Two sphere `A and B` of masses `m_(1) and m_(2)` respectivelly colides. A is at rest initally and `B` is moving with velocity `v` along x-axis. After collision `B` has a velocity `(v)/(2)` in a direction perpendicular to the original direction. The mass `A` moves after collision in the direction.

Options

A. same as that `B`
B. opposite to that of `B`
C. `theta = tan^(-1) (1//2)` to the `x`-axis
D. `theta = tan^(-1) (-1//2)` to the `x`-axis

Answer

D

Explanation

Momentum conservation :
`x : m_(2) v = m_(1) cos theta (i) rArr v cos theta = (m_(2))/(m_(1)) v`
`y : 0 = m_(2) (v)/(2) - m_(1) v sin theta (ii) rArr v sin theta = (m_(2))/(m_(1)) (v)/(2)`
`tan thet = (v sin theta)/( v cos theta) = (1)/(2) rArr theta = tan^(-1) ((1)/(2))`
SInce `A` is moving below the `x`-axis , hence the angle will be `tan^(-1)(-(1)/(2))` to the axis.
Note : A angles measured in the anticlockwise direction are considered `+ ve` and in clockwise , `- ve`.