Question

A(n) 0.49 kg softball is pitched at a speed of20 m/s. The batter hits it back directly at thepitcher at a speed of 25 m/s. The bat acts onthe ball for 0.012 s. What is the magnitude of the average force exerted by the bat on the ball? Answer in units of N.

Answer 1

Given,

The mass of the softball, m=0.49 kg

The speed with which the ball was thrown, u=-20 m/s

The speed of the ball after the batter hits it, v=25 m/s

The time duration for which the bat acts on the ball, t=0.012 s

Here we have considered the direction from the pitcher to the batter as the negative direction and the direction from batter to the pitcher as the positive direction.

The momentum of an object is given by the product of the mass of the object and its velocity.

The change in the momentum of the ball is given by,

`\begin{gathered} \Delta p=mv-mu \\ =m(v-u) \end{gathered}`

On substituting the known values,

`\begin{gathered} \Delta p=0.49(25-(-20)) \\ =22.05kg\cdot(m)/(s) \end{gathered}`

The change in the momentum is equal to the impulse that acts on an object due to the applied force. And the impulse is given by the product of the applied force and the time duration.

Thus,

`\begin{gathered} I=\Delta p \\ =F\cdot t \\ \Rightarrow F=(\Delta p)/(t) \end{gathered}`

Where I is the impulse and F is the magnitude of the average force exerted by the bat on the ball.

On substituting the known values,

`\begin{gathered} F=(22.05)/(0.012) \\ =1837.5\text{ N} \end{gathered}`

Thus the magnitude of the average force exerted by the bat on the ball is 1837.5 N