This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full Document
 Spring '08
 MURTHY
 Physics, Force, Friction, Momentum, m/s, Chaney

Click to edit the document details