A 15.0 block is attached to a very light horizontal spring of force constant 350

Question

A 15.0 block is attached to a very light horizontal spring of force constant 350 and is resting on a smooth horizontal table. (See the figure below (Figure 1) .) Suddenly it is struck by a 3.00 stone traveling horizontally at 8.00 to the right, whereupon the stone rebounds at 2.00 horizontally to the left. A) Find the maximum distance that the block will compress the spring after the collision.(Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

Explanation / Answer

First apply the law of conservation of momentum to the collision: The momentum before collision = 3*8=24 kg-m/s. This equals the momentum after collision: -3*2+15*v=24, where v=speed of block right after collision. Thus: v=(24+6)/15=2 m/s. Next apply the law of conservation of energy to the compression of the spring: K.E. of block (initial energy)= P.E. of spring (final energy) or 1/2 M v^2 = 1/2 k x^2 1/2 * 15 * 2^2 = 1/2 550 * x^2 or 30= 225 x^2 or x^2= 30/225 Hence x=0.36 m=36 cm. Note that you can't apply the law of conservation of energy to a collision unless you know that the collision is elastic.

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