Quick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 6 Momentum and Collisions Quick Quizzes 1. (d). We are given no information about the masses of the objects. If the masses are the same, the speeds must be the same (so that they have equal kinetic energies), and then . If the masses are not the same, the speeds will be different, as will the momenta, and either , or , depending on which particle has more mass. Without information about the masses, we cannot choose among these possibilities. 1pp=22 21pp<1pp>2. (c). Because the momentum of the system (boy + raft) remains constant with zero magnitude, the raft moves towards the shore as the boy walks away from the shore. 3. (c). The total momentum of the car-truck system is conserved. Hence, any change in momentum of the truck must be counterbalanced by an equal magnitude change of opposite sign in the momentum of the car. 4. (a). The total momentum of the two-object system is zero before collision. To conserve momentum, the momentum of the combined object must be zero after the collision. Thus, the combined object must be at rest after the collision. 5. (a) Perfectly inelastic. Any collision in which the two objects stick together afterwards is perfectly inelastic. (b) Inelastic. Both the Frisbee and the skater lose speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved. (c) Inelastic. The kinetic energy of the Frisbee is conserved. However, the skater loses speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved. 6. (a). If all of the initial kinetic energy is transformed, then nothing is moving after the collision. Consequently, the final momentum of the system is necessarily zero. Because momentum of the system is conserved, the initial momentum of the system must be zero meaning that the two objects must have had equal magnitude momenta in opposite directions before the collision. 191192 CHAPTER 6 Answers to Even Numbered Conceptual Questions 2. The glass, concrete, and steel were part of a rigid structure that shattered upon impact of the airplanes with the towers and upon collapse of the buildings as the steel support structures weakened due to high temperatures of the burning fuel. The sheets of paper floating down were probably not in the vicinity of the direct impact, where they would have burned after being exposed to very high temperatures. The papers were most likely situated on desktops or open file cabinets and were blown out of the buildings as they collapsed. 4. No. Only in a precise head-on collision with equal and opposite momentum can both balls wind up at rest. Yes. In the second case, assuming equal masses for each ball, if Ball 2, originally at rest, is struck squarely by Ball 1, then Ball 2 takes off with the velocity of Ball 1. Then Ball 1 is at rest. 6. The skater gains the most momentum by catching and then throwing the Frisbee. 8. Kinetic energy can be written as 22pm. Thus, even through the particles have the same kinetic energies their momenta may be different due to a difference in mass. 10. The resulting collision is intermediate between an elastic and a completely inelastic collision. Some energy of motion is transformed as the pieces buckle, crumple, and heat up during the collision. Also, a small amount is lost as sound. The most kinetic energy is lost in a head-on collision, so the expectation of damage to the passengers is greatest. 12. The less massive object loses the most kinetic energy in the collision. 14. The superhero is at rest before the toss and the net momentum of the system is zero. When he tosses the piano, say toward the right, something must get an equal amount of momentum to the left to keep the momentum at zero. This something recoiling to the left must be the superhero. He cannot stay at rest. 16. The passenger must undergo a certain momentum change in the collision. This means that a certain impulse must be exerted on the passenger by the steering wheel, the window, an air bag, or something. By increasing the time during which this momentum change occurs, the resulting force on the passenger can be decreased. 18. A certain impulse is required to stop the egg. But, if the time during which the momentum change of the egg occurs is increased, the resulting force on the egg is reduced. The time is increased when the sheet billows out as the egg is brought to a stop. The force is reduced low enough so that the egg will not break.Momentum and Collisions 193 Answers to Even Numbered Problems 2. (a) 5.40 N·s (b) –27.0 J 4. (a) 0 (b) 1.1 kg m s⋅ 6. 1.7 kN 8. 41.91 10 N upward×10. (a) 7.50 kg m s westward⋅ (b) 375 N eastward12. (a) 12.0 N·s (b) 6.00 m s (c) 4.00 m s 14. 260 N perpendicular to the wall 16. (a) 6.3 kg m s⋅33.2 10 N× toward the pitcher (b) toward the pitcher 18. 62 s 20. (a) 0.49 m s (b) -22.0 10 m s× 22. 2.48 m sthrowerv = , 22.25 10 m scatcherv−=× 24. (a) B exerts a horizontal force on A. (b) A exerts a horizontal force on B that is opposite in direction to the force B exerts on A. (c) The force on A is equal in magnitude to the force on B, but is oppositely directed. (d) Yes. The momentum of the system (the two skaters) is conserved because the net external force on the system is zero (neglecting friction). (e) 2.22 m/s 26. , no broken bones 3av3.75 10 NF =×28. 25.3 10 m s× 30. 143 m s 32. (a) 20.9 m s East (b) into internal energy 38.68 10 J×34. (a) 2.2 m s toward the right (b) No 36. 40.0 cm s− (10.0-g object), 10.0 cm s+ (15.0-g object) 38. (a) 2.50 m s (b) 43.75 10 J×194 CHAPTER 6 40. (a) 0, 1. (b) 50 m s 1.00 m s, 1.50 m s− (c) 1.00 m s, 1.50 m s 42. (a) 12.4 m s at 14.9° N of E (b) 7.20 % 44. No, his speed was 41.5 mi h . 46. 40.5 g 48. 0.556 m 50. 4minMvgm=A 52. 0.960 m above the level of point B 54. 91 m s 56. (a) 9.90 m s, 9.90 m s− (b) 16.5 m s, 3.30 m s− (c) 13.9 m, 0.556 m 58. 0.980 m 60. (a) 0, 3.00 m sred bluevv== (b) 0.212 m 62. (a) 03022, 3mmvv v v== (b) 35.3° 64. (a) 90.0° (b) ()3.46 m s cue ball , ()2.00 m s target 66. 0.31 m 68. (a) See solution for diagrams. (b) From Newton’s third law, the forces have equal magnitudes and opposite directions. (c) 23ApMv∆=− ,