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Two identical pucks of radius 10 cm are placed in a round arena of radius 1 m. They are positioned 50 cm away from the center of the arena on opposing sides. Assuming no energy losses during sliding and collisions, can you hit one of the pucks such that the two pucks touch exactly once and then never again? If so, how can you do it?

Notes:

  • The collisions conserve energy and momentum.
  • The walls of the arena don't move.
  • As there is no friction the pucks can't exchange angular momentum.
  • Please don't invent additional dimensions. Everybody knows that the universe is two-dimensional.
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  • $\begingroup$ Poincaré recurrence is applicable: en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem $\endgroup$ – Johannes Feb 4 at 15:45
  • $\begingroup$ @Johannes This is perfectly true, but would you please place this comment under the answer instead of here? Otherwise it's a too direct hint for everyone familiar with physics. $\endgroup$ – A. P. Feb 4 at 21:21
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Without actually solving the problem formally I hypothesize that the answer is

No

My reasoning is that

The dynamics of perfectly elastic collisions is reversible in time. So if there existed a dynamical state where a collision occurs followed by no collisions forever, then that would be equivalent to being able to define a state where no collision occurs forever, but then a collision follows. Which doesn't make logical sense to my way of thinking.

I presume that there is probably a simple geometrical conjecture involving rational angles that could confirm the above. Proof that irrational angles don't produce a solution is as follows:

If both the balls bounces off the wall at an irrational angle (as a fraction of $360°$ or $2\pi$ radians), then the balls must collide eventually. Just consider stretching their future collision times along a single time axis. A ball $a$ or $b$ that does not collide with any other will hit the wall at constant time interval $T_a$ or $T_b$ and each collision with the wall will precess some angle $\alpha_a$ or $\alpha_b$, where $–\pi \le \alpha_a \le \pi$, around arena. So at the $n$th collision the ball a will collide with the wall at position $n\alpha_a \, \text{mod} \, 2\pi$ at time $n T_a$. As angle $\alpha_a$ is an irrational fraction of $2\pi$, an $n$ can be chosen such that the ball is arbitrarily close to the zenith ($0°$) position of the arena, and similarly an $m$ can be chosen such that the ball will arbitrarily closely revisit that position after a further $p$ (or $2p$, or $3p…kp$) collisions – where $m$ and $k$ are functions of what you define as arbitrarily close. If both balls are bouncing irrationally then they each get arbitrarily close to the zenith after $p$ and $q$ collisions respectively at time intervals $p T_a$ and $q T_b$. Now identify a fraction $h/k$ which is arbitrarily close to $p T_a / q T_b$ , and after $kp$ and $hq$ collisions each, the two balls will both be arbitrarily close to the zenith within an arbitrarily short time period and will collide. The same argument still holds if one ball is cycling a rational angle.

So the only remaining possibility is that both balls bounce at rational angle increments after the collision and manage to miss each other, which I still think is unlikely, although it is obviously possible to set two balls bouncing so they never collide...

If such a rational solution exists after one initial collision then:

Both the balls bounce rational angles so each repeats its path exactly after some $p$ and $q$ collisions. In this case the repeat period for each ball will be $p T_a$, and $q T_b$ respectively. There is no guarantee that the repeat times are rational or rational ratios of each other, but regardless of that, you can choose some large number of repeat periods $u$ and $v$ for each ball such that $u p T_a$ is arbitrarily equal to $v q T_b$. At the next collision after that time, the two balls will strike the walls at positions and relative times arbitrarily close to those that occurred at the time of their first collision with the wall (after they initially struck each other). At that precise moment (when the pattern repeats), trace the ball's positions backwards and you will observe that their time-reversed paths are arbitrarily close to their initial paths after they struck each other. So tracing it backwards they would strike again (or to put it another way – they couldn't get to the arbitrarily identical positions without colliding with each other immediately beforehand).

I think that is sufficient to confirm the hypothesis in all possible cases other than zero initial velocity. I'm sure things could be formalized but I will stick with the hand-wavy "proof".

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  • $\begingroup$ We could remove the arena walls and your reasoning still applies. $\endgroup$ – deep thought Feb 3 at 23:20
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    $\begingroup$ I agree with your comment completely - but within a finite arena I believe the "logical sense" line still holds in the sense that "I cant't point out the exact problem, but something smells wrong with the consistency of such a dynamic". That is why I only present it as an informal hypothesis. $\endgroup$ – Penguino Feb 3 at 23:23
  • $\begingroup$ Hmmm, yes, finite arena (with puck of non-zero radius). I think you're right actually... $\endgroup$ – deep thought Feb 3 at 23:27
  • $\begingroup$ rot13(engvbany => crevbqvp, bevtvany nethzrag nccyvrf) $\endgroup$ – deep thought Feb 4 at 0:06
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    $\begingroup$ Rational for both pucks doesn't apply to my original argument as then the contact positions of each puck do not precess around the arena(in the sense that there will be many angles on the wall of the arena that a ball will never get arbitrarily close to). Instead, after some p*q collisions, both balls will be in teh exact same places again and will repeat their dance. But that gives me an idea... $\endgroup$ – Penguino Feb 4 at 1:45
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I'm on the same boat with @Penguino, but there's the edge case still to consider:

Can the pucks come to touching distance of each other without actually colliding?

If that were the case, then it could theoretically be possible for the pucks to complete arbitrarily many synchronised loops of some kind without actually colliding, and it would be possible to have a collision in the mix as well.

This corner case seems equivalent to this problem:

enter image description here

What are the constant non-zero speeds (directions given) of these two points, given that they never came closer than ten units of each other?

If there's a solution to this problem, then there could be a solution to the whole, even given @Penguino's excellent argument: after the first collision the pucks must be moving at right angles because of the conservation laws, and they must repeat the position later, (because infinity is quite big), but given an answer, this could be a way to get around @Penguino's argument.

I think the answer must be "nuh-uh, can't do that, $\text{cos}(\theta)\approx1$ for small $\theta$, and infinite speed ain't cool", but I'm not exactly sober, and I'm way past my bedtime even without that, so I'm afraid I'll have to leave it to you guys.

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