If the momentum of a body of mass ' \(m\) ' is increased by \(20 \%\) then its kinetic energy increases by

K.E. \(=\frac{\mathrm{p}^2}{2 \mathrm{~m}}\)
When momentum increases by \(20 \%\), new momentum is \(\mathrm{p}^{\prime}\)
\(\begin{aligned}
\mathrm{p}^{\prime} & =\mathrm{p}+20 \% \mathrm{p} \\
& =\mathrm{p}+0.2 \mathrm{p}=1.2 \mathrm{p}
\end{aligned}\)
New kinetic energy,
\(\begin{aligned}
\mathrm{K} \cdot \mathrm{E}^{\prime} & =\frac{\mathrm{p}^{\prime 2}}{2 \mathrm{~m}} \\
& =\frac{(1.2 \mathrm{p})^2}{2 \mathrm{~m}}=\frac{1.44 \mathrm{p}^2}{2 \mathrm{~m}}
\end{aligned}\)
\(\therefore \quad K \cdot E^{\prime}=1.44 \mathrm{~K} . \mathrm{E}\).
Increase in kinetic energy is calculated by change in kinetic energy.
\(\begin{aligned}
\Delta \mathrm{KE} & =\left(\mathrm{K} \cdot \mathrm{E}^{\prime}-\mathrm{K} \cdot \mathrm{E}\right) \\
& =(1.44 \mathrm{~K} \cdot \mathrm{E}-\mathrm{K} \cdot \mathrm{E}) \\
& =0.44 \mathrm{~K} . \mathrm{E} \\
& \Rightarrow 44 \%
\end{aligned}\)