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

First we need to find maximum height attained by projectile , because collision occurs at that point .
Max height H = `(u_(0)^(2)sin^(2)alpha)/(2g)`
Another object is thrown vertically upwards with velocity `u_(0)` . Let its velocity becomes v after reaching height H , then
`v^(2) =u_(0)^(2) = 22 H `
`= u_(0)^(2) - 2g (u_(0)^(2) sin^(2) alpha)/(2g)`
`rArr " " v^(2) = u_(0)^(2) cos^(2) alpha `
`rArr " " v = u_(0) cos alpha `
As we know projectile also have same velocity `u_(0) cos alpha ` but in horizontal direction , at it.s height point of trajectory . Hence resultant momentum will be at an angle `45^(@)` with horizontal .