This question is asked on Physics SE and MathOverflow by somebody else. I don't think it belongs there, but rather here (for reasons given there in my comments there; edit: now self-removed).



This is it.  The perfectly centered billiards break.  Behold.

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The Collision



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The Trajectories


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Linear and Quadratic Responses


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Stiff Response



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As you can see, the result is very similar to my first answer below. This seems like good evidence that the behavior discussed in my first answer is indeed the limiting behavior in the case where the stiffness goes to infinity.


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While the simplicity of this behavior is appealing, I would guess that "real" billard balls do not have such a stiff force response. Of the models listed here, the intial Hertz-based model is probably the most accurate. Qualitatively, it certainly seems the closest to an "actual" break.


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    $\begingroup$ Also, can you check if changing the form of the force function, say from $2-d$ to $(2-d)^2$, changes the results? $\endgroup$ – user856 Feb 1 '14 at 9:03
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    $\begingroup$ @JimBelk, If changing parameters asymmetrically doesn't change the qualitative behavior of the solution, then it is usually safe. $\endgroup$ – achille hui Feb 1 '14 at 9:57
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    $\begingroup$ Aha! What this shows is that the original question is indeed underdetermined. One cannot simply treat rigid balls as the limiting case as stiffness approaches infinity, for it matters how the limit is approached (in the sense of what the shape of the stiffness function is). $\endgroup$ – user856 Feb 1 '14 at 11:01
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    $\begingroup$ But that's not what "stiffness" means. In Hooke's law $F = k x$, the stiffness is the constant $k$ in front of the $x$, not the unwritten exponent $1$ over the $x$. Infinite stiffness in this billiards-balls scenario would be a force that rises to infinity as soon as $d<2$. All of your examples are extremely stiff; the only difference is how they approach the infinite step function. $\endgroup$ – user856 Feb 1 '14 at 21:11
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    $\begingroup$ @RahulNarain I suppose that's true -- they're all "stiff" in the usual physics sense of the word. Actually, what I call the "high-stiffness" answer could also be interpreted as an answer for the case where there is a very small distance between the balls, which must be overcome for a pair of balls to collide. $\endgroup$ – Jim Belk Feb 1 '14 at 21:17

Note: I'm not deleting this answer since the discussion in the comments is interesting, but I no longer believe that this answer is physically realistic. See my other answer for a much better model.

Later Note: Well, it turns out that this model is realistic if you put certain hypothesis on the bouncing of the balls! See the "stiff response" section in my other answer.

The two balls in the back corners shoot away along rays parallel to the two sides of the triangle. Here is a picture showing the forces, with each force vector emanating from the point of contact.

enter image description here

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    $\begingroup$ Yeah, but ball 2 (balls are numbered 0-15) touches ball 4 and 5. I think the impact will be allocated to 4 and 5 by a ratio of 2:1. (Although there might be a $\pi$ involved.) $\endgroup$ – Řídící Jan 31 '14 at 20:58
  • $\begingroup$ @GlenTheUdderboat I don't think so. The force on ball 2 is entirely from its contact with ball 1. This is a normal force, so it has to be perpendicular to the tangent planes, i.e. parallel to the vector from the center of ball 1 to the center of ball 2. The forces that ball 2 exerts on balls 4 and 5 have to add up to the negative of this force vector, only the only such linear combination is that ball 5 exerts no force on ball 2, and ball 4 exerts the same magnitude of force on ball 2 as ball 1 did. $\endgroup$ – Jim Belk Jan 31 '14 at 21:06
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    $\begingroup$ You almost got me there. :) But you didn't account for ball 2 to move left(ish) after impact. Note that, because of the instantaneity, there is no physical causality or intuition to assume. $\endgroup$ – Řídící Jan 31 '14 at 21:13
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    $\begingroup$ I don't believe your explanation in the comment, Jim. Note that your argument applies just as well to the situation where only balls 1, 2, 4, and 5 were present; your argument would conclude that ball 5 doesn't move at all, which is false. $\endgroup$ – Greg Martin Jan 31 '14 at 21:32
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    $\begingroup$ @GlenTheUdderboat Part of what's going on with this problem is that conservation of momentum and energy do not completely determine the solution. The solution I gave certainly obeys both, because it's what would happen if the six balls not along the sides were removed. $\endgroup$ – Jim Belk Jan 31 '14 at 22:35

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