$n$-way elastic collisions of particles in 1D, for $n \ge 3$

Question

This is my own question, not a HW question.

I took first year physics$\,{\large{-}}\,$enjoyed it and did well in it, but that was a long time ago, and I've forgotten most of it.

To get started, take $n = 3$.

Here's the setup . . .

Three point particles $P_1,P_2,P_3$ are moving on the $x$-axis.

For particle $P_k$,

• $\,m_k$ is its mass
• $\,x_k(t)$ is its position at time $t$
• $\,v_k(t)$ is its velocity at time $t$

Initial data:

• $\,x_1(0) = -1,\;\;v_1(0) = 1$
• $\,x_2(0) = 1,\;\;\;\;\,v_2(0) = -1$
• $\,x_3(0) = 2,\;\;\;\;\,v_3(0) = -2$

Assumptions:

• Velocities can only change as the result of a collision.
• Particles can't pass though each other.
• All collisions are elastic.

It's clear that there will be a $3$-way collision at time $t=1$.

The basic question is, what are the velocities after the collision?

Presumably, it depends on the masses.

The answers given so far assert that the velocities after the collision are not uniquely determined, but I'm not sure those answers are using all of the available information.

Intuitively, I would expect the initial information to be sufficient to determine the motion.

A proposed resolution: In an $n$-way collision, where $n \ge 3$, assume that for each particle $P$ in the collision, the post-collision velocity of $P$ is the same as it would be if $P$ collided with a fictitious particle $Q$, such that the mass of $Q$ is equal to the total mass of the set of complementary particles (the set of particles other than $P$), and the velocity of $Q$ is chosen so that the momentum of $Q$ is equal to the sum of the momentums of the particles in the complementary set.

Thus, for each particle $P_k$, there is a fictitious particle $Q_k$, temporarily replacing all the particles other than $P_k$, whose only purpose is to determine the post-collision velocity of $P_k$. After that calculation, $Q_k$ is discarded.

Call this the "fictitious particle" method.

As a test example for $n=3$, using $m_1=m_2=m_3=1$, together with the data previously specified for this question (at the top of this post), the fictitious particle method yields post-collision velocities for $P_1,P_2,P_3$ of $-\frac{7}{3},-\frac{1}{3},\frac{2}{3}$.

As another test example, using $m_2=2$ and $m_2=m_3=1$, but all other data the same, the fictitious particle method yields post-collision velocities for $P_1,P_2,P_3$ of $-\frac{3}{2},\frac{1}{2},\frac{3}{2}$.

Note that for both of the above test examples, the calculated post-collision velocities preserve the original total momentum, as well as the original total energy.

Is there any problem with this proposed way of modeling $n$-way collisions?

Actually, there is a problem, but I don't have time to discuss it right now.

But I have a new understanding of these $n$-way collisions, based on some calculations I did just a little while ago. It's not the same as the fictitious particle method, but it yields what I think is the true, correct resolution. I need to check it some more, and I won't have time to post the details until Sunday, but if it checks out, I'll post it as an answer.

Answer 1

Because you write that a 3-way collision takes place at $t=1$, I'll assume the initial velocities remain equal. I'll call all the $v$'s before the collision $v_{i k}$, and after $v_{ak}$. From conservation of momentum:

$m_1 v_{i1}+m_2v_{i2}+m_3v_{i3}=m_1v_{a1}+m_2v_{a2}+ m_3v_{a3}$

Putting in the values of the velocities:

$m_1-m_2-2m_3=m_1v_{a1}+m_2v_{a2}+m_3v_{a3}$

From conservation of kinetic energy:

$\frac1 2 m_1+\frac1 2m_2+2m_3=\frac1 2m_1v_{a1}^2+\frac1 2m_2v_{a2}^2+\frac1 2m_3v_{a3}^2$

If you know the values of $m_1$, $m_2$ and $m_3$, I'll think you can see that there is no unique solution, because you have two equations with three variables, in which case you have infinite many solutions.

Answer 2

You can't calculate it. I explain, why.

In the analogous 2-body problem, you have 2 equations (impulse + energy preservation), and 2 unknowns (the outgoing velocities).

Here you still have 2 equations, but 3 unknowns. Thus, the information you have is not enough to calculate the end result. Although you can calculate an 1-dimensional parametrized space of the possible outcomes.

The deeper reason behind that is this: the rigid body, elastic collision, pointlike particles, 1-dimensional rotationless particles, they are all abstractions. They are an idealized approximation of the reality. In the reality, none of them exists, the reality a little bit differs from them.

In the 2-body case, this little difference causes only a little deviation from the calculated result. But in the 3-body case, their minor details significantly change the outcome. Where exactly will be the outcome of your experiment on this set of the 1d parameter space, it depends on these minor details.

Answer 3

Mind your approximations!

Point particles are abstract things, and point collisions also are not so real. What truly happens for real bodies in the real word is that they have shapes and density distributions.

Remember, the conservation laws are not a substitute of the dynamical equations (in your case, Newton's second law). Take a system in three dimensions composed by particles at $x_i(t)$, where $i$ indicates the number of the particle. With $N$ particles, we have $3N$ degrees of freedom, and we can relate them to the forces through Newton's law. We only have $4$ conservation laws (energy and impulse), so we have to solve $3N-4$ differential equations and then we can obtain the last $4$ coordinates through conservation laws.

In case of real bodies, the shape mainly determines the impact. In this case, forces act through collision and exchange of impulse, but you can calculate the final directions with some effort.

There is also a second thing that you can do to have a better model of collisions. Take electromagnetism, where charges attract and repel each other. In the simple case of 2 particles, you can describe the motion through the Coulomb force. If you take charges of the same sign, they behave in the same way as colliding bodies (just with a contact without interactions). In general, to have a complete model for collisions you have to know the forces in the system and solve some differential equations. Point collisions are just an approximation, and in this case the approximation is not sufficient.