If momentum of a body is increased by 20 %,th | vedclass

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Question

(C) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$.Let the initial momentum be $P$ and the initial kine

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$.Let the initial momentum be $P$ and the initial kinetic energy be $K = \frac{P^2}{2m}$.The new momentum $P'$ is increased by $20\%$,so $P' = P + 0.20P = 1.2P$.The new kinetic energy $K'$ is $K' = \frac{(P')^2}{2m} = \frac{(1.2P)^2}{2m} = 1.44 	imes \frac{P^2}{2m} = 1.44K$.The percentage increase in kinetic energy is given by $\frac{K' - K}{K} 	imes 100$.Substituting the values,we get $\frac{1.44K - K}{K} 	imes 100 = 0.44 	imes 100 = 44\%$.

If momentum of a body is increased by $20 \%$,then its kinetic energy increases by $....\%$

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