Question

A particle of mass $m$ is projected with a speed $u$ from the ground at an angle $\theta =\frac{\pi }{3}$ w.r.t. horizontal $x-\text{axis}$ . When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity. The horizontal distance covered by the combined mass before reaching the ground is

• A. $2\sqrt{2}\frac{u^2}{g}$
• B. $\frac{3\sqrt{3}{u}^2}{8g}$
• C. $\frac{5u^2}{8g}$
• D. $\frac{3\sqrt{2}{u}^2}{4g}$

Answer 1

The correct option is B $\frac{3\sqrt{3}{u}^2}{8g}$

Step 1: Draw a free body diagram of the given problem


It is given that the collision is completely inelastic so both the particle will combined. And let velocity of the combined mass be $v$'.

Step 2: Find the common velocity of the particles.
From conservation of linear momentum
$\frac{mu}{2}+mu=2mv$'
$v^{\prime }=\frac{3u}{4}$

Step 3: Find the horizontal distance covered by the combined mass.
The height at which collision happened, $H=\frac{u^2{\sin }^2{60}^{\circ }}{2g}$
Horizontal distance covered by combined mass $x,=v^{\prime }t$ $t$ Time taken by combined mass to reach ground
$t=\sqrt{\frac{2H}{g}}$
Hence,
$x=\frac{3u}{4}\sqrt{\frac{2u^2{\sin }^2{60}^{\circ }}{g2g}}$
$=\frac{3}{4}\frac{\sqrt{3}}{2}\cdot \frac{u^2}{g}=\frac{3\sqrt{3}{u}^2}{8g}$