Question

A particle of mass $m$ is projected from the ground with an initial speed $u_0$at an angle $\alpha$with the horizontal. $A$t 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 $\frac{\pi }{4}$

Velocity of the particle performing projectile motion at highest point
$=V_1=V_{0}cos\alpha$
Velocity particle thrown vertically upwards at the position of collision

$=v_2^2=u_0^2-2g\frac{u^2{\sin }^2\alpha }{2g}=v_0\cos \alpha$

So, from conservation of momentum

$\tan \theta =\frac{m{v}_0\cos \alpha }{m{u}_0\cos \alpha }=1$

$\Rightarrow \theta =\pi /4$