Question

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 sherre after collision.

Answer 1

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)`