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USAPhO 2026, Полуфинал — B2. Say Cheese… It's a Conic

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Всички материали от USAPhO 2026

2026 USA Physics Olympiad Exam

Внимание: Бележка към темата

The supplied archive folder contains the full question packet and the official interleaved key. The cover titled “Instructions for the Student” contains only a reference sheet and acknowledgements; no administrative instructions, timing, point allocations or blank answer sheets are supplied. These values remain unspecified. Printed source inconsistencies are preserved: A3's key changes $C_P,C_V$ to $c_p,c_v$ , prints an empty equation label (0-5), has a plus sign in radial force balance (0-12) followed by a positive pressure gradient (0-13), and gives the exterior pressure without any radial dependence. B2 refers to “part (2)” despite lettered subparts, and its key uses the angle between the asymptote rays as the deflection angle. B3 refers to “part (1)” despite lettered subparts. The original key is transcribed without mathematical corrections.

AAPT UNITED STATES PHYSICS TEAM — AIP 2026; USA Physics Olympiad Exam — Instructions for the Student — REFERENCE SHEET

Fundamental Constants

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$

Useful Approximations

(1+x)^n\approx1+nx+n(n-1)x^2/2\quad\text{for }|nx|\ll1

e^x\approx1+x+x^2/2+x^3/6\quad\text{for }|x|\ll1

\sin\theta\approx\theta-\theta^3/6\quad\text{for }|\theta|\ll1

\cos\theta\approx1-\theta^2/2\quad\text{for }|\theta|\ll1

You may use this sheet for both parts of the exam.

We acknowledge the US Physics Team coaches and other people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Eddie Chen, Kellan Colburn, Evan Erickson, Harry Kroft, Natalie LeBaron, Rishab Parthasarathy.

Условие

You are studying the motion of charged particles constrained to the $xy$ -plane. Particle $\alpha$ , with charge $+1\ \mathrm{C}$ , is fixed at the origin.

For motion under an inverse-square central force, trajectories are conic sections with the fixed particle at a focus. In polar coordinates:

For an attractive interaction (ellipse),

r = \frac{r_0}{1+e\cos\phi}, \qquad e < 1.

For a repulsive interaction (hyperbola),

r = \frac{r_0}{e\cos\phi - 1}, \qquad e > 1.

You have a camera that takes three snapshots of a moving particle at equal time intervals.

(a)

A particle $\beta$ , with charge $-1\ \mathrm{C}$ , moves under the electrostatic force of particle $\alpha$ . In three consecutive snapshots, its positions are

(0,-5\ \mathrm{m}), \quad (3\ \mathrm{m}, 0), \quad (0, 5\ \mathrm{m}).

Assuming the motion is governed only by the Coulomb interaction with $\alpha$ , determine the maximum distance that particle $\beta$ reaches from the origin.

(b)

Now particle $\beta$ is replaced by particle $\gamma$ , which has charge $+2\ \mathrm{C}$ and is free to move in the plane. In three consecutive snapshots, its positions are

(3\ \mathrm{m}, -4\ \mathrm{m}), \quad (2\ \mathrm{m}, 0), \quad (3\ \mathrm{m}, 4\ \mathrm{m}).

Assuming the motion is governed only by the Coulomb interaction with $\alpha$ , determine the angle $\theta$ (measured from the positive $x$ -axis) of the velocity of particle $\gamma$ at large times.

(c)

A family of particles, each identical to particle $\gamma$ (that is, each has charge $+2\ \mathrm{C}$ ), approaches from infinity in the $xy$ -plane. All particles have the same speed $v_\infty$ far from the origin and move along lines parallel to the initial asymptotic direction of particle $\gamma$ from part (2). The particles are injected one at a time, so they do not interact with one another.

For each trajectory, define the impact parameter $B$ to be the perpendicular distance between the initial straight-line path of the particle and the origin.

As a particle passes near the scattering center, its direction changes due to Coulomb repulsion. Let $\alpha$ denote the total deflection angle of the trajectory, i.e., the angle between the incoming and outgoing asymptotic directions.

