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USAPhO 2015, Полуфинал — Задача 5. Question B1

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Всички материали от USAPhO 2015

USA Physics Olympiad Exam — Part A / Part B · 50 т.

Внимание: Бележка към темата

Original question and matching key disagree at A4: the question uses Tmax and X=C; repeated key statement uses400K and says X need not coincide with C, but actual worked key graph/argument still place C/X/Y together. Both native versions are explicitly retained. B1 question says three independent parts although only a/b exist; repeated key says two. Original mathematical/sign/symbol defects remain printed, with evidence notes.

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 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 9-11), and several answer sheets for one of the questions in Part A (pages 13-13). 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, 2015.
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 AAPT ID number, 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,

AAPT ID # Doe, Jamie 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 15, 2015.

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

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ 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 $\vert \theta\vert \ll1$

Условие

Part B

This problem is divided into three parts. It is possible to solve these three parts independently, but they are not equally weighted.

a.

An ideal rocket when empty of fuel has a mass $m_r$ and will carry a mass of fuel $m_f$ . The fuel burns and is ejected with an exhaust speed of $v_e$ relative to the rocket. The fuel burns at a constant mass rate for a total time $T_b$ . Ignore gravity; assume the rocket is far from any other body.

i. Determine an equation for the acceleration of the rocket as a function of time $t$ in terms of any or all of $t$ , $m_f$ , $m_r$ , $v_e$ , $T_b$ , and any relevant fundamental constants.

ii. Assuming that the rocket starts from rest, determine the final speed of the rocket in terms of any or all of $m_r$ , $m_f$ , $v_e$ , $T_b$ , and any relevant fundamental constants.

b.

The ship starts out in a circular orbit around the sun very near the Earth and has a goal of moving to a circular orbit around the Sun that is very close to Mars. It will make this transfer in an elliptical orbit as shown in bold in the diagram below. This is accomplished with an initial velocity boost near the Earth $\Delta v_1$ and then a second velocity boost near Mars $\Delta v_2$ . Assume that both of these boosts are from instantaneous impulses, and ignore mass changes in the rocket as well as gravitational attraction to either Earth or Mars. Don't ignore the Sun! Assume that the Earth and Mars are both in circular orbits around the Sun of radii $R_E$ and $R_M=R_E/\alpha$ respectively. The orbital speeds are $v_E$ and $v_M$ respectively.

Full native circular Earth/Mars orbits and bold transfer ellipse, complete RE/RM radius arrows and labels.

i. Derive an expression for the velocity boost $\Delta v_1$ to change the orbit from circular to elliptical. Express your answer in terms of $v_E$ and $\alpha$ .

ii. Derive an expression for the velocity boost $\Delta v_2$ to change the orbit from elliptical to circular. Express your answer in terms of $v_E$ and $\alpha$ .

iii. What is the angular separation between Earth and Mars, as measured from the Sun, at the time of launch so that the rocket will start from Earth and arrive at Mars when it reaches the orbit of Mars? Express your answer in terms of $\alpha$ .

Решение

Source variant note. The repeated key introduction reads: “This problem is divided into two parts. It is possible to solve these two parts independently, but they are not equally weighted.” The original question's “three parts” wording is retained above; the actual native tasks are a and b.

a. i.

Since there are no external forces on the system,

0=\frac{dp}{dt}=\frac{dm}{dt}v+m\frac{dv}{dt}

which means

a=-\frac1{m(t)}v_e\frac{dm}{dt}=\frac{v_e}{m_r+m_f(1-t/T)}\frac{m_f}{T}.

a. ii.

Rearrange the previous result for

\frac1{v_e}\,dv=-\frac1m\,dm.

Integrating both sides gives

\frac1{v_e}v=\ln\left(\frac{m_r+m_f}{m_r}\right)\quad\Rightarrow\quad v=v_e\ln\left(\frac{m_r+m_f}{m_r}\right).

This result is called the ideal rocket equation.

b. i.

First off, for a circular orbit of radius $R_c$ , we have

\frac{GM_S}{R_c^2}=\frac{v_c^2}{R_c}

where $M_S$ is the mass of the sun, so

v_E=\sqrt{\frac{GM_S}{R_E}},\qquad v_M=\sqrt{\frac{GM_S}{R_M}}.

Now consider an elliptical orbit with minimum radius $R_1$ and maximum radius $R_2$ . Energy and angular momentum give

\frac12v^2-\frac{GM_S}{r}=E,\qquad v_1R_1=v_2R_2.

Combining and eliminating $v_2$ ,

\frac12v_1^2-\frac{GM_S}{R_1}=\frac12v_1^2\left(\frac{R_1}{R_2}\right)^2-\frac{GM_S}{R_2}

which can be solved for $v_1$ ,

\frac12v_1^2\left(1-\left(\frac{R_1}{R_2}\right)^2\right)=GM_S\frac{R_2-R_1}{R_1R_2}.

Setting $R_1=R_E$ and $R_2=R_M$ , we have $\alpha=R_1/R_2$ , so

\frac12v_1^2(1-\alpha^2)=\frac{GM_S}{R_1}(1-\alpha)\quad\Rightarrow\quad v_1=v_E\sqrt{\frac2{1+\alpha}}.

As expected, this is greater than $v_E$ , and the boost is

\Delta v_1=v_E\left(\sqrt{\frac2{1+\alpha}}-1\right).

b. ii.

This is similar to the previous part, except we now eliminate $v_1$ ,

\frac12v_2^2\left(1-(1/\alpha)^2\right)=\frac{GM_S}{R_2}(1-(1/\alpha))\quad\Rightarrow\quad v_2=v_M\sqrt{\frac2{1+1/\alpha}}.

This is less than $v_M$ , so the rocket must receive a second positive boost,

\Delta v_2=v_M\left(1-\sqrt{\frac2{1+1/\alpha}}\right)=v_E\sqrt\alpha\left(1-\sqrt{\frac2{1+1/\alpha}}\right).

where we used $v_M=v_E\sqrt\alpha$ .

b. iii.

Kepler's third law gives the time for the orbital transfer,

\frac T{T_M}=\frac12\left(\frac{(R_E+R_M)/2}{R_M}\right)^{3/2}=\frac12\left(\frac{\alpha+1}{2}\right)^{3/2}.

During this time Mars moves through an angle of

2\pi\frac T{T_M}=\pi\left(\frac{\alpha+1}{2}\right)^{3/2}

while the rocket moves through an angle of $\pi$ , so the angular separation from Earth will be

\theta=\pi\left(1-\left(\frac{\alpha+1}{2}\right)^{3/2}\right).

Оригинал в Архива: USAPhO-2015.pdf · официални решения: USAPhO-2015-Solutions.pdf