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

Options

A. `pi/4`
B. `pi/4+alpha`
C. `pi/4-alpha`
D. `pi/2`

Answer

A

Explanation

(a) Activity B to M for particle upwards
`v_1^2-u_0^2=2(-g) [(u_0^2sin^2 alpha)/(2g)]`
`:. v_1^2=u_0^2(1-sin^2 alpha)=u_0^2 cos^2 alpha`
`:. v_1=u_0 cos alpha….(i)`

Applying conservation of linear momentum in
Y-direction
`2mv sin theta=mv_1=mu_0cos alpha....(ii) [from (i)]`
Applying conservation linear momentum in
X-direction
`2mv cos theta=mu_0cos alpha....(iii)`
on dividing (ii) and (iii) we get
`tan theta=1 :. theta=pi/4`