Moving Beads About To Collide

The system starts with energy $\frac12 m v^2 + \frac12 m v^2 = m v^2$ as well as momentum $m v + m v = 2 m v$ .

Now momentum is conserved, and when the beads collide their velocities in the frame of the ring cancel each other out. Therefore, the total momentum at time of collision is due to the ring carrying the balls for a total mass of $3m$ . Therefore, its velocity, $v'$ , must be $\frac23 v$ .

The kinetic energy of the ring alone is therefore $\frac12 m \left(\frac23 v\right)^2 = \frac29 m v^2$ , so the remaining two beads have a total kinetic energy of $\frac79 m v^2$ . So due to symmetry, each has an energy of $\frac7{18} m v^2 = \frac12 m (v_\text{ball})^2$ . So $v_\text{ball} = \frac{\sqrt7}3 v$ . So $\frac yx = \frac{\sqrt7}2$ , so $\left(\frac yx \right)^2 = \frac74 = \boxed{1.75}$ .

Pick an axis system $Oxy$ such that $Ox$ is in the direction of the original motion of the beads. By symmetry, both beads will always make the same angle with the $x$ -axis. By symmetry again, the ring will move along the $x$ -axis. Since all reaction forces are smooth, the ring does not rotate.

Suppose that at some time in the motion the centre of the ring has moved a distance $u$ from its starting position, and the two beads make an angle of $\theta$ above and below the $x$ -axis. Then the centre of the ring and the two beads have coordinates $(u,0)$ , $(u+a\cos\theta,a\sin\theta)$ and $(u+a\cos\theta,-a\sin\theta)$ respectively. Since all forces acting on the system are either irrelevant (like gravity) or internal, momentum is conserved, and hence $m\left(\begin{array}{c} \dot{u} \\ 0 \end{array}\right) + m\left(\begin{array}{c} \dot{u} - a\sin\theta \dot\theta \\ a\cos\theta \dot\theta\end{array} \right) + m\left(\begin{array}{c} \dot{u} - a\sin\theta\dot\theta \\ -a\cos\theta \dot\theta\end{array} \right)$ is constant. Thus we deduce that $3\dot{u} - 2a\sin\theta\dot\theta$ is constant. Since, initially, $\dot{u} = 0$ , $\theta = \tfrac12\pi$ and $\dot\theta = -\tfrac{v}{a}$ , we deduce that $3\dot{u} - 2a\sin\theta\dot\theta \; = \; 2v$ All forces are conservative, and so energy is also conserved. Thus $\tfrac12m \dot{u}^2 + 2\left[ \tfrac12m(\dot{u} - a\sin\theta\dot\theta)^2 + \tfrac12ma^2\cos^2\theta\dot\theta^2\right]$ is constant, and hence $\tfrac32\dot{u}^2 - 2a \sin\theta \dot{u}\dot\theta + a^2\dot\theta^2 \; = \; v^2$ When $\theta = 0$ , the conservation of momentum equation tells us that $\dot{u} = \tfrac23v$ , and the conservation of energy equation tells us that $a^2\dot\theta^2 = \tfrac13v^2$ , and hence $a\dot\theta = -\tfrac{1}{\sqrt{3}}v$ . Thus $x = \tfrac23v$ , and the velocity of the bead below the $x$ -axis is $\left(\begin{array}{c} \dot{u} - a\sin\theta\dot\theta \\ -a\cos\theta \dot\theta\end{array} \right) \; = \; \left(\begin{array}{c} \tfrac23v \\ \tfrac{1}{\sqrt{3}}v \end{array} \right)$ so that $y^2 = \tfrac49v^2 + \tfrac13v^2 = \tfrac79v^2$ , and hence $y = \tfrac{\sqrt{7}}{3}v$ . We deduce that $\left(\frac{y}{x}\right)^2 \; = \; \left(\frac{\sqrt{7}}{2} \right)^2 \; = \; \boxed{1.75}$ If you prefer, we can consider the equation of motion of the bead above the $x$ -axis, which is $m\left(\begin{array}{c} \ddot{u} - a\sin\theta\ddot{\theta} - a\cos\theta\dot\theta^2 \\ a\cos\theta\ddot\theta - a\sin\theta\dot\theta^2\end{array}\right) \; = \; -R\left(\begin{array}{c}\cos\theta \\ \sin\theta\end{array} \right)$ where $R$ is the magnitude of the reaction of the ring on the bead. This equation reads $m\ddot{u}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) + ma\ddot\theta\left(\begin{array}{c}-\sin\theta \\ \cos\theta \end{array}\right) + (R- ma\dot\theta^2)\left(\begin{array}{c} \cos\theta \\ \sin\theta \end{array}\right) \; = \; \mathbf{0}$ Taking the scalar product of this equation with the vector ${-\sin\theta \choose \cos\theta}$ , we deduce that $-m\sin\theta \ddot{u} + ma\ddot{\theta} = 0$ , and hence $\sin\theta \ddot{u} \; = \; a\ddot\theta$ If we combine this identity with the (derivative of the) conservation of momentum equation, we can derive the differential equation $\dot\theta^2 \; = \; \frac{v^2}{a^2(3 - 2\sin^2\theta)}$ which again tells us that $a\dot\theta = -\tfrac{1}{\sqrt{3}}v$ when $\theta = 0$ . We could, if we wished, solve this differential equation to determine how long it takes for the two beads to collide, namely $\frac{a\sqrt{3}}{v}E\big(\sqrt{\tfrac23}\big) \approx. 2.18444\frac{a}{v}$ where $E$ is the complete elliptic integral of the second kind: $E(k) \; = \; \int_0^{\frac12\pi} \sqrt{1 - k^2\sin^2\theta}\,d\theta$