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.

Find the ratio `m_(1):m_(2)`

Options

A. `1`
B. `sqrt(2)`
C. `1/sqrt(2)`
D. `2`

Answer

C

Explanation

Just before colisions, velocilty of `m_(1)`

`u_(1)=sqrt(2g2a)=2sqrt(ga)`
just after collision `m_(1)` is brought to rest and let velocity of `m_(2)` be `v_(2)`.
From conservation of linear momentum, `m_(2)v_(2)=m_(1)u_(1)`...........i

Now `v_(2)=sqrt(2ga)` so that `m_(2)` can rise up to point `m`. Putting the value of `u_(1)` and `v_(2)` in eqn i we get
`m_(2)sqrt(2ga)=m_(1)2sqrt(ga)impliesm_(1)/m_(2)=1/sqrt(2)`