= Solution
For the scattering calculation, take the heavy target as fixed and define $\kappa=2Ze^2/(4\pi\epsilon_0)>0$. The incoming <kinetic energy> and <angular momentum> are $E=mv^2/2$ and $J=mvb$. In the original sketch, the horizontal dashed incoming asymptote has perpendicular distance $b$ from the nucleus, while $d$ is the <radius> from the nucleus to the point where $\dot r=0$.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-1-scattering.png]
{title=Original repulsive <Rutherford scattering> trajectory with <impact parameter> and closest approach}
At closest approach, the <velocity> is tangential and $v_t=J/(md)=vb/d$. <conservation of energy> gives
$$
\frac12mv^2=\frac{J^2}{2md^2}+\frac\kappa d,
\qquad mv^2d^2-2\kappa d-mv^2b^2=0.
$$
Choose the positive root and put $s=\kappa/(mv^2)$:
$$
\boxed{d_{\rm rep}=s+\sqrt{s^2+b^2}.}
$$
Head-on incidence gives $d_{\rm rep}=2s$, and repulsion always gives $d_{\rm rep}>b$. With an attractive <Coulomb force>, change the sign of the potential term:
$$
\boxed{d_{\rm att}=\sqrt{s^2+b^2}-s.}
$$
For $b>0$, the <attractive and repulsive inverse-square closest approaches> obey $d_{\rm att}d_{\rm rep}=b^2$. For $b=0$ attraction causes a collision with the point centre, so $d=0$ is a singular endpoint, not a regular scattering turning point. Finite nuclear size is excluded by the stated point-force model.
For the repulsive <inverse-cube central force> $\mathbf F=\kappa_3\widehat{\mathbf r}/r^3$, the <torque> vanishes, $\mathbf r\times\mathbf F=0$, and <angular momentum> is still conserved. Its <potential energy> is $+\kappa_3/(2r^2)$, since $-dV/dr=+\kappa_3/r^3$. Thus
$$
\frac12mv^2=\frac{m^2v^2b^2}{2md^2}+\frac{\kappa_3}{2d^2},\qquad
\boxed{d=\sqrt{b^2+\frac{\kappa_3}{mv^2}}.}
$$
The new coefficient is denoted $\kappa_3$ because its units are $\mathrm{N\,m^3}$, unlike the $\mathrm{N\,m^2}$ units of the inverse-square coefficient. Conserving <angular momentum> depends on centrality, not on a particular radial power law.
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