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 400N/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

using momentum conservation on block and stone ,

15 x 0 + 3 x 8 = 15u + 3 x -2

u = 2 m/s

after that using energy conservation on block-spring system,

kx^2/2 = mv^2/2

400x^2/2 = 15 x 2^2/2

x = 0.387 m   or 38.7 cm

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