ch15-p036 - 36. We note that the spring constant is k =...

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block 2 is x = x m cos( ω t + φ ) where φ = π /2 which gives x = x m cos( π /2 + π /2) = x m . This means block 2 is at a turning point in its motion (and thus has zero speed right before the impact occurs); this means, too, that the spring is stretched an amount of 1 cm = 0.01 m at this moment. To calculate its after-collision speed (which will be the same as that of block 1 right after the impact, since they stick together in the process) we use momentum conservation and obtain v = (4.0 kg)(6.0 m/s)/(6.0 kg) = 4.0 m/s. Thus, at the end of the impact itself (while block 1 is still at the same position as before the impact) the system (consisting now of a total mass M = 6.0 kg) has kinetic energy K = 1 2 (6.0 kg)(4.0 m/s) 2 = 48 J and potential energy U = 1 2 kx 2 = 1 2 (1.97 × 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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