A uniform smooth rod (mass `m` and length `l`) placed on a smooth horizontal floor is it by a particle (mass `m`) moving on the floor, at a distance `l/4` from one end elastically `(e = 1)`. The distance travelled by the centre of the rod after the collision when it has completed three revolutions will be

Options

A. `2pil`
B. cannot be determined
C. `pil`
D. none of the above

Answer

A

Explanation

Applying conservation of linear momentum
`mv=mv'=mVimpliesv=v'=V`……….i
Applying conservation of angular momentum about point of collision.
`0=((ml^(2))/12)omega-mV(l/4)implieslomega=3V`………..ii
Applying restituting equation
`(u_(1)-u_(2))=(v_(2)-v_(1))_(n)implies(v-0)=(V-v')` ...........iii
solving eqn i, ii, and iii we get `V=v` and
`omega=(3v)/l`
Time taken to complete three revolutions `(theta=6pi)`
`t=theta/omega-(6pi)(omega=(6pil)/(3v)=(2pil)/v)`
Hence, distance travelled by the centre of the rod is
`s=Vt=v((2pil)/v)=2pil`