Two beads of massesm_(1) and m_(2) are closedthreaded on a smoothcircularloop of wireof radius R. Intially both the beads arein positionA and B in verticalplane . Now , the beadA is pushed slightly so that itslide on the wire and collideswith B . if B risesto the height of the centre ,the centre of the loop Oon the wire and becomes stationaryafterthe collision , provem_(1) : m_(2) = 1 : sqrt(2)

Two beads of massesm_(1) and m_(2) are closedthreaded on a smoothcircu

1.	Two beads of massesm_(1) and m_(2) are closedthreaded on a smoothcircularloop of wireof radius R. Intially both the beads arein positionA and B in verticalplane . Now , the beadA is pushed slightly so that itslide on the wire and collideswith B . if B risesto the height of the centre ,the centre of the loop Oon the wire and becomes stationaryafterthe collision , provem_(1) : m_(2) = 1 : sqrt(2)
Answer» Solution :Let `v_(1)`be the velocityof head A of mass `m_(1)` when it slideon the wireand collideswith 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 lawof 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 valueof `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)`