 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2014/paper-1/3/a/ii/solution

The second decay has

p_{A} = - p_{s}

in the B rest frame. Using

p_{A}^{2} = m_{A}^{2}

and

p_{s}^{2} = 0

gives

∣ p_{A} ∣ = ∣ p_{s} ∣ = \frac{m _{B}^{2} - m _{A}^{2}}{2 m _{B}}, E_{A} = \frac{m _{B}^{2} + m _{A}^{2}}{2 m _{B}} .

This is an ordinary two-body decay in its parent's rest frame.

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