A smooth ball of mass `m` is suspended from a light string of length `1 m`. Another ball of mass `2m` strikes the ball of mass in horizontally with a speed of `u=sqrt(35)m//s`. The coefficient of restitution for the collision is `e`. The string becomes loose, when it makes an angle of `30^(@)` with the horizontal, find the value of `e`.

Explanation

`v_(1)` and `v_(2)` are after collisiion
`v_(2)=((1+e)2mu)/(2m+m)=(2u(1+e))/3`
since tension becomes zero at `B`, so
`mg sin 30^(@)=(mv^2)/rimpliesv=sqrt((2r)/2)`
Applying conservation of energy between `A` and `B`
`1/2mv^(2)+mg(r+rsin30^(@))=1/2mv_(2)^(2)`
`implies e=0.5`