A tiny ball of mass `m` is released from the state of rest over a large smooth sphere of mass `M` and radius `R`, which is at rest on a smooth horizontal surface. If the ball strikes the sphere perfectly inelastically, find their velocities after collision.

Explanation

Velocity attained by ball just before it strikes the surface
`u=sqrt(2gh)`
`:. Sintheta=(R//2)/(R )rArrtheta=30^(@)`
After collision let the velocity of the sphere be `V` and the components of velocity of ball along common normal `OL` and along common tangent be `v_(a)` and `v_(t)` respectively.
In the absence of external force in the horizontal direction, the linear momentum conserved.
`0+0= -MV+mv_(n)sintheta+mv_(t)costheta` (`1`)
Along the common normal at ,
`Vsintheta-(-v_(n))=e[ucostheta-0]`(`2`)
`e=0` for inelastic collision.
`:. Vsintheta= -v_(n)`
`:. v_(n)=-v//2` (`3`)
Along the common tangent velocity components before and after collision remain same.
`:. usintheta=v_(1)rArrv_(t)=(u)/(2)` (`4`)
Using (`3`) and (`4`) in (`1`),
`MV=m(-(V)/(2))(1)/(2)+m((u)/(2))(sqrt(3))/(2)`
`(M+(m)/(2))V=m(usqrt(3))/(4)`
Velocity of sphere `V=(msqrt(6gh))/(2(2M+m)):'u=sqrt(2gh)`
Velocity of ball `v=sqrt(v_(n)^(2)+v_(t)^(2))=sqrt((6ghm^(2))/(4(2M+m)^(2))+(gh)/(2))`