If momentum of a body is increased by 20 % , then its kinetic...

A body of mass $m$, moving with momentum $p$ and kinetic energy $K$ are related as follows,
$K = \dfrac{{p}^{2}}{2m}$.
As the momentum of the body is increased by $20$, then its momentum becomes,
\left(p + \dfrac{20}{100}p\right) = 1 . 2p$<br />and the mass of the body remains unchanged. So, kinetic energy is directly related to the square of momentum. </p><p>Hence, new kinetic energy$\left({K}^{'}\right)$will be,<br />${K}^{'} = \dfrac{{\left(1 . 2p\right)}^{2}}{2m}$</p><p>${K}^{'} = 1 . 44K$.</p><p>So, the percentage change in kinetic energy<br />$= \dfrac{{K}^{'} - K}{K} \times 100$<br />$= \dfrac{1 . 44K - K}{K} \times 100$<br />$= \left(1 . 44 - 1\right) \times 100$<br />$= 44 %.