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Unformatted text preview: chaney (glc568) Impulse, Momentum, Angular Motion murthy (21118) 1 This printout should have 49 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Consider the elastic headon collision between a sledge hammer with 3000 g mass and a golf ball with a 9 g mass. The initial velocity of the sledge hammer is 14 m / s , and the golf ball is initially at rest. 3000 g 9 g 14 m / s Estimate the approximate final speed v 2 of the golf ball. 1. v 2 4 . 45634 m / s 2. v 2 140 m / s 3. v 2 70 m / s 4. v 2 7 m / s 5. v 2 140 m / s 6. v 2 43 . 9823 m / s 7. v 2 112 m / s 8. v 2 56 m / s 9. v 2 28 m / s correct 10. v 2 14 m / s Explanation: v 1 = 14 m / s , v 2 = 0 m / s , m 1 = 3000 g , and m 2 = 9 g . For the present case, v cm = m 1 v 1 + m 2 v 2 m 1 + m 2 v 1 , since m 1 m 2 . Using the expression for the final velocity of the golf ball, we have v 2 = 2 v cm v 2 , where v 2 = 0 (the golf ball is initally at rest). The approximate final speed of the ball is v 2 = 2 v 1 = 2 (14 m / s) = 28 m / s . 002 10.0 points Particle 1 with momentum p 1 strikes Par ticle 2, which is at rest. The particles move in different directions after the collision. The momenta after the collision are p 1 and p 2 . What relationship is true? 1. p 1 x = p 1 x + p 2 x correct 2. p 1 y = p 2 y 3. p 2 1 = ( p 1 ) 2 + ( p 2 ) 2 4. p 1 = p 2 5. p 1 y = p 1 y p 2 y 6. p 1 x = p 1 x p 2 x 7. ( p 1 ) 2 = p 2 1 + ( p 2 ) 2 Explanation: The momentum is conserved. We have p 1 = p 1 + p 2 So in the x direction, p 1 x = p 1 x + p 2 x , and in the y direction, p 1 y = p 1 y + p 2 y . 003 10.0 points Two particles of masses m and 3 m are moving toward each other along the xaxis chaney (glc568) Impulse, Momentum, Angular Motion murthy (21118) 2 with the same speed v . They undergo a head on elastic collision and rebound along the x axis. m v 3 m v Determine the final speed of the heavier object. 1. v 3 m = 3 v 2. v 3 m = 3. v 3 m = 1 3 v 4. v 3 m = 3 2 v 5. v 3 m = 4 v 6. v 3 m = 2 3 v 7. v 3 m = v 8. v 3 m = 2 v 9. v 3 m = 0 correct 10. v 3 m = 1 2 v Explanation: Let : v 1 = v , v 2 = v , m 1 = m and m 2 = 3 m. Basic Concepts: Conservation of momentum Solution: Denote the smaller mass initially moving to the right as m 1 and the larger mass initially moving to the left as m 2 . Since no external forces act on the two masses, even during the collision, the total momentum is conserved....
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This note was uploaded on 12/01/2008 for the course PHYSICS 160 taught by Professor Murthy during the Spring '08 term at Kentucky.
 Spring '08
 MURTHY
 Physics, Momentum

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