Type:
Solution Manual
Resource:
Conceptual Physics
Edition:
13th Edition
Author(s):
Paul G. Hewitt
1 About Science ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 1.1 Scientific Measurements How Eratosthenes Measured the Size of Earth PRACTICING PHYSICS Earth’s Size via Tree/Flagpole Shadows Size of the Moon Distance to the Moon Distance to the Sun Size of the Sun Mathematics—The Language of Science 1.2 Scientific Methods The Scientific Attitude Dealing with Misconceptions 1.3 Science, Art, and Religion FAKE SCIENCE 1.4 Science and Technology RISK ASSESSMENT 1.5 Physics—The Basic Science 1.6 In Perspective
OPENING PHOTOS AND PROFILE All chapters open with photos of educators. Chapter 1 opens with a photo of my wife, Lil, pointing to many circular images of the Sun beneath a tree in front of our residence in St. Petersburg, Florida. The second photo is of my young protégé Einstein Dhayal from India (right) with his friends telling time with solar shadows. The third photo is of my friend and science writer Judith Brand who assisted in the writing of this edition. She shows how lining up a pea with the Moon leads to estimating its distance from Earth. The last photo is of the Vasquez family, all teachers, that begin with Phyllis Vasquez, colleague at City College of San Francisco (CCSF), and her sons in order of age from oldest to youngest: Michael, David, Rodney, John, and Robert, who all took my class. The question suggested by the photo is how many Moons away from Earth is the Moon? This leads to a very nice Moon-Earth activity. The profile is of Greek mathematician Eratosthenes.
INTRODUCTION Like many introductions, much of this introductory chapter can be regarded as a personal essay. In my teaching days I didn’t discuss Chapter 1. Today I would, due to the high interest in Eratosthenes and his intriguing way of measuring Earth’s circumference in about 235 BC. And I would certainly present what I see as a great feature of this edition, the easy-to-do green-pea activity of judging the Moon’s distance from Earth, as Judith Brand illustrates.
SUPPLEMENTS Practicing Physics • Measuring the Size of Planet Earth • Flagpole Physics • Solar Images Next-Time Questions • Scientific Claims • Solar Image
• The Moon, Sun, and a Green Pea • Making Hypotheses
• Pinhole Image of the Sun • With Simply a Pair of Sticks • Flagpole Shadows • Just a Pair of Sticks ..
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• Cone, Ball, and a Cup
• Eratosthenes Sticks
• Two Sticks and a Shadow
Hewitt-Drew-It! Screencast • Eratosthenes (no. 148) Laboratory Manual There are no labs for this chapter.
SUGGESTED PRESENTATION Eratosthenes, Aristarchus, and Sun-Moon Measurements Measurements are a hallmark of science. What more intriguing ones than the sizes and distances of our Sun and Moon. I suggest beginning your course with the measurements of Eratosthenes for Earth, followed by Aristarchus and his measurements of the Moon. Whenever a half moon is viewed in the sky, our line of sight to the Moon and its line of sight to the Sun define the imaginary right triangle in the sky envisioned by Aristarchus (Figure 1.8), which is worthy of intriguing class discussion. Begin by asking your students to come up with two ways to determine the circumference of a pizza pie if they had only one slice. Then progress to the following demonstration. DEMONSTRATION: I’m excited about a new activity in this edition (Flagpole Physics in Practicing Physics) inspired by the first calculation of Earth’s size by Eratosthenes. And that’s to do the reverse: using Earth’s circumference to calculate the distance between far-apart locations—such as far-apart school flagpoles. You can use the second part of the activity, A Model Orange, as a demo. Hold an orange before your students. The orange nicely approximates a sphere. Poke two long toothpicks perpendicular to the surface of the spherical fruit and ask where they’ll meet inside. They’ll intersect at the center of the sphere. Whatever the angle is between the toothpicks, that’s the vertex angle they make with the center of the sphere. The arc between the two sticks makes up a segment of a circle. How many such segments make up a full 360°? That number, multiplied by the distance between the toothpicks at the surface tells you the circumference of the sphere. If the vertex of the segment is, say, 36°, then ten segments form the circumference. Apply this to flagpoles that are very far apart—extreme social distancing. Rather than calculating Earth’s circumference, use its known value of 40,000 km and calculate the distance between school flagpoles. (The first part of the Flagpole Physics activity.) The Next-Time Question Flagpole Shadows also goes well with this material. It’s reproduced following this Suggested Presentation. Measuring Solar Diameter One of my very favorite class assignments is the task of measuring the diameter of the Sun with a ruler or tape measure. This makes sense by first explaining the physics of a pinhole camera. The pinhole image technique is described in Solar Images in Practicing Physics and in Figure 1.9. Hold a meterstick up and tell the class that with a measuring
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device, such as a strip of measuring tape or a simple ruler, they can measure the diameter of the Sun. Call attention to Figure 1.9, then sketch and explain the simple pinhole camera thusly. (More pinhole information is in Chapter 28.) Tell of how a small hole poked in a piece of cardboard will show the image of the Sun when the card is placed in sunlight. You can explain this without referring to Figure 1.9 in the textbook because the figure gives the ratio you wish them to determine. I find this early assignment very successful, in that simple measurements yield a most impressive value—a confidence builder. For those who don’t succeed, or succeed partially, I urge them to try again for full credit. Pinhole Images of the Sun I can’t overestimate the value of spending time on the pinhole image idea. If you’re near a sunlit window you can show how a tiny hole in a piece of card projects an image of the Sun. Most intriguing is doing this with different shaped holes. A small triangle in a piece of card will show not a triangle, but the circular shape of the Sun. The shape of the hole is irrelevant. It just has to be small compared with the distance of the card from the floor. All students find this intriguing. Especially when you tell them of the solar images beneath sunlit trees. They will forever look at the circular spots of light beneath sunlit trees in a different way. And they will be reminded of your lesson. (A question to their friends: When was the first time you noticed that spots of light beneath sunlit trees are pinhole images of the Sun?) That 110 Suns can fit in the space between Earth and the Sun should be intriguing to students. Additionally, 110 Moons can fit in the space between Earth and the Moon! Discussion of the pea held at arm’s length to eclipse the Moon holds high interest. (See The Moon, Sun, and a Green Pea in Practicing Physics.) Scientific Thinking Consider elaborating on the idea about the possible wrongness versus rightness of ideas; an idea that characterizes science. A test for being wrong is not well understood. Expand on the idea that honesty in science is not only a matter of public interest, but is a matter of self-interest. Ideally, there are no second chances for a scientist who’s found to be dishonest. The high standards for acceptable performance in science, unfortunately, do not extend to other fields that are as important to the human condition. For example, consider the standards of performance required of politicians. Distinguish among Hypothesis, Theory, Fact, and Concept Point out that theory and hypothesis are not the same. A hypothesis (an educated guess) proposes an explanation for something observed; it must be testable. A theory applies to a synthesis of a large body of information. The criterion of a theory is not whether it is true or untrue, but rather whether it is useful or not. A theory is useful even though the ultimate causes of the phenomena it encompasses are unknown. For example, we accept the theory of the big bang as the beginning of the observable universe. The theory can be refined, or with new information it can take on a new direction or develop new insights and contribute to the advancing of knowledge. It is important to acknowledge the common misunderstanding of what a scientific theory is, as revealed by those who say, “But it is not a fact; it is only a theory.” Many people have the mistaken notion that a theory is tentative or speculative, while a fact is absolute. Impress upon your class that a fact is not immutable and absolute, but is generally a close agreement by competent observers of a series of observations of the same phenomena. The observations must be testable. Since the activity of science is the determination of the most probable, there are no absolutes.
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Facts that were held to be absolute in the past are seen altogether differently in the light of present-day knowledge. By concept, we mean the intellectual framework that is part of a theory. We speak of the concept of time, the concept of energy, or the concept of a force field. Time is related to motion in space and is the substance of the special theory of relativity. We find that energy exists in tiny grains, or quanta, which is a central concept in the quantum theory. A concept envelops the overriding idea that connects various phenomena. Thus, when we think “conceptually” we envelop a generalized way of looking at things. Dealing with Misconceptions The notion of a self-test for determining whether one’s thinking is on a right or wrong track goes back to many rap sessions with fellow students in my college days. I incorporated the outcome of those insightful chats in the first edition of Conceptual Physics, which I’ve repeated in all editions thereafter. I hope this material elicits spirited discussions among your students today. Prediction Prediction in science is different than prediction in other areas. In the everyday sense, one speaks of predicting what has not yet occurred, like whether or not it will rain next weekend. In science, however, prediction is not so much about what will happen, but about what is happening and is not yet noticed, like what the properties of a hypothetical particle are and are not. A scientist predicts what can and cannot happen, rather than what will or will not happen. In biology, for example, you explain events once you see them. In a sense you’re looking at the historical behavior and then you explain patterns. In physics you’re more likely to predict patterns before they’re seen. Science and Technology In discussions of science and technology and their side effects, a useful statement is that you can never do just one thing. Doing this affects that. Or, You can never change only one thing. Every time you show an equation, it’s evident that changing a variable on one side of the equation changes one or more on the other side. This idea is nicely extended with there is never just one force in discussions of Newton’s third law. Science, Art, and Religion Einstein said, “Science without religion is deaf; religion without science is blind.” The topic of religion in a science textbook is rare. I treat it briefly only to address what is foremost on many students’ minds. Do religion and science contradict each other? Must one choose between them? I hope my very brief treatment presents a satisfactory answer to these questions. My take is that religion and science are compatible when they address their respected realms. Your students need no new examples of the horror that results when extremist thinking dominates. Disputes in science are settled by experiment and better evidence. No wars are fought over science! Your feedback on my inclusion of science/religion in this book will be appreciated.
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Next-Time Question
Answer
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ANSWERS AND SOLUTIONS FOR CHAPTER 1 For brevity, and a space saver in this manual, only the answers and solutions to end-of-chapter questions and problems are included in this manual. Refer to the questions and problems in your textbook. READING CHECK QUESTIONS (COMPREHENSION) 1.1 Scientific Measurements 1. Science is the product of human curiosity about how the world works. It describes order and causes within nature and is an ongoing human activity dedicated to gathering and organizing knowledge about the world. 2. The general reaction has been to forbid new ideas. 3. Alexandria was farther north, at a higher latitude. 4. The center of Earth is the same as the center of any of Earth’s great circles. 5. Like the Sun, the Moon’s diameter is 1⁄110 the distance between Earth and the Moon. 6. The Sun’s diameter is 1⁄110 the distance between Earth and the Sun. 7. At the time of a half moon, he knew that the line from the Moon to Earth intersecting the line from the Moon to the Sun formed a 90° angle. 8. The circular spots are pinhole images of the Sun. 9. The equations are guides to thinking that show the connections between concepts in nature. 1.2 Scientific Methods 10. First, observe; 2. Question; 3. Predict; 4. Test predictions; 5. Draw a conclusion. 11. The answer is stated in the Summary of Terms. 12. Competent scientists must be experts at changing their minds. 13. A scientific hypothesis must be testable. 14. Whereas people who misrepresent something are often given second chances in daily life, second chances ideally are not granted to scientists by the scientific community. 15. If you can state an antagonist’s position to his or her satisfaction, and compare how well the antagonist can state your position, if he or she can’t state your position, the likelihood is that you are correct in your position. 1.3 Science, Art, and Religion 16. To know more than what your beliefs and attitudes are is to expand your education. 17. No, science and religion can work well together, and even complement each other. (Religious extremists, however, may assert that the two are incompatible). 18. One benefit is an open, exploring mind. 1.4 Science and Technology 19. Science is gathering knowledge and organizing it; technology puts scientific knowledge to practical use and provides the instruments for scientists to conduct their investigations. 1.5 Physics—The Basic Science 20. The other sciences build upon physics, and not the reverse. THINK AND DO (HANDS-ON APPLICATION) 21. The ratio of the coin diameter to the coin distance is similar to the ratio of the Sun diameter to the Sun distance, so the numbers of coins and Suns are the same. The number of Suns that would fit between Earth’s surface and the Sun is 110. The Sun is 110 Suns away from us! 22. You’ll find that 110 peas fit in the space, just as 110 Moons fit in the space to the Moon. 23. We see what we’re taught to see! 24. Open ended, as lists will vary.
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THINK AND EXPLAIN (SYNTHESIS) 25. The penalty for fraud is professional excommunication. 26. (a) This is a scientific hypothesis, for there is a test for wrongness. For example, you can extract chlorophyll from grass and note its color. (b) This statement is without a means of proving it wrong and therefore is not a scientific hypothesis. It is speculation. (c) This is a scientific hypothesis. It could be proved wrong, for example, by showing tides that do not correspond to the position of the Moon. 27. What is likely being misunderstood is the distinction between theory and hypothesis. In common usage, “theory” may mean a guess or hypothesis, something that is tentative or speculative. But in science a theory is a synthesis of a large body of validated information (e.g., cell theory or quantum theory). The value of a theory is its usefulness (not its “truth”). 28. For a 45° angle, the lengths of the flagpole and its shadow are the same. 29. Earth’s radius is 8 times 800 km = 6400 km. 30. For a smaller Earth, the shadow would be longer because the angle of the pole would be greater relative to the sunlight. The ratio of the shadow to pole height would be greater than 1 to 8 as in the previous answer. 31. The difference in angles is 6°, which is also the vertex angle at Earth’s center. How many 6° “wedges” circle Earth? It’s 360/6 = 60. So the flagpoles are 40,000 km/60 = 667 km apart. 32. The vertex at the planet’s center is 20° − 15° = 5°. To calculate the circumference, the distance between the stakes must be known, which is 5/360 of a circle. Then, the circumference = (360/5)(distance between stakes).
THINK AND DISCUSS (EVALUATION) 33. To publicly change your mind about your ideas is a sign of strength rather than a sign of weakness. It takes more courage to change your ideas when confronted with counter evidence than to hold fast to them. If your ideas and view of the world are no different after a lifetime of varied experience, then you were either miraculously blessed with unusual wisdom at an early age, or learned nothing. Education is learning what you don’t yet know. Arrogance is thinking you know it all in the later stages of your education, and stupidity is thinking so at the beginning of your education. 34. The examples are endless. Knowledge of electricity, for example, has proven to be extremely useful. The number of people injured by electricity who understood it is far fewer than the number of people injured by it who don’t understand it. A fear of electricity is much more harmful than useful to one’s general health and attitude. 35. All fields of science are based on the freedom to question. Not being able to answer questions is altogether different than not being able to question. Never questioning flies in the face of good science. Since early friendships affect lifelong ones, and unless you value submission to authority, be on the side of science and suggest your relative seek friends elsewhere.
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CHAPTER 1 MULTIPLE-CHOICE PRACTICE EXAM Two ways in which you might use this practice exam: 1. To assess how well your students understand the concepts. 2. To help your students learn the concepts they have yet to understand. 1. The main thrust of equations in this textbook is (a) guides to thinking. (b) recipes for solving algebraic problems. (c) a route to “plugging and chugging.” (d) to challenge students weak in math. Answer: (a) Explanation: Although the equations of physics are central to choices (b–d), the distinguishing feature of equations in Conceptual Physics is to guide thinking, which is of value in its own right and crucial to understanding whatever math problems are solved. 2. The classic scientific method includes (a) guessing. (b) experimenting. (c) predicting. (d) all of the above Answer: (d) Explanation: Most scientists follow all these steps, but other methods such as trial and error or just plain luck can lead to scientific discoveries. 3. If the pillar in Alexandria were taller and the well in Syene were deeper, Eratosthenes’ calculation of Earth’s size would be (a) larger. (b) smaller. (c) no different. (d) impossible to say Answer: (c) Explanation: A crucial measurement is the angle of sunbeams with the pillar in Alexandria, which is 7.2 degrees for a short or tall pillar. Another crucial measurement is the distance between Alexandria and Syene, which has nothing to do with this question. So no difference occurs with a taller pillar and deeper well. 4. If in the dime-Moon experiment you used a larger coin than a dime, your distance from the taped coin would be (a) closer. (b) farther. (c) the same distance. (d) none of these Answer: (b) Explanation: A larger coin would have to be positioned farther away to attain the same 1/110 ratio. Conversely, a coin smaller than a dime would be closer to your eye for the same ratio. 5. Beneath a sunlit tree are pinhole images of the Sun. If the tree that casts the images is taller, then the images of the Sun are (a) larger. (b) smaller. (c) the same size. (d) impossible to say Answer: (a) Explanation: The ratio of the solar image diameter to tree height is about the same as the ratio of the Sun’s diameter to the Sun’s distance from Earth. So taller trees cast larger solar images.
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6. If the diameter of the solar image cast by a pinhole is the same as the diameter of a particular coin, and 110 coins can fit between the image and the pinhole, the Sun must be (a) 55 Sun diameters from Earth. (b) 110 Sun diameters from Earth. (c) not enough information (d) impossible to say Answer: (b) Explanation: The ratio for the coin diameter (same size as solar image) to its distance to the pinhole is the same ratio diameter of the Sun to its distance from Earth. Coincidentally, the same occurs for the Moon. Same 1/110 ratio! 7. Which of these is a scientific hypothesis? (a) The full Moon is a poor time to make decisions. (b) Your truest friends and you have the same astrological sign. (c) The dumpster in your back alley is filled with garbage. (d) none of the above Answer: (c) Explanation: Only choice (c) has a test for wrongness. Look and see! The other statements are speculation. 8. Which of these is a scientific statement? (a) Candied walnuts contain no sugar. (b) Things exist that we will never know about. (c) Matter is filled with undetectable particles. (d) Parts of the universe exist that will never be discovered by humans. Answer: (a) Explanation: Only (a) has a test for wrongness. Choices (b–d) have no tests for wrongness and are therefore unscientific statements. Because a statement may be true does not make it a scientific statement. 9. Reasonable and valid statements that are not testable (a) are a small segment of the scientific method. (b) lie outside the domain of science. (c) have no value. (d) all of the above Answer: (b) Explanation: Statements galore are everywhere, many highly valued and correct. But if not testable, they lie outside the domain of science. Statements that have no test for wrongness, however highly valued, lie outside the domain of science. 10. Science, art, and religion need not contradict one another because (a) all three involve different domains. (b) choosing the right one means no need to heed the other two. (c) choosing religion and art means no need to heed science. (d) choosing science means no need to heed religion and art. Answer: (a) Explanation: Conflicts among science, art, and religion mainly occur when an idea or formulation from one domain is applied to a different domain. When ideas are kept within their own domain, conflicts are minimal. Some students are deeply bothered by what they see as a personal conflict between science and religious faith. They needn’t be.
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2 Newton’s Second Law of Motion ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 2.1 Aristotle on Motion ARISTOTLE (384–322 BC) Copernicus and the Moving Earth 2.2 Galileo’s Experiments Leaning Tower Inclined Planes GALILEO GALILEI (1564–1642) 2.3 Newton’s First Law of Motion PERSONAL ESSAY 2.4 Net Force and Vectors 2.5 The Equilibrium Rule PRACTICING PHYSICS 2.6 Support Force 2.7 Equilibrium of Moving Things 2.8 The Moving Earth
OPENING PHOTOS AND PROFILE The little boy with the Newton’s cradle apparatus in the Part One opener is Ian Evans, son of Bart and Jill Evans, both featured in chapter-opening photos (Chapters 23 and 25). Bart is the workshop coordinator at the Mission Science Workshop, a mini-Exploratorium, devoted to hands-on learning in San Francisco. Jill is a physics instructor at CCSF, my teaching heaven. Photos of Jill appear throughout the book. Chapter opener photos begin with Swedish friends Cedric and Anne Linder investigating rope tensions for the suspended red block. Photo two shows a demonstration of inertia with a blacksmith’s anvil resting on my body, and friend Will Maynez swinging the sledgehammer. Newton’s law of inertia is again illustrated by Ana Miner in photo three. Photo four is of Karl Westerberg of CCSF showing one of my favorite demos with the ball suspended on a string. This is the best demo I know that clearly distinguishes mass and weight. The profile is of Cedric Linder, the first Professor of Physics Education Research in Sweden at Uppsala University.
INTRODUCTION Whereas the study of mechanics customarily begins with kinematics, I begin with a much easier concept for your students—force. In my experience I’ve come to see kinematics as the “black hole of physics instruction”—sucking up class time better devoted to physics. It may be fun to teach, but know that there are NO laws of physics in kinematics. Examples of force provide a better initial hook for student interest than speed and acceleration. So forces in Chapter 2, and motion postponed a bit until Chapter 3. What I hope will be of particular interest to students is the Personal Essay, which relates to events that inspired me to pursue a life in physics—my meeting with Burl Grey on the sign-painting scaffolds of Miami, Florida. Relative tensions in supporting cables are what first caught my interest in physics, and I hope to instill the same interest in your students with my story—also featured on the first of the many Hewitt-Drew-It screencasts, Equilibrium Rule, which nicely introduces vectors. The distinction between mass and weight will await the following chapter, when it’s needed in Newton’s second law. I see the key to good instruction as treating somewhat difficult topics only when they are
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used. For example, I see as pedagogical folly spending the first week on unit conversions, vector notation, graphical analysis, and scientific notation. How much more sensible that the first week in your course is a hook to promote class interest, with these things introduced later if and when they are needed.
SUPPLEMENTS Practicing Physics • Static Equilibrium • The Equilibrium Rule: F = 0 • Vectors and Equilibrium Next-Time Questions • Ball Swing
• Pellet in the Spiral
Hewitt-Drew-It! Screencasts • Equilibrium Rule • Equilibrium Problems • Nellie’s Rope Tensions • Nellie’s Ropes • Force Vectors on an Incline
• Falling Elephant and Feather • Net Force and Vectors • Force Vector Diagrams
Laboratory Manual • Walking the Plank—Equilibrium Rule (Experiment)
SUGGESTED PRESENTATION Newton’s First Law Begin by pointing to an object in the room and stating that if it started moving, one would reasonably look for a cause for its motion. We would say that a force of some kind was responsible, and that would seem reasonable. Tie this idea to the notion of force maintaining motion as Aristotle saw it. State that a cannonball remains at rest in the cannon until a force is applied, and that the force of expanding gases drives the ball out of the barrel when it is fired. But what keeps the cannonball moving when the gases no longer act on it? This leads you into a discussion of inertia. In the everyday sense, inertia refers to a habit or a rut. In physics it’s another word for laziness, or the resistance to change with regard to motion. To demonstrate, I use a 10-cm diameter solid steel sphere, actually a huge ball bearing. I roll the ball along the lecture table to show its tendency to keep rolling. Inertia was first introduced by Galileo—a result of his inclined-plane experiments. DEMONSTRATION: Show that inertia refers also to objects at rest with the classic tableclothand-dishes demonstration. Be sure to pull the tablecloth slightly downward so there is no upward component of force on the dishes! I precede this demo with a simpler version, a simple block of wood on a piece of cloth—but with a twist. I ask what the block will do when I suddenly whip the cloth toward me. After a neighbor check, I surprise the class when they see that the block has been stapled to the cloth! This illustrates Newton’s zeroth law—be skeptical. Then follow up with the classic tablecloth demo. Don’t think either of these demos is too corny. Your students will love both. Of course when we show a demonstration to illustrate a particular concept, there is almost always more than one concept involved. The tablecloth demo is no exception, which also features impulse and momentum (Chapter 6 material). The plates on the table experience two impulses; one that first involves the friction between the cloth and dishes, which moves them slightly toward you. It is brief and very little momentum builds up. Once the dishes are no longer on the cloth, a second impulse occurs due to friction between the sliding dishes and table, which acts in a direction away from you and prevents continued sliding toward you. This second impulse brings the dishes to rest. Done quickly, the brief displacement of
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the dishes is hardly noticed. Is inertia really at work here? Yes, for if there were no friction, the dishes would strictly remain at rest. DEMONSTRATION: Continuing with inertia, fashion a wire coat hanger into an m shape as shown. Two globs of clay are stuck to each end. Balance it on your head, with one glob in front of your face. State that you wish to view the other glob and ask how you can do so without touching the apparatus. Then simply turn around and look at it. It’s like the bowl of soup you turn only to find the soup stays put. Inertia in action! (Of course, like the tablecloth demo, there is more physics here than inertia; this demo can also be used to illustrate rotational inertia and the conservation of angular momentum. In the interest of “information overload,” it may be best that you don’t discuss this.) CHECK QUESTION: How does the law of inertia account for removing snow from your shoes by stamping on the floor, or removing dust from a coat or rug by shaking it? When your foot moves downward, the snow on it tends to continue moving downward when your foot is brought to a sudden halt by the floor. Likewise for shaking dust from a garment. DEMONSTRATION: Do as Marshall Ellenstein does and place a metal hoop atop a narrow jar. On top of the hoop balance a piece of chalk. Then whisk the hoop away and the chalk falls neatly into the narrow opening. The key is grabbing the hoop on the inside, on the side farthest from your sweep. This elongates the hoop horizontally and the part that supports the chalk drops from beneath the chalk. (If you grab the hoop on the near side, the elongation will be vertical and pop the chalk skyward!) Units of Force—Newtons I suggest not making a big deal about the unfamiliar unit of force—the newton. I simply state it is the unit of force used by physicists, and if students find themselves uncomfortable with it, they should simply think of “pounds” in its place. Relative magnitudes, rather than actual magnitudes, are the emphasis of conceptual physics anyway. Do as my influential pal Burl Grey does in Figure 2.11 and suspend a familiar mass from a spring scale. If the mass is a kilogram and the scale is calibrated in newtons, it will read 10 N (more precisely, 9.8 N). If the scale is calibrated in pounds it will read 2.2 pounds. State that you’re not going to waste good time in conversions between units (students can do enough of that in a follow-up course). Plus any conversion they might want is instantly available on the Web. Net Force Discuss the idea of more than one force acting on something, and the resulting net force, which in vector notation we call the resultant. Figure 2.6 captures the essence. The Equilibrium Rule This is the gist of Newton’s first law. It is a rule to guide the answers to many questions, especially about structures. The rule is simple: F = 0. The symbol stands for “vector sum of.” A vector sum takes into consideration positive and negative directions. For example, if a scaffold is in equilibrium, then the vector sum of the forces acting on it “add to” zero. If only upward rope tensions and gravity act on the scaffold, then the combination of upward forces, minus the combination of downward forces, will add to zero. Ups is equal to downs, and the scaffold hangs at rest. For any object in equilibrium, the net force on it must be zero. Sign Painter Skit From my Personal Essay on page 31 in the textbook.
