When the momentum of a body increases by 100% | vedclass

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(C) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$,where $m$ is the mass of the body.Let t

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$300$

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(C) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$,where $m$ is the mass of the body.Let the initial momentum be $p_1 = p$. Then the initial kinetic energy is $KE_1 = \frac{p^2}{2m}$.When the momentum increases by $100\%$,the new momentum $p_2$ becomes:$p_2 = p + 100\% 	ext{ of } p = p + p = 2p$.The new kinetic energy $KE_2$ is:$KE_2 = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4 	imes KE_1$.The percentage increase in kinetic energy is given by:$	ext{Percentage Increase} = \frac{KE_2 - KE_1}{KE_1} 	imes 100\%$$= \frac{4 KE_1 - KE_1}{KE_1} 	imes 100\%$$= \frac{3 KE_1}{KE_1} 	imes 100\% = 300\%$.

When the momentum of a body increases by $100\%$,its $KE$ increases by .............. $\%$

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