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 $P=\sqrt{2m{H}_0}$ and $p=p_{z0}$ givesThe allowed interval obeys $|p+a{x}^2|\le P$. The sign of $\dot{x}$ 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.