A classical mechanics problem by Akshat Sharda

Assume $m_{1}$ collide with $m_{2}$ at the top point of the circle. From conservation of linear momentum, $m v_{0} = m v_{1}' + m v_{2}' \:,$ and from definition of coefficient of restitution,
$v_{1}' = v_{2}' - e\; v_{0}\:,$ we get $v_{2}' = \frac{3}{4} v_{0}$ .

Let $\theta_{2}(t)$ and $\theta_{3}(t)$ be angular positions of $m_{2}$ and $m_{3}$ respectively, relative to vertical line that through common center of two circle. Let $l$ be the length of the spring. Then, $l^{2} = 5R^{2} - 4R^{2} \cos( \theta_{2} - \theta_{3})$ For maximum expansion, we have $\frac{dl^{2}}{dt} = 0$ which imply $\frac{d\theta_{2}}{dt} = \frac{d\theta_{3}}{dt}$ .

Since the track is smooth, then the mechanical energy of system $m_{2}$ and $m_{3}$ is conserved. Thus, $\frac{1}{2}m (v_{2}')^{2} = \frac{1}{2}m(2R)^{2} \left(\frac{d\theta_{2}}{dt}\right)^{2} + \frac{1}{2}m(R)^{2} \left(\frac{d\theta_{3}}{dt}\right)^{2} + \frac{1}{2}k(l-R)^{2}\:.$ And, since the external force in tangent direction is zero, then the angular momentum (respect to common center) is also conserved, $mv_{2}' (2R) = m (2R)^{2} \frac{d\theta_{2}}{dt} + m (R)^{2} \frac{d\theta_{3}}{dt}\:.$

Using those equations, we get the maximum expansion of the spring is given by, $(l-R) = \frac{3}{4}v_{0} \sqrt{\frac{m}{5k}} \:.$