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Условие Отговори Решение

USAPhO 2026, Първи кръг — Задача 17

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Всички материали от USAPhO 2026

2026 $F = ma$ Exam · 1 т.

Указания

25 QUESTIONS - 75 MINUTES

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Use $g = 10\ \mathrm{N/kg}$ throughout, unless otherwise specified.
You may write in this question booklet and the scratch paper provided by the proctor.
This test has 25 multiple choice questions. Select the best response to each question, and use a No.2 pencil to completely fill the box corresponding to your choice. If you change an answer, completely erase the previous mark. Only use the boxes numbered 1 through 25 on the answer sheet.
All questions are equally weighted, but are not necessarily equally difficult.
You will receive one point for each correct answer, and zero points for each incorrect or blank answer. There is no additional penalty for incorrect answers.
You may use a hand-held calculator. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any external references, such as books or formula sheets.
The question booklet, answer sheet and scratch paper will be collected at the end of this exam.
To maintain exam security, do not communicate any information about the questions or their solutions until after February 13, 2026.

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

We acknowledge The US Physics Team coaches and other contributors to this year’s exams:

Tengiz Bibilashvili, Eddie Chen, Orhun Ciftcioglu, Kellan Colburn, Evan Erickson, Harry Kroft, Natalie LeBaron, Mai Qi, Elena Yudovina

Условие

A ladder is leaning against a vertical wall. The ladder is a uniform rod of mass $M$ and length $L$ , and both the wall and the ground are frictionless. The ladder is released from rest from an almost-vertical position and begins to slide. What is the speed of the point of the ladder that is in contact with the floor when it is a horizontal distance $\frac{\sqrt{3}L}{2}$ away from the wall?

(A) $0.46\sqrt{gL}$

(B) $0.51\sqrt{gL}$

(D) $0.61\sqrt{gL}$

(E) $0.66\sqrt{gL}$

Отговори

Решение

Key physical subtlety: the ladder loses contact with the wall before the base reaches $x = \frac{\sqrt{3}}{2}L$ . Thus there are two distinct phases: (i) motion while touching both wall and floor, and (ii) motion after detachment. The multiple-choice options $0.61\sqrt{gL}$ and $0.66\sqrt{gL}$ correspond to the two most natural “no-calculus” estimates at the transition.

Phase I (still touching wall): detachment speed. Let $\theta$ be the angle the ladder makes with the horizontal. While both contacts are maintained, the constraints give

x_B = L\cos\theta, \qquad y_{\mathrm{CM}} = \frac{L}{2}\sin\theta.

Starting from rest near vertical, the drop in the center of mass from $\theta = 90^{\circ}$ to angle $\theta$ is

\Delta U = Mg\left(\frac{L}{2} - \frac{L}{2}\sin\theta\right) = \frac{1}{2}MgL(1 - \sin\theta).

Using the instantaneous center (intersection of normals at the contacts), the kinetic energy can be written

T = \frac{1}{2}I_P\Omega^2, \qquad I_P = I_{\mathrm{CM}} + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2,

and the base speed is $v_B = \Omega(L\sin\theta)$ . Energy conservation gives

\frac{1}{2}\left(\frac{1}{3}ML^2\right)\Omega^2 = \frac{1}{2}MgL(1 - \sin\theta) \quad\implies\quad \Omega^2 = \frac{3g}{L}(1 - \sin\theta),

v_B^2 = \Omega^2L^2\sin^2\theta = 3gL(1 - \sin\theta)\sin^2\theta.

The ladder detaches when the normal force at the wall drops to zero; the standard result for this setup is

\sin\theta_{\mathrm{det}} = \frac{2}{3}.

Substituting into the expression above gives

v_{B,\mathrm{det}}^2 = 3gL\left(1 - \frac{2}{3}\right)\left(\frac{2}{3}\right)^2 = \frac{4}{9}gL \quad\Rightarrow\quad v_{B,\mathrm{det}} = \frac{2}{3}\sqrt{gL} \approx 0.67\sqrt{gL}.

Thus the choice $0.66\sqrt{gL}$ matches the exact detachment speed.

What about the asked position $x = \frac{\sqrt{3}}{2}L$ ? If one (incorrectly) assumes the ladder remains in contact with the wall all the way to $x = \frac{\sqrt{3}}{2}L$ , then $\cos\theta = \frac{\sqrt{3}}{2}$ , i.e. $\theta = 30^{\circ}$ , and the same formula yields

v_B^2 = 3gL(1 - \tfrac{1}{2})(\tfrac{1}{2})^2 = \frac{3}{8}gL \quad\Rightarrow\quad v_B = \sqrt{\frac{3gL}{8}} \approx 0.61\sqrt{gL},

which is exactly the $0.61\sqrt{gL}$ choice.

Why we accept both 0.61 and 0.66. The correct motion does detach earlier, and determining the speed at $x = \frac{\sqrt{3}}{2}L$ requires analyzing the post-detachment evolution (translation + rotation) over time. That level of modeling is beyond what can be resolved reliably by a short algebraic argument in the F=ma format. However:

$0.66\sqrt{gL}$ is an exact milestone speed (at detachment),
$0.61\sqrt{gL}$ is the natural value obtained under the common “stays in contact” simplification,
distinguishing which is closer to the true value at $x = \frac{\sqrt{3}}{2}L$ requires post-detachment dynamics not intended for this exam.

Therefore, we award full credit for either choice $0.61\sqrt{gL}$ or $0.66\sqrt{gL}$ .

Оригинал в Архива: 2026_FMA_exam.pdf · официални решения: 2026_FMA_solutions.pdf