Question

A particle of mass $m$ is projected with a speed $u$ from the ground at angle is $\theta =\frac{\pi }{3}$ w.r.t. horizontal $x-axis$. When it has reached its maximum height, it collides completely in elastically 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. $\frac{3\sqrt{3}{u}^2}{8g}$
• B. $\frac{2\sqrt{2}{u}^2}{g}$
• C. $\frac{5u^2}{8g}$
• D. $\frac{3\sqrt{2}{u}^2}{4g}$

Answer 1

The correct option is A

$\frac{3\sqrt{3}{u}^2}{8g}$

Step 1: Given data and assumptions.

Mass of the particle$=m$

Projectile angle, $\theta =\frac{\pi }{3}$

Initial velocity $=u$

Step 2: Finding the horizontal velocity of the combined mass.

We know that,

According to the law of conservation of momentum,

$P_i=P_f$

Where, $P_i$ is initial momentum, and $P_f$ is final momentum.

$$\begin{gathered} mu+mu\cos \theta =\left(m+m\right)v \\ mu+mu\cos \theta =2mv \end{gathered}$$ [As it collides elastically, a mass will stick to the other mass]

Where, $m$ is the mass of the particle, $u$ is the initial velocity of the particle, and $v$ is the final velocity.

$v=\frac{u\left(1+\cos \left(60\right)\right)}{2}$ $\left[\cos 60°=\frac{1}{2}\right]$

$v=\frac{3}{4}u$ ….$\left(i\right)$

Step 3: Find the time required to reach the maximum height

Since,

Vertical maximum range,

$H=\frac{1}{2}g{t}^2$

Where, $t$ is time to reach the maximum height and $g$ is the acceleration due to gravity.

$t=\sqrt{\frac{2H}{g}}$ ……$\left(ii\right)$

Step 4: Finding the horizontal distance traveled by the combined mass

Since the horizontal range after collision$=vt$

$=\frac{3}{4}u\times \sqrt{\frac{2H}{g}}$ [ From$\left(i\right),\left(ii\right)$]

$=\frac{3}{4}u\times \sqrt{\frac{2u^2{\sin }^{2}60}{2g^2}}$ $\left[\because H=\frac{u^2{\sin }^2\theta }{2g}\right]$

$=\frac{3}{4}{u}^2\frac{\sqrt{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}{g}$

$=\frac{3\sqrt{3}{u}^2}{8g}$

Hence, option A is Correct.