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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

Vedclass · Accepted Answer

$300$

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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 $E_1 = \frac{p^2}{2m}$.Given that the momentum increases by $100\ \%$,the new momentum $p_2 = p + 100\% 	ext{ of } p = p + p = 2p$.The new kinetic energy $E_2 = \frac{p_2^2}{2m} = \frac{(2p)^2}{2m} = 4 \left( \frac{p^2}{2m} ight) = 4E_1$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100\ \%$.Substituting the values: $\frac{4E_1 - E_1}{E_1} 	imes 100\ \% = \frac{3E_1}{E_1} 	imes 100\ \% = 300\ \%$.

If the momentum of an object is increased by $100\ \%$,then its kinetic energy increases by ......... $(\%)$ percent.

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