A player kicks a football of mass 0.5 kg and | vedclass

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(C) According to Newton's second law of motion,the force $F$ is equal to the rate of change of momentum.$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u

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$250$

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(C) According to Newton's second law of motion,the force $F$ is equal to the rate of change of momentum.$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$Given:Mass $m = 0.5 \ kg$Initial velocity $u = 0 \ m/s$Final velocity $v = 10 \ m/s$Time interval $\Delta t = \frac{1}{50} \ s$Substituting the values:$F = \frac{0.5 	imes (10 - 0)}{1/50}$$F = 0.5 	imes 10 	imes 50$$F = 5 	imes 50 = 250 \ N$Therefore,the force acted on the football is $250 \ N$.

$A$ player kicks a football of mass $0.5 \ kg$ and the football begins to move with a velocity of $10 \ m/s$. If the contact between the leg and the football lasts for $\frac{1}{50} \ s$,then the force acted on the football should be ........... $N$.

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