# In the two-body scattering. How to calculate the momentum of final state particles?

For the process two-body scattering $1+2 \to 3 + 4$ , how to calculate the momentum of the final state particles in the COM frame: \begin{align} |\vec p_3| =|\vec p_4| &= \frac{1}{2\sqrt{s}} \sqrt{[s-(m_3+m_4)^2][s-(m_3-m_4)^2]} \\ \end{align} $\sqrt{s}$ is the energy of COM frame, $\sqrt{s} = E_1+E_2$.

As I thought, I begin with the energy conservation and momentum conservation: $$E_1+E_2=E_3+E_4$$ $$|\vec p_3|=|\vec p_4|$$ we have: $$E^2=\vec p^2+m^2$$ then: \begin{align} \Rightarrow \sqrt{s} &= \sqrt{\vec p^2_3+m^2_3} + \sqrt{\vec p^2_4+m^2_4} \\ s &= \vec p^2_3+m^2_3 + \vec p^2_4+m^2_4 + 2\sqrt{(\vec p^2_3+m^2_3)(\vec p^2_4+m^2_4)} \end{align}

After this, I don't know what to do next step. Maybe I'm wrong from the beginning? Thanks a lot!

$$\sqrt{s}=\sqrt{\mathbf{p}_3^2+m_3^2}+\sqrt{\mathbf{p}_4^2+m_4^2}$$
As you are working in COM frame you have: $$\mathbf{p}_3^2=\mathbf{p}_4^2=x(\text{say})$$
\begin{align} \left (\sqrt{s}-\sqrt{x+m_3^2}\right)^2&=\left(\sqrt{x+m_4^2}\right)^2\\ \Rightarrow s+x+m_3^2-2\sqrt{s\left(x+m_3^2\right)}&=x+m_4^2 \\ \Rightarrow 4s\left(x+m_3^2\right)&=\left(s+\left(m_3^2-m_4^2\right)\right)^2\\ \Rightarrow \sqrt{x}&=\sqrt{\frac{\left(s+\left(m_3^2-m_4^2\right)\right)^2}{4s}-m_3^2}\\&=\frac{1}{2\sqrt{s}} \sqrt{[s-(m_3+m_4)^2][s-(m_3-m_4)^2]} \end{align}