NBPhO 2025

Part (i) — 2.0 / 2.0 pts

Scored against the canonical accelerating-frame grading column. Claude actually presents two derivations — the direct lab-frame component approach (the official’s “Solution 2”) and the accelerating-frame energy argument (the official’s canonical solution); either column gives full marks.

Criterion	Points	Result
Idea of switching to the coaccelerating frame	1.0	✓ “Pass to the drone’s (non-inertial) rest frame. There the drone sits at the origin forever, and every body is augmented by a pseudo-acceleration $-\vec{a}_d = -g\hat{x}$“
Ball gains horizontal acceleration $g$	0.3	✓ pseudo-acceleration $-g\hat{x}$ on the ball, combined with $-g\hat{y}$ from real gravity
Effective gravitational field $g\sqrt{2}$ pointing from drone to throwing point	0.2	✓ ”$\vec{g}_{\mathrm{eff}} = -g\hat{y} - g\hat{x}$, $\;\|\vec{g}_{\mathrm{eff}}\| = g\sqrt{2}$, pointing along $-(\hat{x}+\hat{y})/\sqrt{2}$ — that is, exactly along the line from the drone to the ball’s launch point $(-h,-h)$“
Energy conservation or the respective kinematical equation	0.3	✓ $w_{\min}^{2} = 2 g_{\mathrm{eff}} H = 2 \cdot g\sqrt{2}\cdot h\sqrt{2}$
Correct answer	0.2	✓ $w_{\min} = 2\sqrt{gh}$, matching the official

Part (ii) — 3.0 / 4.0 pts

Claude takes the official’s Solution 2 path (free-fall frame plus a velocity-space optimisation), then computes $\tau$, $v$, $u$ via lab-frame algebra rather than the geometric construction. The grading scheme explicitly caps Solution 2 at 3.5 / 4.0 — its two replacement items (0.3 + 0.5 = 0.8 pts) substitute for Solution 1’s three optimisation items (0.3 + 0.5 + 0.5 = 1.3 pts), a structural 0.5 pt discount. The remaining 0.5 pt loss is the “well drawn geometrical construction” item, which Claude bypasses entirely.

Criterion	Points	Result
The idea of switching to the free-falling frame	0.5	✓ “in a frame falling freely with acceleration $-g\hat{y}$, both stone and ball move at constant velocity”
Stating (explicitly or implicitly) that the stone and the ball travel in straight lines	0.5	✓ same passage, plus “the ball must travel from $\vec{B}$ to $\vec{S}$ at the relative velocity $\vec{u}-\vec{v}$ along the straight line $\vec{D}$”
$SC' = vt$ and $BC' = ut$	0.2	✓ implicit through $\vec{r}_b(t) - \vec{r}_s(t) = (\vec{B}-\vec{S}) + (\vec{u}-\vec{v})t$ — in the free-fall frame the ball travels $\vec{u}\tau$ from $B$ and the stone travels $\vec{v}\tau$ from $S$, meeting at the same point
(Sol 2 substitution) Noticing that $\vec{v} - \vec{u}$ is on $SB$	0.3	✓ $\vec{u} = \vec{v} + \vec{D}/\tau$ rearranges to $\vec{v} - \vec{u} = -\vec{D}/\tau$, parallel to $\vec{D}$ along $SB$
(Sol 2 substitution) A valid argument that $u$ is minimal when $\vec{u} \perp \overrightarrow{SB}$	0.5	✓ “the locus of admissible ball velocities is a straight line in velocity space … The minimum-magnitude point on a line is its foot of perpendicular from the origin, so $\vec{u}_{\min} \perp \hat{D}$“
In the free falling frame $C'$ is shifted upwards with respect to $S$, $B$, $C$	0.5	✓ algebraically equivalent: equation (1) $\vec{v}\tau = (\vec{C}-\vec{S}) + \tfrac{1}{2}g\hat{y}\tau^{2}$ rearranges to $\vec{S} + \vec{v}\tau = \vec{C} + \tfrac{1}{2}g\tau^{2}\hat{y}$, identifying the free-fall-frame collision point as $C' = C + \tfrac{1}{2}g\tau^{2}\hat{y}$
$\|CC'\| = \tfrac{1}{2}gt^{2}$	0.1	✓ same equation — the vertical shift between lab-frame $C$ and free-fall-frame $C'$ is $\tfrac{1}{2}g\tau^{2}$
Well drawn geometrical construction (that is correct)	0.5	✗ — Claude has only an ASCII layout of $S$, $B$, $C$ with coordinates; no geometric drawing of $C'$, the perpendicular through $B$, the vertical through $C$, or their intersection. The criterion explicitly rewards a correct drawn construction (“straight lines are straight lines and the 90° angles look like 90° angles”), and unlike problem 2’s “graphical method” item it does not include an “any other valid numerical method” allowance
$v = \sqrt{g\,SC'}/\sqrt{2\,CC'}$ and $u = \sqrt{g\,BC'}/\sqrt{2\,CC'}$	0.2	✓ algebraically equivalent: $v = \sqrt{v_x^{2}+v_y^{2}}$ from eq. (1) and $\vec{u} = \vec{v} + \vec{D}/\tau$ from $(\star)$ — both formulas evaluated component-wise after $\tau$ is fixed by eq. (3)
$v \in [11.8,\,12.7]\,\text{m s}^{-1}$, $u \in [10.5,\,11.4]\,\text{m s}^{-1}$	0.2	✓ $v \approx 12.2\,\text{m s}^{-1}$ and $u \approx 10.9\,\text{m s}^{-1}$, both inside the bands

