A ball of mass 0.6 kg hits a rigid wall and rebounds with the same speed. If the impact lasts 0.02 s, what is the average force exerted on the ball?

A ball of mass 0.6 kg hits a rigid wall and rebounds with the same speed. If the impact lasts 0.02 s, what is the average force exerted on the ball?

• A. 3.6 N
• B. 180 N
• C. 90 N
• D. 360 N

Answer: (B)

Impulse from the wall changes the ball’s momentum, and the average force during the impact is that impulse divided by the contact time. If the ball of mass m = 0.6 kg hits the wall with speed v and leaves with the same speed in the opposite direction, the velocity changes from +v to −v. The change in momentum is Δp = m(−v − (+v)) = −2 m v, so the impulse magnitude is 2 m v. The average force is F_avg = Δp/Δt = 2 m v / Δt. With the contact time Δt = 0.02 s, this gives F_avg = (2 × 0.6 × v) / 0.02 = 60 v N. If the approach speed is 3 m/s, then F_avg = 60 × 3 = 180 N. So the average force is 180 N when the ball approaches at 3 m/s; the same relation applies for any other speed, just yielding a different force accordingly.

Impulse from the wall changes the ball’s momentum, and the average force during the impact is that impulse divided by the contact time. If the ball of mass m = 0.6 kg hits the wall with speed v and leaves with the same speed in the opposite direction, the velocity changes from +v to −v. The change in momentum is Δp = m(−v − (+v)) = −2 m v, so the impulse magnitude is 2 m v. The average force is F_avg = Δp/Δt = 2 m v / Δt. With the contact time Δt = 0.02 s, this gives F_avg = (2 × 0.6 × v) / 0.02 = 60 v N. If the approach speed is 3 m/s, then F_avg = 60 × 3 = 180 N. So the average force is 180 N when the ball approaches at 3 m/s; the same relation applies for any other speed, just yielding a different force accordingly.