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2026 National Astronomy Competition 1 Instructions (Please Read Carefully)Table of Physical Constants Astrophysical Data Other data and conversion factors Trigonometrical identities Prefixes Условие Решение

USAAAO 2026, II кръг — Задача 8. Watch Out!

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2026 National Astronomy Competition · 55 т.

Внимание: Бележка към темата

Questions 3 and 9 refer to designated answer sheets, and question 6 to provided grid paper; these sheets were not obtained in the two native 2026-II PDFs or their two-row same-folder archive catalogue. The separately attributed physical-constants.pdf is an undated common USAAAO archive reference; its identity as the table supplied for this exam is unproven.

Instructions (Please Read Carefully)

2026 National Astronomy Competition

1 Instructions (Please Read Carefully)

The top 15 eligible scorers on the NAC will be invited to the training camp. A third round will be conducted during the camp to select the national team for the IOAA 2026. In order to qualify for camp, you must be a high school student with US citizenship or permanent residency.

The test must be completed within 2.5 hours (150 minutes) and the maximum number of points is 300 points. The point breakdown per problem is as follows.

Problem	P1	P2	P3	P4	P5	P6	P7	P8	P9	Total
Points	10	15	15	20	20	40	40	55	85	300

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. Some problems may require a designated answer sheet; these will be provided. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific calculator (non-programmable and non-graphing) for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on April 20, 2026. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Students must be proctored during the entire duration of the exam.

This exam sheet and your answer sheets (including scratch papers) should be returned to your proctors once the exam ended. Your proctors should upload your answer sheets to the Google Form provided to them.

We acknowledge the following people for their contributions to this year’s exam:

Srihari (Hari) Balaji, Feodor Yevtushenko, Ferdinand, Hagan Hensley, Joe McCarty, Lucas Pinheiro, Vincent Bian, Elizabeth Lee

After reading the instructions, please make sure to sign, affirming that:

All work on this exam has been completed by me.
I took this exam under the supervision of a proctor.
I did not receive any external help beyond the materials provided.
I will not discuss the contents of this exam with anyone until April 20, 2026.
Failure to follow these rules will result in disqualification from the exam.

Student Signature: ____________________ Date: ____________________

Undated common archive reference · Table of Physical Constants

Table of Physical Constants

Quantity	Symbol	Value
Electron rest mass	$m_e$	$9.109\times10^{-31}\,\mathrm{kg}$
Proton rest mass	$M_p$	$1.6726\times10^{-27}\,\mathrm{kg}$
Electronic charge	$e$	$1.6022\times10^{-19}\,\mathrm C$
Speed of light in free space	$c$	$2.9979\times10^8\,\mathrm{m\,s^{-1}}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}\,\mathrm{H\,m^{-1}}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\,\mathrm{F\,m^{-1}}$
Planck’s constant	$h$	$6.626\times10^{-34}\,\mathrm{J\,s}$
Reduced Planck’s constant	$\hbar=h/2\pi$	$1.0546\times10^{-34}\,\mathrm{J\,s}$
	$\hbar c$	$197.33\,\mathrm{MeV\,fm}$
Boltzmann’s constant	$k_B$	$1.3807\times10^{-23}\,\mathrm{J\,K^{-1}}$
Gas constant	$\mathcal R=k_B/m_H$	$8.250\times10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}$
Molar gas constant	$R$	$8.315\,\mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\,\mathrm{mol^{-1}}$
Standard molar volume		$22.414\times10^{-3}\,\mathrm{m^3\,mol^{-1}}$
Unified atomic mass unit ( $\vphantom{C}^{12}\mathrm C$ scale)	$u$	$931.5\,\mathrm{MeV}/c^2=1.660538\times10^{-27}\,\mathrm{kg}$
Mass of hydrogen atom	$m_H$	$1.0078u=1.6735\times10^{-27}\,\mathrm{kg}$
Bohr magneton	$\mu_B$	$9.274\times10^{-24}\,\mathrm{A\,m^2}$ or $\mathrm{J\,T^{-1}}$
Nuclear magneton	$\mu_N$	$5.051\times10^{-27}\,\mathrm{A\,m^2}$ or $\mathrm{J\,T^{-1}}$
Proton magnetic moment	$\mu_p$	$2.7928\mu_N$
Neutron magnetic moment	$\mu_n$	$-1.9130\mu_N$
Bohr radius	$a_0$	$5.292\times10^{-11}\,\mathrm m$
Fine structure constant	$\alpha=e^2/(4\pi\epsilon_0\hbar c)$	$(137.04)^{-1}$
Compton wavelength of electron	$\lambda_C=h/(m_ec)$	$2.4263\times10^{-12}\,\mathrm m$
Rydberg’s constant	$R_\infty$	$1.0974\times10^7\,\mathrm{m^{-1}}$
	$R_\infty hc$	$13.606\,\mathrm{eV}$
Stefan-Boltzmann constant	$\sigma$	$5.671\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$
Radiation density constant	$a=4\sigma/c$	$7.561\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$
Gravitational constant	$G$	$6.673\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$

