Discussion Forum

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 

Answer:

Explanation:

From the momentum conservation equation, we have

   P_i=P_j

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

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