 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/8b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\overset{p}{˙}_{i} = n_{i} H_{1} (p_{1}, p_{2}) sin θ,

n_{2} p_{1} - n_{1} p_{2} is conserved .

Use

F_{2} = (n_{1} q_{1} + n_{2} q_{2}) P_{1} + q_{2} P_{2} .

Then

Q_{1} = θ, Q_{2} = q_{2}, p_{1} = n_{1} P_{1}, p_{2} = n_{2} P_{1} + P_{2},

and

P_{2} = p_{2} - (n_{2} / n_{1}) p_{1}

is the conserved combination. The transformed Hamiltonian depends on

Q_{1}

but not

Q_{2}

. Thus

H

and

P_{2}

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