A 0.5 kg ball is thrown against a wall with a speed of 10 m/s and rebounds with a speed of 6 m/s. The ball is in contact with the wall for 0.02 seconds. What is the average force exerted by the wall on the ball?

Core Concept

The impulse experienced by the ball equals the change in momentum. Initial momentum = 0.5 kg × 10 m/s = 5 kg·m/s (toward wall). Final momentum = 0.5 kg × (-6 m/s) = -3 kg·m/s (away from wall). Change in momentum = -3 kg·m/s - 5 kg·m/s = -8 kg·m/s. Average force = change in momentum / time = -8 kg·m/s / 0.02 s = -400 N. The negative sign indicates direction, so the magnitude is 400 N.