If momentum is increased by 20%,then kinetic | vedclass

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Question

(A) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.Since mass $m$ is constant,we have $E \propto P^2$.L

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.Since mass $m$ is constant,we have $E \propto P^2$.Let the initial momentum be $P_1 = P$ and the final momentum be $P_2 = P + 0.20P = 1.2P$.The ratio of kinetic energies is $\frac{E_2}{E_1} = \left( \frac{P_2}{P_1} ight)^2 = \left( \frac{1.2P}{P} ight)^2 = (1.2)^2 = 1.44$.This implies $E_2 = 1.44 E_1$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100\% = (1.44 - 1) 	imes 100\% = 0.44 	imes 100\% = 44\%$.Therefore,the kinetic energy increases by $44\%$.

If momentum is increased by $20\%$,then kinetic energy ($K$.$E$.) increases by ........... $\%$

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