Disintegrating $\mu$

Let us utilise the momentum four-vector equation for the collision. $\vec{p_\pi}=\left(m_\pi c,0,0,0\right)$ , $\vec{p_\mu}=\left(mc,\gamma_\mu m_\mu v,0,0\right)$ , and $\vec{p_\nu}=\left(m_\nu c,\gamma_\nu m_\nu,0,0\right)$ . Now the conservation law gives $\vec{p_\pi}-\vec{p_\mu}=\vec{p_\nu}$ Squaring both sides, ${p_\pi}^2+{p_\mu}^2-2\vec{p_\pi}\cdot\vec{p_\mu}={p_\nu}^2$ Solving this by plugging the initial values I had stated gives $m_\pi^2+m_\mu^2-2\gamma_\mu m_\pi m_\mu =m_\nu^2$ But the rest mass of the neutrino is very small compared to others, and can be neglected. Yielding: $\gamma_\mu=\frac{{m_\pi}^2+{m_\mu}^2}{2m_\pi m_\mu}$ The distance covered by the muon in the frame of the earth is $L=v\tau \gamma_\mu$ , with $v=c\sqrt{1-1/\gamma^2_\mu}$ , thus giving $L=c\tau\sqrt{\gamma^2_\mu-1}=\frac{1}{2}\cdot\frac{m_\pi^2+m_\mu^2}{m_\pi m_\mu}$ Therefore $AB=2$