Two beads of masses `m_(1) and m_(2)` are closed threaded on a smooth circular loop of wire of radius R. Intially both the beads are in position A and B in vertical plane .
Now , the bead A is pushed slightly so that it slide on the wire and collides with B . if B rises to the height of the centre , the centre of the loop O on the wire and becomes stationary after the collision , prove `m_(1) : m_(2) = 1 : sqrt(2)`

Explanation

Let `v_(1)` be the velocity of head A of mass `m_(1)` when it slide on the wire and collides with head B ,
`1/2 m_(1)v_(1)^(2) =m_(1)g (2R) " " ( :. h = 2R)`
` :. V_(1) = sqrt(4g)R " "……..(1)`
Collision being elastic , so the law of momentum is ,
` m_(1)v_(1) =m_(2)v_(2)`
` :. v_(2) =(m_(1))/(m_(2)) v_(1) =(m_(1))/(m_(2)) (sqrt(4g)R) " ".....(2)`
After , collision bead of mass `m_(2)` reaches to the height R ,
` :. 1/2 m_(2)v_(2)^(2) =m_(2)gR " " ( :. h = R)`
` :. v_(2) = sqrt(2gR) " "......(3)`
Substituting the value of `v_(2)` from equation (3) in euation (2) ,
`sqrt(2gR) =(m_(1))/(m_(2)) sqrt(4gR)`
` :. (m_(1))/(m_(2)) =1/(sqrt(2))`
` :. m_(1) : m_(2) =1: sqrt(2)`