Rest masses of some leptons and hadrons in $\mathrm{MeV}/c^2$ :

$e^\pm$ 0.5110, $\mu^\pm$ 105.66, $\tau^\pm$ 1777, $\pi^0$ 134.98, $\pi^\pm$ 139.57, $\mathrm K^\pm$ 493.7, $\mathrm K^0$ 497.7, $\eta$ 547, $\mathrm D^0$ 1865, $\mathrm D^\pm$ 1869, p 938.3, n 939.6, $\Lambda^0$ 1115.7, $\Sigma^+$ 1189, $\Sigma^0$ 1193, $\Sigma^-$ 1197, $\Xi^0$ 1315, $\Xi^-$ 1321, $\Omega^-$ 1672, $\mathrm Z^0$ $91.187\times10^3$ , $\mathrm W^\pm$ $80.41\times10^3$ .

Quark	Charge	$I_3$	$S$	$C$	$B$	$T$	Mass ( $\mathrm{GeV}/c^2$ )
u	$+\frac23$	$\frac12$	0	0	0	0	$\sim0.003$
d	$-\frac13$	$-\frac12$	0	0	0	0	$\sim0.006$
c	$+\frac23$	0	0	+1	0	0	$\sim1.25$
s	$-\frac13$	0	-1	0	0	0	$\sim0.11$
t	$+\frac23$	0	0	0	0	+1	174.3
b	$-\frac13$	0	0	0	-1	0	4.2

Undated common archive reference · Astrophysical Data and Conversion Factors

Astrophysical Data

Quantity	Symbol	Value
1 astronomical unit	AU	$1.496\times10^{11}\,\mathrm m$
1 parsec	pc	$3.086\times10^{16}\,\mathrm m$
Luminosity of Sun	$L_\odot$	$3.85\times10^{26}\,\mathrm W$
Mass of Sun	$M_\odot$	$1.989\times10^{30}\,\mathrm{kg}$
Radius of Sun	$R_\odot$	$6.96\times10^8\,\mathrm m$
Mass of Earth	$M_E$	$5.9742\times10^{24}\,\mathrm{kg}$
Radius of Earth	$R_E$	$6.3781\times10^6\,\mathrm m$

Other data and conversion factors

Quantity	Symbol	Value
1 ångstrom	Å	$10^{-10}\,\mathrm m$
1 fermi	fm	$10^{-15}\,\mathrm m$
1 barn	b	$10^{-28}\,\mathrm{m^2}$
1 pascal	Pa	$1\,\mathrm{N\,m^{-2}}$
1 standard atmosphere		$1.0132\times10^5\,\mathrm{Pa}$
Standard acceleration due to gravity	$g$	$9.807\,\mathrm{m\,s^{-2}}$
1 electron volt	eV	$1.6022\times10^{-19}\,\mathrm J$
	$\mathrm{eV}/hc$	$8.065\times10^5\,\mathrm{m^{-1}}$
	$\mathrm{eV}/k_B$	$1.1604\times10^4\,\mathrm K$
Wavelength of 1 eV photon		$1.2399\times10^{-6}\,\mathrm m$

Trigonometrical identities

\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)

\cos(\theta+\phi)=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)

\sin\alpha+\sin\beta=2\sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha+\cos\beta=2\cos\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha-\cos\beta=2\sin\frac12(\alpha+\beta)\sin\frac12(\beta-\alpha)

In a triangle ABC, $a/\sin A=b/\sin B=c/\sin C$

and $a^2=b^2+c^2-2bc\cos A$

Prefixes

T = tera = $10^{12}$	c = centi = $10^{-2}$
G = giga = $10^9$	m = milli = $10^{-3}$
M = mega = $10^6$	$\mu$ = micro = $10^{-6}$
k = kilo = $10^3$	n = nano = $10^{-9}$
	p = pico = $10^{-12}$
	f = femto = $10^{-15}$

