Consider the following preliminary problem.

A box of mass $m$ is attached to a box of mass $M$ by a compressed spring, and both are moving at velocity $v_{0}$. The compressed spring has potential energy $U$. What are the velocities of the boxes after the spring is released and $m$ is ejected backward? (Assume the little box loses contact with spring right when the spring loses all of its potential energy.)

enter image description here

The right direction is positive and the left direction is negative.

This can be solved by applying energy and momentum conservation. By comparing the energies before and after the release, we find \begin{align*} E &= \frac{1}{2}(m+M)v_{0}^{2} + U = \frac{1}{2}mv_{1}^{2} + \frac{1}{2}Mv_{M}^{2}, \\ P &= mv_{0} + Mv_{0} = mv_{1} + Mv_{M}. \end{align*} The above is two equations and two unknowns ($v_{M}, v_{1}$). Solving for $v_{M}$ and $v_{1}$ becomes incredibly difficult, but eventually you get

\begin{align*} v_{M} = v_{0} + \sqrt{\frac{2U}{M(1 + \frac{M}{m})}} \qquad\text{and}\qquad v_{1} = v_{0} - \sqrt{\frac{2U}{m(1 + \frac{m}{M})}} \end{align*}

Now the main scenario.

Let's say there are two boxes of masses $m_{1}, m_{2}$ attached to the big box, and the springs have energies $U_{1}, U_{2}$, respectively. Then suppose both springs are released simultaneously. A sketch is given here.

enter image description here

In a realistic scenario, the big box would have some torque imparted on it, and this would no longer be a one-dimensional problem. To counteract this, let's modify the image a bit.

Instead of two boxes attached, let's say there are two pairs of boxes arranged at the back edges of the big box. The back of the box can look like this:

enter image description here

Boxes $1$ and $2$ are of mass $m_{1}/2$ with identical springs, and boxes $3$ and $4$ are of mass $m_{2}/2$ with identical springs. This way symmetry is preserved for any "launching" of the boxes, and we can continue to treat this as a one-dimensional problem.

To simplify this further, let's assume $v_{0} = 0$ (as we should have done in the initial problem).

Then energy and momentum conservation gives us \begin{align*} E &= U_{1} + U_{2} = \frac{1}{2}Mv_{M}^{2} + \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}, \\ P &= 0 = Mv_{M} + m_{1}v_{1} + m_{2}v_{2}. \end{align*}

We have $2$ equations and $3$ unknowns ($v_{M}, v_{1}, v_{2}$). It seems like the solution is not uniquely determined.


  • What exactly is going on here? Why is the solution not uniquely determined? We know in real-life the scenario has a single, uniquely determined outcome.
  • What additional info do you need to solve this problem uniquely?
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    $\begingroup$ To solve it, you need to assume a delay between ejections. Even a little delay like $\epsilon$ will do. Or you have to assume other restrictions. $\endgroup$ – Paradoxy Jul 31 '19 at 23:39
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    $\begingroup$ @Paradoxy Isn't the three body problem related to the inclusion of gravitation between the objects? $\endgroup$ – JMac Jul 31 '19 at 23:39
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    $\begingroup$ @JMac Of course not. It's a lot more general than that. For a simple example, assume that 3 balls with different masses collide at the same time (like at a billiard table or something) You can't find their final velocity, because there are 3 equations (two conservation of momentum in y and x) and one for conservation of energy. While there are 4 unknowns, their final velocity and direction of motion of one ball. And just for more information, if you assume that their mass are the same, you can solve the problem. It's done in halliday physics book $\endgroup$ – Paradoxy Jul 31 '19 at 23:44
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    $\begingroup$ @Paradoxy I just mean in the traditional sense of the term "three-body problem" you're mostly going to find information on how to deal with massive bodies who also interact with gravity. I'm not saying there's no issue here, just that calling it the three-body problem isn't particularly relevant in most contexts (including the one you linked). $\endgroup$ – JMac Jul 31 '19 at 23:49
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    $\begingroup$ you folks both have good points. That is a three-body problem by definition of course, and simultaneously not the three-body problem. Now I'm going to wander off and think about the three simultaneous balls problem... ;-) $\endgroup$ – uhoh Jul 31 '19 at 23:54

For your 3 body problem you will find that one pair of boxes detaches before the other pair of boxes. There are thus two equations for each of momentum and energy conservation at these two transition times. You will find then that you have enough to go on in finding the unknowns.

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