If you haven't watched the linked problem video (the solution isn't revealed in it), you should -- the animations are great. I'll assume in this answer you have a full understanding of what the problem is. Not a tall order.

I could actually give you a picture proof right now -- the solution is that amazing -- but I won't. Let's build our insight up to it, so when you see it, you are ready.

The moment I think of $\pi$, I think of circles. Well, where are the circles here (besides my shoddy drawings of the two balls)? Here's another thing to think about: how do you solve for any result of a collision? You consider

**conservation of momentum and energy**, of course. Aha, that should click in your mind -- conservation of energy is sort of like a circle, it's an ellipse! The condition:

$$\frac12mv_m^2+\frac12Nmv_{Nm}^2=\frac12Nmw^2$$

Is the equation for an ellipse. And conservation of momentum is the equation for a line. But there are

**two sorts of collisions**that can occur in this system: collisions between $m$ and $Nm$, and collisions between $m$ and $\infty$. The former conserves momentum, the latter does not (I mean, the law isn't violated, but the momentum of just the two balls together isn't conserved -- it is

**transferred to the wall**). Respectively under the two collisions, we have:

$$mv_m+Nmv_{nm}=C\\

Nmv_{Nm}=C$$

Now, the

*idea*we have is that when we impose conservation of energy and momentum, we are solving for the

**intersection points**of the ellipse and the line -- one intersection point is the pre-collision configuration of velocities, and the other is the one after collision. So the idea we have in our mind is that of a bunch of lines, each corresponding to a different momentum value (because that is not conserved) intersecting a single ellipse, and we want to

**count the number of intersections**.

The key idea here is that the collisions, as they occur

**in real time**, correspond to the bouncing of an object off the ellipse as it moves across the lines -- first across the slanted blue line, bouncing off the ellipse, then across the red line, then bouncing off the ellipse onto the green line, then the other blue line, and then its last collision.

However, to say this with confidence, we need to be sure that we can really

**map every collision onto an intersection in the velocity space**above, we need the following lemma:

**Lemma:**Among the inter-collision periods, any given configuration of velocities occurs no more than once, i.e. the velocity space configurations are unique (so there is a bijection (well, the surjection is obvious) between the intersections and the collisions).

**Proof:**

*Case 1:*the number of collisions is finite (why do I make these my cases? Because the first argument I thought of requires this assumption, so let's consider the other possibility separately). If a given velocity configuration occurs again, then the system will have to repeat itself (so there will be an infinite number of collisions). Why?

Because the number of collisions from a point in time onwards depends only on the configuration of velocities at that point in time (think about why this is true -- specifically, it does not depend on the distance between the masses, or between the masses and the wall -- is this true when you have more than two masses? But don't we actually have three masses here, counting the wall? So it matters if the masses are mobile. Why? What do mobile things do different that immobile things don't).

*Case 2:*we don't even need the finiteness assumption. We know the velocity (not speed) of $Nm$ is non-decreasing (with sign convention of right being positive). Suppose it stabilises at 0. Then both walls have stabilised at 0, and the collisions must be finitely many. Suppose it doesn't. Then the velocity is continually changing as it hits the smaller ball (if their velocities become equal -- "worldlines become parallel" -- then there must have been only finitely many collisions), so each configuration is unique.

Ok, so we need to find the number of intersections between the ellipse with radii $v$ and $v\sqrt{N}$, and the lines, with slope $-1/N$. Truth be told, I spent a lot of time staring at the diagram at this point, having no way of how to proceed. And so should you.

One thing

*is*easy to see here, which is that the answer is*something*times $\sqrt{N}$, i.e. it goes to infinity at the rate $\sqrt{N}$ as $N$ goes to infinity. Why is this easy to see?Here's how I found the realisation: there are two ways to find $\pi$ in something -- by looking at circles' areas and lengths, or by looking at angles. I had exhausted everything I could looking at lengths (and again, you should too), so let's think about

**angles**. Let's read the angles between the lines.

Well, first, let's scale the diagram so the ellipse becomes a circle -- let's make it a unit circle. This is good, because now the slope of the lines becomes $-1/\sqrt{N}$, which means there's only one crazy diverging infinite term to bother about -- rather than an infinite ellipse going crazy and an infinite number of infinite line segments each going infinitely crazy.

Why is such a scaling okay? Because the number of intersections is invariant under a scaling. There are other things, like distances, that aren't invariant (which is why finding the perimeter of an ellipse is quite hard). What things are invariant? Why?

Each of these angles is clearly $\arctan(1/\sqrt{N})$. The key is to think about the sum of these angles. You might see the trick -- I do quite like it when a funny geometric argument comes out in some bizarre space, a velocity space here, in a physics proof -- and it's "angles in the same segment are equal".

So the sum of those angles approaches the angle subtended by that big chord. What is that big chord? I wonder if it has a name. Well, as $N$ approaches infinity, the chord approaches the

**diameter**, and that angle approaches $\pi/2$. There's another $\pi/2$ from the angles on the other side, so the total of the angles at all intersections is $\pi$.

(Great, another fact from elementary high-school geometry.)

So the total number of angles/intersections/collisions is $\frac{\pi}{\arctan(1/\sqrt{N})}$, which as $N\to\infty$, is the same thing as $\pi\sqrt{N}$.

Which is great. It's just

*great*.

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