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

Question

A 15.0kg block is attached to a very light horizontal spring of force constant 550N/m and is resting on a smooth horizontal table. 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.

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

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= 30/275 Hence x=0.33 m=33 cm.

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