A solid 600-g concrete block is struck by a 50-g rubber ball traveling at 1.75 m/sec. If the ball bounces backward off the block at 0.9 m/sec, what will be the speed of the concrete block?

Question

TheoryPractice · Accepted Answer

0.07 m/sec. To find the speed of the concrete block (v2), use the momentum equation (m1v1 + m2v2 = m1v1 + m2v2).

v2 = (m1v1 + m2v2 – m1v1) / m2

Where the rubber ball is m1, the concrete block is m2, and the initial velocity of the concrete block is 0.

v2 = [(50 g x 1.75 m/sec) + (600 g x 0 m/sec) – (50 g x 0.9 m/sec)] / 600 g

v2 = 0.07 m/sec

A solid 600-g concrete block is struck by a 50-g rubber ball traveling at 1.75 m/sec. If the ball bounces backward off the block at 0.9 m/sec, what will be the speed of the concrete block?

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