Why Conservation of Momentum Only Gives You Mass Ratios

I saw a question online recently from a physics student who was given a conservation of linear momentum problem during lecture. They were concerned that the problem was impossible since they could only find the ratio of masses instead of exact values, but it turned out to be much worse than that. In fact, the problem given to them was unphysical (did not conserve momentum). In the student’s attempt, they had accidentally wrote the problem down incorrectly which incidentally gave them an answer for the ratio of masses.

Thus their online post really, by chance, had two interesting questions baked in:

• Why can we only solve for mass ratios in 2D collision problems?
• Could we generalize a 2D conservation of linear momentum problem to determine if a solution exists?

Admittedly, these don’t help actually solve the given homework problem, but it instead tells us if the homework problem is physically possible and what we can solve for if it is.

Background

Consider an arbitrary conservation of linear momentum problem in 2D for the collision of two objects with masses $m_1$ and $m_2$. We have the following equations for each axis:

Re-writing this, so as to put it into a nice form for a matrix equation, we get

So that if we define

we obtain the matrix equation

Equivalently,

Calling the velocity-difference matrix $V$, this is just

Here we can use some properties of matrices to make general conclusions:

1. Clearly the trivial solution, $m_1=m_2=0$, always solves this system.

2. If $V$ is invertible, i.e. $\det(V)\neq 0$, then there is only one solution. By the first point, that solution must be the trivial solution.

3. If $\det(V)=0$, then the columns of the $V$ matrix, i.e.

are linearly dependent. In other words, they are scalar multiples of one another,$\mathbf{\Delta v_1} = -\lambda \mathbf{\Delta v_2}$. It also means the change in the individual velocity vectors must be antiparallel, scaled multiples of one another! They must be antiparallel since both masses must be larger than zero (try it for yourself). This scalar multiple is in fact the mass ratio, and thus there is not a unique solution of mass pairs that satisfy this condition.

In fact, in the case of 3., there are infinitely many solutions and they form a line in the $(m_1,m_2)$-plane. The slope of this line gives the ratio of the masses. Thus, mathematically, if your problem has solutions other than the trivial solution, you cannot uniquely determine $m_1$ and $m_2$. You can only determine their ratio.

To translate some of this into physical wording: if any objects with any mass are present in the problem, then clearly the trivial solution makes no physical sense, since that would imply the objects have no mass. We are ignoring photons and relativistic momentum, obviously. This means that for a physically interesting problem, we require

Anything else and the problem does not make sense physically.

Broken Example

A ball of mass $m_1$ moves at $2 \text{ m/s}$ in the $\hat{x}$ direction, hits another ball of mass $m_2$ initially at rest. The ball of $m_2$ flies off at an angle of $60^\circ$, below the horizontal, at $4\text{ m/s}$. The ball of $m_1$ is deflected at an angle of $45^\circ$, above the horizontal, at $2 \text{ m/s}$. What is each ball’s mass?

You can probably just look at the velocities and see how this is likely not physical, but we can now rigorously verify this claim.

Applying conservation of momentum in the $x$- and $y$-directions gives

Bringing all terms to the left,

Writing this as a matrix equation,

Define

so that

The determinant is

Substituting the numerical values,

or equivalently,

Since

the matrix $V$ is invertible. Therefore, the only solution is the trivial solution:

which implies

or explicitly,

This means the problem is not physically compatible with conservation of momentum based on the values given.

Takeaway

Very rarely is conservation of momentum viewed from the perspective of linear algebra. Truthfully, this is likely due to students in Physics 1 having not taken linear algebra at that point — which lends itself to a whole other discussion on the best way to approach introductory physics. But that’s for another time.

Viewing the problem in this way, however, gives us a result that is physically consistent and follows from a purely mathematical argument. Specifically, conservation of momentum alone is not enough to uniquely determine two masses given the incoming and outgoing velocities from a collision. Moreover, we can derive a constraint on these velocities, $\det(V)=0$, that if broken yields only the trivial solution. In other words, the only physically interesting examples occur when $\det(V)=0$.

Note that we have been working in what I would call a “mass basis” for this problem. That is, the unknown vector is the vector of masses. This is useful for asking whether the given velocities are compatible with conservation of momentum. However, if the goal is to derive physical quantities such as scattering angles (1), then the velocity basis, or more generally the center-of-mass/velocity-space picture, is much more useful (2,3).

What this simple example does provide is a foreshadowing of a very common occurrence in higher-level physics such as condensed matter theory, quantum mechanics, electrodynamics, and thermodynamics. Very often, we run into a system of vector equations that can be rewritten in matrix form, and the physically nontrivial system occurs when $\det(A)=0$. Such systems are incredibly interesting, but warrant their own space to discuss.

References