JEE Main 2023 — Center of Mass Momentum and Collision Question with Solution

Let initial velocity of bullet be $u$.

Given, mass of bullet $m=10 \mathrm{g}$ and mass of ball $M=200 \mathrm{g}$,

Initially ball is at height $5 \mathrm{m}$ and at rest, the only acceleration acting is due to gravity.

Time of flight of each ball and bullet is $t=\sqrt{\frac{2h}{g}}$.

Now, velocity of ball and bullet as $v_1=\frac{30}{\sqrt{{\displaystyle \frac{2h}{g}}}} \mathrm{and} v_2=\frac{120}{\sqrt{{\displaystyle \frac{2h}{g}}}}$

As the collision is elastic, so applying conservation of linear momentum.

$mu+M\left(0\right)=M{v}_1+m{v}_2$

$\Rightarrow \left(0.01\right)u=\left(0.2\right)\frac{30\sqrt{g}}{\sqrt{2h}}+\left(0.01\right)\frac{120\sqrt{g}}{\sqrt{2h}}$

$\Rightarrow u=300+60=360 \mathrm{m} {\mathrm{s}}^{-1}$