Colliding Blocks

This is a simulation of elastically colliding blocks inspired by a 3Blue1Brown video on YouTube about the connection between the number of collisions and $\pi$. During collisions between blocks, both energy \[ E = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \] and momentum \[ P = m_1 v_1 + m_2 v_2 \] is conserved. However, the blocks can also collide with the wall, which corresponds to $v \mapsto -v$. This changes overall the momentum, but not the energy. In particular, if we rescale the axes, then the energy condition requires the points $\left(\sqrt{m_1}v_1,\sqrt{m_2}v_2\right)$ will lie on a circle of radius $\sqrt{2E}$. Likewise, the momentum condition requires points to lie on a straight line with slope $\sqrt{m_1}/\sqrt{m_2}$. In other words, the angle subtended by a collision point in phase space is \[ \theta = \arctan \sqrt{\frac{m_1}{m_2}}. \] This corresponds to angle $\gamma=2\theta$ as seen from the center of the energy-conservation circle. Therefore, If there is only one wall, the maximal number of collisions $N$ before the two blocks part their ways is the largest integer such that $N\gamma < 2\pi$. That is \[ N = \Bigg\lfloor \frac{\pi}{\arctan \sqrt{m_1/m_2}} \Bigg\rfloor \simeq \Big\lfloor \pi \sqrt{\frac{m_2}{m_1}} \Big\rfloor, \] where, in the last approximation, the arc tangent has been expanded around $m_1/m_2 = 0$ to linear order. This means that if $m_2 = 10^{2n}m_1$, then the number of collisions will be the floor of $\pi 10^n$. For example, $m_2=100m_1$ means we expect $N=31$ collisions, and for $m_2=10^8m_1$ we expect $N=31415$ collisions. These are the first $4=8/2$ digits of $\pi\simeq 3.14159265358979...$