Съдържание

Условие Отговори Решение

USAPhO 2003, Полуфинал — Задача 5. B1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2003

AAPT / AIP United States Physics Team 2003 — 2003 Semi-Final Exam

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.
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	$k = 1/4\pi\varepsilon_0 = 9.0 \times 10^{9}\ \mathrm{N{\cdot}m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T{\cdot}m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol{\cdot}K)}$
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 $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$

Условие

a. In classical mechanics, the kinetic energy $K$ , momentum $p$ , and mass $m$ of a particle are related by $K = \frac{p^2}{2m}$ . Derive the relationship between the total energy $E$ , momentum $p$ , and mass $m$ of a particle in special relativity? [10 т.]

Hint: Express $E^2$ in terms of $p^2$ . In addition to $m$ , the expression may also include $c$ . Relativistic expressions for the total energy $E$ and momentum $p$ of a particle of mass $m$ moving at speed $v$ are

E = \frac{mc^2}{\sqrt{1 - (v/c)^2}} \qquad\qquad p = \frac{mv}{\sqrt{1 - (v/c)^2}}

Particle A (mass $1000\ \mathrm{MeV}/c^2$ ) is at rest at the center of a spherical gamma detector, which completely surrounds it except for a small hole. We accelerate particle B (mass $500\ \mathrm{MeV}/c^2$ ) to a total energy of $700\ \mathrm{MeV}$ , sending it through the hole towards particle A. When the particles collide, a single, excited particle of unknown mass, C*, is produced. C* quickly decays to its unexcited state, C (mass $1300\ \mathrm{MeV}/c^2$ ), emitting a single gamma ray.

b. Find the mass of C*. [10 т.]

c. If the gamma is emitted directly away from the hole, what energy does the detector record? [15 т.]

d. The experiment is repeated, and the detector records an energy of $300\ \mathrm{MeV}$ . At what angles did particle C and the gamma emerge? Assume that C* decays at the center of the detector. Let an angle of 0 be directly away from the hole, and an angle of $\pi$ be directly towards the hole. [15 т.]

Отговори

Решение

a. Squaring the relativistic expressions for $E$ and $p$ ,

E^2 = \frac{m^2c^4}{1 - (v/c)^2} \qquad\qquad p^2 = \frac{m^2v^2}{1 - (v/c)^2}.

Multiplying the $p^2$ equation by $c^2$ and subtracting from $E^2$

E^2 - c^2p^2 = \frac{m^2c^4}{1 - (v/c)^2} - \frac{m^2v^2c^2}{1 - (v/c)^2} = \frac{m^2c^4\left(1 - (v/c)^2\right)}{1 - (v/c)^2} = m^2c^4

E^2 = (cp)^2 + \left(mc^2\right)^2

b. Since A is at rest, $E_A = m_Ac^2 = 1000\ \mathrm{MeV}$ and $p_A = 0$ ;
and

cp_B = \sqrt{E_B^2 - m_B^2c^4} = \sqrt{(700\ \mathrm{MeV})^2 - (500\ \mathrm{MeV})^2} = 490\ \mathrm{MeV}.

By conservation of energy and momentum,

E_{C*} = E_A + E_B = 1000\ \mathrm{MeV} + 700\ \mathrm{MeV} = 1700\ \mathrm{MeV}

cp_{C*} = cp_A + cp_B = 0 + 490\ \mathrm{MeV} = 490\ \mathrm{MeV}

Thus

m_{C*}c^2 = \sqrt{E_{C*}^2 - (cp_{C*})^2} = \sqrt{(1700\ \mathrm{MeV})^2 - (490\ \mathrm{MeV})^2} = 1628\ \mathrm{MeV}.

c. Again, using conservation of energy and momentum,

E_C + E_\gamma = E_{C*}

cp_C + cp_\gamma = cp_{C*}

Since the gamma is massless, $E_\gamma^2 = (cp_\gamma)^2$ . It is emitted along the direction of travel of C*, so we select the positive root, $cp_\gamma = E_\gamma$ . Thus, rearranging,

E_C = E_{C*} - E_\gamma

cp_C = cp_{C*} - E_\gamma

We need a fourth relationship,

E_C^2 - (cp_C)^2 = \left(m_Cc^2\right)^2.

