A uniform rod of length `L` lies on a smooth horizontal table. The rod has a mass `M`. A particle of mass `m` moving with speed `v` strikes the rod perpendicularly at one of the ends of the rod sticks to it after collision.
Find the velocity of the centre of mass `C` and the angular, velocity of the system about the centre of mass after the collision.

Options

A. `(2mv)/(M+m),(3Mv)/((M+2m)L`
B. `(Mv)/(M+m),(6Mv)/((M+4m)L`
C. `(2mv)/ (M+m),(3Mv)/((M+4m)L`
D. `(mv)/(M+m),(6Mv)/((M+4m)L`

Answer

D

Explanation

Velocity of the centre of mass of the rod `+`particle system
`V_(CM)=(mv)/((M+m))`

`vecv_(p,CM)=vecv_(p)-vecv_(c)`
`v_(c)=(mv)/(M+m)` and `v_(p)=v`
`vecv_(p,CM)=(Mv)/(M+m)`
`vecv_(rod,CM)=vecv_(rod)-vecv_(CM)=0-(mv)/(M+m)`
`v_(rod,CM)=-(mv)/(M+m)`
Location of the centre of mass of the system rod + particle from `A`
` r_(1)=(mxx0+Ml//2)/((m+M))`
`=(Ml)/(2(m+M))`
Angular momentum of the particle just before collision about `C`,
`L_(p,c)=mv_(p,c)r_(1)`
`=(M^(2)mv)/(2(M+m)^(2))`
Angular momentum of the rod about the centre of mass of system `C`,
`L_(rod,CM)=Mv_(rod,CM)(l/2-r_(1))`
`=M((mv)/(M+m))((ml)/(2(M+m)))`
`=(Mm^(2)vl)/(2(M+m)^(2))`
Moment of inertia about the vertcal axis passing through `C`,
`I_(c,rod)=I_(0)=M(l/2-r_(1))^(2)`
`I_(c,rod)=(Ml^(2))/12+m((ml)/(2(M+m)))^(2)`
`I_(c,"particle")=mr_(1)^(2)=m((Ml)/(2(M+m)))^(2)`
`I_(c)=(Ml^(2))/12+M((ml)/(2(M+m)))^(2)+m((Ml)/(2(M+m)))^(2)`
`I_(c)=(M(M+4m))/(12(M+m))L^(2)`
As no external forces are acting on the system rod `+` particle, hence the velocity of the centre of mass of the system will remain constant.
`v_(CM)=(mv)/(M+m)`
For angular velocity about `C`
`mvr_(1)=I_(c)omega`
`omega=(mvr_(1))/(I_(c))=(mv((ml)/(2(M+m))))/(((M(M+4m)L^(2))/(12(m+m)))`
`=(6mv)/((M+3m)L)`