Условие

Starting at time $t=0$ , two spaceships A and B begin traveling apart from each other at a relative speed of $v=\beta c$ . Let a given ship’s reference frame be the frame in which it is constantly at position $x=0$ . For a given event, if it occurs at position $x$ and time $t$ in A’s reference frame (call this frame $S$ ), let $x'$ and $t'$ be the position and time, respectively, of the event in B’s reference frame (call this frame $S'$ ). Assume that the spaceships start at the same position, at which point their clocks are initially synchronized.

(a) (5 points) Using only Newtonian mechanics, and neglecting all relativistic corrections, write $x'$ and $t'$ in terms of $x$ , $t$ , and $v$ .

(b) (2 points) Explain the primary physical contradiction that arises in the above system if $\beta$ is significantly larger than 0.

Let us now account for relativistic effects. When switching between reference frames moving at a speed $v$ relative to each other, the relativistically correct transformations are

x'=\gamma(x-vt)

t'=\gamma\left(t-\frac{v}{c^2}x\right)

where

\gamma=\frac{1}{\sqrt{1-v^2/c^2}}.

Here, $x$ , $t$ , $x'$ , and $t'$ are defined as before. You may also find the relativistic momentum ( $\vec p=\gamma m\vec v$ ) and total energy ( $E=\gamma mc^2$ ) formulas helpful.

The spaceships have previously agreed to communicate with each other through light signals. In particular, red light ( $\lambda=700\,\mathrm{nm}$ ) indicates that there is imminent danger ahead. Spaceship B, equipped with better radar system, detects a cluster of asteroids along the path that spaceship A is taking and sends a laser signal in red light with a wavelength of $\lambda$ that lasts a time $\Delta t_0$ . Both $\lambda$ and $\Delta t_0$ are as measured in B’s reference frame, and let the overall emitted power in B’s reference frame be $P$ . Use $\beta=0.4$ for all numeric answers.

You may refer to the following table of color versus approximate wavelength for parts (c)(v) and (c)(vi):

$\lambda$ (nm)	Color
$\leq390$	ultraviolet
$390-420$	purple
$420-480$	blue
$480-560$	green
$560-585$	yellow
$585-610$	orange
$610-720$	red
$\geq720$	infrared

(c) (27 points) By virtue of relativity, both the observed duration of the signal and the observed wavelength of the signal change. We will now examine this.

(i) (5 points) In terms of $\Delta t_0$ and $\beta=v/c$ , how long does B spend emitting the signal, in A’s reference frame? (Denote this as $\Delta t_1$ .)

(ii) (7 points) Due to the finite travel time of light, and given that B is moving away from A, A observes a different length $\Delta t_2$ for the signal. (This is analogous to the non-relativistic Doppler effect.) Calculate $\Delta t_2$ in terms of $\Delta t_0$ and $\beta$ .

(iii) (3 points) In terms of $\lambda$ and $\beta$ , what wavelength $\lambda'$ does A observe for the signal?

(iv) (8 points) Using your answers to (ii) and (iii), calculate the power $P'$ that A observes for the signal in terms of $P$ and $\beta$ . (Assume the entire beam emitted by B hits A.)

(v) (2 points) Evaluate $\lambda'$ numerically. What part of the electromagnetic spectrum does this correspond to?

(vi) (2 points) In order for A to observe red light at a wavelength $\lambda$ , numerically evaluate the wavelength $\lambda''$ that B would have to emit light at. What part of the electromagnetic spectrum does this correspond to?

An asteroid (of mass $5m$ ) is traveling directly toward spaceship A (of mass $m$ ) at a speed of $0.5c$ in spaceship A’s reference frame. From spaceship A’s perspective, the asteroid and spaceship B are diametrically opposite each other, and both are moving in the $+x$ direction. Let $F_0$ be the reference frame in which spaceship A is initially stationary and at the origin. (Again, use $\beta=0.4$ for all calculations, and from now on, express all answers in terms of $m$ , $c$ , and numeric prefactors. Onwards, $p$ and $E$ are not referring to four-vectors, rather, $p$ refers to the momentum along the x-axis, $E$ refers to the total energy (rest mass-energy + relativistic kinetic energy), and the “mass” refers to rest mass.)

