If the momentum of a body is increased by 100 | vedclass

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(D) The kinetic energy $E$ of a body with momentum $P$ and mass $m$ is given by $E = \frac{P^2}{2m}$.Let the initial momentum be $P_1 = P$ and the ini

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$300$

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(D) The kinetic energy $E$ of a body with momentum $P$ and mass $m$ is given by $E = \frac{P^2}{2m}$.Let the initial momentum be $P_1 = P$ and the initial kinetic energy be $E_1 = \frac{P^2}{2m}$.Since the momentum is increased by $100\%$,the new momentum $P_2 = P + 100\% 	ext{ of } P = P + P = 2P$.The new kinetic energy $E_2$ is given by $E_2 = \frac{P_2^2}{2m} = \frac{(2P)^2}{2m} = 4 \left( \frac{P^2}{2m} ight) = 4E_1$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100\%$.Substituting the values,we get $\frac{4E_1 - E_1}{E_1} 	imes 100\% = \frac{3E_1}{E_1} 	imes 100\% = 300\%$.

If the momentum of a body is increased by $100\%$,then the percentage increase in the kinetic energy is ............ $\%$

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