Two beads `A` and `B` of masses `m_(1)` and `m_(2)` respectively, are threaded on a smooth circular wire of radius a fixed in a vertical plane. `B` is stationary at the lowest point when `A` is gently dislodged from rest at the highest point. A collided with `B` at the lowest point. The impulse given to `B` due to collision is just great enough to carry it to the level of the centre of the circle while `A` is immediately brought to rest by the impact.

If `m_(2)` again comes down and collides with `m_(1)` then after the collision

Options

A. `m_(1)` will rise the same height as risen by `m_(2)`
B. `m_(1)` will rise the less height as risen by `m_(2)`
C. `m_(1)` will rise the more height as risen by `m_(2)`
D. `m_(1)` and `m_(2)` will move in opposite directions

Answer

A

Explanation

`m_(1)v_(1)+m_(2)v_(2)=m_(2)u_(2)`
`impliesm_(1)v_(1)+sqrt(2)m_(1)v_(2)=sqrtm_(1)u_(2)`

`impliesv_(1)+sqrt(2)v_(2)=sqrt(2)u_(2)`
`e=1/sqrt(2)=(v_(1)-v_(2))/(u_(2)-u_(1))implies1/sqrt(2)=(v_(1)-v_(2))/u_(2)`
`implies v_(1)-v_(2)=(u_(2))/(sqrt(2))`…………..ii
Solving eqn i and ii we get `v_(1)=u_(2)=sqrt(2ga)`
Both are positive so both move same direction. height raised by `m_(1)`,
`v_(1)^(2)/(2g)=a`, same as that of `m_(2)`