A small particle of mass `m` is released from a height `h` on a large smooth sphere kept on a perfectly smooth surface as shown in the figure. Collision between particle and sphere is perfectly inelastic. Determine the velocities of particle and sphere after collision.

Explanation

Collision takes lace between smooth bodies, therefore, tangential velocity of particle is unchanged.
`v_(r)=usintheta` and `u=sqrt(2gh)`
when `sintheta=(R/2)/R=1/2` or `theta=30^(@)`
`v_(t)=u/2` …………i

From conservation of momentum along `x`-axis,
`P_(i)=P_(f)`
`0+0=MV+mv_(n)sintheta+mv_(t)costheta`..............iii
from the definitation of coefficient of restitution.
`e=(Vsintheta-(-v_(n)))/(ucoshteta-0)=0`
[`e=0,` for a perfectly inelastic collision]
or, `Vsintheta=-v_(n)`
`v_(n)=-V/2` .............iii
On substitution expression for `v_(n)` and `v_(t)` in eqn ii we obtain
`MV=m(-V/2)1/2+m(u/2)(sqrt(3))/2`
`(M+m/4)V=m(usqrt3)/4`
where `u=sqrt(2gh)` or `V=(msqrt(6gh))/((4M+m))`
`v=sqrt(v_(n)^(2)+v_(t)^(2))=sqrt((6m^(2)gh)/(4(4M+m)^(2))+(gh)/2)`