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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.

quasi
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    The fact that the question has not been set as homework doesn't prevent it from being an exercise. Site policy requires that you show your attempt and ask about a conceptual difficulty. – sammy gerbil Feb 18 '17 at 21:49
  • See Is my method here for solving this 3-body collision problem correct? – sammy gerbil Feb 18 '17 at 22:05
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    Another reason this doesn't work (that I don't think has been mentioned in the numerous answers) is that even though each particle collides elastically you will not necessarily collide elastically with your 'fictitious particle'. It is perfectly valid to treat a subset of the particles as a system with a position given by the center of mass. But some energy can be lost or gained within your fictitious particle by changing the internal motion of the particles relative to the center of mass. – octonion Feb 19 '17 at 23:11
  • I'm not claiming the fictitious particle $Q_k$ conserves the energy of the total system. It's just used to simulate the impact on $P_k$. Of course, the calculated post-collision velocities better be such that the original total momentum and original total energy are conserved. Both of the posted examples satisfy that validity check. – quasi Feb 19 '17 at 23:16
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    If you don't make some assumption on the energy of P and Q, how are you getting a unique answer for the velocity of P after the collision? Also how are you going back from knowing the momentum of Q to the momentum of the individual particles? – octonion Feb 20 '17 at 00:34
  • Hi quasi. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Feb 21 '17 at 11:58
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    @Qmechanic: I could be wrong, but I've never seen an exercise with 3-way simultaneous collisions. Moreover, the standard 2-way collision theory doesn't appear to be sufficient to resolve the outcome of a 3-way collision. So I don't think my question fits the criteria for the homework-and-exercises tag. – quasi Feb 21 '17 at 12:21
  • These are three idealized point particles with zero volume. In non-idealized reality the faster-moving $P_3$ would have to "pass through" $P_2$ to hit $P_1$, without first colliding with the "right-hand side" of $P_2$ or sandwiching it between $P_1,P_3$. And the behavior is very sensitive to small variations in velocities and displacements which might alter the sequence of which particle collides with which first, thus changing this into a sequence of two two-particle collisions. Can anyone comment on that? – smci May 13 '17 at 22:48

3 Answers3

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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.

Deschele Schilder
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  • In view of your answer, I edited my question. If possible, could you address my idea of regarding a 3-way collision as a 2-way collision of a combined pair with the remaining particle? – quasi Feb 19 '17 at 00:14
  • Consider [edit]ing this answer to include the content of your two other, now-deleted, answers to this question. – rob Jul 19 '21 at 19:31
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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.

peterh
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  • What are some of the details that can change the outcome? – quasi Feb 19 '17 at 00:09
  • @quasi The difference of the physical reality from the abstractions. For example: 1) actually you won't have a 3-body collision, you will have 2-body collisions in a quick sequence. Their order will significantly alter the result. 2) The bodies aren't rigid and point-like, although, for example, small bearing balls would be quite similar as if they would be. But they will still have an elasticity modulus. Thus, the collision(s) won't happen in a moment, the speed of their parttakers won't change instantly. It would mean infinite acceleration for a moment. – peterh Feb 19 '17 at 00:20
  • @quasi None of these will have a major effect to the in the 2-body case, but they will select the real outcome in the 3-body case (between the infinite possibilities, what you get by solving the 2 equations with the 3 unknowns). – peterh Feb 19 '17 at 00:20
  • Forget the order for the moment -- I'll come back to that. But why can't we simply assume that the bodies are rigid and point-like, for the sake of simplifying the model. Isn't that what's done in most of the early parts of Newtonian mechanics? – quasi Feb 19 '17 at 00:23
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    @quasi You can assume, but you lose 1 real number while you produce that model. You have 3 input parameters, you want 3 output parameters, but the Newtonian physics can calculate only 2 for you, because you ignored some important details as you set up the model. – peterh Feb 19 '17 at 00:28
  • I'll edit my question to give some more details of what I'm troubled about, since there's not enough space allowed for a comment. Thanks for staying with me on this. – quasi Feb 19 '17 at 00:38
  • @quasi What you found accidentally, is a chaotical system. A chaotical system means that a little change in its initial state will significantly alter its end state. In the 2-body collision, if one of the bodies is late with a millisecond, the outcome of the collision will be nearly the same. Thus, it is not chaotic. In the 3-body case, if one of the bodies is in late by a millisecond, the outcome will be totally different. It is a chaotical system. You can assume in the model, that they will collide coincidentally, but real bearing balls won't follow this assumption. – peterh Feb 19 '17 at 00:43
  • @quasi As you changed your view from real bearing balls to the hyphotetical ones, you didn't lose any significant information in a 2-body system. But you did from 3 bodies. – peterh Feb 19 '17 at 00:45
  • Let us continue this discussion in chat. – peterh Feb 19 '17 at 00:47
  • @octonion 1) An example: if one of the bodies is a little bit late, or early, then we have not a 3-body collision, but multiple 2-body collisions, and their order depends on the timings. If they collide in different order, the end result will be in general significantly different. This was my reasoning for the sensitive dependence, why it wouldn't be ok? Note: the masses aren't equal here. – peterh Feb 20 '17 at 02:22
  • @octonion 2) We still have 2 equations and 3 unknowns, thus we have only a parametrized 1d space of the possible outcomes, how can you make it welldefined? – peterh Feb 20 '17 at 02:25
  • Yes my mistake you are correct, if the masses are not equal you can can have more than one result. It is more well defined than you'd expect by counting unknowns since we can perturb it slightly and treat it in terms of two body collisions. If they are equal masses there is one unique answer, but if they are different masses yes there might by a discrete set of possible solutions. – octonion Feb 20 '17 at 03:26
  • @octonion Sorry for the late react, I hope you are still here. My idea is the following: in the model, the collisions are instantenous. In the reality, they aren't, instead there is an elastic transformation of the colliding balls, what then transforms back into kinetical energy. If we have only 2, hard, elastic balls, it is not a problem - the end result is perfectly determined by the two equations we have. But if we have 3 balls, then we have at least 2 collisions, all of them have a - short - time interval, in which the kinetical energy transforms to elastic energy, and then back. – peterh Mar 11 '17 at 23:02
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    @octonion How these time intervals relate to eachother, is not only a discrete thing, but a continuous thing, and it determines the real outcome of the real experiment. – peterh Mar 11 '17 at 23:04
  • I think of these three body collisions as defining a boundary in the space of initial conditions. On either side of the boundary there is a small (in this example just 2?) number of well defined asymptotic solutions defined by the two body collisions. In the realistic case you are talking about where the collisions take a finite time, this boundary might have a thickness, and you can't predict what happens in this thin region. But still if you move a little bit away from the boundary you are in the well defined asymptotic region so I think the idealization has value. – octonion Mar 12 '17 at 00:50
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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.

Salvatore Baldino
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