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. `2 pil`
B. cannot be determined
C. `pil`
D. none of these

Answer

A

Explanation

`mv=mv=mV rArrv=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))_(n)=(v_(2)-v_(1))_(n)rArr(v-0)=(V-v') (iii)`
Solving Eqs. (i),(ii),(iii) we get `V=v` and
`omega=(3v)/l`
Time taken to complete three revolutions
`(theta=6 pi), t=(theta)/(omega)=(6pi)/(omega)=(6pil)/(3v)=(2pi)/v`
Hence distance travelled by the centre of the rod
is `s=vt =v((2pil)/v)=2pil`