A 15.0kg block is attached to a very light horizontal spring of force constant 3

Question

A 15.0kg block is attached to a very light horizontal spring of force constant 325N/m and is resting on a smooth horizontal table. (See the figure below (Figure 1) .) Suddenly it is struck by a 3.00kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00m/s horizontally to the left.

Part 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.)

A 15.0kg block is attached to a very light horizontal spring of force constant 325N/m and is resting on a smooth horizontal table. (See the figure below (Figure 1) .) Suddenly it is struck by a 3.00kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00m/s horizontally to the left. Part 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 = 1/2 600 * x^2
30= 300 x^2 or x^2= 30/300 Hence x=0.31 m=31 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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