Question

A cricket ball of mass 0.20 kg is moving with a velocity of 1.2 m/s. Find the average force applied by the player if he is able to stop the ball in 0.10 s?​

Answer 1

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m = 0.2 kg

u = 1.2 m/s

v = 0 (as the player stops the ball)

t = 0.1 s

\begin{gathered}a = \frac{v - u}{t} \\ = \frac{ 0 - 1.2}{0.1} \\ = - 12\end{gathered}a=tv−u=0.10−1.2=−12

a = -12 m/s²

F = ma

= 0.2×(-12)

= -2.4 N

Impulse = F×t

= -2.4×0.1

= -0.24 N s

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