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

The correct option is A

$\frac{\pi }{4}$

At the instant of collision, momentum of the first particle $=m{u}_{0}cos\alpha$ [Horizontal]
Velocity of second particle at collision $=\sqrt{u_0^2-2g\frac{u_0^{2}si{n}^2\alpha }{2g}}=u_{0}cos\alpha$
Momentum of the second particle $=m{u}_{0}cos\alpha$ [Vertical]
Now, $tan\theta =\frac{m{u}_{0}cos\alpha }{m{u}_{0}cos\alpha }=1\Rightarrow \theta ={45}^{\circ }$