A 0.49 kg softball is pitched at a speed of 18 m/s.
The batter hits it back directly at the pitcher at a speed of 23 m/s. the bat acts on the ball for 0.015 s. What is the magnitude of the average force exerted by the bat on the ball during contact?
The initial velocity of the ball is 18 m/s moving away from the pitcher and towards the batter. The batter strikes the ball and its velocity changes direction and is now moving at 23 m/s towards the pitcher.
The time for the change in velocity is 0.015 sec. Now acceleration is the rate of change in velocity. Here it is (23 - (-18 ))/ 0.015 = 8200/3 m/ s^2.
The force exerted, to result in this acceleration is equal to mass*acceleration. As the ball is 0.49 kg, the force is 0.49 * 8200/3 = 4018/3 N.
The required force is 4018/3 N.