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Step 1: If both painters have the same weight and each stands next to a rope as shown to the left, the supporting force of the ropes will be equal. If spring scales were used, one on each rope, the forces in the ropes would be evident. Ask what the scale reading for each rope would be in this case. The answer is each rope will support the weight of one man and half the weight of the rig— both scales will show equal readings.
Step 2: Suppose one painter walks toward the other as shown to the right, which you draw on the board (or equivalent). Will the tension in the left rope increase? Will the tension in the right rope decrease? Grand question: Will the tension in the left rope increase exactly as much as the decrease in tension of the right rope? You might quip that is so, but how does either rope “know” about the change in the other rope? After neighbor discussion, be sure to emphasize that the answers to these questions lie in the framework of the equilibrium rule: F = 0. Since there is no change in motion, the net force must be zero, which means the upward support forces supplied by the ropes must add up to the downward force of gravity on the two men and the rig. So a decrease in one rope must necessarily be met with a corresponding increase in the other. (This example is dear to my heart. Burl and I didn’t know the answer way back then because neither he nor I had a model for analyzing the problem. We didn’t know about Newton’s first law and the Equilibrium Rule. How different one’s thinking is when one has or does not have a model to guide it. If Burl and I had been mystical in our thinking, we might have been more concerned with how each rope “knows” about the condition of the other—an approach that intrigues many people with a nonscientific view of the world.) Cite other static examples, where the net force is zero as evidenced by no changes in motion. Hold a 1-kg mass at rest in your hand and ask how much net force acts on it. Be sure students distinguish between the 10-N gravitational force on the object and the zero net force on it—as evidenced by its state of rest. (The concept of acceleration is introduced in the next chapter.) When suspended by the spring scale, point out that the scale is pulling up on the object, with just as much force as Earth pulls down on it. Pretend to step on a bathroom scale. Ask how much gravity is pulling on you. This is shown on the scale reading. Then ask what the net force is that acts on you. This is evident by your absence of motion change. Consider two scales, one foot on each, and ask how each scale would read. Then ask how the scales would read if you shifted your weight more on one than the other. We’ll return to support force, called normal force, in following chapters. Inertia and the Moving Earth Stand facing a wall and jump up. Then ask why the wall does not smash into you as Earth rotates under you while you’re airborne. Relate this to the idea of a helicopter ascending over San Francisco, waiting motionless for 5 hours and waiting until Washington, D.C., appears below, then descending. Hooray, this would be a neat way to fly cross-country! Except, of course, for the fact that the “stationary” helicopter remains in motion with the ground below. “Stationary” relative to the stars means it would have to fly as fast as Earth turns (what jets attempt to do).
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Introducing End-of-Chapter Exercises After you’ve assigned exercises, tell your students to be nice to their brains and avoid the easy route of searching on the Web for solutions to Think and Solve and Think and Explain exercises. This is a time for brain exercise. Just as doing pushups strengthens their arms, thinking through solutions strengthens their cognitive skills.
If you want your students to read chapter material before coming to class, reward them for doing so. Give them a quickie quiz at the outset of your class—perhaps at unannounced times. The quiz can be one or two end-of-chapter Reading Check Questions of assigned chapters. Furthermore, you don’t even have to grade the quizzes. Tell your class that you may or may not look at or count their quizzes. The quiz itself is motivation. Nobody wants the indignity of submitting a blank quiz. Not being able to answer simple questions is personally debilitating—enough to motivate students to read assigned material before coming to class. Common sense—reward the behavior you want!
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ANSWERS AND SOLUTIONS FOR CHAPTER 2 READING CHECK QUESTIONS (COMPREHENSION) 2.1 Aristotle on Motion 1. Aristotle classified the motion of the Moon as natural. 2. Aristotle classified the motion of the Earth as natural. 3. Copernicus stated that Earth circles the Sun, and not the reverse. 2.2 Galileo’s Experiments 4. Galileo discovered that falling objects pick up equal speeds whatever their weights. 5. Galileo discovered that moving objects continue in motion without the need of a force. 6. Inertia is the name for the property of matter resisting a change in motion. 2.3 Newton’s First Law of Motion 7. Newton’s law is a restatement of Galileo’s concept of inertia. 8. In the absence of force, a moving body follows a straight-line path. 9. No force is required for continued motion. 2.4 Net Force and Vectors 10. The net force is 80 N to the right. 11. A description of force involves magnitude and direction, and is a vector quantity. 12. The resultant is √2 N. 13. The resultant is 50 N. 2.5 The Equilibrium Rule 14. Yes, although scientists prefer the newton. 15. The net force is zero. 16. The net force is zero. 17. All the forces on something in mechanical equilibrium add vectorially to zero. 18. F = 0. 2.6 Support Force 19. The support force is 12 N. The net force on the book is zero. 20. Weight and support force have equal magnitudes. 2.7 Equilibrium of Moving Things 21. Yes, straight-line motion of the ball at constant speed is dynamic equilibrium. 22. An object in either static or dynamic equilibrium has a zero net force on it. 23. The force of friction is 120 N. 2.8 The Moving Earth 24. They had no comprehension of the concept of inertia. 25. The bird still moves at 30 km/s relative to the Sun. 26. Yes, like the bird of Figure 2.16, you maintain a speed of 30 km/s relative to the Sun, in accord with the concept of inertia. THINK AND DO (HANDS-ON APPLICATION) 27. Consider connecting with your grandparents via traditional handwritten notes. 28. This illustrates Newton’s first law—the law of inertia. 29. Again, the application of the law of inertia. The inertia of the block and books keeps them from suddenly crushing down on you.
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THINK AND SOLVE (MATHEMATICAL APPLICATION) 30. (a) 70 N. (b) 30 N. 31. (a) Zero friction. (b) 100 N. (c) 120 N. 32. Since each scale reads 350 N, Lucy’s total weight is 700 N. 33. 400 N on one scale, 800 N on the other. (2x + x = 1200 N; 3x = 1200 N; x = 400 N) 34. From the equilibrium rule, F = 0, the upward forces are 800 N, and the downward forces are 500 N + the weight of the scaffold. So the scaffold must weigh 300 N. 35. From the equilibrium rule, F = 0, the upward forces are 800 N + tension in the right scale. This sum must equal the downward forces 500 N + 400 N + 400 N. Arithmetic shows the reading on the right scale is 500 N. THINK AND RANK (ANALYSIS) 36. C, B, A 37. C, A, B, D 38. (a) B, A, C, D (b) B, A, C, D 39. (a) A = B = C (no force) (b) C, B, A THINK AND EXPLAIN (SYNTHESIS) 40. Aristotle preferred philosophical logic while Galileo preferred experimentation. 41. The tendency of a rolling ball is to continue rolling—in the absence of a force. The ball slows down likely due to the force of friction. 42. Copernicus and others of his day thought an enormous force would continuously push the Earth to keep it in motion. He was unfamiliar with the concept of inertia, and didn’t realize that once a body is in motion, no force is needed to keep it moving (assuming no friction). 43. Galileo discredited Aristotle’s idea that the rate at which bodies fall is proportional to their weight. 44. Galileo demolished the notion that a moving body requires a force to keep it moving. He showed that a force is needed to change motion, not to maintain motion, with negligible force. 45. Galileo proposed the concept of inertia before Newton was born. 46. Nothing keeps asteroids moving. The Sun’s force deflects their paths but is not needed to keep them moving. 47. Nothing keeps the probe moving. In the absence of a propelling or deflecting force it would continue moving in a straight line. 48. If you pull the cloth upward, even slightly, it will tend to lift the dishes, which will disrupt the demonstration of dishes remaining at rest. The cloth is best pulled horizontally and slightly downward for the dishes to remain at rest. 49. The inertia of a whole roll resists the large acceleration of a sharp jerk and only a single piece tears. If a towel is pulled slowly, a small acceleration is demanded of the roll and it unwinds. This is similar to the hanging ball and string in Figure 2.5. 50. Your body tends to remain at rest, in accord with Newton’s first law. The back of the seat pushes you forward. Without support at the back of your head, your head is not pushed forward with your body, which likely injures your neck. Hence, headrests are recommended. 51. In a bus at rest your head tends to stay at rest. When the bus is rear-ended, it lurches forward and you and your head also move forward. Without a headrest, your body tends to leave your head behind. Hence a neck injury. 52. The law of inertia applies in both cases. When the bus slows, you tend to keep moving at the previous speed and lurch forward. When the bus picks up speed, you tend to keep moving at the previous (lower) speed and you lurch backward. 53. The maximum resultant occurs when the forces are parallel in the same direction—32 N. The minimum occurs when they oppose each other—8 N.
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54. The vector sum of the forces equals zero. That means the net force must be zero. 55. No, if only a single nonzero force acts on an object, its motion will change and it will not be in mechanical equilibrium. Other forces are needed for a zero net force for equilibrium. 56. Yes, it’s in dynamic equilibrium because the net force on it is zero when sliding at constant velocity. 57. No, because the force of gravity still acts. Net force is not zero! 58. No, because the force of gravity still acts, and the player is in motion. Net force is mg. 59. Tension in each rope is half her weight. 60. At the top of its path (and everywhere else along its path) the force of gravity acts to change the ball’s motion. Even though it momentarily stops at the top, the net force on the ball is not zero and therefore it is not in equilibrium. 61. Yes, if the puck moves in a straight line with unchanging speed, the forces of friction are negligible. Then the net force is practically zero, and the puck can be considered to be in dynamic equilibrium. 62. You can say that no net force acts on your friend at rest. The key word is net. When the net force is zero, your friend is in static equilibrium. 63. The scale will read half her weight. In this way, the net force (upward pull of left rope + upward pull of right rope − weight) = 0. 64. In the left figure, Harry is supported by two strands of rope that share his weight (like the little girl in the previous exercise). So each strand supports only 250 N, below the breaking point. Total force up supplied by ropes equals weight acting downward, giving a net force of zero and no acceleration. In the right figure, Harry is now supported by one strand, which for Harry’s wellbeing requires that the tension be 500 N. Since this is above the breaking point of the rope, it breaks. The net force on Harry is then only his weight, giving him a downward acceleration of g. The sudden return of zero velocity changes his vacation plans. 65. Two significant forces act on the book: the force due to gravity and the support force (normal force) of the table. 66. If the upward force were the only force acting, the book indeed would rise. But another force, due to downward gravity, results in the net force being zero. 67. When you stand on a floor, the floor pushes upward against your feet with a force equal to that of gravity, your weight. This upward force (normal force) and your weight are oppositely directed, and since they both act on the same body, you, they cancel to produce a net force on you of zero— hence, you are not accelerated. 68. The gravitational force on you and the scale are a constant. What does fluctuate is the upward support force (normal force) due to your jouncing. Since the reading on a scale is the normal force, your best reading is standing at rest on the scale on a horizontal floor, which matches your weight. 69. Without water, the support force is W. With water, the support force is W + w. 70. The friction on the crate is 200 N, opposite to your 200-N pull. 71. The friction force is 600 N for constant speed. Only then will F = 0. 72. The support force on the crate decreases as the load against the floor decreases. When the crate is entirely lifted from the floor, the support force by the floor is zero. The support force on the workmen’s feet correspondingly increases as the load transfers from the floor to them. When the crate is off the floor and at rest, its weight is transferred to the men, whose normal force is then increased. 73. Emily will not be successful, for her speed will be zero relative to the land. THINK AND DISCUSS (EVALUATION) 74. Aristotle would likely say it stops due to its natural place. Galileo would say it stops due to encountering friction along the way. 75. In both cases the top string supports the weight of the ball. When the lower string is pulled gradually, its tension adds to the top string, and it breaks. But if a pull of the lower string is quick enough, inertia of the ball “keeps it in place” while the snapped lower string breaks. ..
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76. We aren’t swept off because we are traveling just as fast as Earth, like moving along in a fastmoving vehicle. Also, there is no atmosphere through which Earth moves, which would more than blow our hats off! 77. Your friend should learn that inertia is not some kind of force that keeps things like Earth moving, but is the name for the property of things to keep doing what they are doing in the absence of a force. So your friend should say that nothing is necessary to keep Earth moving. Interestingly, the Sun keeps it from following the straight-line path it would take if no forces acted, but it doesn’t keep it moving. Nothing does. That’s inertia. 78. You should disagree with your friend. In the absence of external forces, a body at rest tends to remain at rest; if moving, it tends to remain moving. Inertia is a property of matter that causes it to behave this way, not some kind of force. 79. The tendency of the ball is to remain at rest. From a point of view outside the wagon, the ball stays in place as the back of the wagon moves toward it. (Because of friction, the ball may roll along the cart surface—without friction the surface would slide beneath the ball.) 80. The car has no tendency to resume to its original twice-as-fast speed. Instead, in accord with Newton’s first law, it tends to continue at half speed, decreasing in speed over time due to air resistance and road friction. 81. No, if there were no friction acting on the cart, it would continue in motion when you stop pushing. But friction does act, and the cart slows. This doesn’t violate the law of inertia because an external force indeed acts. 82. An object in motion tends to stay in motion, hence the discs tend to compress upon each other just as the hammerhead is compressed onto the handle in Figure 2.5. This compression results in people being slightly shorter at the end of the day than in the morning. The discs tend to separate while sleeping in a prone position, so you regain your full height by morning. You can notice this if you find an overhead point you can almost reach in the evening, then easily reached in the morning. Try it! 83. No, if there were no force acting on the ball, it would continue in motion without slowing. But air drag does act, along with slight friction with the lane, and the ball slows. This doesn’t violate the law of inertia because external forces indeed act. 84. The normal force is greatest when the table surface is horizontal. It progressively decreases as the angle of tilt increases. As the angle of tilt approaches 90°, the normal force approaches zero. When the table surface is vertical, it no longer presses on the book, which then freely falls. 85. No, the normal force would be the same whether the book was on slippery ice or sandpaper. Friction plays no role unless the book slides or tends to slide along the surface. 86. A stone will fall vertically if released from rest. If the stone is dropped from the top of the mast of a moving ship, the horizontal motion is not changed as the stone is falls—providing air resistance on the stone is negligible. From the frame of reference of the uniformly-moving ship, the stone falls in a vertical straight-line path, landing at the base of the mast. 87. A body in motion tends to remain in motion, so you move with the moving Earth whether or not your feet are in contact with it. When you jump, your horizontal motion matches that of the Earth and you travel with it. Hence the wall does not slam into you. 88. The coin moves along with you when tossed. While airborne it maintains this forward motion, so the coin lands in your hand. If the train slows while the coin is airborne, it will land in front of you. 89. If the train rounds a corner while the coin is in the air, it will land off to the side of you. The coin continues in its horizontal motion, in accord with the law of inertia. 90. Gravitation is a vertical force. On an incline, a component of this force lies along the surface, which produces acceleration if the ball is free to move. But on a horizontal surface, no horizontal component of gravitational force exists, and no change in force occurs.
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CHAPTER 2 MULTIPLE-CHOICE PRACTICE EXAM A distinguishing new feature in this thirteenth edition is the sample multiple-choice practice exams at the end of each chapter. I see these as a must for student learning. Even though answers and explanations are displayed upside down at the bottom of the exam page, consider using class time to highlight the explanations, some of which are explained more fully here. Exams can be learning experiences as well as an assessment of student learning. 1. According to Galileo, inertia is a (a) force like any other force. (b) special kind of force. (c) property of all matter. (d) concept attributed to Aristotle. Answer: (c) Explanation: By definition, inertia is a property of all matter, choice (c). It is not a force or a concept opposite to force. This is a definition of inertia, pure and simple. 2. Galileo’s use of inclined planes allowed him to effectively (a) slow the ball’s changes in speed. (b) reduce the time of the ball’s changes in speed. (c) eliminate major changes in speed. (d) eliminate friction. Answer: (a) Explanation: Galileo lacked suitable time devices (clocks had not yet been invented) so he searched for a pattern to changes in a ball’s downward motion by using small-angle inclines. By successively increasing the angle of a plane to vertical, he found acceleration to be that of free fall. Look ahead to the step-by-step stages of Figure 3.6 in Chapter 3. Hooray for simplicity. 3. A space probe flying in remote outer space continues traveling (a) due to a force acting on it. (b) in a curved path. (c) even if no force acts on it. (d) due to gravity. Answer: (c) Explanation: In accord with Newton’s first law of motion, a body in motion tends to remain in motion in a straight-line path unless acted upon by a force. This question is a simple example of that. With no applied force, the space probe will move continuously. 4. A hockey puck slides along an icy pond. Without any kind of friction, the force needed to sustain sliding is (a) none at all. (b) equal to the weight of the puck. (c) the weight of the puck divided by its mass. (d) the mass of the puck multiplied by 10 m/s2. Answer: (a) Explanation: In accord with Newton’s first law of motion, a body in motion tends to remain in motion in a straight-line path unless acted upon by a force. With no friction to overcome, the puck will slide indefinitely until it encounters a force. The notion that no force is required for sustained motion, although somewhat accepted by the public today, was historically earthshaking.
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5. What is the net force on a box of chocolates when pushed across a table with a horizontal force of 10 N while friction between it and the surface is 6 N? (a) 16 N (b) 10 N (c) 6 N (d) 4 N Answer: (d) Explanation: The net force across the table is simply the sum of the horizontal forces 10 N − 6 N, which is 4 N, choice (d). This simple question should not have been missed. 6. The sum of the forces that act on any object moving at constant velocity is (a) zero. (b) 10 m/s2. (c) equal to its weight. (d) about half its weight. Answer: (a) Explanation: In accord with Newton’s first law of motion, no force is needed for a moving object to continue moving. It will “move of itself.” The key here is “constant velocity”—no acceleration. 7. The equilibrium rule F = 0 applies to (a) objects or systems at rest. (b) objects or systems in uniform motion in a straight line. (c) both. (d) neither. Answer: (c) Explanation: By definition, the equilibrium rule applies to objects at rest, static equilibrium, or in straight-line motion, dynamic equilibrium—both, so the best choice is (c). 8. The tensions in each of the two supporting ropes that support Burl and Paul on opposite ends of a scaffold (a) are equal. (b) depend on the relative weights of Burl and Paul. (c) combine to equal zero. (d) none of the above Answer: (b) Explanation: According to the equilibrium rule, the sum of the upward tensions must be equal and opposite to the combined sums of the weights of Burl plus Paul, and the weight of the scaffold. Each rope tension is not equal because the weights of Burl and Paul are not equal, nor do they combine to zero. What does combine to zero is the sum of upward forces (tensions in the two ropes) and downward forces of the weights the ropes support. 9. A man weighing 800 N stands at rest on two bathroom scales so that one scale shows a reading of 500 N. The reading on the other scale is (a) 200 N. (b) 300 N. (c) 400 N. (d) 800 N. Answer: (b) Explanation: This is a numerical example of F = 0. What number + 500 N = 800 N? Simple math tells you that number must be 800 N − 500 N = 300 N, choice (b). 10. If you leap straight up inside a high-speed train while it gains speed, you land (a) slightly ahead of your original position. (b) at your original position. (c) slightly behind your original position. (d) none of the above Answer: (c) Explanation: Unlike the bird that drops from a tree branch to a worm below where no changes in horizontal motion occur, the train gains speed while you are airborne. Your horizontal component of speed (velocity) during your jump remains the same as your speed when jumping. While you are jumping the floor below gains speed, which means you’ll land behind your original position. Got it?
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3 Linear Motion ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 3.1 Speed Instantaneous Speed Average Speed Motion Is Relative 3.2 Velocity Constant Velocity Changing Velocity 3.3 Acceleration Acceleration on Galileo’s Inclined Planes 3.4 Free Fall How Fast How Far How Quickly “How Fast” Changes HANG TIME 3.5 Velocity Vectors
OPENING PHOTOS AND PROFILE The photo openers begin with my good friend and Conceptual Physics Laboratory Manual author Dean Baird. Photo two is of Sue Johnson (wife of Dan Johnson) who with her racing-shell team won national honors in rowing. Shown in the third photo is Norwegian friend Carl Angell of University of Oslo who rolls a ball through a photo timer. Photo four is of friend and colleague from CCSF, Chelcie Liu, who shows the tracks he made while teaching his daughter Cindy some physics. The tracks have been and are very well used at CCSF and many other schools. This chapter opens with a profile on Dean Baird, a Secondary Science Presidential Award recipient. INTRODUCTION TAKE CARE NOT TO SPEND TOO MUCH TIME ON THIS CHAPTER. Doing so is the greatest pacing mistake in teaching physics! Time spent on kinematics is time not spent on why satellites continually fall without touching Earth, why high temperatures and high voltages (for the same reason) can be safe to touch, why rainbows are round, why the sky is blue, and how nuclear reactions keep Earth’s interior molten. Too much time on this chapter is folly. I strongly suggest making the distinction between speed, velocity, and acceleration, and move quickly to Chapter 4. (I typically spend only one class period on this chapter.) By all means, resist the temptation to do the classic motion problems that involve 90% math and 10% physics! Too much treatment of motion analysis can be counterproductive to maintaining the interest in physics that hopefully began in the previous chapter and moves further in chapters ahead. I suggest you tell your class that you’re skimming the chapter so you’ll have more time for more interesting topics in your course—let them know they shouldn’t expect to master this material, and that mastery will be expected in later material (that doesn’t have the stumbling blocks of kinematics). It’s okay not to fully understand this early part of your course, which interestingly has no laws of physics. Just as wisdom is knowing what to overlook, good teaching is knowing what to omit. If you are getting into more problem solving than is customary in a conceptual course, be sure to look at the student ancillary, Problem Solving in Conceptual Physics. It has ample problems for a lightweight algebra-trigonometry physics course.
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The box on Hang Time at chapter’s end may be especially intriguing to your students if they’re unaware of the short time involved. The hang times of basketball’s greatest legends were less than 1.0 s. Height jumped is less than 1.25 m (4 feet). Those who insist on a hang time of 2 s are way off, for 1 s up would be 16 feet or 4.9 meters—clearly, no way!). Detail is shown in the box on page 56. I’ll stress here the importance of the “check with your neighbor” technique of teaching. Please do not spend your time talking to yourself in front of your class! The procedure of “check with your neighbor” is a routine that keeps you and your class engaged. I can’t emphasize enough its importance! The distinction between velocity and acceleration is an important one and is a prerequisite for the following chapters on mechanics. SUPPLEMENTS The Practicing Physics book of exercises treats the distinction between velocity acquired and distance fallen for free fall via a freely-falling speedometer-odometer. Students do learn from these practice pages, in class or out of class, so whether by paper or digitally, get these to your students. Ideally print and distribute them to your classes. Practicing Physics • Free Fall Speed • Hang Time • Acceleration of Free Fall • Nonaccelerated Motion Next-Time Questions • Relative Speeds
• Bikes and Bee
Hewitt-Drew-It! Screencasts • Free Fall • Sideways Drop • Bikes and Bee • Velocity Vectors
• Ball Toss
Laboratory Manual • Go! Go! Go!—The Basics of Graphing Motion (Experiment) • Sonic Ranger—Graphing Motion in Real Time (Activity) The textbook does not treat motion graphically, but leaves that to the laboratory manual. For this thirteenth edition we use the twelfth edition lab manual. Labs are enhanced with the sonic ranger device, which is conceptual graphing at its best. If not done as lab experiments, consider demonstrating the sonic ranger as part of your presentation. Problem Solving in Conceptual Physics Chapter 3 has abundant and insightful kinematics problems requiring straight-forward algebra, some with sample solutions.
