Two identical discs of mass m and of radius R touch each other and move with the same velocity perpendicularly to the line segment which joins their centres of mass, along the surface of a horizontal smooth tabletop. There is a third disc of mass M and of radius R at rest, at a point on thhe perpendicular bisector of of the line segment joining the centres of mass of the two moving discs as shown in the figure. The two moving discs collide elastically with the third one, which is at rest there is no friction between the rims of the discs. What should the ratio of M/m be in order that after the collision the two discs of mass m move perpendicularly to their intial velocity?

Options

A. 3
B. 2
C. `sqrt(3)`
D. `sqrt(2)`

Answer

A

Explanation

Conservation of momentum of (M+m)
`2mv_(0)=Mv_(1)` .(i)
Newton's second law:
`v_(0)cos30^(@)=v_(1)cos30^(@)+v_(2)cos60^(@)`
`sqrt(3)v_(0)=sqrt(3)v_(1)+v_(2)` .(ii)
For 'm' `v_(0)sin30^(@)=v_(2)sin60^(@)` ....(iii)
`impliesv_(2)=(v_(0))/(sqrt(3)),v_(1)=(2)/(3)v_(0)`
from (i) `(2mv_(0))/(M)=(2v_(0))/(3)`
`((m)/(M))=(1)/(3)implies((M)/(m))=3`