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.)
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.
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:
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.
Questions
- 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?
