In this question we will look for regularities in how a bowling ball falls under
ID: 3156226 • Letter: I
Question
In this question we will look for regularities in how a bowling ball falls under the influence of gravity. Specifically, we will look for a relationship between the number t of seconds the bowling ball remains in free fall, and the distance D it falls in that time. To get some data on this question, you bring a bowling ball to the tallest building in the world --- the Burj Khalifa in Dubai, UAE --- and you drop it from the top. The table below summarizes your measurements of how far the ball had fallen at each second:
Time (seconds) 0 1 2 3 4 5 6
Distance fallen (feet) 0 16 63 145 255 401 578
1. In early 2016, user truescientist224 proposed in a YouTube comment that the distance fallen by objects follows the pattern D=c2t. Use regression to find the 'c' that best fits the data, in the sense of the sum-of-squares measure.
Explanation / Answer
SAMPLE PROBLEM 1 An 8-kg bowling ball rolling at 2 m/s bumps into a padded guardrail and stops. (a) What is the momentum of the ball just before hitting the guardrail? (b) How much impulse acts on the ball? (c) How much impulse acts on the guardrail? SOLUTION: (a) The momentum of the ball is (b) In accord with the impulse– momentum relationship, the impulse on the ball is equal to its change in momentum. The momentum changes from to zero. So 16 kg # m/s = 16 N # s. (Note that the Ft = ¢mv = (16 kg # m/s) - 0 = 16 kg # m/s mv = (8 kg)(2 m/s) = 16 kg # m/s. units and are equivalent.) (c) In accord with Newton’s third law, the force of the ball on the padded guardrail is equal and oppositely directed to the force of the guardrail on the ball. Because the time of the interaction is the same for both the ball and the guardrail, the impulses are also equal and opposite. So the amount of impulse on the ball is SAMPLE PROBLEM 2 An ostrich egg of mass m is thrown at a speed v into a sagging bedsheet and is brought to rest in time t. (a) Show that the average force of egg impact is . (b) If the mass of the egg is 1.0 kg, its speed when it hits the sheet is 2.0 m/s, and it is brought to rest in 0.2 s, show that the average force that acts is 10 N. mv t 16 N # s. N # kg s # m/s (c) Why is breakage less likely with a sagging sheet than with a taut one? SOLUTION: (a) From the impulse–momentum equation, where in this case the egg ends up at rest, and simple algebraic rearrangement gives (b) (c) The time during which the tossed egg’s momentum goes to zero is extended when it hits a sagging sheet. Extended time means less force in the impulse that brings the egg to a halt. Less force means less chance of breakage. = 10 kg # m s2 = 10 N. F = mv t = (1.0 kg)A2.0m s B (0.2 s) F = mv t . ¢mv = mv, Ft = ¢mv, M04_HEWI6954_04_SE_CH03.qxd 10/1/07 3:25 PM Page 63 2008934301 Conceptual Physical Science, Fourth Edition, by Paul G. Hewitt, John Suchocki, and Leslie A. Hewitt. Published by Addison-Wesley. Copyright © 2008 by Pearson Education, Inc. impulses are greater when an object bounces. The impulse required to bring an object to a stop and then to “throw it back again” is greater than the impulse required merely to bring the object to a stop. Suppose, for example, that you catch the falling pot with your hands. You provide an impulse to reduce its momentum to zero. If you throw the pot upward again, you have to provide additional impulse. This increased amount of impulse is the same that your head supplies if the flowerpot bounces from it. The fact that impulses are greater when bouncing occurs was used with great success during the California gold rush. The waterwheels used in gold-mining operations were not very effective. A man named Lester A. Pelton recognized a problem with the flat paddles on the waterwheels. He designed a curved paddle that caused the incoming water to make a U-turn upon impact with the paddle. Because the water “bounced,” the impulse exerted on the waterwheel was increased. Pelton patented his idea, and he probably made more money from his invention, the Pelton wheel, than any of the gold miners earned. Physics can indeed enrich your life in more ways than one. 64 PART ONE PHYSICS FIGURE 3.8 Howie Brand shows that the block topples when the swinging dart bounces from it. When he removes the rubber head of the dart so it doesn’t bounce when it hits the block, no tipping occurs. Impulse FIGURE 3.9 The Pelton wheel. The curved blades cause water to bounce and make a U-turn, which produces a greater impulse to turn the wheel. CHECK POINT 1. In Figure 3.7, how does the force that Cassy exerts on the bricks compare with the force exerted on her hand? 2. How does the impulse resulting from the impact differ if her hand bounces back upon striking the bricks? Were these your answers? 1. In accordance with Newton’s third law, the forces are equal. Only the resilience of the human hand and the training she has undergone to toughen her hand allow this feat to be performed without broken bones. 2. The impulse is greater if her hand bounces back from the bricks upon impact. If the time of impact is not correspondingly increased, a greater force is then exerted on the bricks (and her hand!). M04_HEWI6954_04_SE_CH03.qxd 10/1/07 3:25 PM Page 64 2008934301 Conceptual Physical Science, Fourth Edition, by Paul G. Hewitt, John Suchocki, and Leslie A. Hewitt. Published by Addison-Wesley. Copyright © 2008 by Pearson Education, Inc. 3.3 Conservation of Momentum Only an impulse external to a system can change the momentum of a system. Internal forces and impulses won’t work. For example, consider the cannon being fired in Figure 3.10. The force on the cannonball inside the cannon barrel is equal and opposite to the force causing the cannon to recoil. Because these forces act for the same amount of time, the impulses are also equal and opposite. Recall Newton’s third law about action and reaction forces. It applies to impulses, too. These impulses are internal to the system comprising the cannon and the cannonball, so they don’t change the momentum of the cannon–cannonball system. Before the firing, the system is at rest and the momentum is zero. After the firing, the net momentum, or total momentum, is still zero. Net momentum is neither gained nor lost. Momentum, like the quantities velocity and force, has both direction and magnitude. It is a vector quantity. Like velocity and force, momentum can be canceled. So although the cannonball in the preceding example gains momentum when fired and the recoiling cannon gains momentum in the opposite direction, there is no gain in the cannon–cannonball system. The momenta (plural form of momentum) of the cannonball and the cannon are equal in magnitude and opposite in direction.