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-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 :

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

Answer 1

The Correct Option is A

To solve this problem, let's break down the motion into two parts: the motion of the initially projected particle and the subsequent motion following an inelastic collision at the maximum height.

1. Initial Motion of the Particle:
   • The particle is projected with a speed $u$ at an angle $\theta = \frac{\pi}{3}$ relative to the horizontal.
   • The horizontal component of the velocity $(u_x)$ is given by: $$ u_x = u \cos\theta = u \cos\left(\frac{\pi}{3}\right) = \frac{u}{2}.$$
   • The vertical component of the velocity $(u_y)$ is: $$ u_y = u \sin\theta = u \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}u}{2}.$$
2. Maximum Height Calculation:
   • At maximum height, the vertical velocity becomes zero. Using the kinematic equation, we have: $$ 0 = \left(\frac{\sqrt{3}u}{2}\right)^2 - 2gH $$, where $H$ is the maximum height.
   • Solving for $H$ gives: $$ H = \frac{3u^2}{8g}. $$
3. Time to Reach Maximum Height:
   • The time to reach the maximum height $(t_1)$ is given by: $$ t_1 = \frac{u_y}{g} = \frac{\sqrt{3}u}{2g}. $$
4. Collision at Maximum Height:
   • When the particle reaches maximum height, it collides inelastically with another particle of the same mass $m$ moving with velocity $u\hat{i}$.
   • The velocity after inelastic collision, considering conservation of momentum in the horizontal direction: $$ (mu/2 + mu) = (2m) v' \Rightarrow v' = \frac{3u}{4}. $$
5. Distance Covered by Combined Mass:
   • After the collision, the combined mass moves horizontally with the velocity $v' = \frac{3u}{4}$.
   • The time to fall from maximum height back to the ground $(t_2)$ is equal to the time to rise: $$ t_2 = \frac{\sqrt{3}u}{2g}. $$
   • The horizontal distance $(d)$ covered after collision is: $$ d = v' \cdot t_2 = \left(\frac{3u}{4}\right) \cdot \left(\frac{\sqrt{3}u}{2g}\right) = \frac{3\sqrt{3}}{8} \frac{u^2}{g}. $$

Hence, the horizontal distance covered by the combined mass before reaching the ground is $\frac{3\sqrt{3}}{8} \frac{u^{2}}{g}$.