A ball of mass m is attached to a cord at point of length L, pivoted O, as shown in figure. 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 μ. After collision, the ball comes momentarily to rest at C when cord makes an angle of θ with the vertical and block moves a distance of 3L/2 on rough horizontal floor before stopping. The values of μ and θ are, respectively,

Velocity of the ball just before collision is v0=2gL----(1)Apply1ing momentum conservation along horizontal direction (because momentum is conserved in collision along the line of impact), we get mv0=−mv1+2mv2Applying coefficient of restitution equation, we gete=23=v1+v2v0Solving the above two equations, we getv1=v09 and v2=5v09----(2)As the block moves a distance of 3L/2 before coming to rest, so from work-energy theorem,0−12(2m)v22=−μ×2mg3L2⇒v22=3μgL2581×2gL−3μgL⇒μ=50243For ball, KE is converted to gravitational potential energy after collision, so 0−mv122=−mg×L(1−cos⁡θ)from (1) and (2) we get cos⁡θ=8081