(d) (5 points) In $F_0$ , what are the combined initial momentum $p_i$ and initial total energy $E_i$ of the asteroid and spaceship A?

Spaceship A and the asteroid undergo a completely inelastic collision, with no mass or energy escaping the system. According to relativity, both $p$ and $E$ are conserved.

(e) (11 points) Following the collision, what is the mass $M$ of the resulting object, and what is its velocity $v_0$ in frame $F_0$ ?

(f) (5 points) Switching to the reference frame of spaceship B, what are the mass $M'$ and velocity $v'_0$ of the resulting object? Is it moving toward or away from spaceship B?

The relativistic velocity addition formula is given as follows:

v'=\frac{v+u}{1+vu/c^2}

where

$v'=$ velocity of the object in the stationary frame,
$v=$ velocity of the object in the moving frame,
$u=$ velocity of the moving frame with respect to the stationary frame.

Решение

Solution:

(a) (5 points) According to classical Newtonian physics, time is absolute, so the clocks on $S$ and $S'$ stay synchronized (2 points):

\boxed{t=t'}.

Additionally, we have the following “standard” linear shift for the x-coordinates based on how far ship B has moved (3 points):

\boxed{x'=x-vt}.

Note that these can also be interpreted as the (fictitious) $c\to\infty$ limits of the Lorentz transformations, thus neglecting all complications that result from near-lightspeed motion.

(b) (2 points) The speed of light has to remain constant regardless of the frame of reference. For example, if spaceship A sends a light signal to spaceship B, in B’s frame of reference, according to the Newtonian relationship above, the speed of light would be measured as $c-v$ , where $v$ is the speed at which spaceship B is moving away from A.

(i) (5 points) We can solve for the inverse Lorentz transformations by inverting the given system of equations:

x=\gamma(x'+vt')

t=\gamma\left(t'+\frac v{c^2}x'\right).

In the frame $S'$ , the signal begins at $(x',t')=(0,t_0)$ and ends at $(x',t')=(0,t_0+\Delta t_0)$ . Plugging into the above, we see that the change in $t$ is

\Delta t_1=\gamma\left((t_0+\Delta t_0)+\frac v{c^2}(0)\right)-\gamma\left((t_0)+\frac v{c^2}(0)\right)=\gamma\Delta t_0,

which is just $\boxed{\frac{\Delta t_0}{\sqrt{1-\beta^2}}}$ . Note that this is just the “standard” time dilation formula.

(ii) (7 points) We claim that $\Delta t_2=(1+\beta)\Delta t_1$ (5 points), just as in the standard non-relativistic Doppler effect. To see this, work in A’s frame. Say that B begins emitting the signal at $(x,t)=(vt_1,t_1)$ and finishes emitting the signal at $(x,t)=(v(t_1+\Delta t_1),t_1+\Delta t_1)$ . Keeping in mind that the signal travels at speed $c$ , the difference in start and end times for receiving the signal is

\Delta t_2=[(t_1+\Delta t_1)+\frac1c(v(t_1+\Delta t_1))]-[t_1+\frac1c(vt_1)]=(1+\beta)\Delta t_1,

which simplifies to $\boxed{\Delta t_0\sqrt{\frac{1+\beta}{1-\beta}}}$ (2 points).

Note that this step has nothing to do with relativity and is purely a consequence of the finite travel time of light! Additionally, our combined answer should remind you of the relativistic Doppler shift formula; this is because the phases (peaks and troughs) of electromagnetic waves get stretched through exactly the same mechanism.

(iii) (3 points) According to the relativistic Doppler effect, the frequency of the light wave measured by spaceship A becomes

f=f'\sqrt{\frac{1-\beta}{1+\beta}}

Since $\lambda=c/f$ ,

\lambda'=\boxed{\lambda\sqrt{\frac{1+\beta}{1-\beta}}}.

(iv) (8 points) We combine our results from (ii) and (iii). Specifically, the power decrease is due to a combination of the signal being stretched out in time (resulting in a lower photon collision rate per second with ship A) as well as a redshift of the light itself (resulting in a lower energy per photon received). Since $E=hc/\lambda$ for a photon, we have,

\frac{P'}P=\frac{\Delta t_0}{\Delta t_2}\cdot\frac\lambda{\lambda'}=\left(\sqrt{\frac{1-\beta}{1+\beta}}\right)^2=\frac{1-\beta}{1+\beta}\to P'=\boxed{P\cdot\frac{1-\beta}{1+\beta}}.