SUGGESTED PRESENTATION Your first question: What means of motion has done more to change the way cities are built than any other? Answer: The elevator! Then explain the importance of simplifying. Motion is best understood if you first neglect the effects of air resistance, buoyancy, spin, and the shape of moving objects. Beneath these are simple relationships that might otherwise be masked by “covering all bases,” and these relationships are what Chapter 3 and
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your presentation are about. State that completely neglecting the effects of air resistance not only exposes the simple relationships, but is a reasonable assumption for heavy and compact (dense) objects traveling at moderate speeds; e.g., one would notice no difference between the rates of fall of a heavy rock dropped from the classroom ceiling to the floor below, when falling through either air or a complete vacuum. For a feather and heavy objects moving at high speeds, air resistance does become important, and will be treated in Chapter 4. Mention that there are few pure examples in physics, for most real situations involve a combination of effects. There is usually a “first order” effect that is basic to the situation, but then there are second, third, and even fourth or more order effects that interact also. If we begin our study of some concept by considering all effects together before we have studied their contributions separately, understanding is likely to be difficult. To have a better understanding of what is going on, we strip a situation of all but the first order effect, and then examine that. When that is well understood, then we proceed to investigate the other effects for a fuller understanding. DEMONSTRATION: Drop a sheet of paper and note how slowly it falls because of air resistance. Crumple the paper and note that it falls faster. Air resistance has been reduced. Then drop a sheet of paper and a book, side by side. Of course the book falls faster, due to its greater weight compared to air resistance. (Interestingly, the amount of air drag is greater for the fasterfalling book—an idea you’ll return to in the next chapter.) Now place the paper against the lower surface of the raised horizontally-held book and when you drop them, nobody is surprised to see they fall together. The book has pushed the paper with it. Now hold the paper on top of the book and ask for predictions and neighbor discussion. Then surprise your class by refusing to show it! Tell them to try it out of class! (Good teaching isn’t giving answers, but raising good questions— good enough to prompt wondering. Let students discover that the book will “plow through the air” leaving an air-resistance free path for the paper to follow!) Explain that air resistance will be treated in later chapters. Again, simplifying brings out the concepts better. You can briefly acknowledge the important effects of air drag. In a bicycle race, for example, the lead cyclist carries along a flow of air that creates a “sweet spot” of low air pressure for the cyclist riding close behind. Air resistance on spinning balls changes their course, and so on. In keeping with the adage “Wisdom is knowing what to overlook,” we neglect the effects of air in this chapter to more clearly reveal the connections between distance, time, speed, velocity, and acceleration. Let your students know that the effects of air drag are treated in future chapters. Speed and Velocity Define speed by writing its equation in word form on the board while giving examples—automobile speedometers, etc. Similarly define velocity, noting that a race car driver is interested in his speed, whereas an airplane pilot is interested in her velocity—speed and direction. Mention the difference between a scalar and a vector quantity and identify speed as a scalar and velocity as a vector. Tell your class that you’re not going to make a big deal about distinguishing between speed and velocity, but you are going to make a big deal of distinguishing between velocity and another concept—acceleration. Acceleration Define acceleration by identifying it as a vector quantity, and cite the importance of CHANGE. That’s change in speed, or change in direction. Hence both are acknowledged by defining acceleration as a rate of change in velocity rather than speed. Ask your students to identify the three controls in an automobile that make the auto change its state of motion—that produce acceleration [accelerator, brakes, and steering wheel]. State how one lurches in a vehicle that is undergoing acceleration, especially for circular motion, and state why the definition of velocity includes direction to make the definition of acceleration
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all-encompassing. Talk of how without lurching one cannot sense motion, giving examples of coin flipping in a high-speed aircraft versus doing the same when the same aircraft is at rest. Units for Acceleration Give numerical examples of acceleration in units of kilometershour per second to establish the idea of acceleration. Be sure that your students are working on the examples with you. For example, ask them to find the acceleration of a car that goes from rest to 100 kmh in 10 seconds. It is important that you not use examples involving seconds twice until they taste success with the easier kilometershour per second examples. Have them check their work with their neighbors as you go along. Only after they get the hang of it, introduce meterssecondsecond in your examples to develop a sense for the units ms2. This is treated in the Hewitt-Drew-It! screencasts, Unit Conversion and Acceleration Units. Free Fall If you round 9.8 ms2 to 10 ms2 in your presentation, you’ll more easily establish the relationships between velocity and distance. In lab you can use the more precise 9.8 ms2. CHECK QUESTION: If an object is dropped from an initial position of rest from the top of a cliff, how fast will it be traveling at the end of one second? (You might add, “Write the answer on your notepaper.” And then, “Look at your neighbor’s paper—if your neighbor doesn’t have the right answer, reach over and help him or her—talk about it.” And then possibly, “If your neighbor isn’t very cooperative, sit somewhere else next time!”) After explaining the answer when class discussion dies down, repeat the process asking for the speed at the end of 2 seconds, and then for 10 seconds. This leads you into stating the relationship v = gt, which by now you can express in shorthand notation. After any questions, discussion, and examples, state that you are going to pose a different question—not asking of how fast, but for how far. Ask how far the object falls in one second. Ask for a written response and then ask if the students could explain to their neighbors why the distance is only 5 m rather than 10 m. After they’ve discussed this for almost a minute or so, ask “If you maintain a speed of 60 kmh for one hour, how far do you go?”—then, “If you maintain a speed of 10 ms for one second, how far do you go?” Important point: You’ll appreciably improve your instruction if you allow some thinking time after you ask a question. Not doing so is the folly of too many instructors. Then continue, “Then why is the answer to the first question not 10 meters?” After a suitable time, stress the idea of average velocity and the relation d = vavet. Show the general case by deriving on the board d = 1⁄2gt2. (We tell our students that the derivation is a sidelight to the course—something that will be the crux of a follow-up physics course. In any event, the derivation is not something that we expect of them, but to show that d = 1⁄2gt2 is a reasoned statement that doesn’t just pop up from nowhere.) CHECK QUESTIONS: How far will a freely falling object that is released from rest fall in 2 s? In 10 s? (When your class is comfortable with this, then ask how far in 1⁄2 second.) To avoid information overload, we restrict all numerical examples of free fall to cases that begin at rest. Why? Because it’s simpler that way. We prefer our students to understand simple physics than to be confused about not-so-simple physics! CHECK QUESTION: Consider a rifle fired straight downward from a high-altitude balloon. If the muzzle velocity is 100 ms and air resistance can be neglected, what is the acceleration of the bullet after one second? (If most of your class say that it’s g, you’re on!)
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I suggest not asking for the time of fall for a freely-falling object, given the distance. Why? Unless the distance given is the familiar 5 meters, algebraic manipulation is called for. If one of our teaching objectives were to teach algebra, this would be a nice place to do it. But we don’t have time to present this stumbling block and then teach how to overcome it. We’d rather put our energy and theirs into straight physics! As said, kinematics can be rich with puzzles, graphical analysis, ticker timers, photogates, and algebraic problems. My strong suggestion is to resist these and move quickly into the rest of mechanics, and then into other interesting areas of physics. Getting bogged down with kinematics, with so much physics ahead, is a widespread practice. Please do your class a favor and hurry on to the next chapters. If at the end of your course you have time (ha-ha), then bring out the kinematics toys and have a go at them. One kinematics toy that should be shown, however, is the “Hump and Dip Tracks.” Hump and Dip Tracks Photo 4 at the beginning of the chapter shows an intriguing ballrace demo that elicits a lot of student discussion. The originator, CCSF colleague Chelcie Liu, fashioned the pair of tracks from equal lengths of channel iron, the kind used in installing shelves on a wall. I suggest going further with three tracks, as shown. In doing the demo you can ask two questions: When balls are released simultaneously, which is first to reach the end of the track? Or said another way, which ball wins the race? And a separate but different question, which ball finishes the race with the greatest speed? Note these are different questions with different answers. Your first question involves time. Your second is of speed. You’ll lead your class into distinguishing between speed and time. You’ll also distinguish between average speed and instantaneous speed. Distinctions are the hallmark of good physics, and, good thinking. Our thinking is guided by the definition of speed: Distance covered per amount of time. It’s easy to see that ball C will clearly win the race. Let’s see why. To begin with, ball B takes more time rolling over the hump than moving along a level surface. This means that ball B lags and doesn’t reach the end of its track as quickly as ball A. The opposite occurs for track C, where the ball rolls more quickly in and out of the dip than along a level surface. So ball C with its period of quickness beats both balls A and B and wins the race. This is a great kinematics demonstration, and will be even greater when you later reintroduce it with more detail as a conservation of energy demonstration. Finishing speed is a different story. Careful thought tells us that the speed that ball B loses going up the hump is regained on its downside. This means that it retains the speed it had before encountering the hump. Likewise for ball C, where its gain in speed down into the dip is lost moving up the dip, resulting in no net change of speed. So all three balls finish the race with the same speed. The lesson learned: Equal speeds at the end does not mean equal times! Hang Time As strange as it first may seem, the longest time a jumper can remain in air is less than a second. It is a common illusion that jumping times are more. Even Michael Jordan’s best hang time (the time the feet are off the ground) was 0.9 second. Then d = 1⁄2gt2 predicts how high a jumper can go vertically. For a hang time of a full second, that’s 1⁄2 s up and 1⁄2 s down. Substituting, d = 1⁄2(10 m/s2)(0.5 s)2 = 1.25 m (which is about 4 feet!). So the great athletes and ballet dancers jump vertically no more than 4 feet high! Of course one can clear a higher fence or bar; but one’s center of gravity cannot be raised more than 4 feet in free jumping. In fact very few people can jump 2 feet high! To test this, stand against a wall with arms stretched upward. Mark the wall at the highest point. Then jump, and at the top, again mark the wall.
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For a human being, the distance between marks is at most 4 feet! We’ll return to hang time for running jumps when we discuss projectile motion in Chapter 10. NEXT-TIME QUESTION (NTQ): For projecting or posting. Note the sample reduced page from the Next-Time Questions collection. The original is a full 81⁄2 11 inches, just right for projecting. Next-Time Questions for each chapter are provided with their answers. Consider displaying printed NTQs at some general area outside the classroom—perhaps in a glass case. This display generates general student interest, as students in your class and those not in your class are stimulated to think physics. After a few days of posting then turn the sheets over to reveal the answers. That’s when new NTQs can be displayed. How better a way to adorn your corridors! Because of space limitations, NTQs for other chapters are scarce in this manual. [I’m not displaying the answer to the reproduced one here, but it makes the point that solving this problem requires consideration of time t. Whether or not one thinks about time should not be a matter of cleverness or good insight, but a matter of letting the equation, d = vt, for distance guide thinking. The v is given, but not the time t. Aha! The equation guides you to consider time. Equations are important in guiding our thinking about physics.]
Velocity Vectors Consider Figure 3.13 of the airplane flying in a crosswind. The resulting speed can only be found with vectors. The only vector tools the student needs are the parallelogram rule and perhaps the Pythagorean theorem. Avoid sines and cosines unless your students are studying to be scientists or engineers. We distinguish between physics and the tools of physics. Tools for pre-engineers and scientists only. Physics for everybody! This is a good time for the Hewitt-Drew-It! screencast Velocity Vectors, which treats an airplane flying at different directions relative to the wind. Consider discussing the vector treatment of Appendix D, sailboats sailing faster crosswind than in the direction of the wind.
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ANSWERS AND SOLUTIONS FOR CHAPTER 3 READING CHECK QUESTIONS (COMPREHENSION) 3.1 Speed 1. The two necessary units are distance traveled and time of travel. 2. A speedometer registers instantaneous speed. 3. Average speed v = 160 km/2 h = 80 km/h 4. From d = vt you travel 80 km/h 0.5 h = 40 km. 5. Relative to the chair your speed is zero. Relative to the Sun it’s 30 km/s. 3.2 Velocity 6. Speed is a scalar and velocity is speed and direction, a vector. 7. Yes, a car moving at constant velocity moves at constant speed. 8. The car maintains a constant speed but not a constant velocity because it changes direction as it rounds the corner. 3.3 Acceleration 9. The acceleration is 10 km/h/s. 10. Acceleration is zero, because velocity doesn’t change. 11. You are aware of changes in your speed, but not steady motion. Therefore you are aware of acceleration, but not constant velocity. 12. When motion is in one direction along a straight line, either may be used. 13. Galileo discovered that the ball gained the same amount of speed each specific time interval, which told him the acceleration is constant. 14. Galileo discovered that the greater the angle of incline, the greater the acceleration. 15. When the incline is vertical, the acceleration of the ball is g, that of free fall. 3.4 Free Fall 16. One, only force due to gravity. 17. None. Free fall means falling with no air resistance. 18. Gain in speed is 10 m/s each second. 19. In accord with v = gt, the speed acquired in 4 s is 40 m/s. 20. The unit “second” occurs in velocity, and again in the time. Velocity is divided by time in computing acceleration. Hence the square of “second” in acceleration. 21. The moving object loses 10 m/s for each second moving upward. 22. Galileo found distance traveled is directly proportional to the square of the travel time (d = 1⁄2gt2). 23. The distance of fall in 1 s is 5 m. For a 5-s drop, falling distance is 125 m. 24. Air resistance reduces falling acceleration. 25. 10 m/s is speed, 10 m is distance, and 10 m/s2 is acceleration. 3.5 Velocity Vectors 26. Speed is how fast, a scalar quantity. Velocity as a vector quantity states how fast and which direction. 27. Your path will be the diagonal of a rectangle composed of your speed (velocity) and the velocity of the flowing water. The diagonal of a rectangle is always greater than the length of either of its sides. 28. The diagonal of the square formed by a pair of equal vectors at right angles to each other represents the resultant.
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THINK AND DO (HANDS-ON APPLICATION) 29. Tell Grandma that, in effect, velocity is how fast you’re traveling and acceleration is how quickly how-fast changes. Please don’t tell Grandma that velocity is always the same as speed! 30. A good activity making use of your personal electronics. 31. Another good activity making use of your personal electronics. 32. This can be a social activity, with good physics. 33. This is a follow-up to the previous activity, a good one! 34. Hang times will vary, but won’t exceed 1 second! PLUG AND CHUG (EQUATION FAMILARIZATION) 1.0 m distance d 35. Speed = = = = 2 m/s2 0.5 s time t 36. Average speed =
total distance convered 30 m = = 15 m/s2 2s time interval
37. Acceleration =
change of velocity v 100 km/h = = = 10 km/h s time interval 10 s t
38. Acceleration =
change of velocity v 10 m/s = = = 5 m/s2 2s time interval t
39. Distance covered = vave time = 6.0 m/s 10 s = 60 m 40. Free-fall distance from rest = 1⁄2gt2 = 1⁄2(10 m/s2)(3 s)2 = 45 m THINK AND SOLVE (MATHEMATICAL APPLICATION) 41. (a) Using g = 10 m s2, we see that v = gt = (10 ms2)(10 s) = 100 ms; vave =
vbeginning + vfinal 2
=
0 + 100 m/s = 50 m/s, downward. 2
(b) We can get “how far” from either d = vavet = (50 ms)(10 s) = 500 m, or equivalently, d = 1⁄2gt2 = 5(10)2 = 500 m. (How nice we get the same distance using either formula!) 42. a =
25 m/s − 0 v = = 2.5 m/s2 10 s t
43. (a) The velocity of the ball at the top of its vertical trajectory is instantaneously zero. (b) One second before reaching its top, its velocity is 10 ms. (c) The amount of change in velocity is 10 ms during this 1-second interval (or any other 1-second interval). (d) One second after reaching its top, its velocity is –10 ms—equal in magnitude but oppositely directed to its value 1 second before reaching the top. (e) The amount of change in velocity during this (or any) 1-second interval is 10 ms. (f) In 2 seconds, the amount of change in velocity, from 10 ms up to 10 ms down, is 20 ms (not zero!). (g) The acceleration of the ball is 10 ms2 before reaching the top, when reaching the top, and after reaching the top. In all cases acceleration is downward, toward Earth. 44. Since it starts going up at 30 ms and loses 10 ms each second, its time going up is 3 seconds. Its time returning is also 3 seconds, so it’s in the air for a total of 6 seconds. Distance up (or down) is 1 ⁄2gt2 = 5 m/s2 32 = 45 m. (Or from d = vt, where average velocity is (30 m/s + 0)/2 s = 15 m/s, and time is 3 seconds, we also get d = 15 ms 3 s = 45 m.) ..
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45. From d = ½ gt2 = 5t2, t =
d = 5 m/s
0.6 m = 0.35 s. Doubling for a hang time of 0.746. 5 m/s
(a) Landing speed is 120 km/h − 40 km/h = 80 km/h. (b) Landing speed is then 120 km/h + 40 km/h = 160 km/h (now you see why planes land in a headwind versus a tail wind). (c) In a headwind that matches air speed, the plane would “hover” above the ground as it touches down! So its landing speed is zero. 47. The resultant is zero, the mosquito appears to hover motionless in the air. 48. Speed is
( 2 + 2 ) = 2.8 m/s at an angle of 45° to the mosquito’s intended direction of travel. 2
2
49. The diagonal of any rectangle = 2 times 10 km/h (~14 km/h).
2 the length of either side. So the resultant speed is equal to
THINK AND RANK (ANALYSIS) 50. D, C, A, B 51. C, B = D, A 52. (a) B, A = C (b) A, B, C (c) C, B, A 53. (a) C, B, A (b) A = B = C (10 m/s2) THINK AND EXPLAIN (SYNTHESIS) 54. She’ll be unsuccessful. Her velocity relative to the shore is zero (8 kmh − 8 kmh = 0). 55. The speeds of both are exactly the same, but the velocities are not. Velocity includes direction, and since the directions of the airplanes are opposite, their velocities are opposite. The velocities would be equal only if both speed and direction were the same. 56. Collision speed is 88 km/h − 80 km/h = 8 km/h. 57. Your fine for speeding is based on your instantaneous speed; registered on a speedometer or radar. 58. (a) Yes, because of the change of direction. (b) Yes, because velocity changes. 59. Emily is correct. Jacob is describing speed. Acceleration is the time rate of change in speed—“how fast you get fast,” as Emily asserts. 60. No, you cannot say which car underwent more acceleration without the times involved. 61. The acceleration is zero, for no change in velocity occurs. Whenever the change in velocity is zero, the acceleration is zero. If the velocity is “constant,” or “uniform,” the change in velocity is zero. 62. The greater change in speeds occurs for (30 kmh − 25 kmh = 5 kmh), which is greater than (100 kmh − 96 kmh = 4 kmh). So for the same time, the slower one has more acceleration. 63. Its speed readings would increase by 10 ms each second. 64. Distance readings would indicate greater distances fallen in successive seconds. During each successive second the object falls faster and covers greater distance. 65. No, free-fall acceleration is constant, which accounts for the constant increase of falling speed. 66. Time (in seconds)Velocity (in meters/second)Distance (in meters) 0 0 0 1 10 5 2 20 20 3 30 45 4 40 80 5 50 125 6 60 180 7 70 245 8 80 320 9 90 405 10 100 500 ..
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67. The acceleration of free fall at the end of the fifth, tenth, or any number of seconds is g. Its velocity has different values at different times, but since it is free from the effects of air resistance, its acceleration remains a constant g. 68. Air drag will reduce all free-fall readings. With air resistance, no free fall. 69. In the absence of air resistance, the acceleration will be g no matter how the ball is released. The acceleration of a ball and its speed are entirely different. 70. Whether up or down, the rate of change of speed with respect to time is 10 ms2, so each second while going up the speed decreases by 10 ms. Coming down, the speed increases by 10 ms each second. So with no air resistance, the time rising equals the time falling. 71. With no air resistance, both will strike the ground below at the same speed. Note that the ball thrown upward will pass its starting point on the way down with the same speed it had when starting up. So its trip on downward, below the starting point, is the same as for a ball thrown down with that speed. 72. When air resistance affects motion, the ball thrown upward returns to its starting level with less speed than its initial speed, and also less speed than the ball tossed downward. So the downward thrown ball hits the ground below with a greater speed. 73. If it were not for the slowing effect of the air, raindrops would strike the ground with the speeds of high-speed pellets! 74. If your car is coasting backward and you apply the brakes, you’ll have a negative velocity and a positive acceleration. 75. On the Moon the acceleration due to gravity is considerably less, so hang time would be considerably more (six times more for the same takeoff speed!). 76. As water falls it picks up speed. Since the same amount of water issues from the faucet each second, it stretches as distance increases. It becomes thinner just as taffy that is stretched gets thinner the more it is stretched. When the water is stretched too far, it breaks up into droplets. 77. The speed of falling rain and the speed of the automobile are the same. THINK AND DISCUSS (EVALUATION) 78. Yes, velocity and acceleration need not be in the same direction. A car moving north that slows down, for example, accelerates toward the south. 79. Yes, again, velocity and acceleration need not be in the same direction. A ball tossed upward, for example, reverses its direction of travel at its highest point while its acceleration g, directed downward, remains constant (this idea will be explained further in Chapter 4). Note that if a ball had zero acceleration at a point where its speed is zero, its speed would remain zero. It would sit still at the top of its trajectory! 80. Acceleration occurs when the speedometer reading changes. No change, no acceleration. 81. “The racing car rounded the curve at a constant speed of 100 km/h.” Constant velocity means not only constant speed but constant direction. A car rounding a curve changes its direction of motion. 82. A vertically-thrown ball has zero speed at the top of its trajectory while acceleration remains a constant g. 83. An object moving in a circular path at constant speed is a simple example of acceleration at constant speed because its velocity is changing direction. But something CANNOT accelerate while having constant velocity, which would contradict the notion that constant velocity means no change. 84. Only on the middle hill does speed along the path decrease with time, for the hill becomes less steep as motion progresses. When the hill levels off, acceleration will be zero. On the left hill, acceleration is constant. On the right hill, acceleration increases as the hill becomes steeper. In all three cases, speed increases. 85. The one in the middle. The ball gains speed more quickly at the beginning where the slope is steeper, so its average speed is greater even though it has less acceleration in the last part of its trip.
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86. Free fall is defined as falling only under the influence of gravity, with no air resistance or other nongravitational forces. So your friend should omit “free” and say something like, “Air resistance is more effective in slowing a falling feather than a falling coin.” 87. With no air resistance an object’s acceleration is the same 10 ms2 regardless of its initial velocity. If it is thrown downward, its velocity will be greater, but not its acceleration. 88. Its acceleration would actually be less if the air resistance it encounters at high speed retards its motion. (We will treat this concept in detail in Chapter 4.) 89. When acceleration of the car is in a direction opposite to its velocity, the car is “decelerating,” slowing down. 90. In the absence of air resistance both accelerations are g, the same. Their velocities may be in opposite directions, but g is the same for both. 91. (a) The ball on track B finishes first. (b) Average speed is greater along the lower part. (c) The instantaneous speed at the ends of the tracks is the same because the speed gained on the down-ramp for track B is equal to the speed lost on the up-ramp side. (Many people get the wrong answer for the previous question because they assume that because the balls end up with the same speed that they roll for the same time. Not so.) 92. Refer to the author’s earlier statement. To better understand your study of physics we first learn the most simple case. When we have a good understanding of simple cases, then we proceed to investigate other complexities. Be glad that Galileo was able to free his thinking from real-world friction that allowed his great discoveries in mechanics. 93. Open ended. 94. Open ended.
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CHAPTER 3 MULTIPLE-CHOICE PRACTICE EXAM 1. If an object moves along a straight-line path at constant speed, then it must be (a) accelerating. (b) acted on by a force. (c) both of these. (d) neither of these. Answer: (d) Explanation: Moving in a straight line at a constant speed, constant velocity, means no net force and no acceleration in accord with F = 0. Only if the object were accelerating would a force be acting. Hence the correct choice is (d), and should not be missed. 2. The average speed of a ball rolling along a horizontal track will be increased if the track has a (a) small upward hump along it. (b) downward dip along it. (c) both the same if the hump and dip are opposites of each other. (d) not enough information Answer: (b) Explanation: The ball gains speed moving downward into the dip and loses the gain moving upward. Even so, everywhere in the dip its speed exceeds the speed along the horizontal part of the track. Hence the average speed along the whole track is increased by the presence of the dip. We’ll return to balls on tracks when we treat the conservation of energy. 3. A mosquito flying at 2 m/s encounters a 2-m/s breeze blowing in the opposite direction, giving it a resulting speed over the ground of (a) 0 m/s. (b) 3 m/s. (c) 4 m/s. (d) 6 m/s. Answer: (a) Explanation: This is a “relative-motion” question. The two speeds in opposite directions are relative to the ground. 2 m/s − 2 m/s = 0. The mosquito “hovers” at rest above the ground. And hopefully, not above us! 4. Neglecting air resistance, when you toss a ball upward, by about how much does its upward speed decrease each second? (a) 10 m/s (b) 10 m/s2 (c) The answer depends on the initial speed. (d) None of these Answer: (a) Explanation: Without air resistance, the acceleration of the ball everywhere along its path is g, 10 m/s2, which means the ball’s change in speed is 10 m/s for each second, whatever its initial speed. Note that choice (b) is not a speed, but an acceleration, as evidenced by the second squared. So the correct answer is 10 m/s, choice (a). 5. The acceleration of a block of ice sliding down an inclined plane (a) increases with time. (b) decreases with time. (c) is practically constant. (d) need more information Answer: (c) Explanation: Speed or velocity changes with time, but not acceleration. If the plane is friction-free and without curves or obstructions, acceleration will be constant.
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6. During each second of free fall, the speed of a dropped object (a) increases by the same amount. (b) changes by increasing amounts. (c) remains constant. (d) doubles. Answer: (a) Explanation: In free fall the acceleration of an object is a constant g, which tells us speed constantly changes by 10 m/s each second. So (a) is the correct answer. Choice (c) may incorrectly confuse acceleration g, a constant, with speed v that changes with time. Choice (b) would apply to falling distance, not speed. Choice (d) indicates a wild guess, ouch! 7. A freely falling object has a speed of 40 m/s at one instant. Exactly 1 s later its speed will be (a) the same. (b) 10 m/s. (c) 45 m/s. (d) greater than 45 m/s. Answer: (d) Explanation: A freely-falling body near Earth’s surface gains a speed of 10 m/s for each second of fall. One second after reaching a speed of 40 m/s, the speed would be 50 m/s, which isn’t a choice. Choice (d), says “greater than 45 m/s,” which is true. So choice (d) is the best answer. 8. Mike stands at the edge of a cliff and throws one ball straight up and another ball straight down, both with the same speed. Neglecting air resistance, which ball hits the ground below with the greater speed? (a) The one thrown upward (b) The one thrown downward (c) neither, both hit with the same speed (d) need more information Answer: (c) Explanation: The speeds at ground level are the same, because when the ball thrown upward returns to its starting level, it will have its same initial speed as when thrown, which will also be the same speed as when thrown downward. So either way, both hit the ground below at the same speed, but at different times. Take care not to confuse speed with time. 9. Nellie steps off the edge of a table and lands on the floor below in 1⁄4 second. The height of the tabletop above the floor is about (a) 1⁄3 m. (b) 1⁄2 m. (c) 3⁄4 m. (d) 1 m. Answer: (a) Explanation: Distance fallen from rest at a constant acceleration of 10 m/s2 over a specific time is given by d = 1⁄2gt2 = 1⁄2(10 m/s2)(1⁄4 s)2 = 0.31 m ≈ 1⁄3 m. This is a straight plug of that equation. If missed, it may have been due to the fraction of a meter—unfortunate. 10. The vertical height attained by a basketball player achieving a hang time of a full 1 s is about (a) 0.8 m. (b) 1 m. (c) 1.2 m. (d) 2.5 m. Answer: (c) Explanation: Hang time is calculated from d = 1⁄2gt2. For a full 1-s jump, time up = time down, which is 1⁄2 s either way. This is a round-trip situation, so it is important to see that the distance equation is applied to going up, or, coming down. Then d = 1⁄2gt2 = 1⁄2(10 m/s2)(1⁄2 s)2 = 1.25 m, close enough to 1.2 m, making choice (c) the best answer. This is an activity worth doing, very personal.
