Question

Three identical discs $A,B$ and $C$ rest on a smooth horizontal plane. The disc $A$ is set under motion with velocity $v$ 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. At what value of $\eta$ will the disc $A$ recoil after the collision, stop and move on?

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 ${\nu }^{{}^{\prime }}$ and be directed towards right after the collision. It is also clear from the symmetry of problem that the disc $B$ and $C$ have equal speed (say ${\nu }^{{}^{\prime \prime }}$) in the direction, shown From the condition of the problem,
$\cos \theta =\frac{\eta \frac{d}{2}}{d}=\frac{\eta }{2}$ so, $\sin \theta =\sqrt{4-{\eta }^2}/2\left(1\right)$
Fro the three discs, system, from the conservation of linear momentum in the symmetry direction (towards right)
$m\nu =2m{\nu }^{{}^{\prime \prime }}\sin \theta +m{\nu }^{{}^{\prime }}$ or $\nu =2{\nu }^{{}^{\prime \prime }}\sin \theta +{\nu }^{{}^{\prime }}\left(2\right)$
From the definition of the coefficient of restitution, we have fro the disc $A$ and $B$ (or $C$)
$e=\frac{{\nu }^{{}^{\prime \prime }}-{\nu }^{{}^{\prime }}\sin \theta }{\nu \sin \theta -0}$
But $e=1$, for perfectly elastic collision,
So, $\nu \sin \theta ={\nu }^{{}^{\prime \prime }}-{\nu }^{{}^{\prime }}\sin \theta \left(3\right)$
From $\left(2\right)$ and $\left(3\right)$,
${\nu }^{{}^{\prime }}=\frac{\nu (1-2{\sin }^2\theta )}{(1+2{\sin }^2\theta )}$
$=\frac{\nu ({\eta }^2-2)}{6{\eta }^2}\left\{using\right(1\left)\right\}$
Hence we have, ${\nu }^{{}^{\prime }}=\frac{\nu ({\eta }^2-2)}{6-{\eta }^2}$