A 150 g cricket ball moving at a speed of $12 m s^{- 1}$ is hit by a bat and turned back at a speed of $20 m s^{- 1}$ . The duration of impact is 0.01 s. What is the average force exerted on the ball by the bat?

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Solution

The initial momentum = $\frac{150}{1000} \times 12 = \frac{9}{5} k g - m / s$

Final momentum = $\frac{- 150}{1000} \times 20 = - 3 k g - m / s$

Force $= \frac{Δ P}{Δ t} = \frac{- 3 - \frac{9}{5}}{0.01} = \frac{24 \times 100}{5} 480 N$

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150 g

is moving with a velocity of

12 m / s

and is hit by a bat so that the ball is turned back with a velocity of

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. The force of blow acts for

0.01 s

on the ball. The average force exerted by the bat on the ball is (in

N

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A 150 g cricket ball moving at a speed of 12ms−1 is hit by a bat and turned back at a speed of 20ms−1. The duration of impact is 0.01 s. What is the average force exerted on the ball by the bat?

The initial momentum = 1501000×12=95kg−m/sFinal momentum = −1501000×20=−3kg−m/sForce =ΔPΔt=−3−950.01=24×1005480N

A 150 g cricket ball moving at a speed of $12 m s^{- 1}$ is hit by a bat and turned back at a speed of $20 m s^{- 1}$ . The duration of impact is 0.01 s. What is the average force exerted on the ball by the bat?

The initial momentum = $\frac{150}{1000} \times 12 = \frac{9}{5} k g - m / s$

Final momentum = $\frac{- 150}{1000} \times 20 = - 3 k g - m / s$

Force $= \frac{Δ P}{Δ t} = \frac{- 3 - \frac{9}{5}}{0.01} = \frac{24 \times 100}{5} 480 N$