Alex throws a 0.250-kg rubber ball down onto the floor. The ball’s speed just be

Question

Alex throws a 0.250-kg rubber ball down onto the floor. The ball’s speed just before impact is 5.30 m/s, and just after is 3.71 m/s. If the ball is in contact with the floor for 0.028 s, what is the magnitude of the average force applied by the floor on the ball? Alex throws a 0.250-kg rubber ball down onto the floor. The ball’s speed just before impact is 5.30 m/s, and just after is 3.71 m/s. If the ball is in contact with the floor for 0.028 s, what is the magnitude of the average force applied by the floor on the ball?

Explanation / Answer

Mass = m = 0.25 kg

Velocity before impact = v1 = 5.3 m/s

Velocity after impact = v2 = 3.71 m/s

Time = t = 0.028 s

impulse = change in momentum
impulse = (0.250 kg)(3.71 m/s - (-5.3 m/s))
impulse = 2.25 kg-m/s (upward)

Now

impulse = (force exerted)(time)
2.25 kg-m/s = F(0.028 s)
F = 80.4 N (upward)

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