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Unformatted text preview: Patel (pnp223) – homework 23 – Turner – (92160) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Consider the general case of the collision of a mass m 1 = m with mass m 2 = 2 m along a frictionless horizontal surface. Denote the initial and the final center of mass momenta to be p i cm = p 1 + p 2 and p f cm = p ′ 1 + p ′ 2 and the initial and final kinetic energies to be K i = K 1 + K 2 and K f = K ′ 1 + K ′ 2 . v 1 m v 2 2 m For an elastic collision which pair of state ments is correct? 1. p i CM > p f CM , and K i < K f 2. p i CM > p f CM , and K i > K f 3. p i CM < p f CM , and K i < K f 4. p i CM < p f CM , and K i = K f 5. p i CM < p f CM , and K i > K f 6. p i CM > p f CM , and K i = K f 7. p i CM = p f CM , and K i = K f correct 8. p i CM = p f CM , and K i < K f 9. p i CM = p f CM , and K i > K f Explanation: Momentum and energy are always con served in elastic collisions. 002 (part 2 of 3) 10.0 points Consider a perfectly inelastic headon colli sion with initial velocity for m 1 of v 1 but with m 2 motionless. Find the final speed of the system. 1. v CM = 1 4 v 1 2. v CM = 2 7 v 1 3. v CM = 1 5 v 1 4. v CM = 2 3 v 1 5. v CM = v 1 6. v CM = 3 5 v 1 7. v CM = 1 3 v 1 correct 8. v CM = 3 4 v 1 9. v CM = 2 v 1 10. v CM = 1 2 v 1 Explanation: Let : v 2 = 0 . In a collision the linear momentum is always conserved, so m 1 v 1 + m 2 · 0 = ( m 1 + m 2 ) v f v CM = v f = m 1 m 1 + m 2 v 1 = m 3 m v 1 = v 1 3 . 003 (part 3 of 3) 10.0 points Now consider an elastic headon collision with initial velocity for m 1 of v 1 but with m 2 mo tionless. Find the final speed of m 2 . 1. v f 2 = 1 5 v 1 2. v f 2 = 2 v 1 3. v f 2 = 3 4 v 1 Patel (pnp223) – homework 23 – Turner – (92160) 2 4. v f 2 = 2 3 v 1 correct 5. v f 2 = 1 3 v 1 6. v f 2 = 3 5 v 1 7. v f 2 = 1 2 v 1 8. v f 2 = v 1 9. v f 2 = 2 7 v 1 10. v f 2 = 1 4 v 1 Explanation: Let : v 2 = 0 . In an elastic collision v 1 0 = v f 2 v f 1 and conservation of momentum gives m 1 v 1 + m 2 · 0 = m 1 v f 1 + m 2 v f 2 . Multiplying the first equation by m 1 , m 1 v 1 = m 1 v f 2 m 1 v f 1 m 1 v 1 = m 1 v f 1 + m 2 v f 2 and adding, 2 m 1 v 1 = ( m 1 + m 2 ) v f 2 2 mv 1 = 3 mv f 2 v f 2 = 2 3 v 1 ....
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This note was uploaded on 07/11/2009 for the course PHY 92160 taught by Professor Turner during the Spring '09 term at University of Texas.
 Spring '09
 Turner
 Physics, Work

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