Bouncing Ball: Impulse and Energy Loss
A student investigates a bouncing ball using a force plate and a motion sensor. A 0.050 kg rubber ball is released from rest at a height of 1.20 m above the plate. The motion sensor records the ball's speed just before impact as 4.85 m/s and just after impact as 3.96 m/s. The force plate records a contact time of 0.018 s. The ball rebounds to a height of 0.80 m. In a second trial, the ball is released from 2.40 m and rebounds to 1.60 m. The student claims that the average force exerted by the plate on the ball is equal to the ball's weight, because the ball is only in contact with the plate for a very short time. The student also says that doubling the drop height doubles the rebound height, so the ball loses the same fraction of energy each time. Air resistance is negligible.
1. Calculate the impulse on the ball during contact and the average force exerted by the force plate. Include the direction.
Read the worked answer
Take upward as positive. Just before contact v = −4.85 m/s; just after v = +3.96 m/s. Change in velocity Δv = 3.96 − (−4.85) = 8.81 m/s upward. Impulse J = mΔv = 0.050 × 8.81 = 0.441 N s upward. Average net force on the ball during contact is F_net = J/Δt = 0.441/0.018 = 24.5 N upward. The net force is the vector sum of the plate force and weight: F_net = F_plate − mg. Weight mg = 0.050 × 9.8 = 0.49 N downward. So F_plate = 24.5 + 0.49 = 25.0 N upward. The plate force is about 50 times the weight, so saying it equals the weight is incorrect.
2. Evaluate the student's claim that doubling the drop height doubles the rebound height and that the ball loses the same fraction of energy each time.
Read the worked answer
From 1.20 m the rebound is 0.80 m; from 2.40 m it is 1.60 m. Both give rebound/drop = 0.667, so the data support the claim over this range. The gravitational potential energy at release is mgh, so the energy just before impact is proportional to drop height. The rebound energy is proportional to rebound height. A constant ratio of 2/3 means the ball retains 2/3 of its mechanical energy and loses 1/3 on each bounce. Because v = √(2gh), doubling h increases impact speed by √2, not 2. The result is consistent with a constant fractional energy loss, but it is an empirical result for this ball and surface, not a general law; at much higher speeds air resistance and deformation may change the fraction.
3. The student says the force-plate reading is not the net force. Explain whether the student is correct, and predict the effect of dropping from 2.40 m with the same contact time and the same fractional energy loss.
Read the worked answer
The force plate measures the normal force it exerts on the ball, not the net force. The net force is normal force minus weight. The student is correct: if the plate reading were the net force, the weight would have to be added to get the true net force. From 2.40 m, the impact speed is √2 times larger than from 1.20 m if the same fraction of energy is lost, so the speed after bounce is also √2 times larger. The change in velocity is √2 times larger, so the impulse and average net force are √2 times larger. The new average net force would be about 24.5 × 1.414 = 34.6 N, and the plate force about 34.6 + 0.49 = 35.1 N. The contact time would probably not stay exactly the same in reality, so this is an estimate.