NBPhO 2025

Difficulty hard

Prerequisites

• Hooke's law and the simple harmonic oscillator
• Young's modulus, stress and strain in a thin rod
• Speed of sound in a solid, $c = \sqrt{Y/\rho}$
• Impulse and 1D elastic collisions
• Angular momentum and moment of inertia of point masses
• 2D rigid-body rotation, $\vec{\omega} \times \vec{r}$ for a rotating point

Learning objectives

• Justify the rod-as-massless-spring approximation by comparing $m_\text{rod}/m_\text{ball}$
• Exploit the hierarchy $\tau \ll l/c \ll T$ to decouple contact, wave transit, and breathing
• Estimate contact times from dimensional analysis ($\tau \sim 2r/c$) and from a lumped elastic model
• Decompose post-impact motion into a breathing oscillation along the rod plus rigid rotation about the CoM
• Linearise in the small parameter $\Omega/\omega$ to reduce the two-bounce condition to $\tan^2\alpha = -\frac{\sin s}{s}$
• Use the impulse approximation when one body's response is much faster than the system's natural period

Watch out for

• Half-rod vs full-rod stiffness confusion. Each ball oscillates about the stationary rod midpoint, so the effective spring is a half of the rod ($k = 2YA/l$), not the full rod. Using the full-rod stiffness gives the wrong period by a factor of $\sqrt{2}$.
• The wall's impulse at the first bounce is purely normal to the wall — the tangential component of the ball's velocity is preserved. Reversing the total velocity rather than just its $x$-component gives a wrong angular momentum and therefore a wrong $\Omega$ in parts (iv) and (v).
• A rigid dumbbell would bounce once; the second bounce in part (iii) is what proves the rod is elastic. Missing it (and stopping at "CoM leaves at $+v$") loses the whole point of the problem.
• The answer $v_\text{CoM} = v\sin^2\alpha$ in part (v) is valid only for $\alpha > \alpha_0$. Applying it for $\alpha < \alpha_0$ is wrong because in that regime ball 1 — not ball 2 — delivers the second impulse to the wall, giving a different departure speed.
• Compute the moment of inertia from two point masses ($I = 2m(l/2)^2 = ml^2/2$), not from a distributed-mass rod formula. The balls dominate because $m \gg m_\text{rod}$.