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A cricket ball of mass 150 g moving with a speed of $126 {kmh}^{- 1}$ hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be

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10.5 N

21 N

$1.05 \times 10^{4} N$

$2.1 \times 10^{4} N$

answer is C.

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Detailed Solution

Here, $u = - v = 126 {kmh}^{- 1} = 126 \times \frac{5}{18} = 35 m / s$

Change in momentum of the ball,

$Δ p = m (v - u) = \frac{150}{1000} (- 35 - 35)$

$= \frac{3}{20} (- 70) = - \frac{21}{2} kg - {ms}^{- 1}$

Now, force, $F = \frac{Δ p}{Δ t} = \frac{- 21 / 2}{0.001} N = - 1.05 \times 10^{4} N$

Here, negative sign shows that direction of force will be opposite to the direction of movement of the ball before
hitting.

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