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USAPhO 2014, Полуфинал — Задача 5. Question B1

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Всички материали от USAPhO 2014

USA Physics Olympiad Exam — Part A / Part B · 50 т.

Внимание: Бележка към темата

The printed key is retained with its source defects: A3(a)(i) has a factor 1/2 in the expanded cubic term; A3(b)(ii) switches from question distance d to h; A4(b)(ii) prints r < b before describing the exterior field; B2(a)(i) drops the subscript 0 in one denominator. These are source discrepancies, not silent editorial corrections.

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 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 4 parts: the cover sheet (page 2), Part A (pages 3–5), Part B (pages 7–8), and several answer sheets for two of the questions in part A (pages 10–13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
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 15, 2014.
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.

USA Physics Olympiad Exam — 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. 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 15, 2014.

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

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Условие

Part B

A block of mass $M$ has a hole drilled through it so that a ball of mass $m$ can enter horizontally and then pass through the block and exit vertically upward. The ball and block are located on a frictionless surface; the block is originally at rest.

Original ball and curved hole in block, m/M masses, velocity arrow and frictionless horizontal surface.

a. Consider the scenario where the ball is traveling horizontally with a speed $v_0$ . The ball enters the block and is ejected out the top of the block. Assume there are no frictional losses as the ball passes through the block, and the ball rises to a height much higher than the dimensions of the block. The ball then returns to the level of the block, where it enters the top hole and then is ejected from the side hole. Determine the time $t$ for the ball to return to the position where the original collision occurs in terms of the mass ratio $\beta=M/m$ , speed $v_0$ , and acceleration of free fall $g$ .

b. Now consider friction. The ball has moment of inertia $I=\frac25mr^2$ and is originally not rotating. When it enters the hole in the block it rubs against one surface so that when it is ejected upwards the ball is rolling without slipping. To what height does the ball rise above the block?

Решение

a.

After the collision, the ball and block have the same horizontal velocity $v_1$ . Since the horizontal momentum is conserved,

v_1=\frac m{m+M}v_0.

Let $v_2$ be the vertical component of the velocity of the ball immediately after the collision. Since there are no frictional losses, conservation of energy yields

\frac12mv_0^2=\frac12Mv_1^2+\frac12m(v_1^2+v_2^2).

Since the ball rises to a height much higher than the height of the block, the gravitational potential energy is negligible, so we have ignored it. This assumption also means we can ignore the duration of the collision itself in our calculations below.

Plugging in our result for $v_1$ ,

mv_0^2-(M+m)\left(\frac m{m+M}\right)^2v_0^2=mv_2^2\quad\Rightarrow\quad v_2=\sqrt{\frac M{m+M}}v_0.

The time spent by the ball in the air is

t_1=2v_2/g.

The distance traveled horizontally by the ball while it is in the air is

x=v_1t_1=\frac{2v_1v_2}g=\frac{2v_0^2}g\sqrt{\frac{m^2M}{(m+M)^3}}.

The ball then falls back into the block and is ejected horizontally.

Since energy and momentum are conserved from before the first collision and after the second, the final horizontal velocity $v_3$ of the ball is given by the result for a perfectly elastic collision,

v_3=v_0\frac{m-M}{m+M}

as can be derived from the conservation laws. Note that $v_3$ is positive for $m>M$ . Thus if $\beta\leq1$ , the time $t$ is infinite; the ball never returns to its starting position.

Assuming that $\beta>1$ , the ball must move a distance $x$ towards its original collision point, where we have neglected the time taken for the collisions themselves. Thus the time for the ball to return to its original position horizontally is

t_2=-\frac x{v_3}=\frac{2v_0}g\sqrt{\frac{m^2M}{(M+m)(M-m)^2}}.

The total time since the first collision is

t=t_1+t_2=\frac{2v_0}g\sqrt{\frac M{m+M}}\left(\frac M{M-m}\right)=\frac{2v_0}g\sqrt{\frac\beta{1+\beta}}\left(\frac\beta{\beta-1}\right)

where $\beta>1$ .

b.

As before, we have

v_1=\frac m{m+M}v_0.

Let $v_4$ be the vertical velocity of the ball after the collision; it is less than $v_2$ due to the friction force $f$ . Friction slows the ball down with an impulse given by

\Delta p=f\Delta t=m(v_2-v_4)

while increasing the angular momentum by

\Delta L=\tau\Delta t=rf\Delta t=r\Delta p.

We also know that $\Delta L=I\omega=Iv_4/r$ , so

Iv_4/r=mr(v_2-v_4)\quad\Rightarrow\quad v_4=\frac{v_2}{1+2/5}.

The vertical velocity of the ball will take it to a height

h=\frac{v_4^2}{2g}=\frac{v_0^2}{2g}\frac1{(1+2/5)^2}\frac M{m+M}=\frac{v_0^2}{2g}\frac{25}{49}\frac\beta{1+\beta}.

Оригинал в Архива: USAPhO-2014.pdf · официални решения: USAPhO-2014-Solutions.pdf