Question

A particle of mass $m$ is projected from the ground with an initial speed $u _{0}$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_{0}$. The angle that the composite system makes with the horizontal immediately after the collision is

• A. $\frac{ \pi}{ 4}$
• B. $\frac{ \pi}{ 4} + \alpha$
• C. $\frac{ \pi}{ 4} - \alpha$
• D. $\frac{ \pi}{ 2}$

Answer 1

Answer: (A)

$\frac{ \pi}{ 4}$

Answer 2

Velocity of particle performing projectile motion at highest point
$= v _{1}= v _{0} \cos \alpha$
Velocity of particle thrown vertically upwards at the position of collision
$=v_{2}^{2}=u_{0}^{2}-2 g \frac{u^{2} \sin ^{2} \alpha}{2 g}=v_{0} \cos \alpha$

So, from conservation of momentum
$\tan \theta=\frac{m v_{0} \cos \alpha}{m u_{0} \cos \alpha}=1$
$\Rightarrow \theta=\pi / 4$