The kinetic energy of a body decreases by 36% | vedclass

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(B) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.Since $m$ is constant,$P \propto \sqrt{E}$.If the kineti

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$20$

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(B) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.Since $m$ is constant,$P \propto \sqrt{E}$.If the kinetic energy decreases by $36\%$,the new kinetic energy $E_2$ is $100\% - 36\% = 64\%$ of the initial kinetic energy $E_1$.So,$E_2 = 0.64 E_1$.The ratio of the new momentum $P_2$ to the initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{0.64} = 0.8$.This means $P_2 = 0.8 P_1$,which is $80\%$ of the initial momentum.The decrease in momentum is $100\% - 80\% = 20\%$.

The kinetic energy of a body decreases by $36\%$. The decrease in its momentum is ............... $\%$

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