A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

Answer

C

Explanation

When the solid sphere collides with the wall, the rebounds with velocity v towards left but it continues to rotate in the clockwise direction.
So, the angular momentum
`-mvR-(2/5)mRxxv/R`
After rebounding when pure roling starts lrt the velocity be v and the corresponding angular velocity is `(v'/R)`

So, `mvR-(2/5)mR=mvR+mR(v'/R)`
`mvRx(x(3/5)=mvR=(7/5)`
`v'=(3v)/7`
So the sphere will move with velocity (3v)/7`.