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USAPhO 2022, Полуфинал — Задача 2. Question A2

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Всички материали от USAPhO 2022

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

Both original question and worked-key packets have been read in full. They refer to a separate answer-sheet packet, including a blank plotting grid, which is not present in the retained question/key sources or the matching catalogue entry. All native question data and the actual key graph are retained. Printed variable/sign/wording defects and grading allowances remain source faithful and are documented.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination has two parts. Part A has three questions and lasts for 90 minutes. Part B also 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 4–7), and Part B questions (pages 8–10).
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, but they may be returned to the students after April 19th, 2022.
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 Bibilashvilli, Abijith Krishnan, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the 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, 2022. Violations of this rule may result in disqualification. Below are instructions for writing your solutions.
This year, all of your solutions will be written on 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.
You may have to fit a line to data, which you can plot on the answer sheet with grid lines. You may use a ruler, pencil, pen, or piece of paper as a straightedge.

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 $\lvert x\rvert\ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\lvert\theta\rvert\ll1$	$\cos\theta\approx1-\theta^2/2$ for $\lvert\theta\rvert\ll1$

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

Important Instructions for the Exam Supervisor — Key Packet Pagination

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 4–13), and Part B questions (pages 14–22).

Условие

Death Metal

A droplet of liquid metal has constant mass density and constant surface tension $\sigma$ , which causes it to form into a sphere of radius $R$ . (Throughout this problem, you may neglect gravity.) A thin wire is inserted into the droplet and connected to an electric current source which slowly charges the droplet. There is a critical value of the charge, $Q_0$ , that causes the droplet to split in half. Each half takes half the total charge, $Q_0/2$ , and half the mass of the original droplet. The “ejected half” is repelled far away from the other half, which remains in contact with the wire.

a. For simplicity, assume the droplet splits as soon as the final state (after the droplet has split in half and the two halves are well separated) has a lower total energy than that of the initial, single droplet. What is the value of $Q_0$ ? Give your answer in terms of a dimensionless constant $A$ multiplied by a product of powers of $\sigma$ , $R$ , and the vacuum permittivity $\epsilon_0$ , and give the numeric value of $A$ to three significant figures.

b. As more charge is added to the droplet by the current source, it continues to split in half repeatedly. What is the charge $q_n$ on the $n^{\mathrm{th}}$ ejected droplet? Give your answer in terms of $Q_0$ .

c. In the limit where all of the initial mass of the droplet has been ejected, what is the total work done by the current source? Give your answer in terms of a dimensionless constant $B$ multiplied by a product of powers of $\sigma$ , $R$ , and $\epsilon_0$ , and give the numeric value of $B$ to three significant figures.

Решение

a.

Note that $\sigma$ has units of $\mathrm{energy/length^2}$ , $R$ has units of length, and $\epsilon_0$ has units of $\mathrm{charge^2/(energy\times length)}$ . There is only one combination with dimensions of charge, so we must have

Q_0=A\sqrt{\epsilon_0\sigma R^3}.

To find the value of $A$ , note that a droplet of radius $R$ and charge $Q_0$ has a surface tension energy $4\pi R^2\sigma$ and an electrostatic energy $Q_0^2/2C$ , where $C=4\pi\epsilon_0R$ is the capacitance of a conducting sphere relative to infinity. Thus, the total energy of the drop is

E_0=4\pi\sigma R^2+\frac{Q_0^2}{8\pi\epsilon_0R}.

When the droplet splits, the new halves both have charge $Q_0/2$ and a radius $R_1=R/2^{1/3}$ , since the total volume stays the same. The total energy of each of the halves is thus

E_1=4\pi\sigma R_1^2+\frac{(Q_0/2)^2}{8\pi\epsilon_0R_1}.

Setting $E_0=2E_1$ and solving for $Q_0$ gives

A=8\pi\sqrt{\frac{2^{1/3}-1}{2-2^{1/3}}}\approx14.9.

This model of the instability of a charged liquid drop is not exactly accurate, and was chosen to keep the problem simple. In reality, it isn’t enough for the final state to have lower energy; the repulsive electrostatic force needs to exceed the surface tension force, or else the bubble can’t get to this state of lower energy. Accounting for this requires a different calculation, first performed by Lord Rayleigh in 1882, which gives the “Rayleigh limit” of $A=8\pi\approx25.1$ . Students who successfully derived this result also received full credit for part (a).

b.

Consider the situation just before the $n^{\mathrm{th}}$ split. At this point, the initial droplet has split in half $n-1$ times, so it has a volume that is $2^{n-1}$ times smaller than the original volume, and thus a radius $R_{n-1}=R/2^{(n-1)/3}$ . By part (a), the charge required to split this drop is

Q_n=A\sqrt{\epsilon_0\sigma R_{n-1}^3}=\frac{Q_0}{2^{(n-1)/2}}.

Once the split happens, the $n^{\mathrm{th}}$ ejected droplet takes half of this charge, so

q_n=\frac{Q_n}2=\frac{Q_0}{2^{(n+1)/2}}.

Also note that the radius of this droplet is $R_n=R/2^{n/3}$ .

c.

By dimensional analysis, the total work $W$ done by the voltage source must be

W=B\sigma R^2.

The work $W$ is equal to the difference between the energy $E_f$ of the final state, where there are many small droplets that have been dispersed to far away, and the energy $E_i=4\pi\sigma R^2$ of the initial, uncharged droplet. The final energy includes the surface tension and electrostatic energy of all the small droplets, which have radius $R_n$ and charge $q_n$ found in the previous problem. This means that the final energy is

\begin{aligned}E_f&=\sum_{n=1}^{\infty}\left(4\pi\sigma R_n^2+\frac{q_n^2}{8\pi\epsilon_0R_n}\right)\\&=4\pi\sigma R^2\sum_{n=1}^{\infty}\frac1{2^{2n/3}}+\frac{Q_0^2}{16\pi R_0\epsilon_0}\sum_{n=1}^{\infty}\frac1{2^{2n/3}}\\&=4\pi\sigma R^2\left(1+\frac{A^2}{64\pi^2}\right)\sum_{n=1}^{\infty}\frac1{2^{2n/3}}\\&=4\pi\sigma R^2\left(1+\frac{A^2}{64\pi^2}\right)\frac1{2^{2/3}-1}=4\pi\sigma R^2\times2.30.\end{aligned}

Here we have inserted the results found in part (b), then used the result of part (a) and summed the geometric series. Thus,

W=E_f-E_i\approx(4\pi\sigma R^2)\times1.30=16.3\,\sigma R^2.

Оригинал в Архива: 2022-USAPhO-Exam.pdf · официални решения: 2022-USAPhO-solutions.pdf