A solid spherical ball rolling without slipping collides elastically with an identical ball at rest, as shown in the figure. Assuming that the frictional forces are small enough to have negligible effect during the instant of collision, calculate, the velocity of each ball along enough time after the collision when each ball is again rolling without slipping.

Answer

`5/7V`

Explanation

Before collision `V_(1)=V,V_(2)=0`

`omega_(1)=V/R, omega_(2)=0`
During collision for forces pass through centre of each sphere and friction force is negligibly small so that the angular momentum about the centre of each ball is coserved.
Thus `V_(1)^(')=0, V_(2)^(')=V_(1)=0`
With these linear and angular velocities start to move and eventually start sure rolling. The directions shown in the figure are considered to be positive. The angular momentum of each ball about point of contact is coserved.
For ball 1: `MRV_(1)^(')+Iomega_(1)^(')=MRV_(1)^('')+Iomega_(1)^(')`
`0+I(V/R)=MRV_(1)^('')+I((V_(1)^(''))/R)`
`V_(1)^('')=V/((MR^(2))/I+1)=2/7V`
For ball 2: `MRV_(2)^(')+Iomega_(2)^(')=MRV_(2)^('')+Iomega_(2)^('')`
`MRV=0=MRV_(2)^('')((V_(2)^(''))/R)`
`V_(2)^('')=V/(1+I/(MR^(2)))=5/7V`