Alex throws a 0.10-kg rubber ball down onto the floor.The ball's speed just before impact is 4.5 m/s,and just after is 3.5 m/s.If the ball is in contact with the floor for 0.02 s,what is the magnitude of the average force applied by the floor on the ball?

Question

Quizplus · Accepted Answer

The change in momentum (impulse) is given by the difference in velocity times mass: (0.10 kg)(3.5 m/s - (-4.5 m/s)) = 0.10 kg * 8 m/s = 0.8 kg*m/s. The average force is impulse divided by time: 0.8 kg*m/s / 0.02 s = 40 N.

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