The momentum of a body is increased by 50 %. | vedclass

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(B) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$.Let the initial momentum be $p_i$. The

Vedclass · Accepted Answer

$125$

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(B) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$.Let the initial momentum be $p_i$. The initial kinetic energy is $KE_i = \frac{p_i^2}{2m}$.The momentum is increased by $50 \%$,so the final momentum $p_f = p_i + 0.50 p_i = 1.5 p_i$.The final kinetic energy is $KE_f = \frac{p_f^2}{2m} = \frac{(1.5 p_i)^2}{2m} = \frac{2.25 p_i^2}{2m} = 2.25 KE_i$.The percentage increase in kinetic energy is given by $\frac{KE_f - KE_i}{KE_i} 	imes 100$.Substituting the values: $\frac{2.25 KE_i - KE_i}{KE_i} 	imes 100 = 1.25 	imes 100 = 125 \%$.

The momentum of a body is increased by $50 \%$. The percentage increase in the kinetic energy of the body is $...........\,\%$

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