How far until it won't come back?

First, let's analyze some limiting cases. If the two discs were joined to form a single disc D of mass $2m$ , then by simple calculation we'd expect disc A to recoil with velocity $v_0/3$ toward the left. If disc D had mass $m$ , we'd expect disc A to come to rest, and disc D to move with velocity $v_0$ to the right after the collision. Thus, as the mass of disc D is lowered, the system undergoes a transition whereupon the recoil velocity goes from positive to negative.

Similarly, we expect to find a value $\eta_\textrm{max}$ above which discs B and C fail to provide enough recoil to A for it to continue to recoil. Above this separation we'd expect disc A to continue on toward the right (with diminished velocity) after the collision.

In the system we're considering, the centers of discs B and C make an angle $\theta$ with the center of disc A at the moment of collision. After the collision, discs B and C will take off with velocity in the direction of the displacement vector from the center of A to their respective centers. In the situation shown, with B and C touching, B will take off at an angle of $\SI{30}{\degree}$ with the horizontal, and C will emerge at an angle of $\SI{-30}{\degree}$ with the horizontal.

Since the collision is elastic, we have conservation of energy, in addition to the conservation of momentum. Writing down both conservation relations, we have

$\begin{aligned} mv_L(0) &= mv_L(\infty) + 2mv_R\cos\theta \\ \frac12 mv_L(0)^2 &= \frac12 mv_L(\infty)^2 + mv_R^2 \end{aligned}$

Solving the first relation for $v_R$ , we find $v_R = \frac{mv_L(0) - mv_L(\infty)}{2\cos\theta} ,$ so that the second equation becomes

$\begin{aligned} v_L(0)^2 &= v_L(\infty)^2 + 2v_R^2 \\ v_L(0)^2 - v_L(\infty)^2 &= \frac{\left(v_L(0) - v_L(\infty)\right)^2}{2\cos^2\theta} \\ \left(v_L(0) - v_L(\infty)\right)\left(v_L(0) + v_L(\infty)\right) &= \frac{\left(v_L(0) - v_L(\infty)\right)^2}{2\cos^2\theta} \\ 2\cos^2\theta\left(v_L(0) + v_L(\infty)\right) &= v_L(0) - v_L(\infty) \\ v_L(\infty)\left(1+2\cos^2\theta\right) &= v_L(0)\left(1-2\cos^2\theta\right) \\ v_L(\infty) &= v_L(0) \frac{1-2\cos^2\theta}{1+2\cos^2\theta} \end{aligned}$

Thus, for recoil, we must have $1-2\cos^2\theta < 0$ or $\cos\theta=1/\sqrt{2}$ or $\theta = \pi/4 = \SI{45}{\degree}$ .

This corresponds to an inter-center distance of $\eta_\textrm{max}D = 2D\sin\theta = 2/\sqrt{2}D = \sqrt{2}D$ .

Intuitively we can say that for the largest value of $\eta$ , such that disc A recoils to the left after the collision, the final velocity of A must be 0.

Seeing as the masses are all equal, the conservation of momentum and kinetic energy properties this collision has tells us that:

$\vec{u} = \vec{v}_1+\vec{v}_2$ and $\vec{u}^2=\vec{v}_1^2+\vec{v}_2^2$

where $\vec{u}$ is the inital velocity of disc A, and $\vec{v}_1$ and $\vec{v}_2$ are the final velocities of discs B and C respectively.

However, if we square both sides of the first equation, we have that:

$\vec{u}^2=\vec{v}_1^2+\vec{v}_2^2+2\vec{v}_1\cdot\vec{v}_2$

At first, by comparison with the second equation, this seems like a contradiction: the only way that these two equations could hold would be for $\vec{v}_1\cdot\vec{v}_2=0$ , or in other words, the dot-product between $\vec{v}_1$ and $\vec{v}_2$ is 0. This implies that they are perpendicular.

Thus the angle of separation of $\vec{v}_1$ and $\vec{v}_2$ is $90^\circ$ .

Seeing as we know that the impulse from disc A is transferred normally to the tangents it forms with discs B and C upon contact, we can now form a right-angled triangle between the centers of A, B and C.

Applying Pythagoras to this triangle yields $2D^2=\eta^2 D^2$

From this we can say that $\eta=\sqrt{2}\approx 1.414$