 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s7/1/e/solution

Since

p_{y} = 0

initially and remains constant, the motion stays in a plane of constant

y

. Setting

P = 2 m H_{0}

and

p = p_{z 0}

gives

\overset{z}{˙} = \frac{p + a x ^{2}}{m}, \overset{x}{˙} = \pm \frac{P ^{2} - ( p + a x ^{2} ) ^{2}}{m} .

The allowed interval obeys

∣ p + a x^{2} ∣ \leq P

. The sign of

\overset{x}{˙}

is set by the initial direction and changes at a turning point. The specified

x_{0}

fixes the starting position on the energy orbit, but the data do not specify which sign of the initial transverse velocity is chosen.

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