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A particle of mass m is projected from t...

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

A

`pi/4`

B

`pi/4 +alpha`

C

`pi/4 - alpha`

D

`pi/2`

Text Solution

Verified by Experts

The correct Answer is:
A
From momentum conservation equation, we have,
`p_i = p_f`
`:. M (u_0 cos alpha) hati + m (sqrt(u_0^2 - 2gh)) hatj = (2m ) v ….(i)`
`H = (u_0^2 sin^2 alpha)/(2g) " "…..(ii)`
From Eqs. (i) and (ii) `" " v = (u_0 cos alpha)/(2) hati + (u_0 cos alpha)/(2) hatj`
Since both components of are equal. Therefore, it is making `45^@` with horizontal.
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