Momentum Test v3 PSI AP 1 Physics Name____________________________________ Multiple Choice– Choose the correct answer for each question. No partial credit will be given. 1. A 2 kg toy car is traveling at a speed of 0.5 m/s. For how long must a 0.2 N force be applied to raise the car’s momentum by 1.2 kg-m/s? A) B) C) D) 7s 2s 6s 9s 2. As shown, a baseball of mass m rebounds from a vertical wall with the same speed as it had initially. What is the change in momentum of the baseball in the vertical direction? A) B) C) D) 0 2mvcosθ 2mvcosθ 2mvsinθ 3. The velocity of a moving bicycle is tripled. Which of the following quantities must also be tripled? A) B) C) D) Acceleration Kinetic Energy Momentum Potential Energy 4. A steel ball (1) moving at a constant speed, v on a horizontal frictionless surface, collides obliquely with an identical ball (2) initially at rest. The velocity of the ball (1) before and after the collision is shown in the diagram below. What is the approximate direction of the velocity of ball (2) after the collision? 2 1 A) C) B) D) 1 Questions 5-6: Use the given graph below to answer questions 5-6. The graph represents the force applied over time to an object of mass 2.0 kg that was traveling at a velocity of 0.50 m/s at t = 0 s. 5. What is the change in momentum of the object after 4.0 s? A) B) C) D) 2.0 N-m 3.0 N-m 4.0 N-m 0 N-m 6. What is the speed of the object at t = 2.0 s? A) B) C) D) 1.0 m/s 1.5 m/s 2.0 m/s Not enough information is given. 7. Ball 1 of mass 3m is moving at a velocity of 4v m/s north and strikes Ball 2 of mass 4m which is traveling at a velocity of v north. If the final velocity of ball 1 is 2v north, what is the final velocity of ball 2? A) B) C) D) 3v/2 north 5v/2 north 2v north v/2 north 8. Object 1 of mass 3.0 kg travels east with a speed of 5.0 m/s. Object 2 of mass 4.0 kg travels south with a speed of 3.0 m/s. The two objects collide and stick together. What is the momentum of the two objects (magnitude and the angle they make with the east-west axis) after the collision A) B) C) D) 19 kg-m/s; -40º 19 kg-m/s; -39º 21 kg-m/s; -40º 21 kg-m/s; -39º 9. A physics student is standing on a boat in the middle of a river without a paddle. The student has a very heavy chemistry book on deck. She wants to get to the north shore of the river, so she decides to throw the book off the boat. Which direction should she throw it? Which physics law applies to this situation? A) B) C) D) North; conservation of energy South; conservation of energy North; conservation of momentum South; conservation of momentum 10. A 5 kg block moves with a constant speed of 3 m/s on a horizontal frictionless surface and collides elastically with an identical block initially at rest. The second block then moves and collides with a third, identical block, also initially at rest. This time, it’s a perfect inelastic collision and the two blocks move off together. What is the initial speed of the second block after its collision? What is the initial speed of the second and third block system after their collision? A) B) C) D) 3 m/s; 3/2 m/s 3 m/s; 5/2 m/s 3/2 m/s; 3 m/s 3/2 m/s; 2 m/s Multi-Correct: For each of the questions or incomplete statements below, two of the suggested answers will be correct. For each of these questions, you must select both correct choices to earn credit. No partial credit will be earned if only one correct choice is selected. Select the two that are best in each case and then enter both of the appropriate letters in the corresponding space on the answer sheet. 11. A fisherman is standing in the back of his small fishing boat (the mass of the fisherman is the same as the mass of the boat) and he is a few meters from shore. His is tired of fishing so he starts walking on his boat towards the shore so he can get off the boat. What happens to the boat and the fisherman? Assume there is no friction between the boat and the water. A) B) C) D) The fisherman gets closer to the shore. The fisherman gets further away from the shore. The boat gets closer to the shore. The boat gets further away from the shore. 12. A spring is compressed between two blocks with unequal masses, m 1 and m2, held together by a spring. The objects are initially at rest on a horizontal surface with no friction. The spring is then cut. What is true of the two object system after the spring is cut? A) B) C) D) The net final momentum of the two objects is zero. Both objects will travel with different constant speeds unless they hit another object. The energy of the system increases as the blocks separate. The net final momentum of the two objects is positive. Free Response - Solve this problem, showing all work. Partial credit may be given. A 5.00 kg piece of clay moves with a constant v = 30.0 m/s. The piece of clay collides and sticks to a massive ball of mass 20.0 kg suspended at the end of a 2.00 m string. The combined masses swing up in an arc, reach a maximum height, and return to the original position. a. Calculate the initial momentum of the piece of clay. b. A student predicts that the clay will stick to the ball in an inelastic collision. i. Assuming his prediction is correct, find the net momentum of the clayball system directly after the collision. ii. Find the velocity of the clay-ball system directly after the collision. iii. Find the maximum height that the clay-ball system reaches on its upswing. c. The student conducted an experiment and found that the collision was inelastic, but the clay did not stick to the ball – in fact, it rebounded from the ball with a velocity of -10.0 m/s. i. Find the final velocity of the ball. ii. Find the kinetic energy transferred into other forms of energy during the inelastic collision. Momentum Test v3 PSI AP 1 Physics ANSWER KEY Weight 50% Multiple Choice, 50% Free Response Multiple Choice1 point each, 12 points total 1. C 2. A 3. C 4. D 5. C 6. A 7. B 8. B Free Response – 15 points total a. 2 points p=mv=(5kg)(30m/s) p=150 kg-m.s b. 6 points i. momentum is conserved 150 kg-m/s ii. po=pf 150kgm/s=(m1+m2)vf 150kgm/s=(20+5kg)(vf) vf = 6.00 m/s iii. ½(m1+m2)vf2=(m1+m2)gh h=l(1-cosø) ½(m1+m2)vf2=(m1+m2)gl(1-cosø) vf2/2=gl(1-cosø) vf2/(2gl)=1-cosø cosø=1-vf2/(2gl)=1-((6m/s)2/(2x9.8m/s2x2m)) ø=85.3º h=2(1-cos26) h = 1.84 m c. 7 points i. po=pf Negative is opposite direction of initial path 150=m1v1f+m2v2f 150=-50+m2v2f 200=20v2f v2f =10.0 m/s ii. Ko=1/2mv2=1/2(5)(30)2=2250 J Kf1=1/2mv2=1/2(5)(-20)2=1000 J Kf2=1/2mv2=1/2(10)(12.5)2=781 J ∆K=(Kf1+Kf2)-K0=(1000J+781J)-2250J=-469J KE transferred is 469 J 9. D 10. A 11. A, D 12. A, B

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