(i) Using the conic form of the trajectory, derive a formula for $B(\alpha)$ .

(ii) Using the result of part (2), express $B(\alpha)$ in terms of $r_0$ .

Решение

(a) Because the first and third snapshots are at $(0,-5\ \mathrm{m})$ and $(0,5\ \mathrm{m})$ , while the second snapshot is at $(3\ \mathrm{m},0)$ , and the time intervals are equal, the motion is symmetric about the $x$ -axis. Therefore the point $(3,0)$ must be the point of closest approach, i.e. the pericenter of the orbit.

At pericenter,

r_{\min} = \frac{r_0}{1+e} = 3.

At the two other photographed points, the particle is on the $y$ -axis, so $\phi=\pm\pi/2$ , $r=5$ .

Since $\cos(\pm\pi/2)=0$ , we get

5 = \frac{r_0}{1+e\cdot 0} = r_0.

So $r_0=5$ .

Now use the pericenter condition:

3 = \frac{5}{1+e} \implies 1+e=\frac{5}{3} \implies e=\frac{2}{3}.

Since $e<1$ , the orbit is an ellipse. The maximum distance occurs at apocenter, where $\phi=\pi$ , so

r_{\max} = \frac{r_0}{1-e}.

Thus

r_{\max} = \frac{5}{1-\frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15\ \mathrm{m}.

$\boxed{15\ \mathrm{m}}$

(b) Because the first and third snapshots are at $(3,-4\ \mathrm{m})$ and $(3,4\ \mathrm{m})$ , while the middle snapshot is at $(2,0)$ , and the time intervals are equal, the motion is symmetric about the $x$ -axis. Thus the point $(2,0)$ is the point of closest approach.

At closest approach,

r_{\min} = \frac{r_0}{e-1} = 2.

Hence

r_0 = 2(e-1).

Now use one of the other photographed points, for example $(3,4)$ . Its distance from the origin is

r=\sqrt{3^2+4^2}=5,

and

\cos\phi = \frac{3}{5}.

Therefore

5 = \frac{r_0}{e\cdot\frac{3}{5}-1}.

Substitute $r_0=2(e-1)$ :

5 = \frac{2(e-1)}{\frac{3e}{5}-1}.

Multiply through:

5\left(\frac{3e}{5}-1\right)=2(e-1),

3e-5=2e-2,

e=3.

Then

r_0 = 2(e-1) = 4.

For a hyperbola, the outgoing asymptote is reached when the denominator goes to zero:

e\cos\phi-1=0.

Thus

\cos\phi = \frac{1}{e} = \frac{1}{3}.

So the direction of motion at large times is

\theta = \arccos\left(\frac{1}{3}\right).

\boxed{\theta = \arccos\left(\frac{1}{3}\right) \approx 70.5^{\circ}.}

(c) For a repulsive inverse-square force, the trajectory is a hyperbola

r = \frac{r_0}{e\cos\phi-1}, \qquad e>1.

The asymptotes occur when the denominator vanishes:

e\cos\phi-1=0 \implies \cos\phi=\frac{1}{e}.

Thus the angle between the asymptotes is

\alpha = 2\phi = 2\arccos\left(\frac{1}{e}\right).

Hence

e = \frac{1}{\cos(\alpha/2)}.

Now relate the impact parameter $B$ to the conic parameters. For a hyperbolic trajectory,

B = \frac{r_0}{\sqrt{e^2-1}}.

Substitute $e=1/\cos(\alpha/2)$ :

e^2-1 = \frac{1}{\cos^2(\alpha/2)}-1 = \tan^2(\alpha/2).

Thus

\sqrt{e^2-1} = \tan(\alpha/2),

and therefore

B = \frac{r_0}{\tan(\alpha/2)}.

\boxed{B(\alpha) = r_0\cot\left(\frac{\alpha}{2}\right)}

Оригинал в Архива: 2026-USAPhO.pdf · официални решения: 2026-USAPhO-Solutions.pdf