$\pi$ from collisions

Let $n$ be the number of collisions, $v_n$, $u_n$ be the velocity of $M$, $m$ respectively after $n$ collisions. For fully elastic collisions, $$\left[\begin{array}{c}{v}_n\\ u_n\end{array}\right]=A^{\frac{n}{2}}\left[\begin{array}{c}{v}_0\\ 0\end{array}\right],whereA=\left[\begin{array}{cc}\frac{M-m}{M+m}& \frac{2m}{M+m}\\ \frac{-2M}{M+m}& \frac{M-m}{M+m}\end{array}\right]$$ $A$ can be decomposed as $$\left[\begin{array}{cc}i\sqrt{\frac{m}{M}}& -i\sqrt{\frac{m}{M}}\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{e}^{i\varphi }& 0\\ 0& e^{-i\varphi }\end{array}\right]\left[\begin{array}{cc}-i\sqrt{\frac{M}{m}}& 1\\ i\sqrt{\frac{M}{m}}& 1\end{array}\right],where\cos \varphi =\frac{M-m}{M+m}$$ Therefore $$\begin{array}{c}\left[\begin{array}{c}{v}_n\\ u_n\end{array}\right]=\left[\begin{array}{cc}i\sqrt{\frac{m}{M}}& -i\sqrt{\frac{m}{M}}\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{e}^{i\frac{n}{2}\varphi }& 0\\ 0& e^{-i\frac{n}{2}\varphi }\end{array}\right]\left[\begin{array}{cc}-i\sqrt{\frac{M}{m}}& 1\\ i\sqrt{\frac{M}{m}}& 1\end{array}\right]\left[\begin{array}{c}{v}_0\\ 0\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}{v}_n\\ u_n\end{array}\right]=\left[\begin{array}{c}{v}_{0}cos(\frac{n}{2}\varphi )\\ -\sqrt{\frac{M}{m}}{v}_{0}sin(\frac{n}{2}\varphi )\end{array}\right]\end{array}$$ We want to find $n$ where $v_n>u_n$, which means $$\begin{array}{c}{v}_{0}cos(\frac{n}{2}\varphi )>-\sqrt{\frac{M}{m}}{v}_{0}sin(\frac{n}{2}\varphi ),where{v}_0<0and\frac{\pi }{2}<\frac{n}{2}\varphi <\pi \\ \Rightarrow n>\frac{2(\pi +\arctan (-\sqrt{\frac{m}{M}}))}{\arctan (\frac{2\sqrt{Mm}}{M-m})}\end{array}$$ Let $M=Nm$, then $$n>\frac{2(\pi +\arctan (-\frac{1}{\sqrt{N}}))}{\arctan (\frac{2\sqrt{N}}{N-1})}$$ Set $x=\frac{1}{\sqrt{N}}$, then $$n>\frac{2(\pi +\arctan (-x))}{\arctan (\frac{2x}{1-x^2})}=\frac{\pi }{x}-1+\frac{\pi }{3}x=\pi \sqrt{N}-1+\frac{\pi }{3\sqrt{N}}$$ $$\Rightarrow \pi \sqrt{N}-1<n<\pi \sqrt{N}$$ $$\Rightarrow n=floor(\pi \sqrt{N})$$