If the linear momentum of a body is increased | vedclass

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(B) The kinetic energy $K$ of a body is related to its linear momentum $p$ by the formula $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.Since

Vedclass · Accepted Answer

$125$

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(B) The kinetic energy $K$ of a body is related to its linear momentum $p$ by the formula $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.Since the mass $m$ remains constant,we have $K \propto p^2$.Let the initial momentum be $p_1 = p$ and the final momentum be $p_2 = p + 0.50p = 1.5p$.The ratio of the final kinetic energy $K_2$ to the initial kinetic energy $K_1$ is given by:$\frac{K_2}{K_1} = \left(\frac{p_2}{p_1}ight)^2 = \left(\frac{1.5p}{p}ight)^2 = (1.5)^2 = 2.25$.To find the percentage increase in kinetic energy,we use the formula:$	ext{Percentage increase} = \left(\frac{K_2}{K_1} - 1ight) 	imes 100 \%$.Substituting the value: $(2.25 - 1) 	imes 100 \% = 1.25 	imes 100 \% = 125 \%$.

If the linear momentum of a body is increased by $50 \%$,then the kinetic energy of that body increases by (in $\%$)

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