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USAPhO 2023, Полуфинал — B2. Fast and Furious

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2023 USA Physics Olympiad Exam · 25 т.

Внимание: Бележка към темата

The archive contains the complete question packet and the official interleaved solution packet. The separate official answer sheets referred to in the instructions are absent from this archive folder. Original printed errors are retained: A1(c)'s solution uses $z'_x$ although the question defines $y(x)$ ; A2(b.iii)'s solution writes radius $r+\delta r$ although the question defines $r=r_0+\delta r$ ; A3's key refers to “part (b)” in its part a.iii check, and B1's key calls the rod a “disc”. B2's printed energy expression uses $2.0\times10^7\ \mathrm{J}$ although the question gives energy density in J/kg. The “this paper” references in B2 and B3 have no recoverable URI in the supplied PDF.

AAPT UNITED STATES PHYSICS TEAM — AIP 2023; USA Physics Olympiad Exam

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Important Instructions for the Exam Supervisor

This examination has two parts. Each part has three questions and lasts for 90 minutes.
For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–8), and Part B questions (pages 9–11).
Begin by giving students the instructions and all of the answer sheets. Let the students read the instructions and fill out their information on the answer sheets. They can keep the instructions for both parts of the exam. Also give students blank sheets of paper to use as scratch paper throughout the exam.
Students may bring calculators, but they may not use symbolic math, programming, or graphing features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones or other electronics may not be used during the exam or while the exam papers are present. Students may not use books or other references.
To start the exam, give students the Part A questions, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the Part A questions and answer sheets.
Then give students a 5 to 10 minute break. Then give them the Part B questions, and allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, collect everything, including the questions, the instructions, the answer sheets, and the scratch paper. Students may not keep the exam questions. Everything can be returned to the students after April 19th, 2023.
After the exam, sort each student’s answer sheets by page number. Scan every answer sheet, including blank ones.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Kellan Colburn, Samuel Gebretsadkan, Abi Krishnan, Natalie LeBaron, Kye Shi, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the reference table on the next page, answer sheets, and blank paper for scratch work. Read this page carefully before the exam begins.
You may use a calculator, but its memory must be cleared of data and programs, and you may not use symbolic math, programming, or graphing features. Calculators may not be shared. Cell phones or other electronics may not be used during the exam or while the exam papers are present. You may not use books or other outside references.
When the exam begins, your proctor will give you the questions for Part A. You will have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. If you finish all of the questions, you may check your work, but you may not look at Part B during this time.
After 90 minutes, your proctor will collect the questions and answer sheets for Part A. You may then take a short break.
Then you will work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. When the exam ends, you must return all papers to the proctor, including the exam questions.
Do not discuss the questions of this exam, or their solutions, until after April 19th, 2023. Violations of this rule may result in disqualification.

Below are instructions for writing your solutions.

All of your solutions must be written on the official answer sheets. Nothing outside these answer sheets will be graded. Before the exam begins, write your name, student AAPT number, and proctor AAPT number as directed on the answer sheets.
There are several answer sheets per problem. If you run out of space for a problem, you may use the extra answer sheets, which are at the end of the answer sheet packet. To ensure this work is graded, you must indicate, at the bottom of your last answer sheet for that problem, that you are using these extra answer sheets.
Only write within the frame of each answer sheet. To simplify grading, we recommend drawing a box around your final answer for each subpart. You should organize your work linearly and briefly explain your reasoning, which will help you earn partial credit. You may use either pencil or pen, but sure to write clearly so your work will be legible after scanning.

Reference table of possibly useful information

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$	$(1+x)^n\approx1+nx$ for $\vert x\vert \ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\vert \theta\vert \ll1$	$\cos\theta\approx1-\theta^2/2$ for $\vert \theta\vert \ll1$

Possibly useful integrals

\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)+C

\int\frac{dx}{1-x^2}=\tanh^{-1}(x)+C

\int\frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(x)+C

\int\frac{dx}{1+x^2}=\tan^{-1}(x)+C

\int\frac{dx}{(1+x^2)^{3/2}}=\frac{x}{\sqrt{1+x^2}}+C

\int\sqrt{\frac{1+x}{1-x}}\,dx=-\sqrt{1-x^2}-2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{\sqrt{1+x}}{(1-x)^{3/2}}\,dx=2\sqrt{\frac{1+x}{1-x}}+2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{1/2}}=\sqrt{\frac{1+x}{1-x}}+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{1/2}}=\frac{(2-x)\sqrt{1+x}}{3(1-x)^{3/2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{3/2}}=\frac{1+2x-2x^2}{3(1-x)^{3/2}(1+x)^{1/2}}+C

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

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Условие

A space program wants to accelerate a spaceship of final mass $m = 100\ \mathrm{kg}$ to relativistic speeds to observe distant stars. They have two proposals to evaluate.

a.

Their first proposal is to use traditional rocket propulsion. A rocket of initial mass $m_0$ and final mass $m$ that expels propellant with exhaust speed $u$ relative to the rocket will reach a speed

v = u \ln\left(\frac{m_0}{m}\right).

Suppose the desired final speed is $v_f = 3c/5$ . In the subparts below, neglect relativistic effects and give your answers in the form $10^n$ , where $n$ has at least two significant figures.

i. If the rocket has exhaust speed $u = 3.5\ \mathrm{km/s}$ , what must its starting mass be in kilograms?

ii. If the propellant is exhausted at rate $7.0\ \mathrm{kg/s}$ , how long does the acceleration take, in centuries?

iii. If the energy density of the fuel is $2.0 \times 10^7\ \mathrm{J/kg}$ , how much total energy is required, in Joules?

For the rest of this problem, you should account for special relativity.

b.

