A 24 g steel ball is released from rest and falls vertically onto a steel plate.

Question

A 24 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 3 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
A 24 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 3 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
1410 N
710 N
1170 N
1650 N
940 N

Explanation / Answer

Time of flight = (3-0.0005) / 2 = 1.49975 s

Now

v = gt = 9.8*1.49975 = 14.7 m/s

Change in momentum = 0.024*2*14.7 = 0.7055 kgm/s

Rate of change of momentum = F = 0.7055/0.0005 = 1410.965 N

Hence answer is option A

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