A particle of mass $m$ is projected with a speed $u$ from the ground at an angle $\theta = \dfrac{\pi}{3}$ w.r.t. horizontal ($x$-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $u\hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is:
Answer: (A) $\dfrac{3\sqrt{3}}{8}\dfrac{u^2}{g}$
At the top the projectile moves horizontally with $u\cos60^\circ = \dfrac{u}{2}$, at height
$$H = \frac{u^2\sin^2 60^\circ}{2g} = \frac{3u^2}{8g}$$
Momentum conservation: $m\dfrac{u}{2} + mu = 2mV \Rightarrow V = \dfrac{3u}{4}$ (horizontal).
Time to fall from rest vertically: $t = \sqrt{\dfrac{2H}{g}} = \dfrac{\sqrt{3}}{2}\dfrac{u}{g}$.
$$x = Vt = \frac{3u}{4}\cdot\frac{\sqrt{3}\,u}{2g} = \frac{3\sqrt{3}}{8}\frac{u^2}{g}$$
Solution by Sreeraj P, M.Sc Physics