Three bodies, two collisions I'm struggling with this seemingly simple question regarding elastic collision. I've worked out something but it is not enough.

The problem reads: Consider three small masses A,B and C placed linearly on a frictionless road which weigh $2m$ kg, $m$ kg, and $2m$ kg correspondingly. At $t=0$ the masses are at rest, and then A starts moving towards B with velocity $v$. If the distance between B and C is $L$ meters, prove that B will collide with A again after $12L/7v$ sec.

My take: I know that when mass X (X kg) and moving at velocity $v$ m/sec collides elastically with a resting mass Y (Y kg), then the velocities after the collision are given by
$v_X=(X-Y)v/(X+Y), \  v_Y=(2X)v/(X+Y)$. Applying this for the collision between A and B, I get the velocities
$v_A=v/3, \ v_B=4v/3$.
Applying this again to the collision between B and C, the velocity of B after the collision will be
$v_B'=-4v/9$.
My question is, how to I determine the time until A and B collide for the second time? I mean B will hit C after $3L/4v$ sec, by which time A will have moved $L/4$ meters, and then ...what?
Am I seeing it wrong? Is there another approach I am missing?
 A: Part 1. As you noted, ball A will move with velocity $v_A=\dfrac v3$. Ball B will move at velocity $v_B=\dfrac 43v$. We'd like to know the distance that ball A moves while ball B moves towards C.
Recall that the distance $d=Vt$. We know the speed of ball A. However, we don't know the time elapsed. Luckily, the time can easily be found knowing that ball B will take $t=\dfrac{3L}{4v}$ (by that same distance equation). Therefore, the distance that A travels while B moves towards C is:
$$d=\dfrac{3L}{4v} \dfrac{v}{3}=\dfrac L4$$
Part 2. Find initial distance between balls A and B.
When A and B first collide, the distance between them is obviously zero. Then, A proceeds to move a distance of $L/4$ rightwards, whereas ball B moves rightwards a distance of $L$. Therefore, the distance between the balls is $\dfrac34L$.
Part 3. And lastly, combine it all:
You know:

*

*Distance between balls A and B.

*Ball A has velocity $v_A=\dfrac v3$.

*Ball B has velocity $v_B=-\dfrac 49v$.

A: I think I got it. I didn't put it in terms of distance covered.
Taking t to again be zero at the moment B strikes C, then the distance covered by B will be
$S_B(t)=3L/4-4vt/9$, and the distance covered by A will be $S_A(t)=tv/3$. These are equal when $t=27L/28v$, giving a total time of $12L/7v$ :D
