USAPhO 2012, Полуфинал — Задача 5. Question B1

UNITED STATES PHYSICS TEAM 2012 — Semifinal Exam · 50 т.

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 minute 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 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet.
  • 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 1, 2012.
  • 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.
  • Please provide the examinees with graph paper for Part A.
Указания

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 1, 2012.

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

g=9.8 N/kgg = 9.8\ \mathrm{N/kg}G=6.67×10−11 N⋅m2/kg2G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}
k=1/4πϵ0=8.99×109 N⋅m2/C2k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}km=μ0/4π=10−7 T⋅m/Ak_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}
c=3.00×108 m/sc = 3.00 \times 10^{8}\ \mathrm{m/s}kB=1.38×10−23 J/Kk_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}
NA=6.02×1023 (mol)−1N_{\mathrm{A}} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}R=NAkB=8.31 J/(mol⋅K)R = N_{\mathrm{A}}k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}
σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \mathrm{C}
1eV=1.602×10−19 J1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}h=6.63×10−34 J⋅s=4.14×10−15 eV⋅sh = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}
me=9.109×10−31 kg=0.511 MeV/c2m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}(1+x)n≈1+nx(1 + x)^n \approx 1 + nx for ∣x∣≪1\lvert x\rvert \ll 1
sin⁡θ≈θ−16θ3\sin\theta \approx \theta - \frac{1}{6}\theta^3 for ∣θ∣≪1\lvert\theta\rvert \ll 1cos⁡θ≈1−12θ2\cos\theta \approx 1 - \frac{1}{2}\theta^2 for ∣θ∣≪1\lvert\theta\rvert \ll 1

Условие

A particle of mass mm moves under a force similar to that of an ideal spring, except that the force repels the particle from the origin:

F=+mα2xF = +m\alpha^2x

In simple harmonic motion, the position of the particle as a function of time can be written

x(t)=A cos⁡ωt+B sin⁡ωtx(t) = A\ \cos\omega t + B\ \sin\omega t

Likewise, in the present case we have

x(t)=A f1(t)+B f2(t)x(t) = A\ f_1(t) + B\ f_2(t)

for some appropriate functions f1f_1 and f2f_2.

a. f1(t)f_1(t) and f2(t)f_2(t) can be chosen to have the form erte^{rt}. What are the two appropriate values of rr?

b. Suppose that the particle begins at position x(0)=x0x(0) = x_0 and with velocity v(0)=0v(0) = 0. What is x(t)x(t)?

c. A second, identical particle begins at position x(0)=0x(0) = 0 with velocity v(0)=v0v(0) = v_0. The second particle becomes closer and closer to the first particle as time goes on. What is v0v_0?

Отговори

Решение

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.

Оригинал в Архива: USAPhO-2012.pdf · официални решения: USAPhO-2012-Solutions.pdf