The momentum of an object increases by 50%. B | vedclass

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(B) The relation between kinetic energy $(E)$ and momentum $(p)$ is given by $E = \frac{p^2}{2m}$.Since mass $(m)$ is constant,$E \propto p^2$.Let the

Vedclass · Accepted Answer

$125$

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(B) The relation between kinetic energy $(E)$ and momentum $(p)$ is given by $E = \frac{p^2}{2m}$.Since mass $(m)$ is constant,$E \propto p^2$.Let the initial momentum be $p_1 = 100$ and initial kinetic energy be $E_1$.If the momentum increases by $50\%$,the new momentum is $p_2 = 100 + 50 = 150$.The new kinetic energy is $E_2 = \frac{p_2^2}{2m}$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100$.Substituting the values: $\frac{(150)^2 - (100)^2}{(100)^2} 	imes 100$.$= \frac{22500 - 10000}{10000} 	imes 100 = \frac{12500}{10000} 	imes 100 = 125\%$.Thus,the kinetic energy increases by $125\%$.

The momentum of an object increases by $50\%$. By what percentage will its $K.E.$ increase?

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