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Q.
If kinetic energy of a body is increased by $300 \%$ then percentage change in momentum will be
J & K CETJ & K CET 2017Work, Energy and Power
Solution:
Let $m$ be the mass of the body and $v_{1}$ and $v_{2}$ be the initial and final velocities of the body respectively.
$\therefore $ Initial kinetic energy $=\frac{1}{2} m v_{1}^{2}$
Final kinetic energy $=\frac{1}{2} m v_{2}^{2}$
Initial kinetic energy is increased $300 \%$ to get the final kinetic energy.
$\therefore \frac{1}{2} m v_{2}^{2}=\frac{1}{2}\left(1+\frac{300}{100}\right) m v_{1}^{2}$
$\Rightarrow v_{2}=2 v_{1}$
or $v_{2} / v_{1}=2\,\,\,$...(i)
Initial momentum $=p_{1}=m v_{1}$
Final momentum $=p_{2}=m v_{2}$
$\therefore \frac{p_{2}}{p_{1}}=\frac{m v_{2}}{m v_{1}}=\frac{v_{2}}{v_{1}}=2$
$\therefore p_{2}=2 p_{1}=\left(1+\frac{100}{100}\right) p_{1}$
So momentum has increased $100 \%$.