In the non-relativistic regime,if the momentu | vedclass

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(C) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.Since mass $m$ is constant,we have $K \propto p^2$.Let the i

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$300$

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(C) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.Since mass $m$ is constant,we have $K \propto p^2$.Let the initial momentum be $p_1 = p$ and the final momentum be $p_2 = p + 100\% 	ext{ of } p = 2p$.The ratio of kinetic energies is $\frac{K_2}{K_1} = \left(\frac{p_2}{p_1}ight)^2 = \left(\frac{2p}{p}ight)^2 = 4$.Thus,$K_2 = 4K_1$.The percentage increase in kinetic energy is given by $\frac{K_2 - K_1}{K_1} 	imes 100\% = \frac{4K_1 - K_1}{K_1} 	imes 100\% = 300\%$.

In the non-relativistic regime,if the momentum is increased by $100\%$,the percentage increase in kinetic energy is (in $\%$)

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