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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.

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    $\begingroup$ 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. $\endgroup$ – sammy gerbil Feb 18 '17 at 21:49
  • $\begingroup$ See Is my method here for solving this 3-body collision problem correct? $\endgroup$ – sammy gerbil Feb 18 '17 at 22:05
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    $\begingroup$ 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. $\endgroup$ – octonion Feb 19 '17 at 23:11
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    $\begingroup$ 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? $\endgroup$ – octonion Feb 20 '17 at 0:34
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    $\begingroup$ @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. $\endgroup$ – quasi Feb 21 '17 at 12:21
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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.

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  • $\begingroup$ 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? $\endgroup$ – quasi Feb 19 '17 at 0:14
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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.

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  • $\begingroup$ What are some of the details that can change the outcome? $\endgroup$ – quasi Feb 19 '17 at 0:09
  • $\begingroup$ @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. $\endgroup$ – peterh says reinstate Monica Feb 19 '17 at 0:20
  • $\begingroup$ @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). $\endgroup$ – peterh says reinstate Monica Feb 19 '17 at 0:20
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    $\begingroup$ @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. $\endgroup$ – peterh says reinstate Monica Feb 19 '17 at 0:28
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    $\begingroup$ @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. $\endgroup$ – peterh says reinstate Monica Mar 11 '17 at 23:04
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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.

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I guess (tell me if you don't) you mean these three equations arise (conservation af momentum):

$m_1v_{1i}+(m_2+m_3)(v_{2i}+v_{3i})=m_1v_{1a}+(m_2+m_3)(v_{2a}+v_{3a})$

$m_2v_{2i}+(m_1+m_3)(v_{1i}+v_{3i})=m_2v_{2a}+(m_1+m_3)(v_{1a}+v_{3a})$

$m_3v_{3i}+(m_1+m_2)(v_{1i}+v_{2i})=m_3v_{3a}+(m_1+m_2)(v_{1a}+v_{2a})$

Filling in $m=1$ for all three masses:

$v_{1i}+2(v_{2i}+v_{3i})=v_{1a}+2(v_{2a}+v_{3a})$

$v_{2i}+2(v_{1i}+v_{3i})=v_{2a}+2(v_{1a}+v_{3a})$

$v_{3i}+2(v_{1i}+v_{2i})=v_{3a}+2(v_{1a}+v_{2a})$

Filling in the three initial velocities (1,-1,-2):

$-5=v_{1a}+2(v_{2a}+v_{3a})$

$-3=v_{2a}+2(v_{1a}+v_{3a})$

$-2=v_{3a}+2(v_{1a}+v_{2a})$

Filling in (-7/3,-1/3,2/3):

$-5=-\frac5 3 $

$-3=3$

$-2=-\frac {14} 3$

It's obvious that the final velocities you came up with are not correct. Off course, maybe I made a mistake somewhere. Maybe you can try for yourself if the three equations I came up with for the velocities after the collisions have a solution.

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  • $\begingroup$ The $3$ equations you started with are not how I modeled it. To replace a pair, I considered a fictitious particle whose mass was the sum of the masses of the pair, but whose velocity was chosen to make the momentum of the fictitious particle equal to the total momentum of the pair. $\endgroup$ – quasi Feb 19 '17 at 17:46
  • $\begingroup$ 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 the calculation, $Q_k$ is discarded. $\endgroup$ – quasi Feb 19 '17 at 17:51
  • $\begingroup$ Note that my calculated post-collision velocities $-\frac{7}{3},-\frac{1}{3},\frac{2}{3}$ preserve both the original total momentum, and the original total energy. $\endgroup$ – quasi Feb 19 '17 at 17:55
  • $\begingroup$ But what if the masses are not equal? For example, if $m_1=2$, $m_2=1$ and $m_3=1$? Then the final velocities are -2/3, -1/3 and 2/3. In this case, the momentum is conserved (-1=-1) but the energies not ($3\frac 1 2$ is not equal to $\frac {13} {18}$). $\endgroup$ – descheleschilder Feb 19 '17 at 20:45
  • $\begingroup$ With masses $m_1 =2$, $m_2=1$, and $m_3 = 1$, then, using the fictitious particle method, I get a post-collision velocity of -3/2 for $P_1$, not -2/3. But nevertheless, I do see a problem, since other two post-collision velocities fail the energy conservation requirement, and worse, they imply that particle $P_3$ passes through $P_2$, which is prohibited. I'll have to rethink this. Thanks for alerting me to the issue. $\endgroup$ – quasi Feb 19 '17 at 21:38
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Just for some rest in my head before going to bed:

For $P_1$ you get (letting all the masses be equal to 1):

Conservation of momentum: $1-3(=-2*\frac2 3)=v_{1a}+2v_{fa}$

Conservation of energy: $\frac 1 2 +(-\frac3 2)^2=\frac1 2(v_{1a})^2+(v_{fa})^2$

So from the first equation, we get: $v_{fa}=-1-\frac1 2 v_{1a}$

Putting this in the second equation: $3(v_{1a})^2+4v_{1a}-7=0$

From the abc-formula, we get $v_{1a}=-\frac7 3$ or $v_{1a}=1$

So only $v_{1a}=-\frac7 3$ remains because $v_{1a}=1$ implies that $P_1$ goes through the other two.

The same can be done for $P_2$ and $P_3$ and different masses.

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  • $\begingroup$ Please do not post three different answers to the same question unless they really are three distinct, complete answers. If you just want to add something to your previous answer, edit it instead. $\endgroup$ – ACuriousMind Feb 20 '17 at 0:43
  • $\begingroup$ @ACuriousMind-Okay. I just wanted to know if this was how quasi did it. $\endgroup$ – descheleschilder Feb 20 '17 at 6:50
  • $\begingroup$ @quasi-I guess this was the way you did it. I had a good night sleep! $\endgroup$ – descheleschilder Feb 20 '17 at 6:53
  • $\begingroup$ @quasi-I think in theory your method works fine! But in practice, it's very difficult to let 3 (or n) masses collide at the same time. Two particles collide at the same time by definition. $\endgroup$ – descheleschilder Feb 20 '17 at 7:03

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