 Solution

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

Since

p_{A} = - p_{s}

, the same angle gives

m_{A t}^{2} = m_{A}^{2} + 2 q (E_{A} + p cos θ)

. Therefore

(m_{A t}^{2})_{m i n} = \frac{m _{A}^{2} m _{C}^{2}}{m _{B}^{2}}, (m_{A t}^{2})_{m a x} = m_{A}^{2} + m_{C}^{2} - m_{B}^{2} .

The minimum now occurs when

s

and

t

are antiparallel. The eventwise identity

m_{A t}^{2} + m_{s t}^{2} = m_{C}^{2} + m_{A}^{2} - m_{B}^{2}

independently checks the opposite ordering of the two endpoints.

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