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4 Newton’s Second Law of Motion ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 4.1 Forces 4.2 Friction 4.3 Mass and Weight Mass Resists Acceleration 4.4 Newton’s Second Law of Motion 4.5 When Acceleration Is g—Free Fall 4.6 When Acceleration Is Less Than g—Non–Free Fall PROBLEM SOLVING
OPENING PHOTOS AND PROFILE The photo openers begin with Efrain Lopez illustrating equilibrium on an inclined plane. Matt Anderson teaches by way of his impressive learning glass, a method that is catching on with physics presentations. What more exciting Newton’s second law activity than wingsuit flying, deserving of a photo. The bottom photo is of CCSF’s Jill Evans, who most always employs props in her teaching. The profile for this chapter features Isaac Newton. The art piece of Newton was created by Mary Jew, an artistically talented student in my early classes at CCSF. The art work is a personal gift, as is her portrait of Einstein that opens Chapter 35.
INTRODUCTION Inertia, acceleration, and falling objects, as introduced in Chapters 2 and 3, are further developed in this chapter, and we treat the relationship between force and acceleration. We distinguish between mass and weight without making a big deal about their units of measurement. A brief treatment of units and systems of measurement is provided in Appendix A. The term “amount of matter” was called mass by Newton, who stated that both gravitational and mechanical forces acted on mass, as described by F = ma. Consider using class time to go over the many Check Points in this and other chapters. A feature of this edition is an emphasis on Check Points as a means of learning material.
SUPPLEMENTS Practicing Physics • Mass and Weight • Cart • Converting Mass to Weight • Force and Acceleration • A Day at the Races with a = F/m • Friction • Dropping Masses and Accelerating • Falling and Air Resistance • Force-Vector Diagrams Next-Time Questions • Skidding Truck • Spool Pull • Net Force Half-Way Up • Truck and Car Collision • Galileo • Skydiving Nellie
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• Book Push Against the Wall • Acceleration at the Top • Falling Balls • Skydiver • Acceleration on the Way Up • Balanced Scale • Block Pull • Direction of Friction • Accelerating Nellie • Rocket Acceleration
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Hewitt-Drew-It! Screencasts • Mass/Weight • Acceleration Units • Newton’s Second Law • Skydiver Problem Laboratory Manual • The Weight—Connecting Mass and Weight (Activity) • Pushing Things Around—Force, Mass, and Acceleration (Activity) • Putting the Force Before the Cart—Force, Mass, and Acceleration Simulation (Tech Lab) • The Newtonian Shot—Force and Motion Puzzle (Demonstration) Problem Solving in Conceptual Physics More than 100 problems complement this chapter!
SUGGESTED PRESENTATION Friction Drag a block at constant velocity across your table. Acknowledge the force of friction, and how it must exactly counter your pulling force. Show that pulling force with a spring balance. Now since the block moves without accelerating, ask for the magnitude of the friction force. It must be equal and opposite to your scale reading. Then the net force is zero. While sliding, the block is in dynamic equilibrium. That is, F = 0. CHECK QUESTIONS: (similar to one in the textbook.) Suppose in a high-flying airplane the captain announces to the passengers that the plane is flying at a constant 900 kmh and the thrust of the engines is a constant 80,000 newtons. What is the acceleration of the airplane? [Answer: Zero, because velocity is constant.] What is the combined force of air resistance that acts all over the plane’s outside surface? [Answer: 80,000 N. If it were less, the plane would speed up; if it were more, the plane would slow down.] Continue your activity of pulling the block across the table with a spring balance. Show what happens when you pull harder. Your students see that when the pulling force is greater than the friction force, there is a net force greater than zero, as evidenced by the observed acceleration. Show different constant speeds across the table with the same applied force, which shows that friction is not dependent on speed. Distinguish between static and sliding friction, and show how a greater force is needed to get the block moving from a rest position. Show all this as you discuss these ideas. If you don’t show this demo, turn your attention to my granddaughter Marie Hurrell pushing the table in Figure 4.4 and the details presented in the accompanying Check Point. You distinguish between static and dynamic friction. Note that the Practicing Physics book nicely treats details of friction. Static friction exists only in response to an applied force. Consider the table being pushed by Marie in Figure 4.4. Before being pushed there’s no friction between the legs of the table and the floor. Friction does not hold the table at rest. It occurs only in response to an applied force. This is important information. Stress this important point! Mass and Weight To distinguish between mass and weight compare the efforts of pushing horizontally on a block of slippery ice on a frozen pond versus lifting it. Or consider the weightlessness of a massive anvil in space and how it would be difficult to shake, weight or no weight. And if moving toward you, it would be harmful to be in its way because of its great tendency to remain in motion. The following demo (often used to illustrate impulse and momentum) makes the distinction nicely.
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DEMONSTRATION: Hang a massive ball by a string and show that the top string breaks when the bottom is pulled gradually with more force, but the bottom string breaks when the string is jerked. Ask which of these cases illustrates weight. Interestingly enough, it’s the weight of the ball that makes for the greater tension in the top string. Then ask which of these cases illustrates inertia. When jerked, the tendency of the ball to resist the sudden downward acceleration, its inertia, is responsible for the lower string breaking. This is the best demo I know of for showing the different effects of weight and mass. Mass Resists Acceleration The property of massive objects to resist changes is nicely shown with this follow-up demonstration. DEMONSTRATION: Lie on your back and have an assistant place a blacksmith’s anvil on your stomach. Have the assistant strike the anvil rather hard with a sledgehammer. The principles are the same as the ball and string demo. Both the inertias of the ball and the anvil resist the changes in motion they would otherwise undergo. So the string doesn’t break, and your body is not squashed. (Be sure that your assistant is good with the hammer. When I began teaching, I used to trust students to the task. In my fourth year the student who volunteered was extra nervous in front of the class and missed the anvil entirely—but not me. The hammer smashed into my left hand breaking two fingers. I was lucky I was not harmed more. I am left-handed and learned to write on the chalkboard with my right hand for the rest of the semester.) Relate the ideas of tightening a hammerhead by slamming the opposite end of the handle on a firm surface, with the bones of the human spine after jogging or even walking around. Interestingly, we are similarly a bit shorter at night. Ask your students to find a place in their homes that they can’t quite reach before going to bed—a place that is one or two centimeters higher than their reach. Then tell them to try again when they awake the next morning. Unforgettable, for you are likely instructing them to discover something about themselves they were not aware of! Newton’s Second Law of Motion Briefly review the idea of acceleration and its definition, and state that it’s produced by an imposed force. Write this as a F and give examples of doubling the force and the resulting doubling of the acceleration, etc. Introduce the ideas of net force, with appropriate examples—like applying twice the force to a stalled car gives it twice as much acceleration—three times the force, three times the acceleration. CHECK QUESTION: If one were able to produce and maintain a constant net force of only 1 newton on the Queen Mary 2 ocean liner, what would be its maximum speed? Give multiple choices for an answer: (a) 0 ms; (b) 1 ms; (c) less than 1 ms; (d) about 10 ms; (e) close to the speed of light! In the discussion that follows, the key concept is net force. Point out the enormous applied forces necessary to overcome the enormous water resistance at high speeds, to yield a net force of 1 newton; and the meaning of acceleration—that every succeeding second the ship moves a bit faster than the second before. This would go on seemingly without limit, except for relativistic effects which result in (e) being the correct answer. Newton’s second law can be rearranged to read F = ma. When the force on an object of mass m is due to gravity it is common to call the force weight. Then weight = ma = mg, where g is the acceleration of free fall. So we express the weight of objects as mg. In Chapter 9 we’ll refine our definition of weight. In these
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spacefaring times, the effects of gravity are not apparent, for example with inhabitants in an orbiting space station. We will see that a stricter way of looking at weight involves a support force. Astronauts in orbit are not without gravity—they’re without a support force, which is why they feel weightless. As we’ll see, gravity is alive and well above Earth’s atmosphere. Falling Objects Point out that although Galileo introduced the idea of inertia, discussed the role of forces, and defined acceleration, he never tied these ideas together as Newton did with his second law. Although Galileo is credited as the first to demonstrate that in the absence of air resistance, falling objects fall with equal accelerations, he was not able to say why this is so. The answer is given by Newton’s second law. SKIT: Hold a heavy object like a kilogram weight and a piece of chalk with outstretched hands, ready to drop them. Ask your class which will strike the ground first if you drop them simultaneously. Ask them to imagine you asking a bright youngster, who responds by asking to handle the two objects before giving an answer. Pretend you are the kid judging the lifting of the two objects. “The metal object is heavier than the chalk, which means there is more gravity force acting on it, it will accelerate to the ground before the chalk does.” Write the kids argument in symbol notation on the board, a F. Then go through the motions of asking another child, who responds with a good argument that takes inertia rather than weight into account. This kid says, after shaking the metal and the chalk back-and-forth in his or her hands, “The piece of metal is more massive than the chalk, which means it has more inertia, than the chalk, which tells us it will be harder to get moving than the chalk. So the chalk will race to the ground first, while the inertia of the metal causes it to lag behind.” Write this kid’s argument with, a 1/m. State that a beauty of science is that such speculations can be ascertained by experiment. Drop the weight and the chalk to show that however sound each child’s argument seemed to be, the results do not support either. Then bring both arguments together with a F/m, Newton’s second law. The comic strip “Backyard Physics” on page 38 of this manual makes this point. Relate your skit to the case of falling bricks, Figure 4.12, and the falling boulder and feather, Figure 4.13. Once these concepts are clear, ask how the bricks would slide on a frictionless inclined plane, then illustrate with examples such as the time required for a fully loaded roller coaster and an empty roller coaster to make a complete run. In the absence of friction effects, the times are the same. Cite the case of a massive limousine and a much lighter vehicle moving down a hill in the absence of friction. By now you are fielding questions having to do with air resistance and friction. (Avoid getting into the buoyancy of falling objects—information overload.) DEMONSTRATION: After you have made clear the cases with no friction, then make a transition to practical examples that involve friction—leading off with the dropping of sheets of paper, one crumpled and one flat. Point out that the masses and weights are the same, and the only variable is air resistance. Bring in the idea of net force again, asking what the net force is when the paper falls at constant speed. (If you left the Chapter 3 demo of the falling book and paper on top of it unexplained, reintroduce it here.) The way the book and paper drop demonstration is conducted reveals the difference between a new teacher, most eager to provide the explanation, and a seasoned teacher—who withholds that information until after students have pondered it. Be a professional, and show the demo only after you’re convinced that students have tried it!
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CHECK QUESTIONS: What is the acceleration of a feather that “floats” slowly to the ground? The net force acting on the feather? If the feather weighs 0.001 N, how much air resistance acts upward against it? [Acceleration is zero at terminal speed, and air resistance = weight = 0.001 N.] These questions lead into a discussion of the parachutists in Figure 4.15. When the decrease of acceleration that builds up to terminal velocity is clear, return to the point earlier about the huge and lighter vehicles moving down an incline, only this time in the presence of air resistance. Then ask whether or not it would be advantageous to have a heavy cart or a light cart in a soap-box-derby race. Ask which would reach the finish line first if they were dropped through the air from a high-flying balloon. Then consider the carts on an inclined plane. For your information, the terminal velocity of a falling baseball is about 150 kmh (95 mih), and for a falling Ping-Pong ball about 32 kmh (20 mih). Make the distinction between how fast something hits the ground and with what force it hits. Dropping a pebble on one foot, and a boulder on the other makes this clear. Although they both hit at the same speed, the heavier boulder elicits the ouch! A box titled Problem Solving ends this chapter. How to treat this will depend on the amount of time you want to invest. If many of your students will continue in science, it would be worthwhile to go over these simple problems in detail. It doesn’t take long to assess student interest in what I call “mini-problem solving.” The downside is over emphasis that limits course time for covering “rainbows and beyond.” So far, we’ve regarded a force as a push or a pull. We’ll consider a deeper definition of force in the next chapter. Onward!
Mascot Manny Mouse reminds us of that laws of physics are rules of nature itself. The rule a = F/m, perhaps the most important in mechanics, was the fundamental law that first got humans to the Moon and back. Unlike civil laws that are often ambiguous, the laws of physics are solid, and provide a base for clear and correct thinking about the world.
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ANSWERS AND SOLUTIONS FOR CHAPTER 4 READING CHECK QUESTIONS (COMPREHENSION) 4.1 Forces 1. A push or a pull is the most common type of force. 2. Acceleration and net force are proportional to each other, not equal to each other. 3. Your push and the force of friction have the same magnitude. 4.2 Friction 4. Once sliding, the push has the same magnitude as the force of friction. 5. Static friction is greater than sliding friction for the same object. 6. Friction does not vary with speed. 7. Yes, fluid friction varies with speed. 4.3 Mass and Weight 8. Mass is more fundamental than weight. Weight, not mass, varies with location. 9. The blanks are: mass; weight. 10. The blanks are: kilogram; newton. 11. A quarter-pound hamburger after it is cooked weighs about 1 newton. 12. The weight of a 1-kg brick is about 10 newtons. 13. Breaking the top string is due mainly to the ball’s weight. 14. Breaking the lower string is due mainly to the ball’s mass. 15. Acceleration is inversely proportional to mass, as specified by Newton’s second law. 4.4 Newton’s Second Law of Motion 16. The acceleration produced by a net force on an object is directly proportional to the net force, in the same direction as the net force, and inversely proportional to the mass of the object. 17. In equation form Newton’s second law is a = F/m. 18. The acceleration doubles. 19. The acceleration decreases to one half. 20. The acceleration will be unchanged. 21. The acceleration and net force are in the same direction. 4.5 When Acceleration Is g—Free Fall 22. Only one force, due to gravity, acts on an object in free fall. 23. The ratio of force to mass is g. 24. The ratio of force to mass for both is the same, g. 4.6 When Acceleration Is Less Than g—Non–Free Fall 25. The net force is 10 N. 26. The net force is 6 N; zero. 27. Speed and frontal area both affect the force of air resistance. 28. Acceleration is zero. 29. Yes, a heavier parachutist must fall faster for air resistance to balance weight. 30. The faster one encounters greater air resistance. THINK AND DO (HANDS-ON APPLICATIONS) 31. Relate how Newton followed Galileo, and so on. 32. The coin hits the ground first; when crumpled, both fall in nearly the same time; from an elevated starting point, the coin hits first. That’s because it has less frontal area.
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33. When the paper is on top of the dropped book, no air resistance acts on the paper because the book shields it from the air. So the paper and book fall with the same acceleration! 34. In all three kinds of motion they move in unison, in accord with a = F/m. 35. The spool will roll to the right! There is an angle at which it will not roll but slide. Any angle larger will roll the spool to the left. But pulled horizontally it rolls in the direction of the pull, whether the string is at the bottom or top of the spool spindle. PLUG AND CHUG (EQUATION FAMILIARIZATION) 36. Weight = (50 kg)(10 N/kg) = 500 N 37. Weight = (2000 kg)(10 N/kg) = 20,000 N 38. a =
Fnet
39. a =
Fnet
=
m
=
m
500 N 2000 kg
0.25 N
=
= 0.25 m/s2
kg
120,000 N 300,000 kg
=
0.4 N kg
= 0.4 m/s2
THINK AND SOLVE (MATHEMATICAL APPLICATIONS) 40. The net force is mg downward, 10 N (or more precisely, 9.8 N). 41. The net force is10 N − 2 N = 8 N (or more precisely 9.8 N − 2 N = 7.8 N). 42. Ten kilograms weighs about 100 N on Earth (weight = mg = 10 kg 10 m/s2 = 100 N, or 98 N if g = 9.8 m/s2). On the Moon the weight is 1⁄6 of 100 N = 16.7 N (or 16 N if g = 9.8 m/s2). The mass is 10 kg everywhere. 43. A 1-kg mass weighs 10 N, so 30 kg weigh 300 N. The bag can safely hold 30 kg of apples—if you don’t pick it up too quickly. 44. a = 45. a =
Fnet
=
200 N
=
6.0 m/s
m v
40 kg
=
5N kg
= 5 m/s2
= 5.0 m/s2 1.2 s t v 6.0 m/s F 46. a = net = = = 5.0 m/s2 1.2 s t m 47. Fnet = ma = (12 kg)(7.0 m/s2) = 84 kg·m/s2 = 84 N
48.
1 lb = 0.225 lb 4.45 N
(1 N )
49. The acceleration of each is the same: a =
F m
=
2N 2 kg
=
1N kg
= 1 m/s2. (Incidentally, from the
definition 1 N = 1 kg.m/s2, then 1 N/kg is the same as 1 m/s2.) 50. For the jet: a = 51. (a) a =
v t
=
F m
=
2(30,000 N )
9.0 m/s 0.2 s
30, 000 kg
= 2 m/s2.
= 45 m/s2 (b) F = ma = (100 kg)(45 m/s2) = 4500 N
52. (a) Force onthe bus, Ma. New acceleration =
same force new mass
..
=
Ma
M M + 5
=
5Ma
(5M + M )
=
5 = a 6 6M
5Ma
41
5 5 (b) New acceleration = a = 1.2 m/s 2 = 1.0 m/s2 6 6
(
)
THINK AND RANK (ANALYSIS) 53. (a) D, A = B = C (b) A = C, B = D 54. C, B, A 55. (a) A = B = C (b) C, A, B 56. (a) C, A, B (b) B, A, C THINK AND EXPLAIN (SYNTHESIS) 57. The forces acting horizontally are the driving force provided by friction between the tires and the road, and resistive forces—mainly air resistance. These forces cancel and the car is in dynamic equilibrium with a net force of zero. 58. No, weight is proportional to mass, but not equal to mass. 59. The force you exert on the ball ceases as soon as contact with your hand ceases. 60. Yes, as evidenced by its slowing down. A force opposite to its motion is acting—likely air resistance and friction between the ball and lane. 61. Constant velocity means zero acceleration, so yes, no net force acts on the motorcycle. But when moving at constant acceleration there is a net force acting on it. 62. No, inertia involves mass, not weight. 63. Items like apples weigh less on the Moon, so (a) there are more apples in a 1-pound bag of apples on the Moon. (b) Mass is another matter, for the same quantity of apples are in a 1-kg bag on Earth as on the Moon. 64. Buy by weight in Denver because the acceleration of gravity is less in Denver than in Death Valley. Buying by mass would be the same amount in both locations. 65. Shake the boxes. The box that offers the greater resistance to acceleration is the more massive box containing the sand. 66. When you carry a heavy load there is more mass involved and a greater tendency to remain moving. If a load in your hand moves toward a wall, its tendency is to remain moving when contact is made. This tends to squash your hand if it’s between the load and the wall—an unfortunate example of Newton’s first law in action. 67. Mass is a measure of the amount of material in something, not gravitational pull that depends on its location. So mass is the same for an astronaut and weight becomes zero due to no support force. 68. A massive cleaver is more effective in chopping vegetables because its greater mass contributes to its greater tendency to keep moving as the cleaver chops the food. 69. Neither the mass nor the weight of a junked car changes when it is crushed. What does change is its volume, not to be confused with mass and weight. 70. The scale reading will increase during the throw. Your upward force on the heavy object is transmitted to the scale. 71. The change of weight is the change of mass times g, so when mass changes by 2 kg, weight changes by about 20 N. 72. One kg of mass weighs 2.2 pounds at the Earth’s surface. If you weigh 100 pounds, for example, your mass is (100 lb)/(2.2 kg/lb) = 45 kg. Your weight in newtons, using the relationship weight = mg, is then (45 kg)(10 N/kg) = 450 N. 73. Since the crate remains at rest, the net force on it is zero, which means the force of friction by the floor on the crate will be equal and opposite to your applied force F. 74. (a) Her push is one force, the other is friction with the floor. (b) Neither force is greater, for the table is in dynamic equilibrium. Both forces have the same magnitude.
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(c) If she pushes harder so the sliding table gains speed, then her push is greater than the force of friction and the table is no longer in equilibrium. 75. Sliding down at constant velocity means acceleration is zero and the net force is zero. This can occur if friction equals the bear’s weight, which is 4000 N. Friction = bear’s weight = mg = (400 kg)(10 m/s2) = 4000 N. 76. The car has no tendency to resume its original twice-as-fast speed. Instead, in accord with Newton’s first law, it tends to continue at the speed it had when the engine stopped running, half speed, decreasing further over time due to air resistance and road friction. 77. Acceleration (slowing the car) is opposite to velocity (direction car moves). 78. Agree. Acceleration (slowing the car) is opposite to velocity (direction car moves). 79. Acceleration is the ratio force/mass (Newton’s second law), which in free fall is just weight/mass = mg/m = g. Since weight is proportional to mass, the ratio weight/mass is the same whatever the body weight. 80. Lifting the opponent decreases the force with which the ground supports him, and correspondingly decreases the force of friction he can muster. The reduced friction limits the opponent’s effectiveness. 81. (a) No, air resistance is also acting. Free fall means free of all forces other than that due to gravity. A falling object may experience air resistance; a freely falling object experiences only the force due to gravity. (b) Yes, although getting no closer to Earth, the satellite is falling (more about this in Chapter 10). 82. Velocity changes, but not acceleration. The velocity of the ascending coin decreases while its acceleration remains constant. 83. The only force on a tossed coin, except for air resistance, is mg. So the same mg acts on the coin at all points in its trajectory. 84. The acceleration at the top or elsewhere in free fall is g, 10 m/s2, downward. The velocity of the rock is momentarily zero while the rate of change of velocity is still present. Or better, by Newton’s second law, the force of gravity acts at the top as elsewhere; divide the net force by the mass for the acceleration of free fall. That is, a = Fnet/m = mg/m = g. 85. You explain the distinction between an applied force and a net force. It would be correct to say no net force acts on a car at rest. 86. When you drive at constant velocity, the zero net force on the car results from the driving force that your engine supplies against the friction drag force. You continue to apply a driving force to offset the drag force that otherwise would slow the car. 87. When the apple is held at rest the upward support force equals the gravitational force on the apple and the net force is zero. When the apple is released, the upward support force is no longer there and the net force is the gravitational force, 1 N. (If the apple falls fast enough for air resistance to be important, then the net force will be less than 1 N, and eventually can reach zero if the air resistance builds up to 1 N.) 88. High-speed grains of sand grazing Earth’s atmosphere burn up because of air friction. 89. Both forces have the same magnitude. This is easier to understand if you visualize the parachutist at rest in a strong updraft—static equilibrium. Whether equilibrium is static or dynamic, the net force is zero. 90. When anything falls at constant velocity such as raindrops, air resistance and gravitational force are equal in magnitude. 91. When a falling parachutist opens her chute she slows, which means she accelerates upward. 92. There are usually two terminal speeds, one before the parachute opens, which is faster, and one after, which is slower. The difference is mainly the areas presented to the air in falling. The large area by the open chute results in a slower terminal speed, enough for a safe landing. 93. Just before a falling body attains terminal velocity, there is still a downward acceleration because gravitational force is still greater than air resistance. When the air resistance builds up to equal the
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gravitational force, terminal velocity is reached. Then air resistance is equal and opposite to the gravitational force. 94. The terminal speed attained by the falling cat is the same whether falling 50 stories or 20 stories. Once terminal speed is reached, falling extra distance does not affect the speed. (The low terminal velocities of small creatures enables them to fall without harm from heights that would kill larger creatures.) 95. The sphere will be in equilibrium when it reaches terminal speed—which occurs when the gravitational force on it is balanced by an equal and opposite force of fluid drag. 96. Falling in the air is different than falling in a vacuum. The amount of air resistance the boulder encounters is tiny compared with its weight. Not so for the Ping-Pong ball that will reach terminal velocity before striking the ground below. 97. Air resistance is not really negligible for so high a drop, so the heavier ball does strike the ground first (Figure 4.16). But although a twice-as-heavy ball strikes first, it falls only a little faster, not twice as fast, which followers of Aristotle believed. Galileo recognized the small difference is due to air friction, and both would fall together when air friction is negligible. 98. The heavier tennis ball will strike the ground first for the same reason the heavier parachutist in Figure 4.15 strikes the ground first. Note that although the air resistance on the heavier ball is less relative to the ball’s weight, it is actually greater than the air drag that acts on the lighter ball. Why? Because the heavier ball falls faster, and air drag is greater at greater speed. 99. The answer is given in the equation a = F/m. As fuel is burned, the mass of the rocket becomes less. As m decreases and F remains the same, a increases! There is less mass to be accelerated as fuel is consumed. 100. Air resistance decreases the speed of a moving object. Hence the ball has less than its initial speed when it returns to the level from which it was thrown. The effect is easy to see for a feather projected upward by a slingshot. No way will it return to its starting point with its initial speed! 101. The time rising will be shorter than the time falling. Like a feather shot into the air by a slingshot, its time fluttering to the ground is visibly longer than the time going upward. THINK AND DISCUSS (EVALUATION) 102. The second law states the relationship between force and acceleration. If there is no net force, there is no acceleration. Newton’s first law is consistent with the second law, and can be considered a special case of the second law. 103. If only the upward force were acting, the book indeed would rise. But another force, that due to gravity, results in the net force being zero. 104. Yes, as illustrated by a ball thrown vertically into the air. Its velocity is initially upward, and finally downward, all the while accelerating at a constant downward g. 105. Neither a stick of dynamite nor anything else “contains” force. We will see later that a stick of dynamite contains energy, which is capable of producing forces when an interaction of some kind occurs. 106. No, an object can move in a curve only when a force acts. With no force, its path would be a straight line. 107. The only force that acts on a dropped rock on the Moon is the gravitational force between the rock and the Moon. Because the Martian atmosphere is very thin, the “air” drag on the rock is negligible. 108. A dieting person seeks to lose mass. Interestingly, a person can lose weight by simply being farther from the center of Earth—at the top of a mountain, for example. 109. Friction between the crate and the truck-bed is the force that keeps the crate picking up the same amount of speed as the truck. With no friction, the accelerating truck would leave the crate behind. 110. Note that 30 N pulls three blocks. To pull two blocks then requires a 20-N pull, which is the tension in the rope between the second and third block. The tension in the rope that pulls only the third block is therefore 10 N. (Note that the net force on the first block, 30 N − 20 N = 10 N, is the force needed to accelerate that block, having one-third of the total mass.) ..
