A ball of radius `R=20 cm` has mass `m=0.75 kg` and moment of inertia (about its diameter) `I=0.0125 kg m^(2)`. The balls rolls with velocity `V_(0)=10 ms^(-1)` towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is `e=0.7`, velocity `V` of the ball after the collision is `(g=0ms^(-2))`

Answer

2

Explanation

For vertical forces, `N=mg`
For vertical forces, `muN=ma` or `a=mug=10 mu`.
Taking moments (about `o`) of forces acting on the ball, `muNR=Ialpha` or `alpha=120 mu`
Long after the collision, there will be no sliding or it will be pure rolling. let sliding stop after a time `t` after collision, the final translation velocity, `upsilon=(7-at)` or `upsilon=7-10 mut` and final
angular velocity, `omega=(-omega_(0))+alphat`
or `omegqa=(120 mut-50) rads^(-1)`(clockwise)
But at that instant `upsilon=Romega`
`:. (7-10 mut)=0.2(120 mut-50) :. mut=0.5`