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...

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?

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william1941 | College Teacher | (Level 3) Valedictorian

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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.

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