44
111. The net force on the wagon, your pull plus friction, is zero. So F = 0. 112. When you stop suddenly, your velocity changes rapidly, which means a large acceleration of stopping. By Newton’s second law, this means the force that acts on you is also large. Experiencing a large force is what hurts you. 113. The force vector mg is the same at all locations. Acceleration g is therefore the same at all locations. 114. The force you exert on the ground is greater. The ground must push up on you with a force greater than the downward force of gravity to produce a resulting net force that is upward to accelerate you upward. 115. At the top of your jump your acceleration is g. Let the equation for acceleration via Newton’s second law guide your thinking: a = F/m = mg/m = g. If you said zero, you’re implying the force of gravity ceases to act at the top of your jump—not so! 116. Your second friend is correct about acceleration at the top not being zero. Although the velocity at the top of the ball’s path is zero, acceleration is not zero, as evidenced by continuing change (at the top and everywhere along his trajectory). The force of gravity, mg, also acts everywhere along the trajectory, which produces acceleration a = F/m = mg/m = g. 117. For a decreasing acceleration the increase in speed becomes smaller each second, but nevertheless, there’s greater speed each second than in the preceding second. 118. A sheet of paper presents a larger surface area to the air in falling (unless it is falling edge on), and therefore has a lower terminal speed. A wadded piece of paper has a smaller area and therefore falls faster before reaching terminal speed. 119. In each case the paper reaches terminal speed, which means air resistance equals the weight of the paper. So air resistance will be the same on each! Of course the wadded paper falls faster for air resistance to equal the weight. 120. For low speeds, accelerations are nearly the same because air drag is small relative to the weights of the falling objects. From a greater height, there is time for air resistance to build up and more noticeably show its effects. 121. Nowhere is her velocity upward. The upward net force on Nellie during the short time that air resistance exceeds the force of gravity produces a momentary upward net force and upward acceleration. This produces a decrease in her downward speed, which is nevertheless still downward. 122. If Dick and Jane fall separately, their frontal areas will be about the same but their individual weights will be different. Jane has less weight and will have a smaller terminal speed than Dick. Dick will reach the ground before Jane does. 123. Air resistance depends on both frontal area and speed. Both balls have the same frontal area, but the faster-falling lead-filled ball encounters more air resistance. 124. Open ended.
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CHAPTER 4 MULTIPLE-CHOICE PRACTICE EXAM As mentioned earlier, although answers and explanations are shown at the end of each chapter, going over each test item and emphasizing explanations can be a great part of your classroom delivery. 1. As mass is added to a cart pushed by a constant force, its acceleration (a) increases. (b) decreases. (c) remains constant. (d) gradually reaches zero. Answer: (b) Explanation: If we read Newton’s second law, a = F/m, we see that for a constant force F and increasing mass m, acceleration decreases. Let equations guide your thinking. 2. Compared with a 2-kg block of solid iron, a 4-kg block of solid iron has the same (a) mass. (b) volume. (c) weight. (d) none of the above Answer: (d) Explanation: Density is the same for both blocks, but isn’t an answer choice. So this is a “none of the above” question. If you answered incorrectly, you may have exam anxiety. 3. An object with a very small mass must also have a very small (a) weight. (b) volume. (c) size. (d) all of these. Answer: (a) Explanation: Mass and weight are directly proportional to each other, which makes (a) the correct answer. Volume and size depend on density. It should be obvious that mass and volume, choices (b) and (c) are incorrect choices. 4. If Marie pushes horizontally on a table, but not hard enough to get it moving, the friction force acting on the table is (a) zero. (b) equal to the weight of the table. (c) equal and opposite to her push. (d) none of the above Answer: (c) Explanation: The still-at-rest table is in equilibrium. Therefore, her push must be countered by an equal and opposite force of friction. If she pushes harder, the table will slide. If the table slides with constant velocity, then her push is enough to overcome friction. If she pushes even harder, the table will gain speed—it will accelerate. There’s good physics in this question. 5. A heavy block is suspended at rest by a vertical rope. When you raise the block at a constant velocity, the rope tension (a) is less than its weight. (b) equals its weight. (c) is greater than its weight. (d) is zero. Answer: (b) Explanation: “Constant velocity” is the key to this question. This means the raised block is in dynamic equilibrium, with tension equal to its weight. That is, tension up is equal and opposite to weight down, in accord with F = 0. If the block were accelerating, then a different story.
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6. A heavy rock and a light rock in free fall (zero air resistance) have the same acceleration. The reason the heavy rock doesn’t have a greater acceleration is that the (a) force due to gravity is the same on each. (b) air resistance is always zero in free fall. (c) inertia of both rocks is the same for both. (d) ratio of force to mass is the same for both. Answer: (d) Explanation: This nicely illustrates Newton’s second law—a = F/m, where F/m for both rocks is the same, g. This is NOT because of no air resistance, so choice (b) is not the reason. Neither the force due to gravity nor the inertia is the same for the two rocks, which eliminates choices (a) and (c). As illustrated on page 74 of the textbook, big F divided by big m gives the same acceleration as little F divided by little m, making choice (d) correct. (How Galileo would have loved seeing it that way!) 7. You drop a pillow off the edge of the tallest building on your campus. While the pillow falls, its speed (a) and acceleration both increase. (b) increases and its acceleration decreases. (c) and acceleration both decrease. (d) decreases and its acceleration increases. Answer: (b) Explanation: On your campus the pillow falls through air. Try to visualize it plowing through air as it gains speed. It’s important to note that the rate of gaining speed decreases while it falls, but not speed itself. As speed increases, so does air resistance. That reduces the net force (mg − R) on the pillow, which means decreasing acceleration. In no way can acceleration increase during the fall. (If it falls fast enough, air resistance may decrease to zero and the pillow would reach terminal speed.) In the special case of dropping in a vacuum, the pillow would gain speed at a constant acceleration, g. 8. A 200-N sack of onions falls from an airplane. When the air resistance of fall equals 200 N, the acceleration of the sack is (a) 0 m/s2. (b) 5 m/s2. (c) 10 m/s2. (d) infinite. Answer: (a) Explanation: Only two forces act on the falling sack; weight and air resistance. Since they are equal in magnitude, they cancel to zero and acceleration likewise cancels to zero. 9. The amount of air resistance that acts on a wingsuit flyer (and a flying squirrel) depends on the flyer’s (a) frontal area. (b) speed. (c) area and speed. (d) acceleration. Answer: (c) Explanation: The frontal area of a flyer’s body that encounters air as it moves certainly affects gliding, but so also does speed, which makes (c) the best answer. Acceleration may play a role, but a minor one compared with air and speed. So choice (c) is the best answer. 10. Toss a ball vertically upward and its acceleration at the top of its path is (a) zero. (b) 10 m/s2. (c) between zero and 10 m/s2. (d) dependent on the mass of the ball. Answer: (b) Explanation: Answering (a) is likely confusing acceleration with velocity. Although the velocity at the top of the path is zero, acceleration is not! Let a = F/m guide your thinking. At the top, the force of gravity F still pulls downward on the ball, and the ball certainly has mass m, so F/m = mg/m = g. Not zero! (This question is one that almost all get wrong the first time they encounter it, including the author.)
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5 Newton’s Third Law of Motion ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 5.1 Forces and Interactions 5.2 Newton’s Third Law of Motion Simple Rule to Identify Action and Reaction Defining Your System 5.3 Action and Reaction on Different Masses PRACTICING PHYSICS: TUG OF WAR 5.4 Vectors and the Third Law 5.5 Summary of Newton’s Three Laws
OPENING PHOTOS AND PROFILE The photos begin with Darlene Librero and Paul Doherty, my two dear friends who previously were at the Exploratorium. Darlene goes back to the earliest days at the Exploratorium when she worked with Frank Oppenheimer. Paul was the senior staff scientist since the 1980s, and passed away in 2018. His last book, coauthored with Cody Cassidy, is And Then You’re Dead, about individuals going against Nature’s rules and suffering bizarre consequences. Paul himself died of the all-too-common disease of cancer, just as he was preparing for retirement. He is missed by many. The second photo is of David Kagan, physics professor at Chico State University, whose passion is baseball. David is one of the experts I rely on for physics advice. The third photo is of Toby Jacobson, my protégé who pushes on the pair of scales with wife Bruna, who sat with his mom Susan in my Exploratorium Conceptual Physics class when he was 13 years old. Today, both with physics PhDs, Toby and Bruna continue in their love of physics. The touching photo is of my son Paul and his daughter Gracie. Hooray for Newton’s third law! Gracie again appears in the Part Six opener. The profile honors physicist Paul Doherty.
INTRODUCTION So far in the textbook, a force is seen as a push or a pull. Newton’s third law better defines it—as part of an interaction between one body and another. As the baseball and Dave Kagan’s bat attest, you cannot exert a force on something—unless (pause), that something exerts an equal and opposite force on you. So you can’t hit a ball unless the ball hits back. You can’t exert a force on the floor when you walk, unless the floor exerts the same amount of force back on you, etc. In discussing action and reaction, emphasize the word “unless” and follow it with a pause. Also emphasize the word “between,” as with forces between Earth and the Moon. Of Newton’s laws, his third law is the most misunderstood. It is not “what goes around comes around” nor “what you sow is what you reap.” These are not examples of Newton’s third law. Confront this misconception early so that students will be clear that the third law, like the first and seconds laws, is about force, and that action-reaction forces act simultaneously. This chapter concludes with a focus on vectors. Trigonometry, no. For resultants for pairs of vectors, the parallelogram rule, yes! Vector components are also treated, which will be needed when projectiles are covered in Chapter 10. Treatment of vectors continues in the Practicing Physics book.
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SUPPLEMENTS Practicing Physics • Action and Reaction Pairs • Force and Velocity Vectors • Interactions • Force Vectors and the Parallelogram Rule • Vectors and the Parallelogram Rule • Force-Vector Diagrams • Velocity Vectors and Components • More on Vectors Next-Time Questions • Reaction Forces • Scale Reading • Tug of War 2 • No-Recoil Cannon • Suspended Nellie 2
• Leaning Tower of Pisa Drop • Apple on a Table • Atwood Pulley • Tug of War 1 • Tug of War 3 • Airplane in the Wind • Accelerating Boulder • Suspended Nellie 1 • Force to Slow a Car • Punching Force
Hewitt-Drew-It! Screencasts • Newton’s Third Law • Newton’s Laws Problem Laboratory Manual • The Force Mirror—Quantitative Observations of Force Pairs (Activity) • Blowout!— A Summary of Newton’s Three Laws (Activity) Problem Solving in Conceptual Physics Sample Problems and more, also with optional section on trigonometry instruction.
SUGGESTED PRESENTATION Forces and Interactions Hold a piece of tissue paper at arm’s length and ask if the heavyweight champion of the world could hit the paper with 50 pounds of force. Ask your class to check their answer with their neighbors. Then don’t give your answer. Instead, continue with your presentation. Tell your class, “I can’t touch you, without you touching me in return—I can’t nudge this chair without the chair in turn simultaneously nudging me—I can’t exert a force on a body without that body in turn exerting a force on me.” In all these cases of contact there is a single interaction between two things—contact requires a pair of forces, whether they be slight nudges or great impacts, between two things and, simultaneously. This is Newton’s third law of motion. Call attention to the examples of Figure 5.7. Getting back to the boxer: When hitting a massive punching bag, his gloved fist hits the bag (and dents it) while the bag hits back on the boxer’s fist (and stops its motion). The pair of forces involved in hitting the bag can be huge. But when the boxer hits a piece of tissue paper in midair, the pair of forces is tiny. In both cases, the boxer’s fist can only exert as much force on the bag or the tissue paper as they can exert on the fist. A great Next-Time Question carries this idea further in Punching Force. Newton’s Third Law of Motion Extend your arm horizontally and show the class that you can bend your fingers upward only very little. Show that if you push with your other hand, and thereby apply a force to them, or have a student do the same, they will bend appreciably more. Then walk over to the wall and show that the inanimate wall does the same (as you push against the wall). State that everybody will acknowledge that you are pushing on the wall, but only a few realize the fundamental fact that the wall is simultaneously pushing on you also—as evidenced by your bent fingers.
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Do as Linda E. Roach does and place a sheet of paper between the wall and your hand. When you push on the paper, it doesn’t accelerate—evidence of a zero net force on the paper. You can explain that in addition to your push, the wall must be pushing just as hard in the opposite direction on the paper to produce the zero net force. Linda recommends doing the same with an inflated balloon, whereupon your class can easily see that both sides of the balloon are squashed. CHECK QUESTION: Identify the action and reaction forces for the case of a bat striking a ball. Do the same for a ball striking a bat. As Newton said, which force is called action and which is called reaction is arbitrary. Defining Your System Discuss the different systems of orange and apple as in Figures 5.8–5.11. This is also treated in the Hewitt-Drew-It! Screencast, Newton’s Third Law. Ask students to identify action and reaction parts of the systems of Figures 5.14–5.18. That’s my wife Lil and me in Figure 5.19. And continuing with the same important concept of “you can’t touch without being touched,” my brother Steve and his daughter Gretchen do the same in Think and Explain 39 in the end-of-chapter material. A prior photo of them, when Gretchen was a child, occurred in previous editions. Again, my son Paul and his daughter Gracie do the same in photo four of the chapter opening photos. Action and Reaction on Different Masses Discuss walking on the floor in terms of the single interaction between you and the floor, and the pair of action and reaction forces that make up this interaction. Contrast this to walking on frictionless ice, where no interaction occurs. Ask how one could get off a pond of frictionless ice. Make the answer easy by saying one has a massive brick in hand. By throwing the brick there is an interaction between the thrower and the brick. The reaction to the force on the brick, the recoiling force, sends one to shore. Or without such a convenient brick, one has clothing. And one has air in the lungs. One could blow air in jet fashion. Exhale with the mouth facing away from shore, but be sure to inhale with the mouth facing toward the shore. CHECK QUESTION: Identify the force that pushes a car along the road. Interestingly, the force that pushes a car is provided by the road itself. How? The tires push on the road, action, and the road pushes on the tires, reaction. Since the tires connect to the car, we can say roads push cars along. Tell your class to imagine you’re at a car race and say to other spectators, “Hey, look at that road pushing the cars along!” The photo of the pushed bricks in the road in Figure 5.20 is clear evidence that the bricks have been pushed, as they push the tires of automobiles! This nicely elicits class discussion. Another point to make is time. If you pull on a shopping cart today and the cart pulls on something else tomorrow, the two pulls do not constitute an action-reaction pair. An action-reaction pair of forces occurs at the same time. When you pull on a cart, the cart pulls back on you simultaneously. These simultaneous forces constitute a single interaction. Most people say that the Moon is attracted to Earth by gravity. Ask most people if Earth is also attracted to the Moon, and if so, which pulls harder, Earth or the Moon? You’ll get mixed answers. Physicists think differently than most people on this topic. Rather than saying the Moon is attracted to Earth by gravity, a physicist would say there is an attractive gravitational force between Earth and the Moon. Asking if the Moon pulls as hard on Earth as Earth pulls on the Moon is similar to asking if the distance between New York and Los Angeles is the same as the distance between Los Angeles and New York.
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ANSWERS AND SOLUTIONS FOR CHAPTER 5 READING CHECK QUESTIONS (COMPREHENSION) 5.1 Forces and Interactions 1. A pair of forces are required for an interaction. 2. No, your push is part of an interaction with the wall. You push it; it pushes back. 5.2 Newton’s Third Law of Motion 3. He can’t exert as much force on the tissue paper because the paper can’t react with as much force. 4. Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. To every action there is always an opposed reaction. 5. Reaction force is the stake against the hammer. 6. At the same time. Action and reaction forces occur simultaneously. 7. The force of friction on the apple’s feet provides the net force on the orange-apple system. 8. Yes, an external net force is required to accelerate the system. 9. If two opposite and equal forces act on the ball, the net force on it is zero and it will not accelerate. 5.3 Action and Reaction on Different Masses 10. Yes, you pull upward with the same amount of force on Earth. 11. The different accelerations are due to different masses. 12. The force that propels a rocket is the exhaust gases pushing back on the rocket. 13. A helicopter gets its lifting force by pushing air downward, in which case the reaction is the air pushing the helicopter upward. 14. No, you cannot touch without being touched with the same amount of force. 5.4 Vectors and the Third Law 15. Vector resolution is the process of determining the components of a vector. 16. The magnitude of the normal vector decreases. 17. The friction force has the same magnitude, with the sum of all forces being zero. 18. Moving upward, the vertical component of velocity decreases. The horizontal component remains constant, in accord with Newton’s first law. 5.5 Summary of Newton’s Three Laws 19. Inertia; acceleration; action and reaction. 20. Newton’s third law focuses on interactions. THINK AND DO (HANDS-ON APPLICATIONS) 21. Your hand will be pushed upward, a reaction to the air it deflects downward. 22. No, each finger will experience the same amount of force. 23. Of course this can be done without a smartphone. Answers will vary. 24. String tensions should be equal when the weight is suspended at the middle. For steeper angles, string tensions should be greater than for less-steep angles. 25. Open ended. PLUG AND CHUG (EQUATION FAMILARIZATION)
(100 + 100 ) = 141 km/h 27. R = (4.0 + 3.0 ) = 5 units 26. R =
2
2
2
2
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THINK AND SOLVE 28. R =
(200 + 80 ) = 215 km/h 2
29. (a) a =
2
25 m/s v = = 500 m/s2 0.05 s t
(b) F = ma = (0.003 kg)(500 m/s2) = 1.5 N, which is about 1⁄3 pound. (c) By Newton’s third law, the same amount, 1.5 N.
Thewall pushes on you with 30 N. a = 30. 31. a =
F , where F = m
(3.0 + 4.0 ) 2
2
THINK AND RANK (ANALYSIS) 32. A = B = C 33. (a) A, B,C (b) B, C 34. B, A, C 35. (a) B, A, C (b) C, B, A 36. B, A, C 37. A
30 N F = = 0.5 m/s2 m 60 kg 5N F = 5 N. So a = = = 2.5 m/s2. m 2.0 kg
THINK AND EXPLAIN (SYNTHESIS) 38. (a) Action: hammer hits nail. Reaction: nail hits hammer. (b) Action: Earth pulls down on the book. Reaction: book pulls up on Earth. (c) Action: helicopter blade pushes air downward. Reaction: air pushes blade upward. (In these examples, action and reaction may be reversed—which is called which is unimportant.) 39. In accord with Newton’s third law, Steve and Gretchen are touching each other. One may initiate the touch, but the physical interaction can’t occur without contact between both Steve and Gretchen. Indeed, you cannot touch without being touched! 40. No, for each hand pushes equally on the other in accord with Newton’s third law—you cannot push harder on one hand than the other. 41. In either order: (1) Earth’s pull on apple (action), and apple’s pull on Earth (reaction). (2) Hand pushes apple upward (action), and apple pushes hand downward (reaction, the normal force on the ball). 42. Only one pair acts while falling in absence of air drag: Earth pulls down on apple, and apple pulls upward on Earth. 43. (a) Action: Earth pulls you downward. Reaction: you pull Earth upward. (b) Action: you pat tutor’s back. Reaction: tutor’s back pats you. (c) Action: wave hits shore. Reaction: shore hits wave. 44. (a) Action: bat hits ball. Reaction: ball hits bat. (b) Action: Earth pulls down on ball in flight. Reaction: ball pulls up on Earth. Also, action: air pushes on ball, and reaction: ball pushes on air. 45. When the ball exerts a force on the floor, the floor exerts an equal and opposite force on the ball— hence bouncing. The force of the floor on the ball provides the bounce. 46. The billions of force pairs are internal to the book, and exert no net force on the book. An external net force is necessary to accelerate the book. 47. The friction on the crate is 200 N, which cancels your 200-N push on the crate to yield the zero net force that accounts for the constant velocity (zero acceleration).
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48. No, although P and −P are equal and opposite forces, they do not make an action-reaction pair, because they both act on the same object, the book—a no-no for action and reaction. 49. (a) Yes, for the man is in equilibrium. (b) No, because the two forces act on the same body. (c) Action: Earth pulls down on man (by gravity). Reaction: man pulls up on Earth. Here we see the action-reaction pair of forces acting on different objects—man and Earth. 50. The four action-reaction pairs of forces (in any order). 1. You push book; book pushes you. 2. Table exerts force of friction on book; book exerts force of friction on table. 3. Weight of book pushing down on table; table pushing up on weight of book (the normal force) 4. Book pulled down by Earth; Earth pulled upward by book. 51. If action is mg pulls down on ball (by gravity) then (a), reaction is ball pulls upward on Earth with equal and opposite mg. (b) At the top of its path only mg acts on the ball, with acceleration = mg/m = g. (Not zero!) 52. When you pull up on the handlebars, the handlebars simultaneously pull down on you. This downward force is transmitted to the pedals. 53. When the climber pulls the rope downward, the rope simultaneously pulls the climber upward—the direction desired by the climber. 54. When the barbell is accelerated upward, the force exerted by the athlete is greater than the weight of the barbell (the barbell, simultaneously, pushes with greater force against the athlete). When acceleration is downward, the force supplied by the athlete is less. 55. When you push the car, you exert a force on the car. When the car simultaneously pushes back on you, that force is on you—not the car. You don’t cancel a force on the car with a force on you. For cancellation, the forces have to be equal and opposite and act on the same object. 56. The strong man can exert only equal forces on both cars, just as your push against a wall equals the push of the wall on you. Likewise for two walls, or two freight cars. Since their masses are equal, they will undergo equal accelerations and move equally. 57. As in the preceding exercise, the force on each cart will be the same. But since the masses are different, the accelerations will differ. The twice-as-massive cart will undergo only half the acceleration of the less massive cart. 58. In accord with Newton’s third law, the force on each will be of the same magnitude. But the effect of the force (acceleration) will be different for each because of the different masses. The more massive truck undergoes less change in motion than the Civic. 59. Both will move toward each other. Scott’s pull on the rope is transmitted to Kjell, causing him to accelerate toward Scott. By Newton’s third law, the rope pulls back on Scott, causing him to accelerate toward Kjell. 60. The winning team pushes harder against the ground. The ground then pushes harder on them, producing a net force in their favor. 61. The tension in the rope is 250 N. With no acceleration, each must experience a 250-N force of friction via the ground. This is provided by pushing against the ground with 250 N. 62. No, this cannot be done. The net force on the rope is zero, meaning a tension greater than zero is the same on both ends, in accord with Newton’s third law. 63. The forces on each are the same in magnitude, and their masses are the same, so their accelerations will be the same. They will slide equal distances of 6 meters to meet at the midpoint. 64. The writer apparently didn’t know that the reaction to exhaust gases does not depend on a medium for the gases. A gun, for example, will kick if fired in a vacuum. In fact, in a vacuum there is no air drag and a bullet or rocket operates even better. 65. The resultant speed is indeed 5 m/s. The resultant of any pair of 3-unit and 4-unit vectors at right angles to each other is 5 units. This is confirmed by the Pythagorean theorem; a 2 + b2 = c2 gives 32 + 42 = 52. (Or, 32 + 4 2 = 5.)
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66. Again, from the Pythagorean theorem; a2 + b2 = c2 gives 32 + 42 = 52. (Or, 32 + 4 2 = 5.) So the boat has a speed of 5 m/s. 67. Again, from the Pythagorean theorem; a2 + b2 = c2 gives 1202 + 902 = 1502. (Or, 120 2 + 90 2 = 150.) So the groundspeed is 150 km/h. 68. (a) The other vector is upward as shown.
(b) It is called the normal force. 69. (a) As shown.
(b) Yes, because the stone is in equilibrium. 70. (a) As shown.
(b) Upward tension force is greater, resulting in an upward net force. 71. As shown.
72. The acceleration of the stone at the top of its path, or anywhere where the net force on the stone is mg, is g, downward. 73. (a) Weight and normal force only. (b) As shown.
74. (a) As shown.
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(b) Note the resultant of the two normal forces is equal and opposite to the stone’s weight. 75. Vector f will have the same magnitude as the vector sum of mg and N. If f is less, then a net force acts on the shoe and it accelerates down the incline. 76. The magnitudes of mg and N will be equal. 77. When the rope is vertical, S is zero. If the rope were vertical, S would be at an angle such that its vertical component would be equal and opposite to mg. 78. No horizontal force acts on the ball so the initial horizontal component of velocity remains constant as the ball moves through the air in accord with Newton’s first law of inertia. 79. A hammock stretched tightly has more tension in the supporting ropes than one that sags. The tightly stretched ropes are more likely to break. 80. The tension will be greater for a small sag. That’s because large vectors in each side of the rope supporting the bird are needed for a resultant that is equal and opposite to the bird’s weight (easily seen via a parallelogram). 81. By the parallelogram rule, the tension is less than 50 N.
