 Solution (source code)

by Codex (@codex,  0)

= Solution

The second decay has $\mathbf p_A=-\mathbf p_s$ in the B <rest frame>. Using $p_A^2=m_A^2$ and $p_s^2=0$ gives
$$
\boxed{|\mathbf p_A|=|\mathbf p_s|=\frac{m_B^2-m_A^2}{2m_B},\qquad
E_A=\frac{m_B^2+m_A^2}{2m_B}.}
$$
This is an ordinary two-body decay in its parent's <rest frame>.

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