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USAPhO 2023, Полуфинал — A1. Circus Act

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2023 USA Physics Olympiad Exam · 25 т.

Внимание: Бележка към темата

The archive contains the complete question packet and the official interleaved solution packet. The separate official answer sheets referred to in the instructions are absent from this archive folder. Original printed errors are retained: A1(c)'s solution uses $z'_x$ although the question defines $y(x)$ ; A2(b.iii)'s solution writes radius $r+\delta r$ although the question defines $r=r_0+\delta r$ ; A3's key refers to “part (b)” in its part a.iii check, and B1's key calls the rod a “disc”. B2's printed energy expression uses $2.0\times10^7\ \mathrm{J}$ although the question gives energy density in J/kg. The “this paper” references in B2 and B3 have no recoverable URI in the supplied PDF.

AAPT UNITED STATES PHYSICS TEAM — AIP 2023; USA Physics Olympiad Exam

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Important Instructions for the Exam Supervisor

This examination has two parts. Each part has three questions and lasts for 90 minutes.
For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–8), and Part B questions (pages 9–11).
Begin by giving students the instructions and all of the answer sheets. Let the students read the instructions and fill out their information on the answer sheets. They can keep the instructions for both parts of the exam. Also give students blank sheets of paper to use as scratch paper throughout the exam.
Students may bring calculators, but they may not use symbolic math, programming, or graphing features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones or other electronics may not be used during the exam or while the exam papers are present. Students may not use books or other references.
To start the exam, give students the Part A questions, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the Part A questions and answer sheets.
Then give students a 5 to 10 minute break. Then give them the Part B questions, and allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, collect everything, including the questions, the instructions, the answer sheets, and the scratch paper. Students may not keep the exam questions. Everything can be returned to the students after April 19th, 2023.
After the exam, sort each student’s answer sheets by page number. Scan every answer sheet, including blank ones.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Kellan Colburn, Samuel Gebretsadkan, Abi Krishnan, Natalie LeBaron, Kye Shi, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the reference table on the next page, answer sheets, and blank paper for scratch work. Read this page carefully before the exam begins.
You may use a calculator, but its memory must be cleared of data and programs, and you may not use symbolic math, programming, or graphing features. Calculators may not be shared. Cell phones or other electronics may not be used during the exam or while the exam papers are present. You may not use books or other outside references.
When the exam begins, your proctor will give you the questions for Part A. You will have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. If you finish all of the questions, you may check your work, but you may not look at Part B during this time.
After 90 minutes, your proctor will collect the questions and answer sheets for Part A. You may then take a short break.
Then you will work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. When the exam ends, you must return all papers to the proctor, including the exam questions.
Do not discuss the questions of this exam, or their solutions, until after April 19th, 2023. Violations of this rule may result in disqualification.

Below are instructions for writing your solutions.

All of your solutions must be written on the official answer sheets. Nothing outside these answer sheets will be graded. Before the exam begins, write your name, student AAPT number, and proctor AAPT number as directed on the answer sheets.
There are several answer sheets per problem. If you run out of space for a problem, you may use the extra answer sheets, which are at the end of the answer sheet packet. To ensure this work is graded, you must indicate, at the bottom of your last answer sheet for that problem, that you are using these extra answer sheets.
Only write within the frame of each answer sheet. To simplify grading, we recommend drawing a box around your final answer for each subpart. You should organize your work linearly and briefly explain your reasoning, which will help you earn partial credit. You may use either pencil or pen, but sure to write clearly so your work will be legible after scanning.

Reference table of possibly useful information

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$	$(1+x)^n\approx1+nx$ for $\vert x\vert \ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\vert \theta\vert \ll1$	$\cos\theta\approx1-\theta^2/2$ for $\vert \theta\vert \ll1$

Possibly useful integrals

\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)+C

\int\frac{dx}{1-x^2}=\tanh^{-1}(x)+C

\int\frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(x)+C

\int\frac{dx}{1+x^2}=\tan^{-1}(x)+C

\int\frac{dx}{(1+x^2)^{3/2}}=\frac{x}{\sqrt{1+x^2}}+C

\int\sqrt{\frac{1+x}{1-x}}\,dx=-\sqrt{1-x^2}-2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{\sqrt{1+x}}{(1-x)^{3/2}}\,dx=2\sqrt{\frac{1+x}{1-x}}+2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{1/2}}=\sqrt{\frac{1+x}{1-x}}+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{1/2}}=\frac{(2-x)\sqrt{1+x}}{3(1-x)^{3/2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{3/2}}=\frac{1+2x-2x^2}{3(1-x)^{3/2}(1+x)^{1/2}}+C

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

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Условие

In this problem we consider a small ball bouncing back and forth between two points. In all parts below, the acceleration of gravity is $g$ , collisions are perfectly elastic, air resistance is negligible, and the impact points are at the same height. The diagrams are not drawn to scale.

a.

