Three identical discs A, B, and C (figure) rest on a smooth horizontal plane. The disc A is set in motion with vector v after which it experiences an elastic collision simultaneously with the discs B and C. The distance between the centres of the latter discs prior of the collision is `eta` times greater than the diameter of each disc.
Find the velocity of the disc A after the collision. At what value of `eta` will the disc A recoil after the collision, stop, move on?

Explanation

From the symmetry of the problem, the velocity of the disc A will be directed either in the initial direction or opposite to it just after the impact. Let the velocity of the disc A after the collision be `v^'` and be directed towards right after the collision. It is also clear from the symmetry of problem that the discs B and C have equal speed (say `v^('')`) in the directions, shown. From the condition of the problem,
`cos theta=(etad/2)/(d)=eta/2` so, `sin theta=sqrt(4-eta^2)//2` (1)
For the three discs, system, from the conservation of linear momentum in the symmetry direction (towards right)
`mv=2mv^('')sin theta+mv^'` or, `v=2v^('')sin theta+v^'` (2)
From the definition of the coefficient of restitution, we have for the discs A and B (or C)
`e=(v^('')-v^'sintheta)/(vsin theta-0)`
But `e=1`, for perfectly elastic collision,
So, `vsin theta=v^('')-v^'sintheta` (3)
From (2) and (3),
`v^'=(v(1-2sin^2theta))/((1+2sin^2theta))`
`=(v(eta^2-2))/(6-eta^2)` {using (1)}
Hence we have,
`v^'=(v(eta^2-2))/(6-eta^2)`
Therefore, the disc A will recoil if `eta lt sqrt2` and stop if `eta=sqrt2`.
One can write the equations of momentum conservation along the direction perpendicular to the initial direction of disc A and the conservation of kinetic energy instead of the equation of restitution.