Past exam of the physics 1 paper of the University of Oxford / 2021 / 3

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.