(2.5 points for accounting for $\frac{P'}P=\frac{\Delta t_0}{\Delta t_2}$ . 2.5 points for accounting for $\frac{P'}P=\frac\lambda{\lambda'}$ . 1 point for combining to form $\frac{P'}P=\frac{\Delta t_0}{\Delta t_2}\cdot\frac\lambda{\lambda'}$ . 2 points for the correct final answer.)

(v) (2 points) We calculate (1 point)

\lambda'=(700\,\mathrm{nm})\sqrt{\frac{1+0.4}{1-0.4}}\approx\boxed{1070\,\mathrm{nm}}.

Spaceship A doesn’t see the red light, as this is in the (1 point) $\boxed{\text{infrared}}$ part of the spectrum.

(vi) (2 points) Again, according to the Doppler shift formula,

\lambda=\lambda''\sqrt{\frac{1+\beta}{1-\beta}},

giving (1 point)

\lambda''=(700\,\mathrm{nm})\sqrt{\frac{1-0.4}{1+0.4}}\approx\boxed{460\,\mathrm{nm}},

which is (1 point) $\boxed{\text{blue}}$ light.

(d) (5 points) In spaceship A’s frame of reference, $v_{\text{asteroid}}=0.5c$ , and the spaceship is at rest. Therefore, the total initial momentum is (2 points)

P_i=\gamma_{\text{asteroid}}\cdot5m\cdot(0.5c)=\boxed{\frac5{\sqrt3}mc}.

In spaceship A’s frame of reference, $E_{\text{spaceship A}}=mc^2$ and

E_{\text{asteroid}}=\gamma_{\text{asteroid}}(5m)(c^2)=\frac{10}{\sqrt3}mc^2.

Therefore, (3 points)

E_i=E_{\text{spaceship A}}+E_{\text{asteroid}}=\boxed{\left(1+\frac{10}{\sqrt3}\right)mc^2}.

(e) (11 points) We assume that in spaceship A’s frame of reference, spaceship B moves at $+0.4c$ relative to the spaceship A. (From now on, algebraic expressions for the coefficients start to get messy, so we begin to carry coefficients numerically.)

For the resulting mass $M$ , Lorentz factor $\gamma$ , and final velocity $v_0$ , according to conservation of momentum,

M\gamma v_0=\frac5{\sqrt3}mc,

and according to conservation of energy,

M\gamma c^2=mc^2\left(1+\frac{10}{\sqrt3}\right),

Dividing the momentum equation by the energy equation gives:

\frac PE=\frac{v_0}{c^2}=\frac{\frac5{\sqrt3}mc}{mc^2\left(1+\frac{10}{\sqrt3}\right)}.

(3 points for using $\frac PE=\frac{v_0}{c^2}$ . 4 points for correct $v_0$ .)

Simplifying,

v_0\approx\boxed{0.426c}.

Note that both algebraic and numerical answers for these coefficients are acceptable.

From here, we can calculate

M=\frac{E\sqrt{1-(v_0/c)^2}}{c^2}\approx\frac{mc^2\left(1+\frac{10}{\sqrt3}\right)\cdot\sqrt{1-0.426^2}}{c^2}\approx\boxed{6.13m}.

(3 points for using $M=\frac{E\sqrt{1-(v_0/c)^2}}{c^2}$ . 1 point for $M\approx6.13m$ .)

(f) (5 points) Trivially, $M'=M\approx\boxed{6.13m}$ (2 points) by basic relativistic principles, so we focus on computing $v'_0$ via the relativistic velocity addition formula. In the frame of spaceship B, the observed combined velocity is

v'_0=\frac{v_0+v_A}{1+\frac{v_0v_A}{c^2}}

where $v_A=-0.4c$ and $v_0=0.4262c$ (we carry more decimal places in the solution for clarity). Substituting the values gives (2 points)

v'_0\approx\boxed{0.0316c}.

The combined mass of the asteroid and spaceship A is slowly traveling $\boxed{\text{toward}}$ spaceship B in its frame of reference (1 point).

Оригинал в Архива: usaaao2026.pdf · официални решения: usaaao2026_sol.pdf · допълнителни данни: physical-constants.pdf