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(A) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = 4K_1$.Since kinetic energy $K = \frac{P^2}{2m}$,where $P$ is moment

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$100$

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(A) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = 4K_1$.Since kinetic energy $K = \frac{P^2}{2m}$,where $P$ is momentum and $m$ is mass,we have $P = \sqrt{2mK}$.Therefore,the ratio of final momentum $P_2$ to initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{4K_1}{K_1}} = \sqrt{4} = 2$.This implies $P_2 = 2P_1$.The percentage change in momentum is given by $\frac{P_2 - P_1}{P_1} 	imes 100$.Substituting the values,we get $\frac{2P_1 - P_1}{P_1} 	imes 100 = \frac{P_1}{P_1} 	imes 100 = 100\%$.

If the kinetic energy of a moving body becomes four times its initial kinetic energy,then the percentage change in its momentum will be $...\%$

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