A 15.0kg block is attached to a very light horizontal spring of force constant 4

Question

A 15.0kg block is attached to a very light horizontal spring of force constant 475N/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

We can find the block's velocity after impact using conservation of momentum (p)

p(i) = p(f) p(i) = 3.00kg*8.0m/s = 24.0kg-m/s

p(f) = 3.00*(-2.00) + 15*V Setting these equal gives -6.00 + 15V = 24

So V = (24 + 6)/15 = 2.00m/s

Now this kinetic energy is converted to potential energy in the spring

So 1/2*m*v^2 = 1/2*k*x^2 Or x = sqrt(m*v^2/k) = sqrt(15*2^2/475) = 0.35541 m

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