A cricket ball of mass 0.15 kg is moving with a velocity of 12 ms - 1 and is hit by a bat so that the ball is turned back with a velocity of 20 ms - 1. If the force of blow acts for 0.01 s, find the average force exerted on the ball by the bat.

Given that

Initial velocity u=12 ms−1

Final velocity v=20 ms−1

Time for which force is acting, t = 0.01 s

Find out

We have to determine the average force exerted on the ball by the bat

Solution

Change in velocity =20 – (- 12) = 20 + 12 = 32 ms−1

The -ve sign is taken because initial and final velocities are in the opposite direction
Time for which force is acting, t = 0.01 s
∴acceleration a=change in velocity / time

= 32/0.01

= 3200 ms-2

From Newton’s second law of motion, we note that

F=ma

0.15 x 3200 = 480 N

Answer

The average force exerted on the ball by the bat= 480N

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