Q: A particle of mass m is projected from the ground with an initial speed u_{0} at an angle α 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 after the collision is

(a) π/4

(b) $\frac{\pi}{4} + \alpha$

(c) $\frac{\pi}{4} -\alpha$

(d) π/2

Ans: (d)

Sol: Maximum height attained by the particle is

$\large H = \frac{u_0^2 sin^2 \alpha}{2 g}$ …(i)

Velocity of the particle at height H which was thrown vertically

$\large v = \sqrt{u_0^2 – 2 g H}$

Now , applying conservation of linear momentum

$\large m u_o cos\alpha \hat{i} + m(\sqrt{u_0^2 – 2 g H})\hat{j} = 2 m \vec{v’} $

$\large m u_o cos\alpha \hat{i} + m(\sqrt{u_0^2 – 2 g (\frac{u_0^2 sin^2 \alpha}{2 g})})\hat{j} = 2 m \vec{v’} $

$\large m u_o cos\alpha \hat{i} + m u_o cos\alpha \hat{j}= 2 m \vec{v’}$

$\large \vec{v’} = \frac{u_0 cos\alpha}{2}\hat{i} + \frac{u_0 cos\alpha}{2}\hat{j}$

$\large tan\theta = \frac{u_0 cos\alpha /2}{u_0 cos\alpha /2}$

tanθ = 1 ⇒ θ = 45°