# Three connected balls

I'm stuck solving this problem with three small balls of masses $m$, $2 m$ and $3 m$ on a smooth table, connected by two equal, light inextensible strings as shown, and initially at the vertices of an equilateral triangle. The strings are initially taut, the two larger masses are at rest and the smallest mass moves to right with some initial speed.

Eventually $m$ comes to a position where the string to $2 m$ gets taut again, and $m$ exerts an impulse on $2 m$, which in turn exerts some impulse on $3 m$. Mass $m$ and initial speed is known, which leaves me with six unknowns, or three 2D velocities after the pull. But I can only muster five equations: two from the conservation of momentum, one from conservation of energy, one from knowing that the impulse on $3 m$ is in the horizontal direction, and one from knowing that the velocity difference of $m$ is in the direction to $2 m$ at the moment of the pull. What am I missing?

## Update

Following the answer by @Farcher below, the two-stage calculation yields velocities:

$\vec{v_1} = (2 v_0/3, -\sqrt{3} v_0/3)$

$\vec{v_2} = (-v_0/30, \sqrt{3} v_0/6)$

$\vec{v_3} = (2 v_0/15, 0)$

Where $v_0$ is the initial speed of $m$. Total kinetic energy is conserved, and the center of mass is undisturbed by the event as the second animation shows. Eventually $m$ distances too much from $2 m$ and should interact with it again.

• Conservation of total angular momentum? – Martin Ueding Jun 15 '16 at 8:13
• @MartinUeding That's likely the case, but I can't write it down. Let's say the pivot is at $m$ at the moment of the pull. I don't know the direction of where the $2 m$ will go and thus its torque distance. – BoLe Jun 15 '16 at 8:22
• Taking the center of mass of the whole system as the pivot point will give a conserved quantity. – Martin Ueding Jun 15 '16 at 8:34
• @BoLe Energy can't be assumed to be conserved within the Kinetic Energies of balls. – Dvij Mankad Jun 15 '16 at 10:18
• @Bole You can show your equations. Of course there's no potential energy since the strings are 'inextensible'. – philip_0008 Jun 15 '16 at 13:22

• The interaction of mass $m$ and mass $2m$ along the line of the string joining them and this should give you the velocity of mass $2m$.
• The interaction of mass $2m$ (with the velocity found in the first stage) and mass $3m$ along the line of the string.
• I did that intuitively first. But shouldn't one after the two stages you describe plug the calculated velocity of $2 m$ into the next interaction with $m$ and iterate the stages in this way? That doesn't converge though. – BoLe Jun 15 '16 at 16:09
• In this two step analysis mass $m$ gives mass $2m$ some momentum and then that momentum which mass $2m$ has is shared with mass $3m$ and it is the final momentum that mass $3m$ which will give you the answer to the question. To save time on the algebra you might find this Wikipedia article of use? en.wikipedia.org/wiki/… – Farcher Jun 15 '16 at 16:21