Consider a ball bouncing between two inclined planes, which each make an angle $\theta < 90^{\circ}$ to the horizontal. The ball has speed $v_0$ at the impact points, which are separated by a distance $D$ .

i. The ball can bounce back and forth along the same path, as shown. Original same-path bounce, both theta labels and distance D.

For what values of $\theta$ is this motion possible? For these values, what is $v_0$ ?

ii. The ball can also take one trajectory while traveling to the right, and a separate trajectory when traveling back. Let $\phi \neq 0$ be the angle between the paths at the impact points. Original two-path bounce, both theta labels and distance D.

For what values of $\theta$ and $\phi$ is this motion possible? For these values, what is $v_0$ ?

b.

Now suppose the ball bounces within an hemispherical well of radius of curvature $R$ . As in part a.ii, it alternates between two distinct paths, with flight times $t_1$ and $t_2 \neq t_1$ . Original hemispherical well, complete two paths and radius R.

Find all of the possible values of $R$ , in terms of $t_1$ and $t_2$ .

c.

Finally, suppose the well has a sinusoidal shape, described by $y(x) = -L\sin(2x/L)$ . The ball takes two distinct paths with flight times $t_1$ and $t_2 \neq t_1$ , and the horizontal distance between the impact points is less than $\pi L$ . Find all of the possible values of $L$ , in terms of $t_1$ and $t_2$ .

Решение

a.i.

To go back and forth on the same path, the ball must impact the plane normally, which means that $\theta$ is the angle of its velocity to the vertical. The range of the ball is

D = \frac{v_0^2\sin\big(2(90^{\circ}-\theta)\big)}{g} = \frac{v_0^2\sin 2\theta}{g}

which implies

v_0 = \sqrt{\frac{gD}{\sin 2\theta}}.

Evidently, the motion is possible for any $\theta$ .

a.ii.

Official key two distinct paths in red and blue.

The two paths have the same initial speed and the same range, which means the initial angles of the velocity to the horizontal must be $\pi/4 \pm \phi/2$ . Since the initial and final angles to the normal are equal in the collision, the angle of the planes to the horizontal must be $\theta = 45^{\circ}$ . Given this, $\phi$ can take any value in the range $0<\phi<90^{\circ}$ .

Considering the range yields

D = \frac{v_0^2\sin\big(2(45^{\circ}+\phi/2)\big)}{g} = \frac{v_0^2\cos\phi}{g}

from which we find

v_0 = \sqrt{\frac{gD}{\cos\phi}}.

As expected, this reduces to the answer above as $\phi \to 0$ , and becomes infinite as $\phi \to 90^{\circ}$ .

Official key hemispherical well with red and blue paths and R.

From the previous question we know that the slope of the wall $\theta = \frac{\pi}{4}$ at the collision points. So $D = R\sqrt{2}$ .

$x$ displacement with $t_1$ is

v_0\cos\left(\frac{\pi}{4}-\phi/2\right)t_1 = R\sqrt{2} \qquad (A1-1)

$y$ -component of velocity in the upper point is 0

v_0\sin\left(\frac{\pi}{4}+\phi/2\right) - g\frac{t_2}{2} = 0

transforms into

v_0\cos\left(\frac{\pi}{4}-\phi/2\right) = g\frac{t_2}{2} \qquad (A1-2)

Dividing (A1-1) by (A1-2) we get

R = \frac{gt_1t_2}{2\sqrt{2}}.

Only one $R$ satisfies this, so the answer is unique.

c. The slope of the wall $\theta=\frac{\pi}{4}$ at the collision points. So $z_x' = -1$ for one end and there is a symmetric point on the other point of collision. $-2\cos\frac{2x}{L}=-1$ has solutions $x=\pm\frac{\pi L}{6}$ . So we got two possible values of $L$ .

i. Case of $x=\frac{\pi L}{6}$ corresponds to $D=\frac{\pi L}{2}-\frac{\pi L}{3}=\frac{\pi L}{6}$

v_0\cos\left(\frac{\pi}{4}-\phi/2\right)t_1 = \frac{\pi L}{6}

and

v_0\cos\left(\frac{\pi}{4}-\phi/2\right) = g\frac{t_2}{2}.

Combining these two we get

L_a = \frac{3gt_1t_2}{\pi}

ii. Case of $x=-\frac{\pi L}{6}$ corresponds to $D=\frac{\pi L}{2}+\frac{\pi L}{3}=\frac{5\pi L}{6}$

v_0\cos\left(\frac{\pi}{4}-\phi/2\right)t_1 = \frac{5\pi L}{6}

and

v_0\cos\left(\frac{\pi}{4}-\phi/2\right) = g\frac{t_2}{2}.

Combining these two we get

L_b = \frac{3gt_1t_2}{5\pi}

Оригинал в Архива: 2023-USAPhO-Exam.pdf · официални решения: 2023-USAPhO-solutions.pdf