A cricket ball of mass 150g moving with a velocity20m/s strikes the bat and rebound back straight forward to the bowler with a speed 30m/s. If the time of contact of the ball with the bat be 0.01s then what will be the force exerted by the bat on the ball?
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Explanation:
impulse=F×t= Change in Momentum
⇒ Change in Momentum =mv−mu=0.15(12)−0.15(−20)=4.8
⇒F×0.001=4.8
F=4800N
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