A body P of mass m moving along the positive x-direction with velocity u collides with a body of mass M initially at rest. After the collision body P moves along the positive y-direction and body Q moves along a direction making an angle `theta` below the x-axis. The collision is assumed to be elastic.
The angle `theta` is given by

Options

A. `theta = tan^(-1)(M/m)`
B. `theta = tan^(-1)(m/M)`
C. `theta = tan^(-1)((M - m)/(M + m))^(1//2)`
D. `theta = tan^(-1)((M + m)/(M - m))^(1//2)`

Answer

C

Explanation

Conservation of x and y components of momentum gives
`m u = MV cos theta .... (1)`
` 0= mv - MV sin theta`
` mv = MV sin theta ...... (2)`
From conservation of KE we have
`1/2 m u^2 = 1/2 mv^2 + 1/2 M V^2 .....(3)`
Using eqn. (1) and (2) in eqn. (3) we get
`sin theta = ((M - m)/(2M))^(1//2) and cos theta = ((M + m)/(2M))^(1//2)`