\left(E_{C*} - E_\gamma\right)^2 - \left(cp_{C*} - E_\gamma\right)^2 = \left(m_Cc^2\right)^2

E_{C*}^2 - 2E_{C*}E_\gamma + E_\gamma^2 - (cp_{C*})^2 + 2cp_{C*}E_\gamma - E_\gamma^2 = \left(m_Cc^2\right)^2

E_{C*}^2 - (cp_{C*})^2 + 2cp_{C*}E_\gamma - 2E_{C*}E_\gamma = \left(m_Cc^2\right)^2

Since

E_{C*}^2 - (cp_{C*})^2 = \left(m_{C*}c^2\right)^2,

the prior equation becomes,

\left(m_{C*}c^2\right)^2 - 2E_\gamma\left(E_{C*} - cp_{C*}\right) = \left(m_Cc^2\right)^2

E_\gamma = \frac{\left(m_{C*}c^2\right)^2 - \left(m_Cc^2\right)^2}{2\left(E_{C*} - cp_{C*}\right)} = \frac{(1628\ \mathrm{MeV})^2 - (1300\ \mathrm{MeV})^2}{2(1700\ \mathrm{MeV} - 490\ \mathrm{MeV})} = 397\ \mathrm{MeV}

d. Let the gamma be emitted at an angle $\theta_\gamma$ to the axis, and C at an angle $\theta_C$ . We measure the two angles on opposite sides of the axis so that both are in the interval $[0, \pi]$ .
From conservation of energy,

E_C = E_{C*} - E_\gamma = 1700\ \mathrm{MeV} - 300\ \mathrm{MeV} = 1400\ \mathrm{MeV};

thus

cp_C = \sqrt{E_C^2 - \left(m_Cc^2\right)^2} = \sqrt{(1400\ \mathrm{MeV})^2 - (1300\ \mathrm{MeV})^2} = 520\ \mathrm{MeV}

And again,

cp_\gamma = E_\gamma = 300\ \mathrm{MeV}.

Then from conservation of momentum,

p_\gamma\sin\theta_\gamma - p_C\sin\theta_C = 0

p_\gamma\cos\theta_\gamma + p_C\cos\theta_C = p_{C*}

Rearranging,

\left(p_C\sin\theta_C\right)^2 + \left(p_C\cos\theta_C\right)^2 = \left(p_\gamma\sin\theta_\gamma\right)^2 + \left(p_{C*} - p_\gamma\cos\theta_\gamma\right)^2

p_C^2 = p_{C*}^2 + p_\gamma^2 - 2p_{C*}p_\gamma\cos\theta_\gamma

\cos\theta_\gamma = \frac{p_{C*}^2 + p_\gamma^2 - p_C^2}{2p_{C*}p_\gamma} = \frac{(490\ \mathrm{MeV/c})^2 + (300\ \mathrm{MeV/c})^2 - (520\ \mathrm{MeV/c})^2}{2(490\ \mathrm{MeV/c})(300\ \mathrm{MeV/c})} = 0.204

\theta_\gamma = \cos^{-1}0.204 = 1.365\ \mathrm{rad}.

Then,

\sin\theta_C = \frac{p_\gamma}{p_C}\sin\theta_\gamma = \frac{300\ \mathrm{MeV/c}}{520\ \mathrm{MeV/c}}\sin 1.365 = 0.565

Thus,

\theta_C = 0.601\ \mathrm{rad}.

Оригинал в Архива: Semifinal 2003.pdf · официални решения: Semifinal 2003 Solutions.pdf