If the momentum increases by 100 %,what is th | vedclass

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(C) The kinetic energy $E$ is related to momentum $P$ by the formula $E = \frac{P^2}{2m}$.Let the initial momentum be $P_1 = P$ and the initial kineti

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$300$

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(C) The kinetic energy $E$ is related to momentum $P$ by the formula $E = \frac{P^2}{2m}$.Let the initial momentum be $P_1 = P$ and the initial kinetic energy be $E_1 = \frac{P^2}{2m}$.Since the momentum increases 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 increases by $100 \%$,what is the percentage increase in kinetic energy?

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