82. Both forces are equal in magnitude. The two forces are tension in the vine and that due to gravity, both equal and opposite in keeping with the zero net force of equilibrium. 83. By the parallelogram rule, the tension is greater than 50 N.
THINK AND DISCUSS (EVALUATION) 84. The four examples are of cause and effect, not pairs of forces between different objects. Newton’s third law refers to forces acting at the same time. The four examples are misconceptions of Newton’s third law. 85. Yes, in accord with Newton’s third law! Don’t confuse the size of the force with the different consequences of that same-sized force. The fate of the fly is not comparable to the slowing of the flyswatter. 86. No, she could only exert a force on the wall if the wall simultaneously exerts a force of equal strength back on her. For example, if the wall suddenly turned to vapor, the interaction couldn’t occur, and she’d topple over. 87. To climb upward means pulling the rope downward, which moves the balloon downward as the person climbs. You can’t pull down on something without that something pulling up. 88. No, for the force the apple exerts on the orange must be met with an equal and opposite force between the apple’s feet and the supporting surface. No force on ice means no ability to exert the force shown. 89. Going straight across can be represented with a vector pointing directly to the opposite side of the river. The flow of the river can be represented by a vector perpendicular to the swimming vector. The resultant of the two vectors is a diagonal line drawn from the two starting points. 90. Agree with your first friend. However strong her push, the push from the tree is the same, in accord with Newton’s third law. 91. No, for in action-reaction fashion, the tail also wags the dog. How much depends on the relative masses of the dog’s body and its tail. ..
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92. The answer is given in the equation a = F/m. As fuel is burned, the mass of the rocket becomes less. As m decreases but F remains the same, a increases! There is less mass to be accelerated as fuel is consumed. 93. Action: your foot against the ball. Reaction: the ball against your foot, whether or not these forces are affected by low pressure in the ball. Whatever the pressure, both forces have the same magnitude in accord with Newton’s third law. 94. Yes, it’s true. Earth can’t pull you downward without you simultaneously pulling Earth upward. The acceleration of Earth is negligibly small, and unnoticed due to its enormous mass. 95. The scale will read 100 N, the same as if one end were tied to a wall instead of to the 100-N hanging weight. Although the net force on the system is zero, the tension in the rope within the system is 100 N, as would show on the scale reading. 96. The tires cannot exert any more force on the surface than the surface can exert on the tires. Sand, gravel, and snow do not provide the needed reaction to the tires acting on the surface. Road material flies backward because tires push backward. 97. Yes, a baseball exerts an external force on the bat, opposite to the bat’s motion. This external force decelerates the oncoming bat. 98. The rapid deceleration of the speeding ball on the player’s glove produces the force on the player’s glove. In this sense, deceleration produces force (cause and effect can sometimes be a matter of interpretation). 99. The forces do not cancel because they act on different things—one acts on the horse, and the other acts on the wagon. It’s true that the wagon pulls back on the horse, which prevents the horse from running as fast as it could without the attached wagon. But the force acting on the wagon (the pull by the horse minus friction) divided by the mass of the wagon, produces the acceleration of the wagon. To accelerate, the horse must push against the ground with more force than it exerts on the wagon and the wagon exerts on it. So tell the horse to push backward on the ground. 100. Tension would be the same if one end of the rope were tied to a tree. If two horses pull in the same direction, tension in the rope (and in the strongman) is doubled.
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CHAPTER 5 MULTIPLE-CHOICE PRACTICE EXAM 1. Earth exerts a force on the Moon. The Moon exerts a force on Earth. These two forces make up (a) a single interaction. (b) two interactions. (c) multiple interactions. (d) none of the above Answer: (a) Explanation: In nature no single force exists in isolation. Every force has a reaction counterpart force. Earth cannot pull on the Moon unless the Moon pulls with an equal and opposite force on Earth. The Earth-Moon pair of forces make up a single interaction. 2. When you push an ice cube with a 0.5-N force, the ice cube (a) accelerates at 10 m/s2. (b) resists being pushed with its own 0.5-N force. (c) will likely not move. (d) pushes on you with a 0.5-N force. Answer: (d) Explanation: This involves Newton’s laws. Since the mass of the ice cube isn’t given, choice (a) that might involve Newton’s second law, a = F/m, is eliminated. Choice (c) is nonsense. Newton’s third law makes sense, you pushing on ice and ice pushing on you, which unlike confusing choice (b) makes (d) the clearest and best answer. 3. A karate chop delivers a force of 3000 N to a board that breaks. The force that the board exerts on the hand during this event is (a) less than 3000 N. (b) 3000 N. (c) greater than 3000 N. (d) More information is needed. Answer: (b) Explanation: This is straight-forward example of Newton’s third law (A acts on B, B reacts on A), or in this case, hand acts on board, board reacts on hand. Without this knowledge, choices (a, c, and d) might be popular. We also distinguish between a force and the effect of that force. Although the board wouldn’t be hurt, the same force on the hand may hurt a lot. 4. The force that propels a rocket is provided by (a) gravity. (b) Newton’s laws of motion. (c) its exhaust gases. (d) the atmosphere against which the rocket pushes. Answer: (c) Explanation: Think Newton’s third law: Rocket pushes on exhaust gases, exhaust gases push on rocket, making choice (c) a better answer than (a, b). Choice (d) is commonly thought to be correct by those who lack an understanding of Newton’s third law. 5. A flower pot rests on a table. If the action force is Earth pulling downward on the pot, the reaction force is the (a) table pushing upward on the pot. (b) pot pulling upward on Earth. (c) pot pulling upward on both the table and Earth (d) none of the above Answer: (b) Explanation: Follow the rule: A acts on B, B acts on A. The table is not part of the action-reaction pair of forces, so choice (a) is bogus. Also the pot pulling up on the table choice (c) is bogus. Only choice (b) is correct. Note how central the (A on B), (B back on A) rule is!
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6. We cannot exert a force on a wall (a) if the wall resists our push. (b) unless we put our mind to it. (c) unless the wall simultaneously exerts the same amount of force on us. (d) all of the above Answer: (c) Explanation: More (A on B), (B back on A) again! When you push on a wall you interact with the wall. There then exists an action-reaction pair of forces: You pushing wall; wall pushing you, simultaneously, as in choice (c). 7. Girls wearing gym shoes win in a tug-of-war against boys wearing socks on a smooth floor because the girls are able to (a) exert more tension on their rope. (b) exert more force against the slippery floor. (c) enjoy a psychological advantage over boys. (d) none of the above Answer: (b) Explanation: In a tug-of-war, rope tension is the same for both sides. What is different is the force the participants exert on the floor. Gym shoes certainly can exert more force on a slippery floor than socks, so the girls win the contest. Eyebrows raised because gender is mentioned here? Hopefully not. 8. A huge GMC truck and a Honda Civic traveling at the same speed collide head-on. The collision force is greatest on the (a) Honda. (b) GMC. (c) same on both (d) cannot be answered without more information Answer: (c) Explanation: A single interaction occurs when the vehicles collide; the force of the truck on the car and the force of the car on the truck. Both action and reaction forces are equal in magnitude. Equal forces does not mean equal damage! When you swat a fly, the force on the fly and the equal and opposite force on your flyswatter may be equal, but the accelerations of each are drastically different. Likewise with the two vehicles. 9. A boxer’s punch hits with more force when striking a (a) heavier opponent. (b) lighter opponent. (c) same either way. (d) none of the above Answer: (a) Explanation: Think of the boxer hitting a heavy punching bag and hitting a paper napkin. For the same punch, the force against the bag greatly exceeds the force on the napkin. So it is with boxing opponents. Force is greater when interacting with greater mass. Many confronted with this question for the first time get it wrong! 10. The speed of an airplane caught in a right-angle crosswind (a) decreases. (b) remains unchanged. (c) increases. d) cannot be estimated. Answer: (c) Explanation: Think vectors: Two velocity vectors at right angles to each other produce a resultant greater than either of the two vectors, just as the diagonal of a rectangle is longer than either of its two sides. The airplane will travel faster in a right-angle crosswind. In crosswinds that are not at right angles, the resulting speed can be less. The two motions in this question are best represented with vectors.
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6 Momentum ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 6.1 Momentum 6.2 Impulse 6.3 Impulse Changes Momentum Case 1: Increasing Momentum Case 2: Decreasing Momentum Over a Long Time Case 3: Decreasing Momentum Over a Short Time 6.4 Bouncing 6.5 Conservation of Momentum CONSERVATION LAWS 6.6 Collisions PROBLEM SOLVING 6.7 More Complicated Collisions
OPENING PHOTOS AND PROFILE Physics star Derek Muller rises to the occasion in the first of the photos that opens this chapter. And the profile is of Derek also. The second photo is of friend from school days, physics teacher Howie Brand. The physics he shows applies nicely to the Pelton wheel. The last photo is of my grandson Alex on his first skateboard. We’ll see him again, a bit older, with a one-wheel skateboard in Figure 8.48.
INTRODUCTION This chapter begins where Chapter 5 leaves off. Newton’s second and third laws lead directly to momentum and its conservation. We emphasize the impulse-momentum relationship with many selected examples to engage student interest. In your classroom, I suggest using the exaggerated symbol technique as shown in Figures 6.5, 6.6, and 6.8. Draw a comparison between momentum conservation and Newton’s third law in explaining examples such as rocket propulsion. You might point out that either of these is fundamental—i.e., momentum conservation may be regarded as a consequence of Newton’s third law, or equally, Newton’s third law may be regarded as a consequence of momentum conservation. Interesting fact: The time of contact for a tennis ball on a racquet is about 5 milliseconds, whether or not a player “follows through.” The idea that follow-through in tennis, baseball, or golf appreciably increases the duration of contact seems useful in getting the point of extended time across, but it is not supported by recent studies. Follow-through is more important in applying maximum force to supply the impulse. The swinging ball apparatus (Newton’s cradle) shown in the sketch is popular for demonstrating momentum conservation. But any thorough analysis of it ought to be postponed to the next chapter when energy is treated. This is because the question is often raised, “Why cannot two balls be raised and allowed to swing into the array, and one ball emerge with twice the speed?” Be careful. Momentum would indeed be conserved if this were the case. But the case with different numbers of balls emerging never happens. Why? Because energy would not be conserved. For the two balls-one-ball case, the KE after would be twice as much as the KE before impact. KE is proportional to the square of the speed, and the conservation of both momentum and KE cannot occur unless the numbers of balls for collision and ejection are the same. Consider postponing this demo until the next chapter. We’ll return to Newton’s cradle in Chapter 26, where it serves as an analogy for photons of light passing through glass—the photon that emerges, like
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the balls in Newton’s cradle, is not the photon that initiated the cascade. Fascinating physics—at a somewhat later time, that I consider a highlight to the textbook! A system is not only isolated in space, but in time also. When we say that momentum is conserved when one pool ball strikes another, we mean that momentum is conserved during the brief duration of interaction when outside forces can be neglected. After the interaction, friction quite soon brings both balls to a halt. So when we isolate a system for purposes of analysis, we isolate it both in space and in time. You may want to assign an “Egg Drop” experiment. Students design and construct a case to hold an egg that can and will be dropped from a three- or four-story building without breaking. The design cannot include means to increase air resistance, so all cases should strike the ground with about the same speed. By requiring the masses of all cases to be the same, the impulses of all will be the same upon impact. The force of impact, of course, should be minimized by maximizing the time of impact. Or do as Peter Hopkinson does (Think and Explain 68), and simply have students toss eggs into cloth sheets, suspended so the eggs don’t hit the floor after impact. Either of these projects stir considerable interest, both for your students and others who are not yet taking your class. In 2009, 40-year-old Paul Lewis from the UK survived a 10,000-foot skydiving fall after his parachute failed to open. Amazingly, he landed on the roof of an aircraft hanger that broke his fall and flexed sufficiently to reduce impact. That’s wonderful Ft = ∆mv in action! An air track is useful in demonstrating conservation of momentum. An economy air track is available from Arbor Scientific (P4-2710). This chapter is important in its own right, and it serves as a foundation for the concept of energy in the next chapter.
SUPPLEMENTS Practicing Physics • Changing Momentum
• Systems
Next-Time Questions • Car Crash • Ball Catch
• Ice Sail Craft • Lovebug Splat
Hewitt-Drew-It! Screencasts •Momentum • Fish-Lunch Problem •Freddy-Frog Momentum Problem
• Conservation of Momentum
Laboratory Manual • Bouncy Board—Impact Time and Impact Force (Demonstration) • Things That Go Bump—Sticking and Bouncing in Collisions (Demonstration) Problem Solving in Conceptual Physics There are many problems on impulse, momentum, and the impulse-momentum relationship.
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SUGGESTED PRESENTATION Momentum Begin by stating that there is something different between a huge truck and a skateboard—they each have a different inertia. And that there is also something different about a moving huge truck and a moving skateboard—they have different momenta. Define and discuss momentum as inertia in motion. CHECK QUESTION: After stating that a huge truck will always have more inertia than an ordinary skateboard, ask if a huge truck will always have more momentum than a skateboard. [Only when mv for the truck is greater than mv for the skateboard.] Call attention to the supertanker shown in Figure 6.2, and discuss why such huge ships normally cut off their power when they are 25 or so kilometers from port. Because of their huge momentum (due mostly to their huge mass), about 25 km of water resistance are needed to bring them to a halt. Impulse and Momentum Derive the impulse-momentum relationship. In Chapter 3 you defined acceleration as a = Δv/t (really t, but you likely used t as the “time interval”). Later in Chapter 4 you defined acceleration in terms of the force needed, a = F/m. Now simply equate a = a, or F/m = Δv/t. With simple rearrangement you have Ft = mv (as in the footnote in the textbook on page 110). Then choose your examples in careful sequence: First, those where the objective is to increase momentum—pulling a slingshot or arrow in a bow all the way back, the effect of a long cannon for maximum range; driving a golf ball. Second, those examples where small forces are the objective when decreasing momentum—pulling your hand backward when catching a ball, driving into a haystack versus a concrete wall, falling on a surface with give (thick mat) versus a rigid surface. Then lastly, those examples where the objective is to obtain large forces when changing momentum—such as karate and boxing. Point of confusion: It’s often thought that in boxing, one “follows through,” whereas in karate one “pulls back.” But this is not so—a karate expert does not pull back upon striking the target. He or she strikes in such a way that the hand is made to bounce back, yielding up to twice the impulse to the target (just as a ball bouncing off a wall delivers nearly twice the impulse to the wall than if it stuck to the wall). CHECK QUESTION: Why is falling on a wooden floor in a roller rink less dangerous than falling on the concrete pavement? [Superficial answer: Because the wooden floor has more “give.”] Emphasize that this is the beginning of a complete answer—one that is prompted if the question is reworded as follows: Why is falling on a floor with more give less dangerous than falling on a floor with less give? [Answer: Because the floor with more give allows a greater time for the impulse that reduces the momentum of fall to zero. The greater time occurs because the change in momentum means less force.] The loose coupling between railroad cars (Think and Discuss 100) makes a good classroom topic. Discuss the importance of loose coupling in bringing a long train initially at rest up to speed, and its importance in braking the train as well. In effect, the time factor in the impulse is extended. The force needed to produce motion is therefore decreased. (I compare this to taking a series of courses in proper sequence, rather than all at once when for sure one’s wheels would spin.) Conservation of Momentum Distinguish between external and internal forces and progress to the conservation of momentum. Stress that no change in momentum can occur in the absence of an external net force.
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DEMONSTRATION: Show momentum conservation with an air-track performance. Doing so can be the focus of your classroom presentation. Defining Your System Momentum is not conserved in a system that experiences an external net force. The momentum of a system is conserved only when no external impulse is exerted on the system. As the example of the girl jumping from Earth’s surface suggests, momentum is always conserved if you make your system big enough. The momentum of the universe is without change. The numerical example of lunchtime for the fish in the Problem Solving box on page 117 should clarify the vector nature of momentum—particularly for the case of the fish approaching each other. Reinforcing this should be helpful—as with Think and Solve 43, and Think and Rank 51. Bouncing When discussing bouncing, tell how Lester Pelton, the inventor of the Pelton water wheel, made a fortune from applying some simple physics to the old paddle wheels. Fortunately, he patented his ideas and was one of the greatest financial beneficiaries of the California Gold Rush era. Bouncing does not necessarily increase impact force. That depends on impact time. Point out that bouncing involves some reversing of momentum, which means greater momentum change, and hence greater impulse. If the greater impulse is over an extended time (bouncing from a circus net), impact force is small. If over a short time (plant pot bouncing from your head), impact force is large. (Damage from an object colliding with a person may depend more on energy transfer than on momentum change, so in some cases damage can be greater in an inelastic collision without bouncing). DEMONSTRATION: In the second photo that opens this chapter, Howie Brand illustrates the effect of bouncing when swinging a dart against a wooden block. If the dart bounces, then the block tends to tip over. If it doesn’t bounce, the block remains upright. (Note that if the dart securely sticks to the block, it may change the center of gravity of the block to favor nontipping, but without bouncing, the block will remain upright anyway. Bouncing, which increases the impact force, is key here.) DEMONSTRATION: Fashion a pendulum using the “happy-unhappy” rubber balls and let them swing into an upright board. When the less-elastic ball makes impact with very little bounce, the board remains upright. But when the more-elastic ball makes impact, it undergoes a greater change in momentum as it bounces. This imparts more impulse to the board, and it topples. This is akin to Howie Brand’s swinging dart demo. Think and Discuss 106–108 may need your elaboration if you wish to go this deep in your presentation of bouncing. Simply removing the sail, as 108 suggests, is the option used by propeller-driven aircraft. Consider suddenly producing a sail in the airstream produced by the propeller of an airplane. The result would be a loss of thrust, and if bouncing of the air occurred, there would be a reverse thrust on the craft. This is precisely what happens in the case of jet planes landing on the runway. Metal “sails” move into place behind the engine in the path of the ejected exhaust, which cause the exhaust to reverse direction. The resulting reverse thrust appreciably slows the aircraft.
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Collisions If you’re lucky enough to have an air track or air table, share with your students. Help them advance from the stage of play to being able to explain their observations in terms of momentum conversation. Physics can be fun, and delightful when learning occurs.
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ANSWERS AND SOLUTIONS FOR CHAPTER 6 READING CHECK QUESTIONS (COMPREHENSION) 6.1 Momentum 1. The mouse has more momentum since only it is moving. 2. Momentum is inertia in motion. 6.2 Impulse 3. Impulse is force time, not merely force. 4. Impulse can be increased by increasing force or increasing time of application. 5. More speed is imparted because of increased impulse due to longer time for the force on the cannonball to act. 6.3 Impulse-Momentum Relationship 6. The impulse-momentum relationship is derived from Newton’s second law. 7. Both. To increase momentum, increase impulse by using a large force or an extended time. 8. A deployed air bag extends the time you stop in an accident. Less force results if momentum is decreased over an extended time. 9. When the momentum of impact is quick, less time means more force. 10. By rolling with the punch, more time of impact occurs, which means a less forceful punch. 6.4 Bouncing 11. Choice (c) represents the greatest change in momentum. 12. Choice (c) also represents the greatest impulse. 13. The buckets cause the impacting water to bounce, increasing the impulse given to the wheel. 6.5 Conservation of Momentum 14. No, only external forces produce changes in momentum, so pushing on the dash of a car while sitting in it does not produce a net impulse. 15. No, the internal forces within a baseball do not supply a net force to produce a net impulse. 16. Yes, the statement is correct. 17. To say a quantity is conserved is to say its magnitude before an event is the same as its magnitude after the event. Momentum in a collision, for example, is the same before and after in the absence of external forces. 18. Only a vector quantity can cancel to zero. Momentum, velocity, and force are vectors that can cancel one another. 6.6 Collisions 19. Momentum is conserved in both an elastic and an inelastic collision. 20. Car B will have the speed of car A before the collision. 21. After collision, the cars will move at half the initial speed of car A. 22. The momentum of the large fish remains the same, but at a lower speed due to its increased mass. 23. The fish undergoes an inelastic collision. 6.7 More Complicated Collisions 24. The combined wreck slides at a north-east angle. 25. The angle would be 45° to the east-bound car. 26. The momentum after would be equal to the net momentum before. THINK AND DO (HANDS-ON APPLICATION) 27.–30. Open ended.
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PLUG AND CHUG (EQUATION FAMILARIZATION) 31. Momentum: p = mv = (9 kg)(2 m/s) = 18 kg.m/s. 32. p = mv = (40 kg)(4 m/s) = 160 kg.m/s 33. I = (10 N)(3 s) = 30 N.s 34. I = (10 N)(5 s) = 50 N.s 35. I = ∆mv = (9 kg)(2 m/s) = 18 kg.m/s = 18 N.s 36. I = ∆mv = (50 kg)(4 m/s) = 200 kg.m/s = 200 N.s mvbefore 3 m/s 37. From mvbefore + 0 = (m + m)vafter; vafter = = = 1.5 m/s. 2m 2 THINK AND SOLVE (MATHEMATICAL APPLICATIONS) 38. Momentum p = (10 kg)(6 m/s) = 60 kg.m/s, the amount of impulse to stop it. That’s 60 N.s. (Note that units N.s = kg.m/s.) mv (1000 kg )(20 m/s) = 2000 N . 39. From Ft = ∆mv, F = = t 10 s 75 kg mv ( )(25 m/s) = 18,750 N . 40. From Ft = ∆mv, F = = t 0.1 s 41. From the conservation of momentum, Momentumdog = momentumJean + dog. (15 kg)(3.0 m/s) = (40.0 kg + 15 kg)v 45 kgm/s = (55 kg)v, so v = 0.8 m/s. 42. Let m be the massof the freight car, and 4m the mass of the diesel engine, and v the speed after both have coupled together. Before collision, the total momentum is due only to the diesel engine, (4m)(5 km/h), because the momentum of the freight car is 0. After collision, the combined mass is (4m + m), and combined momentum is (4m + m)v. By the conservation of momentum equation: Momentumengine = momentumengine + freight car (4m)(5 km/h) + 0 = (4m + m)v 20m km / h v= = 4 km/h. (Note that you don’t have to know m to solve the problem.) 5m 43. Momentumbefore = momentumafter (5 kg)(1 m/s) + (1 kg)v = 0 5 m/s + v = 0 v = −5 m/s. So if the little fish approaches the big fish at 5 m/s, the momentum after will be zero. 44. By momentum conservation, asteroid mass 800 m/s = Superman’s mass v. Since the asteroid’s mass is 1000 times Superman’s, (1000m)(800 m/s) = mv. Therefore v = 800,000 m/s. This is nearly 2 million miles per hour! 45. Momentum conservation can be applied in both cases. (a) For head-on motion the total momentum is zero, so the wreckage after collision is motionless. (b) As shown in Figure 6.17, the total momentum is directed to the northeast—the resultant of two perpendicular vectors, each of magnitude 20,000 kgm/s. It has magnitude 28,200 kgm/s. The 28,200 kg m/s speed of the wreckage is this momentum divided by the total mass, v = = 14 m/s. 2000 kg p 46. From the impulse-momentum relationship Ft = ∆p, solving for F we get F = . t
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47.
Ft t
=
mv t
v = ma, which reads F = ma; another of the delightful connections in physics! t
= m
THINK AND RANK (ANALYSIS) 48. (a) B, D, C, A (b) B, D, C, A 49. (a) B = D, A = C (b) D, C, A = B 50. (a) A, B, C (b) A, B, C 51. C, A, B
(c) C, B, A
(d) A, B, C
THINK AND EXPLAIN (SYNTHESIS) 52. If the velocity of the speeding bullet is huge, and the massive supertanker is slow, a product of mass velocity for the bullet could exceed the same for the supertanker. 53. The momentum of a supertanker is enormous, which means enormous impulses are needed for changing motion—which are produced by applying modest forces over long periods of time. Due to the force of water resistance, over time it coasts 25 kilometers to sufficiently reduce the momentum. 54. When you are brought to a halt in a moving car, an impulse, the product of force and time, reduces your momentum. During a collision, padded dashboards increase the time of impact while reducing the force of impact. The impulse equals your change in momentum. 55. Air bags lengthen the time of impact thereby reducing the force of impact. 56. The extra thickness extends the time during which momentum changes and reduces impact force. 57. Stretching ropes extend the time during which momentum decreases, thereby decreasing the jolting force of the rope. Note that bringing a person to a stop more gently does not reduce the impulse. It only reduces the force. 58. The steel cord will stretch only a little, resulting in a short time of stop and a large force. Ouch! 59. Bent knees will allow more time for momentum to decrease, reducing the force of landing. 60. Sliding extends the time of the impulse, which reduces the force of impact. 61. The deep knee bend and forward roll increases the time of coming to a stop, and thereby reduces the force of impact. 62. The time during which the stopping force acts is different for the different situations. Stopping time is least on concrete and most on water, hence the different impact speeds. So there are three concepts: speed at impact, time of impact, and force of impact—which are all related by the impulse-momentum relationship. 63. An extended hand allows more time for reducing the momentum of the ball to zero, resulting in a smaller force of impact on your hand. 64. The time during which the ball stops is small, producing a greater force. 65. Crumpling allows more time for reducing the momentum of the car, resulting in a smaller force of impact on the occupants. 66. The blades impart a downward impulse to the air and produce a downward change in the momentum of the air. The air at the same time exerts an upward impulse on the blades, providing lift. (Newton’s third law applies to impulses as well as forces.) 67. Its momentum is the same (its weight might change, but not its mass). 68. The egg hitting the sagging sheet has a longer impact time, which decreases the force that would otherwise break it. 69. The large momentum of the spurting water is met by a recoil that makes the hose difficult to hold, just as a shotgun is difficult to hold when it fires birdshot. 70. Doing so would be a bad idea because the gun would recoil with a speed ten times the muzzle velocity. The target may endure less harm than the shooter. 71. Impulse is force time. The forces are equal and opposite, by Newton’s third law, and the times are the same, so the impulses are equal and opposite.
