JEE Main 2022 — Work Power Energy Question with Solution

A body of mass $m$, moving with momentum $p$ and kinetic energy $K$ are related as follows,
$K=\frac{p^2}{2m}$.
As the momentum of the body is increased by $20\%$, then its momentum becomes,
$\left(p+\frac{20}{100}p\right)=1.2 p$
and the mass of the body remains unchanged. So, kinetic energy is directly related to the square of momentum. 

Hence, new kinetic energy $\left(K^{\text{'}}\right)$ will be,
$K^{\text{'}}=\frac{{\left(1.2 p\right)}^2}{2m}$

$K^{\text{'}}=1.44 \mathrm{K}$.

So, the percentage change in kinetic energy
$=\frac{K^{\text{'}}-K}{K}\times 100$ 
$=\frac{1.44K-K}{K}\times 100$
$=\left(1.44-1\right)\times 100$
$=44\%$.