10_P50InstructorSolution

# 10_P50InstructorSolution - 10.50. Model: We assume the...

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10.50. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Visualize: Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-and- after pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the ball’s rebound velocity is bB fb ib 80 g ( ) ( ) (5.0 m/s) 3.33 m/s 120 g mm vv −− == = + The ball’s speed is 3.3 m/s. (b) An elastic collision gives the block speed B fB 24 0 g ( ) ( ) (5.0 m/s) 1.667 m/s 120 g m = + To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the

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## This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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10_P50InstructorSolution - 10.50. Model: We assume the...

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