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72. The momentum of recoil of Earth is 10 kg m/s. Again, this is not apparent because the mass of the Earth is so enormous that its recoil velocity is imperceptible. (If the masses of Earth and person were equal, both would move at equal speeds in opposite directions.) 73. The momentum of the falling apple is transferred to Earth. Interestingly, when the apple is released, Earth and the apple move toward each other with equal and oppositely directed momenta. Because of Earth’s enormous mass, its motion is imperceptible. When the apple and Earth hit each other, their momenta are brought to a halt—zero, the same value as before. 74. There is usually greater speed and therefore greater impact on a catcher’s mitt than the mitts of other players. That’s why extra padding is used to prolong the time of the impulse to stop the ball and lessen the catching force. 75. The lighter gloves have less padding, and less ability to extend the time of impact, and therefore result in greater forces of impact for a given punch. 76. In jumping, you impart the same momentum to both you and the canoe. This means you jump from a canoe that is moving away from the dock, reducing your speed relative to the dock, so you don’t jump as far as you expected to. 77. The swarm will have a net momentum of zero if the swarm stays in the same location; then the momenta of the many insects cancel and there is no net momentum in any given direction. 78. To get to shore, the person may throw keys, coins or an item of clothing. The momentum of what is thrown will be accompanied by the thrower’s oppositely-directed momentum. In this way, one can recoil towards shore. (One can also inhale facing the shore and exhale facing away from the shore.) 79. If no momentum is imparted to the ball, no oppositely directed momentum will be imparted to the thrower. Going through the motions of throwing has no net effect. If at the beginning of the throw you begin recoiling backward, at the end of the throw when you stop the motion of your arm and hold onto the ball, you stop moving, too. Your position may change a little, but you end up at rest. No momentum given to the ball means no recoil momentum gained by you. 80. Regarding Question 78: If one throws clothing, the force on the clothes will be paired with an equal and opposite force on the thrower. This force can provide recoil toward the shore. Regarding Question 79: According to Newton’s third law, whatever forces you exert on the ball, first in one direction, then in the other, are balanced by equal forces that the ball exerts on you. Since the forces on the ball give it no final momentum, the forces it exerts on you also give no final momentum. 81. Both recoiling carts have the same amount of momentum. So the cart with twice the mass will have half the speed of the less massive cart. That is, (2m)(v/2) = mv. 82. Momentum is not conserved for the ball itself because an impulse is exerted on it (gravitational force time). So the ball gains momentum. Only in the absence of an external force does momentum not change. If the whole Earth and the rolling ball are taken together as a system, then the gravitational interaction between Earth and the ball are internal forces and no external impulse acts. Then the change of momentum of the ball is accompanied by an equal and opposite change of momentum of Earth, which results in no change in momentum. 83. A system is any object or collection of objects. Whatever momentum such a system has, with no external forces, that momentum remains unchanged—that’s what the conservation of momentum is about. 84. For the system that consists of only the ball, momentum changes, and is therefore not conserved. But for the larger system of ball + Earth, momentum is conserved for the impulses acting are internal impulses. The change of momentum of the ball is equal and opposite to the change of momentum of the recoiling Earth. 85. For the system made up of ball + Earth, momentum is conserved for the impulses acting are internal impulses. The momentum of the falling apple is equal in magnitude to the momentum of the Earth toward the apple.
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86. If the system is the stone only, its momentum certainly changes as it falls. If the system is enlarged to include the stone plus Earth, then the downward momentum of the stone is canceled by the equal but opposite momentum of Earth “racing” up to meet the stone. 87. Yes, because you push upward on the ball you toss, which means the ball pushes downward on you, which is transmitted to the ground. So the normal force increases as the ball is thrown (and goes back to equal mg after the ball is released). Likewise, in catching the ball you exert an upward force while stopping it, which is matched by a downward force by your feet on the ground, which increases the normal force. 88. By Newton’s third law, the force on the bug is equal in magnitude and opposite in direction to the force on the car windshield. The rest is logic: Since the time of impact is the same for both, the amount of impulse is the same for both, hence they undergo the same change in momentum. The change in momentum of the bug is evident due to its large change in speed. The same change in momentum of the considerably more massive car is not evident, for the change in speed is correspondingly very small. Nevertheless, the magnitude of m∆V for the bug is equal to M∆v for the car! 89. In accord with Newton’s third law, the forces on each are equal in magnitude, which means the impulses are likewise equal in magnitude, so both undergo equal changes in momentum. 90. The magnitude of force, impulse, and change in momentum will be the same for each. The Mini Cooper undergoes the greater deceleration because its mass is less. 91. Cars brought to a rapid halt experience a change in momentum, and a corresponding impulse. But greater momentum change occurs if the cars bounce, with correspondingly greater impulse and therefore greater damage. Less damage results if the cars stick upon impact than if they bounce. 92. The direction of momentum is to the left, for the momentum of the 0.8-kg car is greater. By magnitude, net momentum = (0.5)(1) − (0.8)(1.2) = − 0.46. 93. The combined momentum is √2 times the magnitude of that of each blob before collision. 94. Momentum conservation is being violated. The momentum of the boat before the hero lands on it will be the same as the momentum of boat + hero after. The boat will slow down. If, for example, the masses of the hero and boat were the same, the boat should be slowed to half speed; mvbefore = (2m)(2/v)after. From an impulse-momentum point of view, when the hero makes contact with the boat, he is moved along with the boat by a friction force between his feet and the boat surface. The equal and opposite friction force on the boat surface provides the impulse that slows the boat. (Here we consider only horizontal forces and the horizontal component of momentum.) 95. Yes, you exert an impulse on a ball that you throw and also when you catch it. Since you change its momentum by the same amount in both cases, the impulse you exert in both cases is the same. To catch the ball and then throw it back again at the same speed requires twice as much impulse. On a skateboard, you’d recoil and gain momentum when throwing the ball, you’d also gain the same momentum by catching the ball, and you’d gain twice the momentum if you did both—catch and then throw the ball at its initial speed in the opposite direction. 96. More impulse occurs when bouncing, which is appreciably greater than when the dart sticks. If it takes one unit of impulse to stop the dart, it takes additional impulse to “bounce it back again.” Ideally, bouncing would involve twice the impulse. THINK AND DISCUSS (EVALUATION) 97. The ramp offers a truck in distress a gradual stop so that the impulse to stop momentum is longer time and therefore less force of impact. 98. Impulse to stop the momentum of the skateboard or heavy truck is given by the product of force and time. Stopping the skateboard in a split second could be a larger impulse than stopping the heavy truck in a long time. 99. When a boxer hits his opponent, the opponent contributes to the impulse that changes the momentum of the punch. When punches miss, no impulse is supplied by the opponent—all effort that goes into reducing the momentum of the punches is supplied by the boxer himself. This tires ..
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the boxer. This is very evident to a boxer who can punch a heavy bag in the gym for hours and not tire, but who finds by contrast that a few minutes in the ring with an opponent is a tiring experience. 100. Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well.) 101. The internal force of the brake brings the wheel to rest. But the wheel, is attached to the tire which makes contact with the road surface. It is the force of the road on the tires that stops the car. 102. If the rocket and its exhaust gases are treated as a single system, the forces between rocket and exhaust gases are internal, and momentum in the rocket-gases system is conserved. So any momentum given to the gases is equal and opposite to momentum given to the rocket. A rocket attains momentum by giving momentum to the exhaust gases. 103. When two objects interact, the forces they exert on each other are equal and opposite and these forces act simultaneously, so the impulses are equal and opposite. Therefore their changes of momenta are equal and opposite, and the total change of momentum of both objects is zero. 104. Let the system be the car and Earth together. As the car gains downward momentum during its fall, Earth gains equal upward momentum. When the car crashes and its momentum is reduced to zero, Earth stops moving upward, also reducing its momentum to zero. 105. This exercise is similar to the previous one. If we consider Bronco to be the system, then a net force acts and momentum changes. In this case, momentum is not conserved. If, however we consider the system to be Bronco and the world (including the air), then all the forces are internal forces and momentum is conserved. Momentum is conserved only in systems not subject to external forces. 106. The craft moves to the right because there are two horizontal impulses that act on the craft: One is that of the wind against the sail, and the other is that of the fan recoiling from the wind it produces. These impulses are oppositely directed, but are they equal in magnitude? No, because of bouncing. The wind bounces from the sail and produces a greater impulse than if it merely stopped. This greater impulse on the sail produces a net impulse in the forward direction, toward the right. We can see this in terms of forces as well. In the sketch there are two force pairs to consider: (1) the fan-air force pair, and (2) the air-sail force pair. Because of bouncing, the forces of the air-sail pair are greater. The net force on the craft is forward, to the right. The principle described is applied in thrust reversers used to slow jet planes after they land. Also, you can see that after the fan is on, there is a net motion of air to the left, so the boat, to conserve momentum, will move to the right. 107. If the air is brought to a halt by the sail, then the impulse against the sail will be equal and opposite to the impulse on the fan. There will be no net impulse and no change in momentum. The boat will remain motionless. Bouncing counts! 108. Removing the sail and turning the fan around is the best means of propelling the boat! Then maximum impulse is exerted on the boat. If the fan is not turned around, the boat is propelled backward, to the left. (Such propeller-driven boats are used where the water is very shallow, as in the Florida Everglades.) 109. Bullets bouncing from the steel plate experience a greater impulse. The plate will be moved more by bouncing bullets than by bullets that stick. 110. In terms of force: When Freddy lands on the skateboard he is brought up to the skateboard’s speed. This means a horizontal force provided by the board acts on Freddy. By action-reaction, Freddy exerts a force on the board in the opposite direction—which slows the skateboard. In terms of momentum conservation: Since no external forces act in the horizontal direction, the momentum
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after the skateboard catches Freddy is equal to the momentum before. Since mass is added, velocity must decrease. 111. Agree with the first friend because after the collision the bowling ball will have a greater momentum than the golf ball. Before collision, the momentum of the system of two balls is all in the moving golf ball. Call this +1 unit. Then after collision, the momentum of the rebounding golf ball is nearly −1 unit. The momentum (not the speed!) of the bowling ball will have to be nearly +2 units. Why? Because only then is momentum conserved. Momentum before is +1 unit: momentum after is (+2 − 1) = +1. 112. We assume the equal strengths of the astronauts means that each throws with the same speed. Since the masses are equal, when the first throws to the second, both the first and second move away from each other at equal speeds. Say the thrown astronaut moves to the right with velocity V, and the first recoils with velocity −V. When the third makes the catch, both she and the second move to the right at velocity V/2 (twice the mass moving at half the speed, like the freight cars in Figure 6.14). When the third makes her throw, she recoils at velocity V (the same speed she imparts to the thrown astronaut), which is added to the V/2 she acquired in the catch. So her velocity is V + V/2 = 3V/2 to the right—too fast to stay in the game. Why? Because the velocity of the second astronaut is V/2 − V = −V/2 to the left—too slow to catch up with the first astronaut who is still moving at −V. The game is over. Both the first and the third astronauts got to throw to the second astronaut only once! 113. Impulse is greater for reflection, which is in effect, bouncing. The vanes therefore recoil more from the silvered sides. The vanes in the sketch therefore rotate clockwise as viewed from above. (This rotation is superseded by a counter rotation when air is present, which is the case for most radiometers. The black surface absorbs radiation and is heated, which warms the nearby air. The surface is pushed away from the warmed air resulting in a recoil that spins the vanes counterclockwise.) 114. Yes, the bat transfers some of its momentum to the ball. If the batted ball has more speed after being batted, the bat must correspondingly have reduced speed. 115. Their masses are the same; half speed for the coupled particles means equal masses for both the colliding and the target particles. This is like the freight car collision of Figure 6.14. 116. If a ball does not hit straight on, then the target ball flies off at an angle (say, to the left,) and has a component of momentum perpendicular to the ball’s initial momentum. To offset this, the striking ball cannot be simply brought to rest, but must fly off in the other direction (say, the right). It will do this in such a way that its sideways component of momentum is equal and opposite to that of the target ball. This means the total sideways momentum is zero—what it was before the collision. (Inspect Figure 6.19 and see how the sideways components of momentum cancel to zero.) 117. Because no external force acts, momentum before the burst equals momentum after. The bursting firecracker sends the balls scattering in directions such that momentum vectors before and after are the same. 118. The chunks have equal and opposite momenta, with the smaller-mass chunk having greater speed (mV = −Mv). 119. Both the cannonball and the recoiling cannon have the same momentum. For the recoiling cannon, most of its momentum is in mass, hence the large size of m. For the cannonball, most momentum is in velocity, hence the large size of v.
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CHAPTER 6 MULTIPLE-CHOICE PRACTICE EXAM 1. When riding your bicycle at twice your normal speed your momentum is (a) almost twice. (b) twice as much. (c) four times as much. (d) none of the above Answer: (b) Explanation: This is a reading test for the momentum equation, mv. Twice the speed v means twice the momentum mv. Sweet and simple. Test taking requires careful reading of the question asked! 2. An iron ball and wooden ball of the same size are dropped simultaneously from a high cliff. Ignore air drag. When hitting the ground below, the iron ball has a greater (a) speed. (b) acceleration. (c) momentum. (d) all of the above Answer: (c) Explanation: With no air drag, both will fall with the same acceleration and experience the same increase in speed. So choices (a) and (b) are wrong. What is different is mass. The iron ball with greater mass will hit the ground with greater momentum. 3. A long-barrel cannon fires a faster cannonball than a short cannon because the ball is given more (a) force. (b) impulse. (c) both of these (d) none of the above Answer: (b) Explanation: For the same force on the cannonball while in the barrel, the longer barrel means more time that the force acts, which gives the cannonball a greater impulse. So we see that the force acts for a longer time. That’s greater impulse. 4. A freight train rolls along a track with considerable momentum. If it rolls at the same speed but has twice the mass, its momentum is (a) zero. (b) doubled. (c) quadrupled. (d) unchanged. Answer: (b) Explanation: This is a straight-forward reading of the equation for momentum, mv. With other variables unchanged, doubling the mass doubles the momentum. 5. When freight cars collide, momentum is conserved in (a) elastic collisions. (b) inelastic collisions. (c) both of these (d) none of the above Answer: (c) Explanation: A feature of momentum conservation is that it holds for both elastic and inelastic collisions. Hence (c) is the best answer. 6. A rolling freight car collides with an identical freight car at rest. After coupling they roll with a speed of 2 m/s. From this information, the initial speed of the rolling freight car was (a) 4 m/s. (b) 5 m/s. (c) 6 m/s. (d) 8 m/s. Answer: (a) Explanation: The total momentum before and after the collision is the same. That means total mv before (which is only mv of the initially rolling car) equals total mv after (which is twice the mass multiplied by 2 m/s). With twice the mass after the collision, the initial speed of the oncoming car must have been twice the initial speed of 2 m/s. That’s 4 m/s. By equation mv + 0 = (2m)(2 m/s); Solving for v we get 4 m/s.
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7. A school bus on a highway brakes to a stop. More braking force is required if the bus has (a) more mass. (b) more momentum. (c) less stopping distance. (d) all of the above Answer: (d) Explanation: Certainly a more massive bus is harder to stop. Or saying the same thing, a bus with more momentum is harder to stop. And having to stop in a shorter distance would require more force. These are common sense responses, all in keeping with the impulse-momentum relationship: Ft = ∆mv. 8. Whether a vehicle comes to an abrupt stop by crashing into a haystack or a brick wall, the impulse is (a) greater with the haystack. (b) greater with the brick wall. (c) both the same. (d) no way to say Answer: (c) Explanation: This asks for a clear understanding of “impulse.” Impulse is not force. Impulse is not time. Impulse is the product of both force and time, which will equal the change in momentum of the vehicle that comes to rest whether hitting the haystack or the wall. So the change in momentum (not force) is the same for both. If the momentum change is the same for both, then the impulse is also the same for both. Careful reading and thinking are required here. 9. A big fish swims upon and swallows a small fish at rest. Immediately after lunch, the fattened big fish has a change in (a) speed. (b) momentum. (c) both (d) neither Answer: (a) Explanation: This is an inelastic collision whereby the big fish gains mass and slows during lunch so choice (a) is correct. Not choice (b) because momentum is conserved, not changed. Conservation of momentum tells us that the total momentum before the inelastic collision is equal to the total momentum after. Momentum before is entirely with the big fish, since the small fish was at rest. Likewise, momentum after is still with the big fish, which has gained mass and lost speed. Hence answer (c) is wrong—the momentum of the big fish is the same before and after lunch—no change. 10. An unfortunate bug splatters on the windshield of a moving car. Which of these is the same in magnitude for the bug and the windshield? (a) force. (b) impulse. (c) change in momentum. (d) all of the above Answer: (d) Explanation: Recall Newton’s third law: the force of the car windshield on the bug is the same amount of force as that of the bug on the windshield—equal and opposite. It’s easy to see that the amount of time for the splatter is the same from both points of view. If force and time are the same for the bug and the car windshield, then the impulse is the same for both. And according to Ft = ∆mv, both undergo the same change in momentum. So it’s all of the above. Careful thinking is being tested in this problem.
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7 Energy ____________________________________________ Conceptual Physics Instructor’s Manual, Thirteenth Edition 7.1 Work 7.2 Power Mechanical Energy 7.3 Potential Energy 7.4 Kinetic Energy 7.5 Work-Energy Theorem 7.6 Conservation of Energy CIRCUS PHYSICS JUNK SCIENCE 7.7 Machines 7.8 Efficiency 7.9 Major Sources of Energy Recycled Energy
OPENING PHOTOS AND PROFILE Swedish physics teacher Christine Lindstrom heads the photo openers, followed by Neil deGrasse Tyson, whose profile is in Chapter 30. Astrophysicist Tyson’s keen wit conveys the personal benefits of science in people’s day to day lives. Cheers to Neil deGrasse Tyson! He is the superstar of science, as the new Carl Sagan, both for his devotion to astrophysics, his insight, and the message he conveys about science to the public. Photo three is of friends who started a small physics products company in 1981, which I’ve watched grow since then. Today Vernier Software and Technology, based in Beaverton, Oregon, is a major supplier of quality equipment for classroom and lab physics. Photo four is of my wife’s and my niece Serena Sinn applying physics to sports. The profile of Émilie du Châtelet cites the controversy in England and continental Europe about the “oomph” of objects in contact. Émilie helped settle the controversy by distinguishing between velocity in momentum (v) and velocity squared in kinetic energy (v2)—a fascinating story.
INTRODUCTION The section Major Sources of Energy, at the end of the chapter, advocates using the heat energy generated by power plants to heat homes and other buildings. New York City and Copenhagen are two cities that do this quite successfully. Rooftop energy is contrasted in Figures 7.25 and 7.26 where water and sunlight are successfully captured and employed. For decades I have been advocating dry-rock geothermal power (Figure 7.27) in Conceptual Physics. It has been only marginally accepted in Australia. The downsides of increased earthquake activity caused by fracturing doesn’t advance the idea. Not mentioned in the textbook are biofuels and the prospects for algae-based fuel oil. Algae can be farmed in special ponds, and the oils they produce can be refined into conventional transportation fuels that can be distributed using existing infrastructures. And like plants, algae absorb carbon dioxide. Given that the energy market is $1 trillion a year globally, biofuels may become a substantial player. David Willey points out that only about 60% to 75% of the PE of a drawn bow goes into the KE of an arrow, depending on the type of bow. The rest of the energy heats the bow. Similarly, only about 30% of the energy of firearms is transferred to the projectiles they fire. The rest heats the firearm. This poses a problem with sustained fire in machine guns, where the bore thermally expands so bullets no longer take the rifling and tip over and over in flight just as though fired from a smooth bore. Hence the water cooling for the machine guns of yesterday’s wars.
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Share this with your students: The energy they get when eating grains and vegetables is “packaged sunshine!”
SUPPLEMENTS Practicing Physics • Work and Energy • Conservation of Energy
• Momentum and Energy • Energy and Momentum
Next-Time Questions • Bumpy Tracks • Cannonball • Skidding Distance • Bouncing Bee
• Ball Toss • Long Cannon • Oomph • Whapped Baseball
Hewitt-Drew-It! Screencasts • Work and Potential Energy • Work-Energy Theorem • Conservation of Energy • Ballistic Pendulum • Machines and Energy
• Roller Coaster • Falling Balls • Hammer Whap
• Potential and Kinetic Energy • Energy of Acrobats
Laboratory Manual • An Uphill Climb—Work on an Inclined Plane (Experiment) • The Fountain of Fizz—Physics in the Soda Pop Geyser (Demonstration) • Dropping the Ball—Energy Conservation During Free Fall (Experiment) Problem Solving in Conceptual Physics Many problems about energy
SUGGESTED PRESENTATION Begin by standing on a chair against a wall with an extended heavy pendulum bob held at the tip of your nose. Say nothing. Release the bob and let it swing out, then back to your nose. Don’t flinch. Then comment on your confidence in physical laws and lead into a distinction between potential and kinetic energy. Point out that where the bob is lowest, it’s moving the fastest, and where it’s highest it doesn’t move at all. The bob transforms energy of motion to energy of position in cyclic fashion. Allow the pendulum to swing to-and-fro while you’re talking. Its motion decays. Why? Then point out the transformation of energy from the moving bob to the molecules of air that are encountered, and to the molecules in the bending string or wire at the pivot point. The energy of the pendulum will end up as heat energy. I quip that on a very hot day, somebody, somewhere, is swinging a giant pendulum to-and-fro. Work Define work and compare it to impulse of the previous chapter. In both cases, the effect of exerting a force on something depends on how long the force acts. In the previous chapter, how long was meant as time, and we spoke of impulse. In this chapter, how long is meant as distance, and we speak of work. Cite the examples of the drawn slingshot and the long-barreled cannon, where the added length produces greater speed. We described this greater speed in terms of greater momentum. Now we describe this greater speed in terms of greater energy—that is, greater KE.
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CHECK QUESTION: Is work done when a weightlifter (Figure 7.3) holds a barbell stationary above her head? [Yes and no. With each contraction of her heart, a force is exerted through a distance in her blood and so work is done on the blood. But this work is not done on the barbell.] Power A watt of power is the work done in vertically lifting a 1-newton mass one meter in one second. Stress that power is the transfer or transformation of energy per unit of time. Potential Energy Return to your pendulum: With the pendulum at equilibrium, show how the force necessary to pull it sideways (which varies with the angle made by the string) is very small compared to the force necessary to lift it vertically (its weight). Point out that for equal elevations, the arced path is correspondingly longer than the vertical path—with the result that the product of the applied force and distance traveled—the work done—is the same for both cases. (Without overdoing it, this is a good place for your students to know a bit about integral calculus—how calculus is required to add up the work segments that continuously increase in a nonlinear way.) Then discuss the work needed to elevate the ball in Figure 7.6. CHECK QUESTIONS: Does a car hoisted for lubrication in a service station have PE? How much work will raise the car twice as high? Three times as high? How much more PE will it have in these cases? You can give the example of dropping a bowling ball on your toe—first from a distance of a couple of centimeters above your toe, then to various distances up to 1 m. Each time, the bowling ball would do more work on your toe because it would transfer more gravitational potential energy when released. Your toe would certainly note the increased kinetic energy! Kinetic Energy Relate force distance = KE to examples of pushing a car, and then to braking a car as treated in the textbook. You may do Think and Solve 47 (about skidding distance as a function of speed) at this point. To a close approximation, skidding force is independent of speed. Hence the change in KE is approximately equal to change in skidding distance. When the car’s brakes are applied, the car’s kinetic energy is changed into internal energy in the brake pads, tire, and road as they become warmer. You may or may not at this point preview future material by relating the idea of the KE of molecules and the idea of temperature. State that in a substance, molecules having the same temperature have the same average KE. If the masses of the molecules are the same, then it follows that the speeds of the molecules are the same. But what if the masses are different—for example, in a sample of gas composed of light and heavy molecules at the same temperature? Which molecules would move faster? (If you shook a container of billiard balls mixed with Ping-Pong balls so that both kinds of balls had the same kinetic energy, which would move faster in the container? If an elephant and a mouse run with the same kinetic energy, which means that both will do the same amount of work if bumping into the door of a barn, can you say which of the two is running faster?) You might consider the demonstration of inhaling helium and talking at this point—particularly if you are not including the chapters on sound in your course design. Relate the higher temperature due to the faster moving helium molecules to the higher temperature in a bugle when faster moving air is blown through it. Work-Energy Theorem When discussing whether or not work is done, be sure to specify done on what. If you push a stationary wall, you may be doing work on your muscles (that involve forces and distances in flexing), but you do no work on the wall. Key point: If work is done on something, then the energy of that something changes.
