Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 4 / b

Multiply the equation of motion by $\dot{x}$ and integrate. The conserved mechanical energy isThe initial conditions $x=\ell$, $\dot{x}=0$ give $E=-C/\ell$, soInitially $\ddot{x}=-C/(m{\ell }^2)<0$, so the particle starts moving towards the origin. For $0<x<\ell$, factor its kinetic energy asThe quadratic function $x(\ell -x)$ has maximum ${\ell }^2/4$ at $x=\ell /2$. Consequently kinetic energy is strictly positive throughout $0<x<\ell$ exactly when $k{\ell }^3<8C$. Thus the collision threshold isFor ${\ell }_1<\ell <{\ell }_2$ there is no intervening turning point, and the particle collides with the fixed particle. The collision takes finite time: near zero, $|\dot{x}|\sim \sqrt{2C/(mx)}$, so the remaining time is bounded by a constant times ${\int }_0^{\delta }\sqrt{x}dx$.

For $\ell >{\ell }_2$, the first turning point encountered isThe particle oscillates periodically between $x_T$ and $\ell$, around the larger stable equilibrium, without collision. The zeros of $E-V$ at the two endpoints of this allowed region are simple, so travel to each turning point takes finite time. At the boundary $\ell ={\ell }_2$, $x=\ell /2$ is a double zero of $E-V$ and an unstable equilibrium; the inward trajectory approaches it only as $t\to \infty$.