A crane drops a 0.250-kg steel ball onto a steel plate.The ball's speeds just before impact and after are 4.00 m/s and 3.70 m/s,respectively.If the ball is in contact with the plate for 0.035 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:$$J=mv_f-mv_i$$$$J=(0.250	ext{ kg})(3.70	ext{ m/s})-(0.250	ext{ kg})(4.00	ext{ m/s})=-0.075	ext{ kg$\cdot$m/s}$$The average force exerted by the ball on the plate is:$$F=\frac{J}{t}=\frac{-0.075	ext{ kg$\cdot$m/s}}{0.035	ext{ s}}=\boxed{2.14	ext{ N}}$$

A Crane Drops a 0