block 2 is x= xmcos(ωt+ φ) where φ =π/2 which gives x= xmcos(π/2 + π/2) = –xm. 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 energyK= 12(6.0 kg)(4.0 m/s)2= 48 J and potential energyU=12kx2=12(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.