PhysicsOlympiad Eval
Benchmarking SOTA LLMs on elite Physics Cup 2026 and EuPhO 2026 problems. Tests first-principles reasoning, complex modeling, and spatial logic on non-pattern-matching scenarios. Text-only format.
Prompt
A small brick is squeezed between two parallel planes in zero-gravity. The planes are perpendicular to the axis $z$. The lower plane always stays at rest, whilst the upper one is moving with constant velocity $\vec{v}$ along the axis $x$. Initially the velocity of the brick $\vec{u}_0$ is perpendicular to $\vec{v}$ and lies along the axis $y$. The friction between the planes and the brick is dry, with the same coefficient of kinetic friction at both surfaces. Given the initial acceleration of the brick $a_0$, find the displacement $l_y$ of the brick along the axis $y$ after a long time. Obtain an expression for $l_y$ for arbitrary speeds $v$ and $u_0$, and evaluate the answer for $v = u_0 = 1\ \mathrm{m/s}$ and $a_0 = 1\ \mathrm{m/s^2}$ with four digits.
Answer guidance
[+6] Final numerical answer 0.6631 m (accept 0.663â0.6632) [+3] Closed-form l_y in terms of v, u_0, a_0 (any algebraically equivalent form) [+3] Writes acceleration as -a(r1/|r1| + r2/|r2|) with r1 = u, r2 = u - v [+3] Shows D = |r1| - |r2| is a conserved quantity [+2] Derives ds/dt = -4a(s²-v²)/(s²-D²) with s = |r1|+|r2| [+2] Uses that r1 - r2 = v has no y-component, so both slips share the same u_y [+2] Gets u_y as the triangle altitude: u_y = sqrt((s²-v²)(v²-D²))/(2v) [+2] Converts the time integral to an s-integral via dt/ds [+2] Eliminates a using a_0 = 2a¡cos(Ď_0/2) with cos Ď_0 = u_0/R [+1] Identifies s â v as t â â (slip vanishes) [+1] Reports four digits as asked [-4] Numeric answer with no derivation [-3] Treats the two friction forces as parallel / collapses them into one [-2] Claims the motion stops in finite time with the brick at rest in the lab frame
OpenAI
OpenAI
Google