= Solution
{c}
Multiply the <equation of motion> by $\dot x$ and integrate. The conserved <mechanical energy> is
$$
\frac12m\dot x^2+V(x)=E,\qquad V(x)=\frac k2(\ell-x)^2-\frac Cx.
$$
The <initial conditions> $x=\ell$, $\dot x=0$ give $E=-C/\ell$, so
$$
\boxed{\frac12m\dot x^2=-\frac k2(\ell-x)^2+C\left(\frac1x-\frac1\ell\right).}
$$
Initially $\ddot x=-C/(m\ell^2)<0$, so the <particle> starts moving towards the origin. For $0<x<\ell$, factor its <kinetic energy> as
$$
E-V(x)=\frac{\ell-x}{2\ell x}\left[2C-k\ell x(\ell-x)\right].
$$
The <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 is
$$
\boxed{\ell_2=\left(\frac{8C}{k}\right)^{1/3}=\left(\frac{2q^2}{\pi\epsilon_0k}\right)^{1/3}.}
$$
For $\ell_1<\ell<\ell_2$ 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 is
$$
\boxed{x_T=\frac{\ell+\sqrt{\ell^2-8C/(k\ell)}}2.}
$$
The <particle> \b[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 $t\to\infty$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv-q4-collision-threshold.png]
{title=The energy barrier distinguishes collision, asymptotic approach, and bounded oscillation}
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