Mass-energy relation Questions

In this case, $$ \begin{matrix} E_{before} = m_πc^2, & \bold{p}_{before} = 0, \\ \\ E_{after}= E_μ + E_v, & \bold{p}_{after} = \bold{p}_μ +\bold{p}_v \end{matrix} $$ Conservation of momentum requires that ,$$\bold{p}_v = -\bold{p}_μ $$ Conservation of energy says that, $$E_μ + E_v = m_{π}c^2 \\$$ Now, using \(E=pc\), \(E_v = |\bold{p}_v|c,\) whereas, $$\\$$ using \(E^2 - p^2c^2 = m^2c^4, |\bold{p}_μ| = \frac{\sqrt{{E_μ}^2 - {m_μ}^2c^4 }}{c} \\\) $$\\$$ so, $$E_μ + \sqrt{{E_μ}^2 - {m_μ}^2c^4 } = m_πc^2,$$ $$ \Rightarrow E_μ = \frac{({m_μ}^2+{m_π}^2)c^2}{2m_π} $$

Here pion is coming to rest and disintegrates into muon and neutrino with the given rest mass. We would apply:- <ol> <li>conservation of momentum</li> <li>conservation of energy</li> </ol> from the conservation of momentum, we get that magnitude of the momentum of pion and neutrino are the same [$p_π = p_μ + p_n$]. Hence let $p$ be the momentum of both. $\\$ Now, let's apply the conservation of energy:- $$\tag{1} E_π = E_μ + E_n$$ where $E_π$ is the energy of π-meson, $E_μ$ is the energy of muon and $E_n$ is the energy of the neutrino. $$\tag{2} \implies m_πc^2 = (T+ m_μc^2) +pc $$ we used:- $E^2 = p^2c^2 + (m_0c^2)^2$ and $E = T + m_0c^2$, where $T$ is the kinetic energy, also $$\tag{3} m_πc^2 = \sqrt{p^2c^2+ m_μ^2c^4} +pc $$ equation (1) could be written as (2) and (3), from equation (2) we get:- $$\tag{4} T = m_πc^2 - m_μc^2 - pc $$ now we are required to find $pc$ from equation (3) to get $T$ for muon. $\\$ using equation (3) and squaring it:- $$ \begin{matrix} (m_πc^2-pc)^2 = [\sqrt{p^2c^2 + m_μ^2c^4}]^2 \\ \implies m_π^2c^4 + p^2c^2 - 2m_πc^2pc = p^2c^2 + m_μ^2c^4 \\ \implies m_π^2c^4 - m_μ^2c^4 = 2m_πc^2pc \end{matrix}$$ $$ \tag{5} pc = \frac{m_π^2c^4 - m_μ^2c^4}{2m_πc^2} $$ now using equation (5) in equation (4) we get the desired answer. $$T = m_πc^2 - m_μc^2 - \frac{m_π^2c^4 - m_μ^2c^4}{2m_πc^2} $$ which gives us $$ T_μ = \frac{(m_π - m_μ)^2 c^2}{2m_π} $$