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USAPhO 2017, Полуфинал — Задача 4. Question A4

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Всички материали от USAPhO 2017

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

Внимание: Бележка към темата

The source gives different exam-security dates on its supervisor and student pages. Printed key inconsistencies at A1, A4(b), B1(c) and B2(f)/(g) are retained with their native wording and mathematics; no silent corrected formula is substituted. Exact original PDFs and detailed source-defect evidence accompany this transcription.

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

This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-6), Part B (pages 8-10). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2017.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN - Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. - After you have completed Part A you may take a break. - Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. - Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. - Start each question on a new sheet of paper. Put your proctor’s AAPT ID, your AAPT ID, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, Doe, Jamie student AAPT ID # proctor AAPT ID # A1 - 1/3 - A hand-held calculator may be used. 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, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. - Questions with the same point value are not necessarily of the same difficulty. - In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 8, 2017.

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

$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$
$m_p$	$1.673\times10^{-27}\ \mathrm{kg}=938\ \mathrm{MeV}/c^2$
$\ln(1+x)$	$\approx x$ for $\vert x\vert \ll1$

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

Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-13), Part B (pages 15-23). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!

Условие

Relativistic particles obey the mass energy relation

E^2=(pc)^2+(mc^2)^2

where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.

A proton with mass $m_p$ and energy $E_p$ collides head on with a photon which is massless and has energy $E_b$ . The two combine and form a new particle with mass $m_\Delta$ called $\Delta$ , or “delta”. It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.

a. Determine $E_p$ in terms of $m_p$ , $m_\Delta$ , and $E_b$ . You may assume that $E_b$ is small.

b. In this case, the photon energy $E_b$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06\ \mathrm{mm}$ . Determine the energy of the photons, writing your answer in electron volts.

c. Assuming this value for $E_b$ , what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by $m_pc^2=938\ \mathrm{MeV}$ and the mass of the $\Delta$ is given by $m_\Delta c^2=1232\ \mathrm{MeV}$ .

The following relationships may be useful in solving this problem:

velocity parameter	$\beta=v/c$
Lorentz factor	$\gamma=1/\sqrt{1-\beta^2}$
relativistic momentum	$p=\gamma\beta mc$
relativistic energy	$E=\gamma mc^2$
relativistic doppler shift	$f/f_0=\sqrt{(1-\beta)/(1+\beta)}$

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Решение

a.

Solution 1. We can solve the problem exactly, approximating only in the last step. This certainly isn't necessary; we do this to illustrate a useful technique. We set $c=1$ throughout, and transform to an inertial frame where the proton is initially at rest. Before the collision,

E_p=m_p,\qquad E_\gamma=|p_\gamma|.

For the $\Delta$ particle, we have the usual relativistic relation

E_\Delta^2=p_\Delta^2+m_\Delta^2.

By energy-momentum conservation,

E_p+E_\gamma=E_\Delta,\qquad p_\gamma=p_\Delta.

Combining these results gives

(m_p+E_\gamma)^2=E_\gamma^2+m_\Delta^2\quad\Rightarrow\quad E_\gamma=\frac{m_\Delta^2-m_p^2}{2m_p}.

Now we need to transform back to the original frame, where the energy of the photon is $E_b$ . We can use the Lorentz transformation for this, but it's a little easier to realize that $E=hf$ for photons, and apply the Doppler shift. Then

\alpha\equiv\frac{E_b}{E_\gamma}=\sqrt{\frac{1-\beta}{1+\beta}}

where $\beta$ is the velocity parameter of the proton in the inertial frame where the photon has energy $E_b$ . Solving for $\beta$ in terms of the energy ratio $\alpha$ ,

\beta=\frac{1-\alpha^2}{1+\alpha^2}.

To calculate the proton energy, we need the Lorentz factor,

\gamma=\frac1{\sqrt{1-\beta^2}}=\frac{1+\alpha^2}{2\alpha}.

Then the proton energy in the original frame is

E_p=\gamma m_p=\frac{m_p}2\left(\alpha+\frac1\alpha\right)=\frac{m_p}2\left(\frac{2m_pE_b}{m_\Delta^2-m_p^2}+\frac{m_\Delta^2-m_p^2}{2m_pE_b}\right)

which is the exact answer.

At this point, we can approximate. The second term is much larger than the first, so

E_p\approx\frac{m_\Delta^2-m_p^2}{4E_b}

which was the required answer for this problem.

Solution 2. We now show another method that approximates throughout. We'll do everything in the lab frame, so the symbols in this solution don't mean the same things they did in solution 1. In the lab frame, energy-momentum conservation gives

p_p-p_b=p_\Delta,\qquad E_p+E_b=E_\Delta.

Squaring both expressions and dropping $E_b^2$ terms, since $E_b$ is small,

p_p^2-2p_pp_b\approx p_\Delta^2,\qquad E_p^2+2E_pE_b\approx E_\Delta^2.

Subtracting these equations gives

m_p^2+2E_pE_b+2p_pE_b=m_\Delta^2\quad\Rightarrow\quad E_p+p_p=\frac{m_\Delta^2-m_p^2}{2E_b}.

Since $E_b$ is small, this quantity must be large. But this means the protons are ultrarelativistic, so $E_p\approx p_p$ , giving

E_p\approx\frac{m_\Delta^2-m_p^2}{4E_b}

as desired.

b.

Plugging in the numbers gives

E=\frac{hc}{\lambda}=0.00112\ \mathrm{eV}.

c.

Restoring the factors of $c$ and plugging in the numbers gives

E_p=1.4\times10^{20}\ \mathrm{eV}.

This is known as the GZK bound for cosmic rays.

Exam Statistics

Question	A1	A2	A3	A4	B1	B2	Total
Mean	11	3	5	6	12	15	53
Standard Deviation	7	4	5	7	10	14	29
Maximum	25	21	25	25	42	50	163
Upper Quartile	18	4	9	8	18	26	73
Median	10	2	5	5	10	12	49
Lower Quartile	4	0	0	0	3	1	30
Minimum	0	0	0	0	0	0	5

Some Trivia:

California had 121 test takers
New Jersey had 42
Texas had 38
Florida, Illinois, Massachusetts, Maryland, New York, and Virginia each had between one dozen and two dozen test takers
Alabama, Alaska, Colorado, Idaho, Iowa, Louisiana, Montana, North Dakota, South Dakota, and Wyoming did not have a test taker this year.

Оригинал в Архива: 2017-USAPhO-exam-2.pdf · официални решения: USAPhO-2017-Solutions.pdf