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Distinguish between the energy one expends in doing things, and the work that is actually done on something. CHECK QUESTION: When a car slows down due to air resistance, does its KE decrease? [Most certainly!] Which is greater, 1 joule or 1 newton? Whoops! [The comparison is silly, for they’re units of completely different things—work and force.] An idea about the magnitude of 1 joule is that it’s the work done in vertically lifting a large apple (approximately 1 N) one meter. Energy Conservation Discuss Figures 7.9 and 7.11 and then return to your pendulum. Explain how the kinetic energy and hence, the speed of the bob at the bottom of its swing is equal to the speed it would have if dropped vertically through the same height. CHECK QUESTION: Refer to Figure 7.6 in “inclines” (a) and (b): How does the speed of the ball compare at ground level when released from equal elevations? [It is impressive that the speeds will be the same. The lesser acceleration down the sloped ramp is compensated by a longer time.] Return to the situation and ask how the times to reach the bottom compare and be prepared for an incorrect response, “The same!” (NOT true!) Quip and ask if the colors and temperatures will also be the same. Straight-forward physics can be confusing enough! DEMONSTRATION: Preview electricity and magnetism and bring out the horseshoe magnet hand-cranked generator that lights up the lamp shown ahead in Figure 25.21 (Karen Jo Matsler producing light). Have student volunteers attest to the fact that more work is needed to turn the crank when the lamp is connected than when it is not. Then relate this to Think and Discuss 100, 101, and 119 (about the car burning more fuel with lights and air conditioner on). When gasoline combines with oxygen in a car’s engine, the chemical potential energy stored in the fuel is converted mainly into molecular KE (internal energy). Some of this energy in effect is transferred to the piston and some of this causes motion of the car. Electric cars are not something new. Electric cars were more popular than gasoline-driven cars in the late nineteenth and early twentieth century. They could go all day on a single charge and move a driver around a city with ease. They required no hand crank to start and had no gears to shift. But back then speed limits were set below 20 mph to accommodate horse-drawn carriages. After World War I these limits were increased, and gasoline-powered cars began to dominate. When most cars go electric, we’ll be doing a full circle! Two-Track Ball Race Return to the race of balls on humped and dipped tracks that you discussed in Chapter 3. The conservation of energy clearly shows that when the PE is the same at the outset of race along tracks A and B, the KE of the balls at the ends of the tracks will be equal to the initial PE and will be the same at the end of each track. There is an important lesson to be learned here. Equal final speeds does not mean equal times. Importantly, have your students explain why ball C wins the race. Problems involving motion are most easily solved via energy. Pose a problem that involves motion to a physicist and chances are high that energy conservation will be invoked—not kinematics or dynamics!
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Perhaps the single most important idea a student can take away from your course is an appreciation of the conservation of energy—energy that moves from one place to another remains constant. It may branch along the way or convert to other forms, but all is accountable. This way of thinking is very good physics. Scams that sell energy-making machines rely on funding from deep pockets and shallow brains! Efficiency It should be enough that your students become acquainted with the idea of efficiency, so I don’t recommend setting the plow too deep for this topic. The key idea to impart is the concept of useful energy. To say that an incandescent lamp of old is 10% efficient is to say that only 10% of the energy input is converted to the useful form of light. All the rest goes to heat. But even the light energy converts to heat upon absorption. So all the energy input to an incandescent lamp is converted to heat. This means that it is a 100% efficient heater (but not a 100% device for emitting light)! Much more efficient light sources, the now common LEDs, are treated in Chapters 23 and 30. Solar Power Solar power is now employed worldwide, and growing. If fossil fuels had never been discovered, solar power would have been popular eons ago. But it’s here now, and we celebrate its growing usage. The latest information about solar power found on the Web is fodder for class discussions. In all discussions of energy transfer, instill the wonder of the conservation of energy. Consider, for example, the transfer of numbers of joules from the Sun to numbers of joules to the consumer. Numbers matter. Hydrogen as a Gas and as a Paste A great activity is duplicating Figure 7.24, separating the oxygen and hydrogen molecules from water. Energy is required to separate the gases. By energy conservation, that energy will be released when the gases recombine to again produce water. The choice fuel for rockets is tanks of oxygen and hydrogen that combine to produce water while releasing awesome amounts of energy when they combine. Emphasize that the same energy required to initially separate the hydrogen and oxygen from water is the same energy they will release, only faster. Better than gaseous hydrogen is hydrogen as a goop, similar to toothpaste. The goop can be stored in a canister and will give up its hydrogen as needed. Unlike a battery, the paste does not gradually lose stored energy if left on a shelf. Hydrogen in this form is ideal for scooters and drones. You simply swap canisters when one is drained. Watch for this! Dark Energy Briefly mentioned in the textbook is the current serious speculation about dark energy, which is postulated to be speeding up the expanding universe. You may want to discuss this current finding, which may be one of the most important discoveries in science in the past decades. SUGGESTED OUT-OF-CLASS ASSIGNMENT: After you’ve shown the swinging ball apparatus, Newton’s cradle, in class (available from Arbor Scientific. P1-6001), consider assigning Think and Discuss 113 as a next-time question to be discussed in the next class. There’s lots of food for thought here. It’s interesting to note that when physicists tackle mechanics problems that involve motion, the routes they first turn to are impulse-momentum and conservation of energy. Both are very helpful physics lenses for viewing situations where changes in motion occur.
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ANSWERS AND SOLUTIONS FOR CHAPTER 7 READING CHECK QUESTIONS (COMPREHENSION) 7.1 Work 1. Force multiplied by distance is called work. 2. No work is done in pushing on a stationary wall, as in Figure 7.4. 3. Energy enables an object to do work. 4. Yes, twice as much work means twice as much energy. 5. It is the same, for the product of each is the same; (50 kg)(2 m) = (25 kg)(4 m). 7.2 Power 6. True, 1 W = 1 J/s. 7. Power = work/time = Fd/t = (12 N)(2 m)/3 s = 8 W 8. The same power when both are raised in the same time. 9. Twice the power for the heavier sack raised in the same time. 7.3 Potential Energy 10. True, gravitational PE relates to a reference level. 11. Its PE would be twice when twice as high. 12. It has the same PE when raised the same vertical height. 7.4 Kinetic Energy 13. True, KE is energy of motion. 14. Four times as much (as 22 = 4). 7.5 Work-Energy Theorem 15. The evidence is a gain in energy. 16. Four times as much work; 4 times as much stopping distance (as 22 = 4). 17. ∆KE = work done = (100 N − 70 N)(10 m) = (30 N)(10 m) = 300 Nm = 300 J 18. Speed has little or no effect on friction. 7.6 Conservation of Energy 19. Its gain in KE will equal its decrease in PE, 10 kJ. 20. Immediately before hitting the ground the initial PE of the peach becomes KE. When it hits the ground its energy becomes heat energy. 21. The source of the energy of sunshine is fusion power in the Sun. 7.7 Machines 22. A machine can multiply input force or input distance, but NEVER input energy. 23. As force is increased, distance is decreased by the same factor. 24. The end moving 1⁄3 as far can exert 3 times the input force, 150 N. 7.8 Efficiency 25. Efficiency would be 100%. 26. Efficiency will be 60%. 7.9 Major Sources of Energy 27. Recycled energy is the reemployment of energy that otherwise would be wasted. 28. The Sun is the source of these energies. 29. Radioactivity is the source of geothermal energy. 30. Like electricity, hydrogen is a carrier of energy, not a source, because it takes energy to separate hydrogen from molecules.
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THINK AND DO (HANDS-ON APPLICATIONS) 31. Where the “pendulum bob” is slowest, its PE is maximum. Where it’s fastest, its KE is maximum and PE is lowest. 32. The temperature of the sand is more after shaking than before. You do work on the sand in shaking it, which increases its temperature. 33. Heat energy is transferred to both the ground and the ball when bouncing. So with reduced KE, the ball doesn’t reach the initial height after bouncing. 34. Some of the basketball’s energy is transferred to the tennis ball by compression. During decompressing, the basketball pushes the tennis ball upward, while the tennis ball pushes the basketball downward. So PE of the bounced basketball is less and PE of the tennis ball is more, but both add to equal the original PEs of the balls before being dropped. PLUG AND CHUG (EQUATION FAMILARIZATION) 35. W = Fd = (4.0 N)(1.4 m) = 5.6 N.m = 5.6 J 36. W = Fd = (500 N)(2.0 m) = 1000 N.m = 1000 J, which is also the gain in PE. 37. P =
80 J W = = 40 W 2.0 s t
38. P =
(500 N )(4.0 m) Fd W = = = 1000 W 2.0 s t t
39. PE = mgh = (4.0 kg)(10 N/kg)(2.0 m) = 80 N.m = 80 J 40. PE = mgh = (1000 kg)(10 N/kg)(5 m) = 50,000 N.m = 50,000 J
41. KE = 1⁄2mv2 = 1⁄2(1.0 kg)(3.0 m/s)2 = 4.5 kg(m/s)2 = 4.5 J 42. KE = 1⁄2mv2 = 1⁄2(100 kg)(10 m/s)2 = 5000 kg(m/s)2 = 5000 J 43. W = ∆KE = ∆1⁄2mv2 = 1⁄2(3.0 kg)(4.0 m/s)2 = 24 J 44. From W = ∆KE; ∆KE = Fd = (5000 N)(500 m) = 2,500,000 J 40 J energy output 45. Efficiency = = = 0.40 or 40% 100 J energy input THINK AND SOLVE (MATHEMATICAL APPLICATION) 46. Work = ∆E = ∆mgh = 250 kg 10 N/kg 6.0 m = 15,000 J 47. Atthree times the speed, it has 9 times (32) the KE and will skid 9 times as far—135 m. Since the frictional force is about the same in both cases, the distance has to be 9 times as great for 9 times as much work done by the pavement on the car. 48. (a) You do F d = 100 N 10 m = 1000 J of work. (b) Because of friction, net work on the crate is less. Therefore ∆KE = net work = net force distance = (100 N − 70 N)(10 m) = 300 J. (c) So the difference, 1000 J − 300J = 700 J of “missing” energy goes into heating the crate and floor. 49. PE + KE = Total E; KE = 10,000 J − 1000 J = 9000 J. d 50. From F d = F , we see F = 3F = 150 N. 3 51. Your input work is 50 N 1 m = 50 J, so 200 N h = 50 J, h = 50/200 = 0.25 m. 52. (F d)in = (F d)out (100 N 10 cm)in = (? 1 cm)out
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So we see that the output force and weight held is 1000 N (less if efficiency < 100%). 53. Power =
(50 N )(6.0 m) 100 J Fd = = = 100 watts 3.0 s 1s t
54. Restating the question: Is (302 − 202) greater or less than (202 − 102)? We can see that (30 2 − 202) = (900 − 400) = 500, which is considerably greater than (20 2 − 10 2) = (400 − 100) = 300. So KE changes more for a given ∆v at the higher speed. 55. The initial PE of the banana is transformed to KE as it falls. When the banana is about to hit the water, all of its initial PE becomes KE. From PE0 = KEf mgh = 1⁄2mv2 v2 = 2gh v = 2gh . 56. From speed v =
2gh and v2 = 2gh; KE = 1⁄2mv2 = 1⁄2(0.0025 kg)(2)(10 m/s2)(381 m) = 95.3 J
THINK AND RANK (ANALYSIS) C, B, A 57. (a) B, A, C (b) C, B, A (c) 58. (a) C,B = D, A (b) C, B = D, A (c) A, B = D, C 59. (a) D, B, C, E, A (b) D, B, C, E, A (c) A, E, C, B, D 60. B = C, A (same as two supporting ropes) THINK AND EXPLAIN (SYNTHESIS) 61. The top hill determines the energy the roller coaster has in total, which spreads throughout successive shorter hills. If successive hills were higher, the coaster wouldn’t have enough energy to reach their tops. 62. You do no work because you haven’t exerted more than a negligible force on the backpack in the direction of motion. Also, the energy of the backpack hasn’t changed. No change in energy means no work done. 63. Stopping a lightly loaded truck of the same speed is easier because it has less KE and will therefore require less work to stop. (An answer in terms of impulse and momentum is also acceptable.) 64. Although no work is done on the wall, work is nevertheless done on internal parts of your body (which generate heat). 65. More force is required to stretch the strong spring, so more work is done in stretching it the same distance as a weaker spring. 66. Work done by each is the same, for they reach the same height. The one who climbs in 30 s exerts more power because work is done in a shorter time. 67. (a) The PE of the drawn bow as calculated would be an overestimate (in fact, about twice its actual value) because the force applied in drawing the bow begins at zero and increases to its maximum value when fully drawn. It’s easy to see that less force and therefore less work is required to draw the bow halfway than to draw it the second half of the way to its fully-drawn position. (b) So the work done is not maximum force distance drawn, but average force distance drawn. 68. When a rifle with a long barrel is fired, more work is done as the bullet is pushed through the longer distance. A greater KE is the result of the greater work, so of course, the bullet emerges with a greater velocity. (Note that the force acting on the bullet is not constant, but decreases with increasing distance inside the barrel.) 69. Your friend does twice as much work (4 1⁄2) > (1 1). 70. Agree, because speed itself is relative to the frame of reference (Chapter 3). Hence 1⁄2mv2 is also relative to a frame of reference. 71. The KE of the tossed ball relative to occupants in the airplane does not depend on the speed of the airplane. The KE of the ball relative to observers on the ground below, however, is a different matter. KE, like velocity, is relative.
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72. You’re both correct, with respect to the frames of reference you’re inferring. KE is relative. From your frame of reference she has considerable KE for she has a great speed. But from her frame of reference her speed is zero and KE also zero. 73. The KE of a pendulum bob is maximum where it moves fastest, at the lowest point; PE is maximum at the uppermost points. When the pendulum bob swings by the point that marks half its maximum height, its PE and KE have the same value. 74. If the ball is given an initial KE, it will return to its starting position with that KE (moving in the other direction!) and hit the instructor. (The usual classroom procedure is to release the ball from the nose at rest. Then when it returns it will have no KE and will stop short of bumping the nose.) 75. There is 200 J of potential energy that doesn’t go into increasing her kinetic energy. This goes into heat energy—heating her bottom and the slide. 76. The design is impractical. Note that the summit of each hill on the roller coaster is the same height, so the PE of the car at the top of each hill would be the same. If no energy were spent in overcoming friction, the car would get to the second summit with as much energy as it starts with. But in practice there is considerable friction, and the car would not roll to its initial height and have the same energy. So the maximum height of succeeding summits should be lower to compensate for friction. 77. You agree with your second classmate. The coaster could just as well encounter a low summit before or after a higher one, so long as the higher one is just the correct amount lower than the initial summit to compensate for energy dissipation by friction. 78. A Super Ball will bounce higher than its original height if thrown downward, but if simply dropped, no way. Such would violate the conservation of energy. 79. When a Super Ball hits the floor, some of its energy is transformed to heat and sound. This means it will have less kinetic energy after the bounce than just before and will not reach its original level. 80. (a) Kinetic energy is a maximum as soon as the ball leaves the hand. (b) Potential energy is a maximum when the ball has reached its highest point. 81. Sufficient work occurs because with each pump of the jack handle, the force she exerts acts over a much greater distance than the car is raised. A small force acting over a long distance can do significant work. 82. When the mass is doubled with no change in speed, both momentum and KE are doubled. 83. When the velocity is doubled, the momentum is doubled and the KE is increased by a factor of four. Momentum is proportional to speed, KE to speed squared. 84. Both have the same momentum, but the faster 1-kg one has the greater KE. 85. The momentum of the car is equal in magnitude but opposite in direction in the two cases—not the same since momentum is a vector quantity. 86. Zero KE means zero speed, so momentum is also zero. 87. Yes, if we’re talking about only you, which would mean your speed is zero. But a system of two or more objects can have zero net momentum, yet have substantial total KE. 88. Not at all. For two objects of the same KE, the one of greater mass has greater momentum. (The mathematical relationship is p2 = 2m KE.) 89. Scissors and shears are levers. The applied force is normally exerted over a short distance for scissors so that the output force is exerted over a relatively long distance (except when you want a large cutting force like cutting a piece of tough rope, and you place the rope close to the “fulcrum” so you can multiply force). With metal-cutting shears, the handles are long so that a relatively small input force is exerted over a long distance to produce a large output force over a short distance. 90. Energy is transformed into less useful forms in an inefficient machine, and is “lost” only in the loose sense of the word. In the strict sense, it can be accounted for and is therefore not lost. 91. An engine that is 100% efficient would not be warm to the touch, nor would its exhaust heat the air, nor would it make any noise, nor would it vibrate. This is because all these are transfers of energy, which cannot happen if all the energy given to the engine is transformed to useful work. (Actually, an engine of 100% efficiency is not even possible in principle. We discuss this in Chapter 18.) ..
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92. Your friend is correct, for changing KE requires work, which means more fuel consumption and decreased air quality. 93. In accord with energy conservation, (a) a person who takes in more energy than is expended stores what’s left over as added chemical energy in the body—which in practice means more fat. (b) One who expends more energy than is taken in gets extra energy by “burning” body fat. (c) An undernourished person who performs extra work does so by consuming stored chemical energy in the body—something that cannot long occur without losing health—and life. THINK AND DISCUSS (EVALUATION) 94. Both will have the same speed. This is easier to see in terms of energy conservation because both balls convert the same PE to KE. Think energy when solving all motion problems! 95. Without the use of a pole, the KE of running horizontally cannot easily be transformed to gravitational PE. But bending a pole stores elastic PE in the pole, which can be transformed to gravitational PE. Hence the greater heights reached by vaulters with very elastic poles. 96. Yes, a car burns more gasoline when its lights are on. The overall consumption of gasoline does not depend on whether or not the engine is running. Lights and other devices are run off the battery, which “runs down” the battery. The energy used to recharge the battery ultimately comes from the gasoline. 97. Except for the very center of the plane, the force of gravity acts at an angle to the plane, with a component of gravitational force along the plane—along the block’s path. Hence the block goes somewhat against gravity when moving away from the central position, and moves somewhat with gravity when coming back. As the object slides farther out on the plane, it is effectively traveling “upward” against Earth’s gravity, and slows down. It finally comes to rest and then slides back and the process repeats itself. The block slides back and forth along the plane. From a flat-Earth point of view the situation is equivalent to that shown in the sketch.
98. Solar energy is merely energy from the Sun. Solar power, like power in general, is the rate at which energy is transferred. Solar power is therefore the same from hour to hour, whereas the amount of solar energy depends on the amount of time energy is transferred. 99. If KEs are the same but masses differ, then the ball with smaller mass has the greater speed. That is, 1 ⁄2Mv2 = 1⁄2mV2. Likewise with molecules, where lighter ones move faster on the average than more massive ones. (We will see in Chapter 15 that temperature is a measure of average molecular KE— lighter molecules in a gas move faster than same-temperature heavier molecules.) 100. A machine can multiply force or multiply distance, both of which can be of value. 101. Your friend may not realize that mass itself is congealed energy, so you tell your friend that much more energy in its congealed form is put into the reactor than is taken out from the reactor. About 1% of the mass that undergoes fission is converted to energy of other forms. 102. The work that the rock does on the ground is equal to its PE before being dropped, mgh = 100 joules. The force of impact, however, depends on the distance that the rock penetrates into the ground. If we do not know this distance we cannot calculate the force. (If we knew the time during which the impulse occurs we could calculate the force from the impulse-momentum relationship— but not knowing the distance or time of the rock’s penetration into the ground, we cannot calculate the force.) 103. When we speak of work done, we must understand work done on what, by what. Work is done on the car by applied forces that originate in the engine. The work done by the road in reacting to the backward push of the tires is equal to the product of the applied force and the distance moved, not the net force that involves air resistance and other friction forces. When doing work, we think of ..
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applied force; when considering acceleration, we think of net force. Actually, the frictional forces of the internal mechanisms in the car, and to some extent the road itself, are doing negative work on the car. The zero total work explains why the car’s speed doesn’t change. 104. When air resistance is a factor, the ball will return with less speed (as discussed in Chapter 4). It therefore will have less KE. You can see this directly from the fact that the ball loses mechanical energy to the air molecules it encounters, so when it returns to its starting point and to its original PE, it will have less KE. This does not contradict energy conservation, for energy is transformed, not destroyed. 105. Both balls strike the ground with the same speed, whether thrown upward or downward. Their initial energies, PE + KE, are the same so they will have the same energy reaching the ground. This means they will strike with the same speed in both cases. Another way to look at this is to consider Chapter 3, Figure 3.8; in the absence of air resistance, the ball thrown upward will return to its starting level with the same speed as the ball thrown downward. Both hit the ground at the same speed (but at different times). We assume negligible air resistance, for if air resistance is a factor, then the ball thrown upward will lose more energy to the air in its longer path and strike with somewhat less speed. 106. Tension in the string supporting the 10-kg block is 100 N (which is the same all along the string). So block B is supported by two strands of string, each 100 N, which means the mass of block B is twice that of block A. So block B has a mass of 20 kg. 107. The other 15 horsepower is supplied by electric energy from the batteries (which are ultimately recharged using energy from gasoline). 108. In a conventional car, braking converts KE to heat. In a hybrid car, braking charges up the batteries. In this way, braking energy can soon be transformed to KE. 109. If an object has KE, then it must have momentum—for it is moving. But it can have potential energy without being in motion, and therefore without having momentum. And every object has “energy of being”—stated in the celebrated equation E = mc2. So whether an object moves or not, it has some form of energy. If it has KE, then with respect to the frame of reference in which its KE is measured, it also has momentum. 110. (a) In accord with Newton’s second law, the component of gravitational force that is parallel to the incline in B produces an acceleration parallel to the incline. (b) In accord with the work-energy theorem, that parallel force component multiplied by the distance the ball travels is equal to the change in the ball’s KE. 111. The physics is similar to that of the ball on the horizontal lane—no force component in the direction of motion. (a) Tension in the string is everywhere perpendicular to the arc of the pendulum, with no component of tension force parallel to its motion. (b) In the case of gravity, a component of gravitational force on the pendulum exists parallel to the arc, which does work and changes the KE of the pendulum. (c) When the pendulum is at its lowest point, however, there is no component of gravitational force parallel to motion. At that instant of motion, gravity does no work (as it doesn’t when the pendulum hangs at rest when the string is vertical). 112. On the hill there is a component of gravitational force parallel to the car’s motion. This component of force does work on the car. But while on the level, there is no component of gravitational force parallel to the direction of the car’s motion, so the force of gravity does no work in this case. 113. There is more to the “swinging balls” problem than momentum conservation, which is why the problem wasn’t posed in the previous chapter. Momentum is certainly conserved if two balls strike with momentum 2mv and one ball pops out with momentum m(2v). That is, 2mv = m2v. We must also consider KE. Two balls would strike with 2(1⁄2mv2) = mv2. The single ball popping out with twice the speed would carry away twice as much energy as was put in: 1⁄2m(2v)2 = 1⁄2m(4 v2) = 2mv2. So popping out with twice its initial energy is clearly a conservation of energy no-no! The exercise is not in Chapter 6 because KE was not discussed in Chapter 6.
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114. In the popular sense, conserving energy means not wasting energy. In the physics sense, energy conservation refers to a law of nature that underlies natural processes. Although energy can be wasted (which really means transforming it from a more useful to a less useful form), it cannot be destroyed. Nor can it be created. Energy is transferred or transformed, without gain or loss. That’s what a physicist means in saying energy is conserved. 115. The rate at which energy can be supplied is more central to consumers than the amount of energy that may be available, so a “power crisis” more accurately describes a short-term situation where demand exceeds supply. (In the long term, the world may be facing an energy crisis when supplies of fuel are insufficient to meet demand.) 116. Once used, energy cannot be regenerated, for it dissipates into less useful forms in the environment—inconsistent with the term “renewable energy.” Renewable energy refers to energy derived from renewable resources—trees, for example. 117. As the world population continues to increase, energy production must also increase to provide decent standards of living. Without peace, cooperation, and security, global-scale energy production likely decreases rather than increases.
My nephew John Luna in 1974.
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CHAPTER 7 MULTIPLE-CHOICE PRACTICE EXAM 1. If Fast Freda doubles her running speed, she also doubles her (a) momentum. (b) kinetic energy. (c) both of these (d) none of these Answer: (a) Explanation: This is a straight-forward reading of the equation and its symbols for momentum, mv. With no mention of other variable changes, doubling the speed doubles the momentum. Kinetic energy, however, would be quadrupled, since it varies with the square of speed v. 2. A model car with three times as much speed as another has kinetic energy that is (a) the same. (b) twice. (c) three times. (d) none of these Answer: (d) Explanation: KE varies with speed squared. So 3 times the speed squared is 9. The model car would have 9 times the KE, which isn’t among the choices (a–c), so (d) is the answer. 3. Which of these equations best illustrates the usefulness of automobile air bags? (a) F = ma (b) Ft = ∆mv (c) KE = 1⁄2mv2 (d) Fd = ∆KE Answer: (b) Explanation: Air bags reduce injury by extending the brief time of stopping with a corresponding reduction of force. The concepts of force and time typify impulse, Ft. This means that choice (b) is the best answer. The other equations only remotely involve air bags. 4. Which of these equations is best for calculating the distance a fast-moving bike will skid when its brakes are locked? (a) F = ma (b) Ft = ∆mv (c) KE = 1⁄2mv2 (d) Fd = ∆KE Answer: (d) Explanation: A key point in this question involves distance moved. The only equation involving distance is the work-energy equation (d), which says that the skidding force times the skidding distance equals the change in the bike’s KE. The other three equations only indirectly involve skidding distance. 5. The power required to raise a 100-kg crate a vertical distance of 2 m in 4 s is (a) 200 W. (b) 500 W. (c) 800 W. (d) 2000 W. Answer: (b) Explanation: Thinking is guided by the definition and equation for power. Power is the rate at which work is done or energy is transferred. P = W/t = Fd/t. In this case the work being done, Fd, is equal to the gain in PE of the crate. So P = W/t = PE/t, where PE, mgh = (100 kg)(10 m/s2)(2 m) = 2000 J, which when divided by 4 s = 500 W. That’s (2000 J)/4 s = 500 W.
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