Solution to 2008 Problem 40

Recall that
\begin{align}E^2 = p^2 c^2 + m^2 c^4 \label{eqn:1}\end{align}

The two decay products must be back to back in order to conserve three momentum (the initial three momentum is 0). Let (E/c,0,0,p) be the four-vector of the massless particle. Equation (1) with m replaced by 0 implies that E/c = p. So, the four-momentum of the massless particle is (p,0,0,p). By three-momentum conservation, the three-momentum of the massive particle must be (0,0,-p). From equation (1), the four-momentum must then be (\sqrt{p^2 + m^2c^2},0,0,-p). The initial energy is M c^2, so by energy conservation,
\begin{align*}Mc = \sqrt{p^2 + m^2c^2} + p\end{align*}
Squaring both sides and solving for p yields
\begin{align*}p = \boxed{\frac{M^2 - m^2}{2M}}\end{align*}
Therefore, answer (B) is correct.

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