 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/h/1/ii/solution

The third particle is free, so

\frac{d}{d t} (m_{3} q_{3} \times \overset{q}{˙}_{3}) = q_{3} \times m_{3} \overset{q}{¨}_{3} = 0.

Thus particle three's angular momentum is conserved.

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