NBPhO 2026

Part (i) — 1.0 / 1.0 pts

The Claude solution leads with Method 1 (velocity projection) and then verifies via Method 2 (implicit differentiation), so it satisfies both grading checklists. Scoring against Method 1.

Criterion	Points	Result
Recognising that the velocity components of the rope endpoints along the rope direction must match (rope inextensibility)	0.4	✓ “The rope is inextensible. Pulling its free end at speed $v$ means the bar-to-weight segment shortens at rate $v$, so the rate at which the weight approaches the bar, measured along the rope, equals $v$“
Identifying the projection $u\sin\alpha$ for the weight’s velocity along the rope	0.3	✓ explicit dot-product $\vec u\cdot\hat e = u\sin\alpha = v$ with the unit vector $\hat e = (-\sin\alpha, +\cos\alpha)$ written out
Final answer $u = v/\sin\alpha$	0.3	✓ boxed $u = v/\sin\alpha$, cross-checked by implicit differentiation of $L^2 = x^2 + H^2$

Part (ii) — 1.0 / 1.0 pts

Criterion	Points	Result
Differentiating $L\dot{L} = x\dot{x}$ to obtain $\ddot{x} = (v^2 - u^2)/x$	0.5	✓ second derivative gives $\dot L^2 + L\ddot L = \dot x^2 + x\ddot x$, with the explicit observation that $\dot L = -v$ constant forces $\ddot L = 0$, yielding $\ddot x = (v^2 - u^2)/x$
Substituting $u$ and $x$ to obtain final answer $a = (v^2/H)\cot^3\alpha$	0.5	✓ substituting $u = v/\sin\alpha$ and $x = H\tan\alpha$ gives the boxed $\lvert a\rvert = (v^2/H)\cot^3\alpha$, with the minus sign on $\ddot x$ explicitly noted as “weight accelerates toward the bar”

Part (iii) — 3.0 / 3.0 pts

Criterion	Points	Result
Identifying that the only horizontal force on the weight is $T\sin\alpha$	0.3	✓ explicit force enumeration (gravity, floor normal, tension along $\hat e = (-\sin\alpha, +\cos\alpha)$) singling out the horizontal component as $-T\sin\alpha$
Writing Newton’s second law horizontally: $T\sin\alpha = M\lvert\ddot{x}\rvert$	0.3	✓ “horizontal: $-T\sin\alpha = M\ddot x$” with sign convention stated
Substituting the expression for $\lvert\ddot{x}\rvert$ from part ii)	0.2	✓ $-T\sin\alpha = -Mv^2\cos^3\alpha/(H\sin^3\alpha)$ written in one line
Isolating $T$ to obtain $T = Mv^2\cos^3\alpha/(H\sin^4\alpha)$	0.2	✓ identical form to the official
Correctly setting up the vertical force balance $T\cos\alpha + N = Mg$	0.5	✓ “vertical: $T\cos\alpha + N - Mg = 0$” with $\ddot y = 0$ floor constraint stated
Correct lift-off condition $N = 0$, i.e., $T\cos\alpha_0 = Mg$	0.5	✓ unilateral-contact reasoning given verbatim (“the floor can push up but not pull down… lift-off is the angle $\alpha_0$ at which $N$ first reaches zero”), then $N = 0$ imposed
Algebraic manipulation to $\tan^4\alpha_0 = v^2/(gH)$	0.5	✓ $Mg = Mv^2\cos^4\alpha_0/(H\sin^4\alpha_0) \Rightarrow \tan^4\alpha_0 = v^2/(gH)$ in one step
Final answer $\alpha_0 = \arctan\sqrt[4]{v^2/(gH)}$	0.5	✓ boxed $\alpha_0 = \arctan[(v^2/gH)^{1/4}]$, identical to the official

Overall score: 5.0 / 5.0 pts — full marks

All three boxed answers — $u = v/\sin\alpha$, $\lvert a\rvert = (v^2/H)\cot^3\alpha$, $\alpha_0 = \arctan\sqrt[4]{v^2/(gH)}$ — match the official key character-for-character.

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads three structural ideas — the bar acts as a frictionless redirection point, $\dot L = -v$ constant forces $\ddot L = 0$ (the most easily forgotten fact), and the natural dimensionless group is the Froude-like ratio $\mathrm{Fr} \equiv v^2/(gH)$ that organises the lift-off behaviour. Part (i) is delivered twice (velocity-projection in vector form and implicit differentiation as a consistency check) and is followed by a physics remark explaining the $u \to \infty$ singularity as $\alpha \to 0$ in terms of the weight’s velocity becoming orthogonal to the rope. Part (ii) gives both end-point limit checks ($\alpha \to \pi/2$ where $\lvert a\rvert \to 0$ because the weight’s velocity is fully aligned with the rope; $\alpha \to 0$ where $\cot^3\alpha$ diverges in lockstep with $u$) and a dimensional check, plus a falling-ladder analogy that frames the result as a generic feature of related-rates problems. Part (iii) supplies a genuinely cleaner re-derivation of the lift-off condition: dividing the two Newton equations at the lift-off instant gives $\tan\alpha_0 = \lvert a\rvert/g$ — a one-line characterisation that recovers $\tan^4\alpha_0 = v^2/(gH)$ on substitution and bypasses the $T$-isolation step entirely. This is followed by three regimes of $\mathrm{Fr}$ (slow pull, $\mathrm{Fr} = 1$ where $\alpha_0 = 45°$ and $v$ equals the free-fall speed accumulated over $H$, fast pull), an explanation of the sluggish $\mathrm{Fr}^{1/4}$ scaling decomposed into one factor of two from $T \propto 1/\sin^4$ and one from the vertical projection $\cos\alpha$, an explicit refutation of the wrong shortcut “lift-off is when $T = Mg$ with the rope vertical”, and a sanity check computing the lift-off speed $u_0^2 = v^2 + v\sqrt{gH}$ to confirm the two regimes interpolate as expected.

Where the official solution is sharper. On every dimension of the grading scheme the Claude write-up is at least as detailed and accurate as the official, and at the lift-off step it is arguably cleaner thanks to the $\tan\alpha_0 = \lvert a\rvert/g$ reformulation. The one feature the official offers that Claude does not is Method 2 of part (ii) — introducing $\omega = -\dot\alpha$ as the angular rate of the rope direction, expressing $u = H\omega/\cos^2\alpha$, and differentiating $u = v/\sin\alpha$ directly to recover the same $\lvert a\rvert = (v^2/H)\cot^3\alpha$. This is an alternative geometric route rather than a sharper one (and is equally weighted in the scheme to Method 1, which Claude takes), so it does not cost points; it is the only sense in which the official surfaces a perspective the Claude solution leaves implicit.