Another option is to use a spaceship with constant mass $m$ , propelled by light produced by lasers on Earth, with total power $P = 6 \times 10^{12}\ \mathrm{W}$ . The light evenly impacts a sail on the spaceship, and reflects off the sail directly back towards the Earth. Neglect the orbital motion of the Earth, and give all your answers in the frame of the Earth.

i. What is the force on the spaceship when the spaceship has speed $v$ ?

ii. How long will it take to accelerate the spaceship to speed $v_f = 3c/5$ , in seconds? You may use the integrals on the reference sheet.

iii. At the moment the spaceship reaches this speed, how much total energy has been used to power the lasers, in Joules?

The following results from relativity may be helpful:

The Lorentz factor is defined as $\gamma = 1/\sqrt{1 - v^2/c^2}$ .
An object of mass $m$ and velocity $\mathbf{v}$ has momentum $\mathbf{p} = \gamma m \mathbf{v}$ and energy $E = \gamma m c^2$ . The force is defined by $\mathbf{F} = d\mathbf{p}/dt$ .
The momentum and energy of light are related by $E = pc$ .
In a frame $S'$ with velocity $v\hat{\mathbf{x}}$ relative to a frame $S$ , the energy and momentum are

E' = \gamma(E - v p_x), \quad p_x' = \gamma(p_x - vE/c^2).

Решение

Question B2

Fast and Furious

a. i. Solution. Plugging in the numbers gives

m_0 = m e^{v_f/u} = 100\ \mathrm{kg} \cdot e^{0.6\times 3\times 10^8 / 3.5\times 10^3} \approx 10^{22337}\ \mathrm{kg}

Equivalently, the exponent is $n = 2.2\times 10^4$ .

ii. Solution. The result is

t = \frac{m_0 - m}{7.0\ \mathrm{kg/s}} \approx 10^{22336}\ \mathrm{s} \approx 10^{22326}\ \text{centuries}.

Equivalently, the exponent is $n = 2.2\times 10^4$ .

iii. Solution. The energy required is

E = (m_0 - m)(2.0\times 10^7\ \mathrm{J}) = 10^{22344}\ J.

Equivalently, the exponent is $n = 2.2\times 10^4$ . When working with such absurdly large numbers, exponents essentially always stay unchanged.

For the rest of this problem, you should account for special relativity.

b. i. Solution. Let $\beta = v/c$ , where $v$ is the ship's speed, and consider a piece of the beam with total momentum $dp_x$ , in the Earth's frame. In the ship's frame, this momentum is $dp'_x = \gamma(1-\beta)dp_x$ , and after the collision the momentum simply flips sign, $dp'_{xf} = -dp'_x$ . Thus, transforming back to the Earth's frame, the final momentum is $dp_{xf} = -\gamma^2(1-\beta)^2 dp_x$ . The change of the spaceship's momentum, still in the Earth's frame, is the difference

dP_x = -(dp_{xf} - dp_x) = \frac{2}{1+\beta}\, dp_x.

To find the force on the spaceship, we need to find the rate at which the beam impacts the spaceship. Accounting for the spaceship's motion, it is $dp_x = \frac{P}{c}(1-\beta)dt$ , so

F = \frac{dP_x}{dt} = \frac{2P}{c}\frac{1-\beta}{1+\beta}.

Alternative solution: In the Earth's frame, if the photons in the incident beam have frequency $f_i$ , then they are reflected with frequency

f_f = \sqrt{\frac{1-\beta}{1+\beta}}\sqrt{\frac{1-\beta}{1+\beta}}f_i = \frac{1-\beta}{1+\beta}\, f_i

where we applied the relativistic Doppler shift formula twice, since the photons are first absorbed by the moving rocket and then reemitted by it. Applying conservation of momentum, and using the fact that the momentum of a photon is related to its energy by $E = hf = pc$ , we have

F = \frac{dN}{dt}\frac{h}{c}(f_i+f_f) = \frac{dN}{dt}\frac{h f_i}{c}\frac{2}{1+\beta}

where $dN/dt$ is the rate at which photons collide with the sail. It is related to the rate at which photons are emitted from the source on Earth, $dN_{\rm em}/dt$ , by

\frac{dN}{dt} = (1-\beta)\frac{dN_{\rm em}}{dt}.

Finally, since the power of the laser is $P = (dN_{\rm em}/dt)(hf_i)$ , we have

F = \frac{2P}{c}\frac{1-\beta}{1+\beta}.

ii. Solution. In the Earth's frame, the relativistic momentum of the spaceship obeys

dP_x = mc\, d\left(\frac{\beta}{\sqrt{1-\beta^2}}\right) = mc\frac{d\beta}{(1-\beta^2)^{1.5}}

Combining this with our expression for the force gives

\frac{2P}{mc^2}dt = \frac{d\beta}{(1-\beta)^2\sqrt{1-\beta^2}}.

Integrating both sides and using an integral on the reference sheet gives

\frac{2P}{mc^2}t = \int_0^{0.6}\frac{d\beta}{(1-\beta)^2\sqrt{1-\beta^2}} = \frac{5}{3}.

Therefore, the time is

t = \frac{5mc^2}{6P} = \frac{4.5\times 10^{19}}{3.6\times 10^{13}} = 1.3\times 10^6\ \mathrm{s}

iii. Solution. The energy is just $5mc^2/6$ from the result in the previous problem, so we get $7.5\times 10^{18}\ \mathrm{J}$ . For reference, the US consumes roughly $10^{16}\ \mathrm{J}$ of electricity per day. For more discussion of this propulsion method, see this paper.

Оригинал в Архива: 2023-USAPhO-Exam.pdf · официални решения: 2023-USAPhO-solutions.pdf