If the kinetic energy of a body is increased | vedclass

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(A) Let the initial kinetic energy be $E_1 = E$.Final kinetic energy $E_2 = E + 300\% 	ext{ of } E = E + 3E = 4E$.We know that the relationship betwe

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$100\%$

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(A) Let the initial kinetic energy be $E_1 = E$.Final kinetic energy $E_2 = E + 300\% 	ext{ of } E = E + 3E = 4E$.We know that the relationship between momentum $P$ and kinetic energy $E$ is $P = \sqrt{2mE}$,which implies $P \propto \sqrt{E}$.Therefore,the ratio of final momentum $P_2$ to initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{\frac{4E}{E}} = \sqrt{4} = 2$.This means $P_2 = 2P_1$.The percentage change in momentum is given by $\frac{P_2 - P_1}{P_1} 	imes 100\% = \frac{2P_1 - P_1}{P_1} 	imes 100\% = 100\%$.Thus,the momentum increases by $100\%$.

If the kinetic energy of a body is increased by $300\%$,then the percentage change in momentum will be:

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