Past exam of the physics 1 paper of the University of Oxford / 2017 / 10 / c

Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is $\dot{y}=y^{\prime }(x)\dot{x}$, so its kinetic energy, gravitational potential energy and Lagrangian areConsequently the canonical momentum conjugate to the horizontal coordinate isIt is not simply the horizontal component $m\dot{x}$ of the mechanical momentum, because a change in $x$ also moves the cart vertically. Differentiate carefully in the Euler-Lagrange equation:Subtracting yieldsThe term involving $y^{\prime \prime }$ accounts for the changing track geometry. An independent check is conservation of energy:

If $y^{\prime \prime }=0$ over the subsequent segment, its slope is a constant $s$ and the equation of motion reduces toThere is uniform horizontal acceleration, not generally zero acceleration. Starting that segment at $(t_0,x_0,{\dot{x}}_0)$,For a downhill straight track with $s=-\tan \theta$, this gives $\ddot{x}=g\sin \theta \cos \theta$, agreeing with b. Only $s=0$ gives constant horizontal velocity.

The condition $y^{\prime \prime }\to 0$ must describe an actually straight limiting segment, with bounded velocity, for that subsequent constant-acceleration conclusion. If $y^{\prime \prime }$ vanishes only at an isolated inflection point, the boxed expression describes the instantaneous acceleration there; it does not make the rest of the track straight. Likewise, a vanishing $y^{\prime \prime }$ alone does not make the product $y^{\prime \prime }{\dot{x}}^2$ negligible in a limit where $\dot{x}$ diverges.