If kinetic energy of a body is increased by $300 \%$ then percentage change in momentum will be\

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Professor PC Thomas Classes · Answer

(a) : Let $m$ be the mass of the body and $v_{1}$ and $v_{2}$ be the initial and final velocities of the body respectively.\ $	herefore \quad$ 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.\ $	herefore \quad \frac{1}{2} m v_{2}^{2}=\frac{1}{2}\left(1+\frac{300}{100}ight) m v_{1}^{2}$\ $\Rightarrow \quad v_{2}=2 v_{1}$ or $v_{2} / v_{1}=2$\ Initial momentum $=p_{1}=m v_{1}$\ Final momentum $=p_{2}=m v_{2}$\ $	herefore \frac{p_{2}}{p_{1}}=\frac{m v_{2}}{m v_{1}}=\frac{v_{2}}{v_{1}}=2$ or $	herefore \quad p_{2}=2 p_{1}=\left(1+\frac{100}{100}ight) p_{1}$\ So momentum has increased $100 \%$.