Question

Three identical discs $A$, $B$ and $C$ (figure shown above) rest on a smooth horizontal plane. The disc $A$ is set in motion with velocity $\overrightarrow{v}$ after which it experiences an elastic collision simultaneously with the discs $B$ and $C$. The distance between the centres of the latter discs prior to the collision is $\eta$ times greater than the diameter of each disc. Find the velocity of the disc $A$ after the collision?

Answer 1

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 $v^{\prime }$ 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 $v^{\prime \prime }$) in the directions, shown in figure below. From the condition of the problem,
$\cos \theta =\frac{\eta \frac{d}{2}}{d}=\frac{\eta }{2}$ so, $\sin \theta =\frac{\sqrt{4-{\eta }^2}}{2}$ (1)
For the three discs, system, from the conservation of linear momentum in the symmetry direction (towards right)
$mv=2m{v}^{\prime \prime }\sin \theta +m{v}^{\prime }$ or, $v=2v^{\prime \prime }\sin \theta +v^{\prime }$ (2)
From the definition of the coefficient of restitution, we have for the discs $A$ and $B$ (or $C$)
$e=\frac{v^{\prime \prime }-v^{\prime }\sin \theta }{v\sin \theta -0}$
But $e=1$, for perfectly elastic collision,
So, $v\sin \theta =v^{\prime \prime }-v^{\prime }\sin \theta$ (3)
From (2) and (3),
$v^{\prime }=\frac{v(1-2{\sin }^2\theta )}{(1+2{\sin }^2\theta )}$
$=\frac{v({\eta }^2-2)}{6-{\eta }^2}$ {using (1)}
Hence we have,
$v^{\prime }=\frac{v({\eta }^2-2)}{6-{\eta }^2}$
Therefore, the disc $A$ will recoil if $\eta <\sqrt{2}$ and stop if $\eta =\sqrt{2}$.