Discussion Forum



1)

A particle of mass m is projected from the ground with an initial speed u0  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 u0, 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:

Option A

Explanation:

From the momentum conservation equation, we have

   Pi=Pj

  $\therefore m(u_{0}\cos\alpha)\hat{i}+m(\sqrt{u_0^2-2gH)}\hat{j}$=(2m)v   .......(i)

 842021229_m3.PNG

   $H=\frac{u_0^2\sin^{2}\alpha}{2g}$      ........(ii)

 From Eqs.(i)   and (ii)

     $v=\frac{u_{0}\cos\alpha}{2}\hat{i}+\frac{u_{0}\cos\alpha}{2}\hat{j}$

  Since both components of v are equal. Therefore it is making 45°  with horizontal.