NBPhO 2025

Part (i) — 2.0 / 2.0 pts

Criterion	Points	Result
Explaining that oscillation is symmetric around centre of the rod (invoking Newton’s third law suffices)	0.5	✓ “By symmetry of the dumbbell about its midpoint, in the centre-of-mass frame the centre of the rod is stationary during the breathing oscillation”
Expressing stiffness of half-rod	0.5	✓ $k = YA/(l/2) = 2YA/l$, with the factor-of-two pitfall called out explicitly
Mass of the ball $m = \tfrac{4}{3}\pi r^{3}\rho$	0.2	✓ stated and evaluated as $\approx 3.27\times 10^{-2}\,\text{kg}$
Realising that the system can be treated as a spring	0.3	✓ a priori check $m_\mathrm{rod}/m \approx 0.02$ justifies “the rod is a massless elastic spring connecting two heavy balls”
Oscillation period $T = 2\pi\sqrt{m/k}$	0.3	✓ $T = 2\pi\sqrt{ml/(2YA)}$
Final answer	0.2	✓ $T \approx 6.4\times 10^{-4}\,\text{s} \approx 0.64\,\text{ms}$, matching the official

Part (ii) — 2.0 / 2.0 pts

Scored against the “Solution 1” (compression-wave) grading column, which is the route Claude takes.

Criterion	Points	Result
Realise compressed ball is essentially a compression wave	0.5	✓ “a compression wave is launched into the ball at the speed of sound… the wave traverses the ball, reflects from the far surface, and returns”
Formula for speed of sound	0.5	✓ $c = \sqrt{Y/\rho} \approx 5.06\times 10^{3}\,\text{m s}^{-1}$
Relation between time, radius and speed	0.5	✓ $\tau \sim 2r/c = 2r\sqrt{\rho/Y}$
Final answer	0.5	✓ $\tau \approx 4\,\mu\text{s}$, matching the official

Part (iii) — 2.0 / 2.0 pts

Criterion	Points	Result
2 hits	0.4	✓ Phase 1 first contact and Phase 3 second contact at $t \approx T/2$
Velocity of front ball flips almost instantaneously	0.4	✓ “near-vertical jump from $-v$ to $+v$ over time $\tau\ll T$” at both contacts
Centre of mass stays at rest	0.4	✓ $\vec v_{\mathrm{CoM}} = 0$ stated explicitly between bounces
Sinusoidal movement of front ball	0.4	✓ $v_{F,x}(t) = v\cos(\omega t)$ over $0 < t < T/2$
Constant velocity $-v$ of front ball after second hit	0.4	✓ — the grading scheme has a sign typo (should read $+v$); Claude’s ”$t > T/2$: flat at $+v$” matches the official’s own narrative which says “both balls move away from the wall with the same velocity $+v$“

Part (iv) — 2.0 / 2.0 pts

Criterion	Points	Result
Realise it behaves as in previous question (balls at velocity $-v$ and $v$, CM at rest), but it now also rotates and oscillates around centre of mass	0.5	✓ “front ball is at $-\tfrac{l}{2}\hat n_{0}$ moving at $(+v,0)$, and the rear at $+\tfrac{l}{2}\hat n_{0}$ moving at $(-v,0)$” + “CoM goes from velocity $-v\hat x$ to rest” + explicit “Decomposition: rotation + breathing”
Expression for the angular speed of rotation	0.3	✓ $\Omega = 2v\sin\alpha/l$ derived from $\vec v_F - \vec v_R = \vec\Omega\times(\vec r_F - \vec r_R)$
Expression of the amplitude of oscillations	0.2	✓ $u_F(t) = (v\cos\alpha/\omega)\sin(\omega t)$, amplitude $v\cos\alpha/\omega$
Realise the difference in interaction is whether the first ball bounces once or twice	0.2	✓ “two (or more) collisions” for $\alpha < \alpha_0$ vs “no second collision of the front ball occurs” for $\alpha > \alpha_0$
Formula for the distance of the front ball to the wall over time	0.3	✓ $x_F(t) \approx \tfrac{l}{2}\sin\alpha\cdot\Omega t + u_F(t)\cos\alpha$, derived by linearising $\cos(\alpha+\Omega t)$
Realise that if the distance is over 0 for all $t > 0$ the first ball does not hit the wall twice	0.2	✓ “the equation has a positive solution iff $\tan^{2}\alpha \le 0.217$… For $\tan^{2}\alpha > 0.217$ the curve never reaches the line and no second collision occurs”
Finding the critical angle given this condition	0.3	✓ $\alpha_0 = \arctan\sqrt{0.217} \approx 25^{\circ} \approx 0.436\,\text{rad}$, matching the official