Overall score: 5.0 / 6.0 pts

Full marks on (i); on (ii), 0.5 pt lost structurally for using Solution 2’s optimisation path (capped at 3.5 vs Solution 1’s 4.0) and 0.5 pt lost for the absent geometric construction.

Numerical answers match the official key on both parts: $w_{\min} = 2\sqrt{gh}$ in part (i) (exact symbolic match), $v \approx 12.2\,\text{m s}^{-1}$ vs official $\approx 12.3$, and $u \approx 10.9\,\text{m s}^{-1}$ matching the official’s $10.9$ exactly — both within the keyed tolerance bands.

Commentary

Where this solution goes beyond the grading scheme. A unifying overview frames both parts around a single move — eliminate gravity by changing reference frame — and quotes the timescale and geometry up front so each later step is a corollary rather than a fresh derivation. Part (i) is solved twice: first via the direct lab-frame component approach (the official’s Solution 2), where Claude derives $w_x = w_y$ as a symmetry of the launch point $(-h,-h)$ rather than an extra constraint imposed by minimisation — a distinction explicitly called out as a pitfall — and minimises via AM–GM ($\frac{h}{t} + \frac{gt}{2} \ge \sqrt{2gh}$) rather than the official’s discriminant or completing-the-square; and a second time via the accelerating-frame energy argument, with the boxed answer $w_{\min} = 2\sqrt{gh}$ verified to agree. An “educational remark” explains why the launch point $(-h,-h)$ produces such a clean answer (it is anti-parallel to $\vec{g}_{\mathrm{eff}}$, collapsing the 2D problem to a vertical throw) and gives the general formula $w^{2}_{\min}=g(L+\Delta y)$ for non-special launch points. Three consistency checks close the part — dimensions, the $h\to 0,\,h\to\infty$ limits, and a comparison to the fixed-target answer at zero drone-acceleration showing $g(L+\Delta y) > 4gh$ as it must be (the running drone makes the problem easier). Part (ii) develops the optimisation in velocity space (“foot of perpendicular from the origin to a line”), with the obtuse-angle condition $\ell C_x > h C_y$ for $\tau^{2} > 0$ given a clean geometric reading (“the boy must stand ‘in the way’ of the stone”). Three independent numerical checks verify the result: $\vec{u}\cdot\vec{D}\approx 0$ to two decimals; the launch angle $56.2^\circ$ matching $90^\circ$ minus the angle of $\vec{BS}$; and the shortcut $u_{\min} = v_\perp = \sqrt{v^{2}-v_\parallel^{2}}$ recovering the same $10.9\,\text{m s}^{-1}$. The “fixed-target trap” is called out explicitly with a numerical comparison ($u_{\rm fixed} \approx 10.7\,\text{m s}^{-1}$, numerically close but conceptually wrong). Most usefully, an explicit error budget on the figure reading — $\delta(\tau^{2})/\tau^{2} \sim 9\%$ from $\delta \sim 0.2\,\text{m}$ on each measured length — justifies rounding to two significant figures, where the official quotes three sig figs without comment.

Where the official solution is sharper. The decisive gap is the geometric construction itself. The official’s Solution 1 — overlay $C'$ on the free-fall-frame picture, drop the perpendicular from $B$ to $SB$, and find $C'$ at the intersection with the vertical through $C$ — is visually the most direct realisation of the result, and the official supplies a labelled figure (/figures/nbpho-2025/sol05-fig1.png). Claude’s lab-frame algebraic route (eq. (1) dotted with $\vec{D}$, combined with the perpendicularity condition (2), gives eq. (3) for $\tau^{2}$) is correct and arguably more general, but it never makes the construction visible, and the grading scheme rewards the construction explicitly. The official’s compact closed forms $v = \sqrt{g\,SC'}/\sqrt{2\,CC'}$ and $u = \sqrt{g\,BC'}/\sqrt{2\,CC'}$ also expose the geometric content of the answer — speed equals the relevant length divided by the geometric mean of $g$ and the vertical drop — in a way that Claude’s vector $v_x^{2}+v_y^{2}$ does not. On part (i), the official’s Solution 2 framing “the highest point of the trajectory has to be as low as possible” (once $v_{x,0} = v_{y,0}$ is established) is a sharp pedagogical observation that Claude’s AM–GM minimisation obscures, even though the two routes give the same $4gh$. Finally, the official offers three independent solution paths for part (ii) (geometric, velocity-space, calculus); Claude blends the first two but does not enumerate them as alternatives, so a reader sees one route rather than the problem’s full structure.