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)/(2) - alpha`
D. `(pi)/(2)`

Answer

A

Explanation

Maximum height of first particle `H_("max") = (u_(0)^(2) sin^(2)alpha)/(2g)`
Speed of `2^(nd)` particle at height `H_("max")` given as `v_(y)^(2) = u_(0)^(2) - 2gH_("max") = u_(0)^(2) - u_(0)^(2) sin^(2) alpha rArr v_(y) = u_(0)cos alpha`
By Momentum Conseravtion `vec(p)_(f) = vec(p)_(i) rArr 2mvec(v)_(f) = mv_(0) cos alphahat(i) + mv_(0) cos alpha hat(j)`
`rArr v_(f) = (v_(0) cos alpha)/(2)(hat(i) + hat(j))`
`rArr` Angle with horizontal immediately after the cosllision `= (pi)/(4)`