The KE of a body is increased by 44 %. What i | vedclass

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(B) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.Let the initial k

Vedclass · Accepted Answer

$20$

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(B) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = K_1 + 0.44 K_1 = 1.44 K_1$.Let the initial momentum be $P_1$ and the final momentum be $P_2$.Since $P \propto \sqrt{K}$,we have $\frac{P_2}{P_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{1.44} = 1.2$.Therefore,$P_2 = 1.2 P_1$.The percentage increase in momentum is $\frac{P_2 - P_1}{P_1} 	imes 100 = (1.2 - 1) 	imes 100 = 20 \%$.

The $KE$ of a body is increased by $44 \%$. What is the percentage increase in its momentum?

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