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Условие Отговори Решение

USAPhO 2002, Полуфинал — Задача 3. A3

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Всички материали от USAPhO 2002

2002 Semi-Final Exam · 25 т.

Указания

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.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

A1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_o = 9.0 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_B = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_A = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_Ak_B = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV}/c^2$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Условие

Fundamental quanta of length, time, and mass are the Planck length $\lambda_{\mathrm{P}}$ , the Planck time $t_{\mathrm{P}}$ , and the Planck mass $m_{\mathrm{P}}$ . Use the fact that they depend only on the Newton’s gravitational constant $G$ , Planck’s constant $h$ , and speed of light in a vacuum $c$ and no other constant to derive expressions for them and then calculate their numerical values.

Отговори

Решение

The Planck length $\lambda_{\mathrm{P}}$ , the Planck time $t_{\mathrm{P}}$ , and the Planck mass $m_{\mathrm{P}}$ depend only on the Newton’s gravitational constant $G$ , Planck’s constant $h$ , and speed of light in a vacuum $c$ and no other constant. Use dimensional analysis to obtain the equations. Let

[T] represent the dimension of time

[L] represent the dimension of length

[M] represent the dimension of mass

Since $G$ has units $\mathrm{N{\bullet}m^2/kg^2 = (kg{\bullet}m/s^2)\ m^2/kg^2 = m^3kg^{-1}s^{-2}}$ its dimensions are $[L]^3[M]^{-1}[T]^{-2}$ for $G$

$h$ has units $\mathrm{J\ s = (kg{\bullet}m^2/s^2)s = m^2kg^1s^{-1}}$ its dimensions are $[L]^2[M]^1[T]^{-1}$ for $h$

$c$ has units $\mathrm{m/s = m^1s^{-1}}$ its dimensions are $[L]^1[M]^0[T]^{-1}$ for $c$

Let

\lambda_p = G^ah^bc^d.

Then analyzing the dimensions

[L]^1[M]^0[T]^0 = [L]^{3a}[M]^{-a}[T]^{-2a}[L]^{2b}[M]^{b}[T]^{-b}[L]^{d}[M]^0[T]^{-d} = [L]^{3a+2b+d}[M]^{-a+b}[T]^{-2a-b-d}.

Equating exponents of [L] $\quad 1 = 3a+2b+d$

Equating exponents of [M] $\quad 0 = -a+b$

Equating exponents of [T] $\quad 0 = -2a-b-d$

Solving these equations $\quad b = a$ and $d = -3a$ then $1 = 3a + 2a - 3a = 2a$

So $a = 1/2$ , $b = 1/2$ , and $d = -3/2$ and

\lambda_p = G^{1/2}h^{1/2}c^{-3/2} = \left(\frac{Gh}{c^3}\right)^{1/2} = \left(\frac{\left(6.67\times10^{-11}\mathrm{N\cdot m^2/kg^2}\right)\left(6.63\times10^{-34}\mathrm{J\cdot s}\right)}{\left(3.0\times10^{8}\mathrm{m/s}\right)^3}\right)^{1/2} = 4.05\times10^{-35}\mathrm{m}

Let

t_p = G^ah^bc^d.

Then analyzing the dimensions

[L]^0[M]^0[T]^1 = [L]^{3a+2b+d}[M]^{-a+b}[T]^{-2a-b-d}.

Equating exponents of [L] $\quad 0 = 3a+2b+d$

Equating exponents of [M] $\quad 0 = -a+b$

Equating exponents of [T] $\quad 1 = -2a-b-d$

Solving these equations $\quad b = a$ , $d = -5a$ , and $1 = -2a - a + 5a = 2a$

So $a = 1/2$ , $b = 1/2m$ and $d = -5/2$

t_p = G^{1/2}h^{1/2}c^{-5/2} = \left(\frac{Gh}{c^5}\right)^{1/2} = \left(\frac{\left(6.67\times10^{-11}\mathrm{N\cdot m^2/kg^2}\right)\left(6.63\times10^{-34}\mathrm{J\cdot s}\right)}{\left(3.0\times10^{8}\mathrm{m/s}\right)^5}\right)^{1/2} = 1.35\times10^{-43}\mathrm{s}

Let

m_p = G^ah^bc^d.

Then analyzing the dimensions

[L]^0[M]^1[T]^0 = [L]^{3a+2b+d}[M]^{-a+b}[T]^{-2a-b-d}.

Equating exponents of [L] $\quad 0 = 3a+2b+d$

Equating exponents of [M] $\quad 1 = -a+b$

Equating exponents of [T] $\quad 0 = -2a-b-d$

Solving these equations $\quad a = -1/2$ , $b = 1/2$ , and $d = 1/2$

m_p = G^{-1/2}h^{1/2}c^{1/2} = \left(\frac{hc}{G}\right)^{1/2} = \left(\frac{\left(6.63\times10^{-34}\mathrm{J\cdot s}\right)\left(3.0\times10^{8}\mathrm{m/s}\right)}{\left(6.67\times10^{-11}\mathrm{N\cdot m^2/kg^2}\right)}\right)^{1/2} = 5.46\times10^{-8}\mathrm{kg}

Автори

The 2002 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Warren Turner – Physics Teacher at Massachusetts Academy of Math and Science, Worcester, MA; Robert Shurtz – Physics Teacher at the Hawken School, Gates Mills, OH; Boris Zbarsky –undergraduate student at MIT, Cambridge, MA; Andrew Lin – undergraduate student at Yale University, New Haven, CT. Questions A2, A4, B1, and B2 were suggested by Leaf Turner and Questions A1 and A3 by Mary Mogge.

Оригинал в Архива: Semifinal 2002.pdf · официални решения: Semifinal 2002 Solutions.pdf