Question

A particle of mass $m$ is projected with a speed of $u$ from the ground at an angle $\theta =\frac{\pi }{3}$ w.r.t. horizontal $\left(\text{x-axis}\right)$. 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:

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

Answer 1

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

Image for the given problem,



Using the principle of conservation of linear momentum for horizontal motion, we have

$p_i=p_f$

$mu+mu\cos {60}^{\circ }=2mv$

$\therefore v=\frac{3u}{4}$

For vertical motion,

$h=0+\frac{1}{2}g{T}^2\Rightarrow T=\sqrt{\frac{2h}{g}}$

Using maximum height formula,

$h=\frac{u^2{\sin }^2{60}^0}{2g}=\frac{u^2{\left(\frac{\sqrt{3}}{2}\right)}^2}{2g}=\frac{3u^2}{8g}$

Let $R$ is the horizontal distance travelled by the body.

$R=vT+\frac{1}{2}\left(0\right)\left(T{)}^2\right(\text{For horizontal motion, a=0})$

$=vT=\frac{3u}{4}\times \sqrt{\frac{2h}{g}}$

$\Rightarrow R=\frac{3u}{4}\times \sqrt{\frac{2\times \frac{3u^2}{8g}}{g}}=\frac{3\sqrt{3}{u}^2}{8g}$

Hence, $\left(\text{A}\right)$ is the correct answer.