JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Conservation of Momentum

At the top of its flight the first particle moves horizontally with $u\cos 60^{\circ }=\frac{u}{2}$, while the incoming particle of equal mass moves with $u$ along $x$. The perfectly inelastic collision conserves horizontal momentum: $m\frac{u}{2}+mu=2mv$, so the combined mass leaves with $v=\frac{3u}{4}$. It starts from the maximum height $H=\frac{u^2{\sin }^{2}60^{\circ }}{2g}=\frac{3u^2}{8g}$ and takes $t=\sqrt{\frac{2H}{g}}=\frac{u\sqrt{3}}{2g}$ to fall, so the horizontal distance covered is $vt=\frac{3u}{4}\cdot \frac{u\sqrt{3}}{2g}=\frac{3\sqrt{3}}{8}\frac{u^2}{g}$.