A crane drops a 0.30 kg steel ball onto a steel plate.The ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s,respectively.If the ball is in contact with the plate for 0.030 s,what is the magnitude of the average force that the ball exerts on the plate during impact?

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Quizplus · Accepted Answer

The impulse on the ball is equal to the change in its momentum:$$J = mv_f - mv_i$$$$J = (0.30 	ext{ kg})(4.2 	ext{ m/s}) - (0.30 	ext{ kg})(4.5 	ext{ m/s})$$$$J = -0.09 	ext{ kg m/s}$$The average force on the ball is equal to the impulse divided by the time of impact:$$F = \frac{J}{t}$$$$F = \frac{-0.09 	ext{ kg m/s}}{0.030 	ext{ s}}$$$$F = 3.0 	ext{ N}$$

A Crane Drops a 0