Question

A pitcher throws a 0.19-kg baseball ball with a velocity of 42 m/s. It is struck...

A pitcher throws a 0.19-kg baseball ball with a velocity of 42 m/s. It is struck by the bat with an average force of 700 N, which results in a velocity of 28 m/s in the opposite direction from the original velocity. How long were the bat and ball were in contact?

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Answer #1

Given,

Mass of the baseball bll, m = 0.19 kg

Initial velocity, v1 = 42 m/s

Final velocity, v2 = -28 m/s

Average force, F = 700 N

We know,

Impulse = change in momentum

Ft = p

Ft = | mv2 - mv1 |

t = | mv2 - mv1 |/F

t = (m|v2 - v1|)/F

t = ( 0.19 x | -28 - 42|)/700

t = 0.019 s

Answer : Average time = 0.019 s = 19 ms

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