Part (v) — 2.0 / 2.0 pts

Criterion	Points	Result
Realise that the dumbbell rotates around its centre of mass (after first collision)	0.2	✓ $\Omega = 2v\sin\alpha/l$ “(counterclockwise), unchanged” between contacts
Realise that the longitudinal oscillations have decayed by the time of the second collision	0.4	✓ “the rod’s breathing dies away to heat… Rod stress: zero (oscillations dissipated)“
Expression for velocity of ball $v\sin\alpha$	0.5	✓ $\vec v_R(t_2^-) = v\sin\alpha(-\sin\alpha, -\cos\alpha)$, magnitude $v\sin\alpha$
Expression for the component of velocity of ball in direction of surface normal $v\sin^2\alpha$	0.5	✓ $v_{R,x} = -v\sin^2\alpha$, reversed by the impulse to $+v\sin^2\alpha$
Realise the component of velocity of second ball in direction of surface normal is also $v\sin^2\alpha$	0.2	✓ rigid rotation gives $\vec v_F(t_2) = (v\sin^2\alpha, v\sin\alpha\cos\alpha)$, the front ball’s $v_x$ component matches by construction
Realise the speed of the centre of mass $v\sin^2\alpha$	0.2	✓ $\vec v_\mathrm{CoM}^\mathrm{after} = (v\sin^2\alpha, 0)$, magnitude $v\sin^2\alpha$

Overall score: 10.0 / 10.0 pts — full marks

The five requested numerical answers all match the official key: $T \approx 0.64\,\text{ms}$ in Part (i), $\tau \approx 4\,\mu\text{s}$ in Part (ii), the two-bounce trace with $T/2$ separation in Part (iii), $\alpha_0 = \arctan\sqrt{0.217} \approx 25^{\circ}$ in Part (iv), and $v_\mathrm{CoM} = v\sin^2\alpha$ in Part (v).

Commentary

Where this solution goes beyond the grading scheme. The opening “Overview” frames everything around the timescale hierarchy $\tau \ll l/c \ll T$ with all three numerics stated up-front, so each later impulse-approximation step is justified by name rather than rederived in place. Part (i) carries an a posteriori validation that $\omega l \approx 0.20\,c$ stays well below the rod’s lowest standing-wave frequency, closing the loop on the massless-rod approximation. Part (ii) explicitly anchors the $\omega\tau \approx 0.04\,\text{rad}$ smallness and $\tau < l/c$ no-transit-during-contact arguments that the rest of the solution leans on. Part (iii) appends an energy-and-momentum table verifying that the wall delivers $4mv$ split evenly across the two bounces with KE preserved at every step, plus an “educational remark — why two bounces, not one” that contrasts the elastic dumbbell against a rigid one and gives the $T \to 0$ limit. Part (iv) explicitly identifies two small parameters of the same order ($\Omega/\omega$ and $v\cos\alpha/(\omega l)$) before linearising, and includes a numerical refinement showing $\alpha_0 = 24.992^{\circ}$ from the exact root of $s\cos s = \sin s$. Part (v) carries an energy-budget breakdown ($\cos^2\alpha$ to heat in the first decay; explicit rotational, translational, and second-decay heat fractions written out individually); a derivation of the new spin rate $\Omega' = 2v\sin\alpha\cos^2\alpha/l$ for the post-second-bounce motion, with the master equation $\sin s/s = -\tan^2\alpha$ shown to recur and to have no solution in the regime $\alpha > \alpha_0$ — a non-trivial self-consistency check that the rear ball does not re-bounce; and a “symmetry-broken bounce” remark connecting the $\sin^2\alpha$ scaling to the slippage loss in a glancing tennis-racket collision. Multiple limits ($\alpha \to 0$, $\alpha \to \pi/2$, threshold $\alpha \to \alpha_0^+ \approx 0.179\,v$) are worked out in each part as sanity checks.

Where the official solution is sharper. Two places. (1) Part (i) closing form: the official packages the period as $T = 4\pi(r/d)\sqrt{2\rho rl/(3Y)}$, which exposes the ball-to-rod aspect ratio $r/d$ as the controlling dimensionless parameter — a reader sees immediately that the period scales linearly with the bluntness ratio of the dumbbell. Claude’s algebraically equivalent $T = 2\pi\sqrt{8r^3 l\rho/(3Yd^2)}$ has the same content but obscures the geometric meaning. (2) Part (v) economy: the official derivation is dramatically shorter — it computes only the magnitude $v\sin\alpha$ of the rear ball’s rigid-body velocity, projects onto the wall normal to get $-v\sin^2\alpha$, reverses, and asserts the symmetry that gives both balls the same $v_x$, all in one paragraph. Claude’s explicit vector decomposition with $\hat n(t_2) = (-\cos\alpha, \sin\alpha)$ and the $y$-component cancellation argument is correct and arguably more rigorous on the front-ball step, but it is several times longer for the same conclusion. Note also a sign typo in the official grading scheme — “Constant velocity $-v$ of front ball after second hit” should read $+v$, as the official’s own solution body confirms; Claude’s solution gets this right and the typo does not affect scoring.