* They cancel to zero for the system as a whole. If no net force or net impulse acts on a system, the momentum of that system cannot change. When momentum, or any quantity in physics, does not change, we say it is conserved. The idea that momentum is conserved when no external force acts is elevated to a central law of mechanics, called the law of conservation of momentum, which states: In the absence of an external force, the momentum of a system remains unchanged. For any system in which all forces are internal—as, for example, cars colliding, atomic nuclei undergoing radioactive decay, or stars exploding—the net momentum of the system before and after the event is the same. CHAPTER 3 MOMENTUM AND ENERGY 65 * Here we neglect the momentum of ejected gases from the exploding gunpowder, which can be considerable. Firing a gun with blanks at close range is a definite no-no because of the considerable momentum of ejecting gases. More than one person has been killed by close-range firing of blanks. In 1998, a minister in Jacksonville, Florida, dramatizing his sermon before several hundred parishioners, including his family, shot himself in the head with a blank round from a .357-caliber Magnum. Although no slug emerged from the gun, exhaust gases did—enough to be lethal. So, strictly speaking, the momentum of the bullet (if any) the momentum of the exhaust gases is equal to the opposite momentum of the recoiling gun. Bowling Ball and Conservation of Momentum Conservation of Momentum: Numerical Example FIGURE 3.10 The net momentum before firing is zero. After firing, the net momentum is still zero, because the momentum of the cannon is equal and opposite to the momentum of the cannonball. INTERACTIVE FIGURE FIGURE 3.11 A cue ball hits an eight ball head-on. Consider this event in three systems: (a) An external force acts on the eight-ball system, and its momentum increases. (b) An external force acts on the cue-ball system, and its momentum decreases. (c) No external force acts on the cue-ball eight-ball system, and momentum is conserved (simply transferred from one part of the system to the other). + 8-ball system (a) (b) (c) cue-ball system cue-ball + 8-ball system The Physical Science Place M04_HEWI6954_04_SE_CH03.qxd 10/1/07 3:25 PM Page 65 2008934301 Conceptual Physical Science, Fourth Edition, by Paul G. Hewitt, John Suchocki, and Leslie A. Hewitt. Published by Addison-Wesley. Copyright © 2008 by Pearson Education, Inc. CHECK POINT 1. Newton’s second law states that if no net force is exerted on a system, no acceleration occurs. Does it follow that no change in momentum occurs? 2. Newton’s third law states that the force a cannon exerts on a cannonball is equal and opposite to the force the cannonball exerts on the cannon. Does it follow that the impulse the cannon exerts on the cannonball is equal and opposite to the impulse the cannonball exerts on the cannon? Were these your answers? 1. Yes, because no acceleration means that no change occurs in velocity or in momentum Another line of reasoning is simply that no net force means there is no net impulse and thus no change in momentum. 2. Yes, because the interaction between both occurs during the same time interval. Because time is equal and the forces are equal and opposite, the impulses, Ft, are also equal and opposite. Impulse is a vector quantity and can be canceled. COLLISIONS The collision of objects clearly illustrates the conservation of momentum. Whenever objects collide in the absence of external forces, the net momentum of both objects before the collision equals the net momentum of both objects after the collision. This is true no matter how the objects might be moving before they collide. When a moving billiard ball makes a head-on collision with another billiard ball at rest, the moving ball comes to rest and the other ball moves with the speed of the colliding ball. We call this an elastic collision; ideally, the colliding objects rebound without lasting deformation or the generation of heat (Figure 3.12). But momentum is conserved even when the colliding objects become entangled during the collision. This is an inelastic collision, characterized by deformation, or the generation of heat, or both. In a perfectly inelastic collision, the objects stick together. Consider, for example, the case of a freight car moving along a track and colliding with another freight car at rest (Figure 3.13). If the freight cars are of equal mass and are coupled by the collision, can we predict the velocity of the coupled cars after impact? net momentumbefore collision = net momentumaftercollision (mass * velocity). 66 PART ONE PHYSICS a b c FIGURE 3.12 Elastic collisions of equally massive balls. (a) A green ball strikes a yellow ball at rest. (b) A head-on collision. (c) A collision of balls moving in the same direction. In each case, momentum is transferred from one ball to the other. INTERACTIVE FIGURE M04_HEWI6954_04_SE_CH03.qxd 10/1/07 3:25 PM Page 66 2008934301 Conceptual Physical Science, Fourth Edition, by Paul G. Hewitt, John Suchocki, and Leslie A. Hewitt. Published by Addison-Wesley
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