A cricket ball of mass 50 g having velocity | vedclass

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(B) Given:Mass $m = 50 \ g = 50 	imes 10^{-3} \ kg = 0.05 \ kg$Initial velocity $u = 50 \ cm \ s^{-1} = 50 	imes 10^{-2} \ m \ s^{-1} = 0.5 \ m \ s^

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$0.05$

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(B) Given:Mass $m = 50 \ g = 50 	imes 10^{-3} \ kg = 0.05 \ kg$Initial velocity $u = 50 \ cm \ s^{-1} = 50 	imes 10^{-2} \ m \ s^{-1} = 0.5 \ m \ s^{-1}$Final velocity $v = 0 \ m \ s^{-1}$Time $t = 0.5 \ s$Initial momentum $p_i = m 	imes u = 0.05 \ kg 	imes 0.5 \ m \ s^{-1} = 0.025 \ kg \ m \ s^{-1}$Final momentum $p_f = m 	imes v = 0.05 \ kg 	imes 0 \ m \ s^{-1} = 0 \ kg \ m \ s^{-1}$According to Newton's second law,the force $F$ is the rate of change of momentum:$F = \frac{\Delta p}{t} = \frac{p_f - p_i}{t}$$F = \frac{0 - 0.025 \ kg \ m \ s^{-1}}{0.5 \ s}$$F = -0.05 \ N$The magnitude of the force applied to stop the ball is $0.05 \ N$.

$A$ cricket ball of mass $50 \ g$ having velocity $50 \ cm \ s^{-1}$ is stopped in $0.5 \ s$. The force applied to stop the ball is (in $N$)

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