If momentum of a body increases by 20%, then the percentage increase in its kinetic energy is

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Solution

The correct option is B 44
We have,
$K . E . = \frac{P^{2}}{2 m}$
$Δ K . E .$ = $\frac{P_{f}^{2}}{2 m}$ - $\frac{P_{i}^{2}}{2 m}$
= $\frac{(\frac{6 P}{5})^{2}}{2 m}$ - $\frac{P^{2}}{2 m}$
= $\frac{36 P^{2} - 25 P^{2}}{50 m}$
= $\frac{11 P^{2}}{50 m}$
Now,
% $C h a n g e$ = $\frac{Δ K E}{K E_{i}}$
= $\frac{\frac{11 P^{2}}{50 m}}{\frac{P^{2}}{2 m}} \times 100$
= $\frac{11}{25} \times 100$
= $44$ %

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