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 / f

Dividing the two velocities givesHere $\dot{z}=a{x}^2/m\ge 0$: the orbit always advances toward increasing $z$, except for instantaneous zeros at each midplane crossing $x=0$. The transverse motion oscillates between $-X$ and $X$. At these outer turning points $\dot{x}=0$ and the trajectory is tangent to the $z$ direction; at the midplane $\dot{x}\ne 0$ while $\dot{z}=0$, so the tangent is along $x$. The sign of the initial $\dot{x}$ only chooses the phase or reflected trace.

With $p=-P$, the allowed range is $|x|\le X_2$, where $X_2=\sqrt{2P/a}$. But an orbit starting with $x\ne 0$ never crosses the midplane in finite time, becauseOne positive-side separatrix, with $u=\omega (t-t_c)$ and $\omega =\sqrt{2Pa}/m$, isThe negative-side orbit is its reflection in $x=0$. For the positive branch, $x$ rises from zero asymptotically, reaches $X_2$ at $u=0$, then returns asymptotically to zero. Thus the infimum is zero but it is not attained in finite time on this nontrivial branch. Since $\dot{z}=(P/m)(2{\mathrm{sech}}^{2}u-1)$, the two axial-velocity zeros occur at $x=\sqrt{P/a}$. Motion is toward decreasing $z$ near either asymptotic midplane tail and toward increasing $z$ near the outer excursion. These directions are marked in part i.

The exactly midplane initial state $x=0$, $\dot{x}=0$ is a separate straight trajectory with $\dot{z}=-P/m$; it must not be described as a finite-time crossing of the separatrix.