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USAPhO 2011, Полуфинал — B2

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Всички материали от USAPhO 2011

AAPT UNITED STATES PHYSICS TEAM AIP 2011 — Semifinal Exam · 50 т.

Внимание: Бележка към темата

Printed source errors are preserved: A1’s solution labels both quoted volumes V₀ and says “we used the fat that”; A2’s solution reverses its graph-variable description; A4 prints ω = f/(2π). The solutions repeat the question statements and common instructions; they are shown once here. No individual subpart point values are printed.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

Important Instructions for the Exam Supervisor

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 3-4), 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 March 31, 2011.
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, 2011.

Possibly Useful Information

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

$g = 9.8\ \mathrm{N/kg}$	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$	$k_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$
$c = 3.00 \times 10^{8}\ \mathrm{m/s}$	$k_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
$N_{\mathrm{A}} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$	$R = N_{\mathrm{A}} k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}$
$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$	$h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$
$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$	$(1+x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $\lvert\theta\rvert \ll 1$	$\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $\lvert\theta\rvert \ll 1$

Условие

A particle is constrained to move on the inner surface of a frictionless parabolic bowl whose cross-section has equation $z = kr^2$ . The particle begins at a height $z_0$ above the bottom of the bowl with a horizontal velocity $v_0$ along the surface of the bowl. The acceleration due to gravity is $g$ .

Left: cross-section of the bowl, a parabola symmetric about a vertical axis labelled z, touching a horizontal axis labelled r at the origin, with the curve labelled z = kr². Right: a perspective wireframe drawing of the parabolic bowl with an elliptical rim, two horizontal rings and several meridian lines.

a. For a particular value of horizontal velocity $v_0$ , which we will name $v_h$ , the particle moves in a horizontal circle. What is $v_h$ in terms of $g$ , $z_0$ , and/or $k$ ?

b. Suppose that the initial horizontal velocity is now $v_0 > v_h$ . What is the maximum height reached by the particle, in terms of $v_0$ , $z_0$ , $g$ and/or $k$ ?

c. i. Suppose that the particle now begins at a height $z_0$ above the bottom of the bowl with an initial velocity $v_0 = 0$ .

Assuming that $z_0$ is small enough so that the motion can be approximated as simple harmonic, find the period of the motion in terms any or all of the mass of the particle $m$ , $g$ , $z_0$ , and/or $k$ .

c. ii. Assuming that $z_0$ is not small, will the actual period of motion be greater than, less than, or equal to your simple harmonic approximation above? (You need not calculate the new value explicitly, but you should show some work to defend your answer.)

Отговори

Решение

a. Let the radius of the bowl at height $z_0$ be $r_0$ and let the angle made by the bowl’s surface to the horizontal at that height be $\theta$ .

Two forces act on the particle: the normal force and gravity. If the particle moves in a horizontal circle, the vertical components of these forces cancel, and the horizontal component of the normal force is the centripetal force. Then

N\sin\theta = \frac{mv_h^2}{r_0}, \quad N\cos\theta = mg.

Combining these,

\tan\theta = \frac{v_h^2}{gr_0}.

Now $\tan\theta$ is the slope of the bowl $dz/dr = 2kr_0$ , so

2kr_0 = \frac{v_h^2}{gr_0}

Using $z_0 = kr_0^2$ and solving for $v_h$ gives

v_h = \sqrt{2gz_0}

b. Let the maximum height be $z$ , let the radius of the bowl at this point be $r$ , and let the speed of the particle at this point be $v$ . By conservation of energy,

\frac{1}{2}mv_0^2 + mgz_0 = \frac{1}{2}mv^2 + mgz.

The two forces acting on the particle never exert a torque in the $z$ -direction, so the $z$ -component of the angular momentum is conserved. Furthermore, at both the initial and final heights, the velocity of the particle is perpendicular to the $z$ -axis, so

mv_0 r_0 = mvr \quad \Rightarrow \quad v = v_0\frac{r_0}{r}.

Using the equation of the bowl,

v = v_0\sqrt{\frac{z_0}{z}}.

Using this to eliminate $v$ in the energy conservation equation, we find a quadratic in $z$ ,

z^2 - \left(\frac{{v_0}^2}{2g} + z_0\right)z + \frac{{v_0}^2}{2g}z_0 = 0.

The roots of this quadratic sum to $z_0 + v_0^2/2g$ , and there is a root $z = z_0$ corresponding to the initial condition, so the desired root is

z = \frac{v_0^2}{2g}.

The two roots are equal when $v_0 = v_h$ , providing an alternate solution to part (a).

i. We present a force-based approach and an energy-based approach. In each case, let $r$ be the radial position of the particle, so that $z = kr^2$ is the height of the particle above the bottom of the bowl.

Let the angle of the bowl’s surface to the horizontal be $\theta$ . Because $z_0$ is small,

\sin\theta \approx \theta \approx \tan\theta = \frac{dz}{dr} = 2kr

and $\cos\theta \approx 1$ . Moreover, since $z_0$ is small, the centripetal acceleration is negligible, so we can consider only the tangential acceleration. Since the force tangential to the bowl is $mg\sin\theta$ , this is $a = g\sin\theta$ . The radial acceleration $a_r$ is

a_r = -a\cos\theta = -g\cos\theta\sin\theta.

Then in the small- $z$ approximation,

a_r \approx -g\tan\theta = -2krg.

This is simple harmonic motion with $\omega = \sqrt{2kg}$ and hence period

T = \frac{2\pi}{\sqrt{2kg}}.

The energy-based approach begins with the total energy

E = \frac{1}{2}mv^2 + mgz.

The velocity $v$ is given by

v^2 = \left(\frac{dr}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2.

Because $z$ is small, $\frac{dz}{dt} \ll \frac{dr}{dt}$ , and we conclude that

E = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 + mgkr^2.

By conservation of energy,

0 = \frac{dE}{dt} = m\frac{dr}{dt}\frac{d^2r}{dt^2} + 2mgkr\frac{dr}{dt}

which implies that

0 = \frac{d^2r}{dt^2} + 2krg

which is the same equation as before.

ii. The period is greater than the simple harmonic period. In the force-based approach, we found

a_r = -g\cos\theta\sin\theta

and approximated it as

a_r = -g\tan\theta.

Since $\cos\theta\sin\theta < \tan\theta$ , this is an overestimate, so the period is actually larger.

In the energy-based approach, we dropped a positive term in the formula for the speed $v$ as expressed in terms of $\frac{dr}{dt}$ . Therefore we overestimated $\frac{dr}{dt}$ , and again the particle takes longer to reach the origin in reality than it does in the approximation.

Оригинал в Архива: USAPhO-2011.pdf · официални решения: USAPhO-2011-Solutions.pdf