If the linear momentum is increased by 50%,th | vedclass

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(C) Let the initial linear momentum be $P_1 = P$.The new linear momentum is $P_2 = P_1 + 0.5P_1 = 1.5P_1 = \frac{3}{2}P_1$.The relationship between ki

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(C) Let the initial linear momentum be $P_1 = P$.The new linear momentum is $P_2 = P_1 + 0.5P_1 = 1.5P_1 = \frac{3}{2}P_1$.The relationship between kinetic energy $E$ and linear momentum $P$ is given by $E = \frac{P^2}{2m}$,which implies $E \propto P^2$.Therefore,the ratio of the new kinetic energy $E_2$ to the initial kinetic energy $E_1$ is $\frac{E_2}{E_1} = \left( \frac{P_2}{P_1} ight)^2$.Substituting the values,$\frac{E_2}{E_1} = \left( \frac{1.5P_1}{P_1} ight)^2 = (1.5)^2 = 2.25$.This means $E_2 = 2.25 E_1 = E_1 + 1.25 E_1$.The percentage increase in kinetic energy is $\frac{E_2 - E_1}{E_1} 	imes 100\% = 1.25 	imes 100\% = 125\%.$

If the linear momentum is increased by $50\%$,the kinetic energy will increase by ............. $\%$

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