There are two pendulums with bobs having indencital size and mass. The pendulum `A` is released from rest in the position as shown in the figure. If the maximum angle formed by cord `BO'` with vertical in the subsequent motion of sphere `B` is equal to the angle `theta_(0)` If the coefficient of restitution between sphere `A` and sphere `B` is `l`. find

a. the velocities of sphere `A` and sphere `B` just after collisions
b. the ratio of lengths of pendulums `l_B//l_A`.

Explanation

Velocity of `A` just before collision
`v_(0)=sqrt(2gl_(A)(1-costheta_(A))`………..i
Collision of `A` and `B`
Using C.O.L.M. , `m.v_(0)+0=mv_(A)+mv_(B)`
`v_(A)+v_(B)=v_(0)`……ii
Using Newton's restitution law
`v_(B)-v_(A)=v_(0)`
`v_(B)-v_(A)=ev_(0)`………ii
Solving eqn i and ii `v_(A)=(v_(0))/2(1-e)` and `v_(B)=(v_(0))/2(1+e)`
As `B` swings angle `thetaB`, hence
`v_(B)=sqrt(2gh_(B)(1-costheta_(B)))` ............iii
`and (v_(0))/2(1+e)=sqrt(2gl_(B)(1-costheta_(B)))`
`((2gl_(A)(1-costheta_(A)))/2(1+e) sqrt(2gl_(B)(1-costheta_(B)))`
`implies (l_(B))/(l_(A))=((1+e)/2)^(2)`