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(A) The relation between kinetic energy $(KE)$ and momentum $(p)$ is given by $KE = \frac{p^2}{2m}$,which implies $p = \sqrt{2m \cdot KE}$.Let the ini

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$100$

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(A) The relation between kinetic energy $(KE)$ and momentum $(p)$ is given by $KE = \frac{p^2}{2m}$,which implies $p = \sqrt{2m \cdot KE}$.Let the initial kinetic energy be $KE_1 = KE$ and the initial momentum be $p_1 = \sqrt{2m \cdot KE}$.The kinetic energy increases by $300\%$,so the new kinetic energy is $KE_2 = KE + 300\% 	ext{ of } KE = KE + 3KE = 4KE$.The new momentum is $p_2 = \sqrt{2m \cdot KE_2} = \sqrt{2m \cdot 4KE} = 2 \sqrt{2m \cdot KE} = 2p_1$.The percentage increase in momentum is given by $\frac{p_2 - p_1}{p_1} 	imes 100\%$.Substituting the values: $\frac{2p_1 - p_1}{p_1} 	imes 100\% = \frac{p_1}{p_1} 	imes 100\% = 100\%$.

The kinetic energy of a particle increases by $300\%$. The percentage increase in its momentum is ........ $\%$.

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