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Unformatted text preview: Solution for Longanswer Homework 8 Energy problems Solution to Longanswer Homework Problem 8.1() Problem: A car with inertia 1200kg is backing out of a parking space at 5 m s . An unobservant driver of a 1800kg pickup truck is coasting through the parking lot at a speed of 3 m s , and runs straight into the rear bumper of the car. (a)What is the change in internal energy of the system, if the speed of the pickup is 1 . 5 m s backwards after they collide? (b)What is the coefficient of restitution for this collision? Solution to Part (a) Conserving momentum: m c v c,i + m t v t,i = m c v c,f + m t v t,f v c,f = m c v c,i + m t ( v t,i v t,f ) m c v c,f = (1200kg) ( 5 m s ) + (1800kg) ( 3 m s ( 1 . 5 m s )) 1200kg v c,f = 1 . 75 m s Energy is conserved, so E + K = 0 E = K = K i K f = 1 2 m c ( v 2 c,i v 2 c,f ) + 1 2 m t ( v 2 t,i v 2 t,f ) E = 1 2 (1200kg) parenleftbigg parenleftBig 5 m s parenrightBig 2 parenleftBig 1 . 75 m s parenrightBig 2 parenrightbigg + 1 2 (1800kg) parenleftbigg parenleftBig 3 m s parenrightBig 2 parenleftBig 1 . 5 m s parenrightBig 2 parenrightbigg E = 19237 . 5J Notice this is positive, which means the internal energy of the system increases during the collision. Grading Key: Part (a) 3 Points Solution to Part (b) The coefficient of restitution is given by: e = v rel,f v rel,i So e = v c,f v t,f v c,i v t,i = 1 . 75 m s ( 1 . 5 m s ) 5 m s 3 m s e = 0 . 40625 Grading Key: Part (b) 2 Points...
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This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.
 Spring '08
 Stewart
 Physics, Energy, Inertia, Work

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