Inside a smooth spherical shell of the radius `R` a ball of the same mass is released from the shown position (Fig.) Find the distance travelled by the shell on the horizontal floor when the ball comes to the lowest point of the shell.

Explanation

As initial momentum of the system in x-direction is zero , and there is no net external force in x-direction the momentum of system remains zero in x-direction and thus the cetner of mass of the system undergoes zero displacement in x-direction
`m_(1)vecs_(1)+m_(2)vecs_(2)=0`
When the ball comes to the lowest position , shell moves backwards say by a distance `x`.
Displacement of ball in x-direction=Displacement of ball w.r.t.
shell+displacement of shell.
Displacement of shell`=(-x)`
`therefore` displacement of ball is x-direction is`((3R)/(4)+(-x))`
`m((3R)/(4)-x)-mx=0`
`therefore" "x=(3R)/(8)`
If we do not consider that the shell moves back ward, we can take its forward displacement to be x,
`therefore` displacement of ball in x-direction`=(3R)/(4)+x`
`m((3R)/(4)+x)+mx=0`
`x=-(3R)/(8)` (-ve sign indicates its backward motion)