A ball of mass `m` is attached to a cord of length `L`, pivoted at point `O`, as shown in Fig. The ball is released from rest at point A, swings down and makes an inelastic collision with a block of mass `2m` kept on a rough horizontal floor. The coefficient of restitution of collision is `e = 2//3` and coefficient of friction between block and surface is After collision, the ball comes momentarily to rest at `C` when cord makes an angle of `theta` with the vertical and block moves a distance of `3L//2` on rough horizontal floor before stopping. The values of `mu` and `theta` are, respectively,

Options

A. `50/243, cos^(-1)(80/81)`
B. `50/81, cos^(-1)(80/81)`
C. `2/81, cos^(-1)(80/243)`
D. `2/243, cos^(-1)(80/243)`

Answer

A

Explanation

Velocity of the ball just before collision `v_(0) =sqrt(2gL)`
Applying momentum conservation along horizontal direction (because momentum is conserved in collision along the line of impact), we get
`mv_(0)=-mv_(1)+2mv_(2)`

Applying coefficient of restitution equation, we get
`e=2/3=(v_(1)+v_(2))/v_(0)`
Solving the above two equations we get
`v_(1)=v_(0)/9` and `v_(2)=(5v_(0))/9`
As the block moves a distance of `3L/2` before coming to rest, so from work energy theorem,
`0-1/2(2m)v_(2)^(2)=-muxx2mg(3L)/2impliesv_(2)^(2)=3mugL`
`25/81xx2gL=3mugLimpliesmu=50/243`
For ball `KE` is converted to gravitational potential energy after collision, so
`0-(mv_(1)^(2))/2=-mgxxL(1-costheta)`
`costheta=80/81`