[User]
4. You are also required to show your detailed work for each question in a reasonably neat and coherent
     way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of
     these pages.
                                                 0

Solution:
4. One detector is either inside or outside the circle, and other detector is at R.

                                         S                     v
                                                  φ

                                                       D
         When source is approaching the stationary detector D is at an angle φ, the frequency
         detected by D is
                                                   f0
                                           f=
                                               1 − c cos φ
                                                   v

         Here f0 is the frequency emitted by the source, and c is the speed of the sound. If we
         consider fourth possibility, we should have observed a sharp change in the plot when
         source is crossing the detector. Hence, we rule out this case.
         When the detector Dout is outside the circle (see figure below):

INPhO 2020 Page 10        Questions & Summary Answers                        Last four digits of Roll No.:

                                            S

                                     R                   φo
                                                                   v
                                O                                                      Dout
                                                                  Do

                                                             f0
                                            f out =
                                                        1 − c cos φo
                                                            v

         For maximum and minimum frequency, cos φ = ±1

                                      Ao

                                 R

                                     αo
                                                    0
                             C2 O

                                      A0o
                                                                  Do
                                                                                Dout

        HB                           at Ao , fmax
                                              out

                                     at A0o , fmin
                                               out
                                                  =

                                                   =
                                                       f0
                                                     1− c
                                                       f0
                                                          v =

                                                     1 + vc
                                                            =
                                                                f0
                                                              1 − ωR
                                                                f0
                                                                   c

                                                              1 + ωR
                                                                   c

         If both the detectors are outside, maximum and minimum frequencies detected by them
                                                                                (4.1)

                                                                                (4.2)

         are same which is not the case if you observe the graph given in the question. Hence, we
         rule out the first possibility.
         When detector Din is inside the circle (see figure below):
                                                              S

                                                                   φi
                                                          βi
                                                    R                           v

                                                              γi
                                                O                      Din
                                                         Di

INPhO 2020 Page 11         Questions & Summary Answers

                                                             f0
                                                f in =                                          (4.3)
                                                         1 − c cos φi
                                                             v

                                          | cos φi | = | sin βi |                               (4.4)
                                             sin γi    sin βi
                                                     =                                          (4.5)
                                               R          Di
                                                       Di
                                             cos φi =       sin γi                              (4.6)
                                                        R
                                                               f0
                                                f in =                                          (4.7)
                                                       1 − c R sin γi
                                                             v Di

                                                                   Ai

                                                     R        βi φi

                                                              γi               v
                                                     αi
                                                O                   Din
                                                         Di

                                                              0
                                      C2
         f in is maximum (minimum) when sin γi is 1 (−1).

                                       at Ai , fmax
                                                in
                                                    =
                                                          f0
                                                       1− cωDi
                                                                   A0i

                                                                    out
                                                                 < fmax                         (4.8)

               HB                      at A0i , fmin
                                                 in
                                                     =
                                                          f0
                                                       1 + ωDc
                                                               i
                                                                    out
                                                                 > fmin                         (4.9)

         If both the detectors are inside, again there are two possibilities: they are at the same
         distances or at the different distances from the center. In the former case, their observed
         peak frequencies will be same which is clearly not evident from the given graph. In the
         latter case, their maximum frequencies will be different but the time difference between
         the maximum and minimum frequencies will also be different. However from the graph,
         the time difference between maximum and minimum frequencies for detectors 1 and 2 is
                                           t(fmin ) − t(fmax ) = 9 s

         This is possible only if one detector is inside and other detector is outside. Also, detector
         may not be colinear with the center.

                                 S
                                  •                                       •1
                                            O
                                            •

                                             •2
         In this part, marking (anywhere) 1 outside and 2 inside the circle with a correct
         justification will be given full credit.

INPhO 2020 Page 12          Questions & Summary Answers                   Last four digits of Roll No.:
    (c) Obtain the frequency f0 of the source.                                                            [3]

         f0 =

          Solution: From equations (4.1) and (4.2),

                                                  f0
                                                       = 1300 Hz                              (4.10)
                                                1 − vc
                                                  f0
                                                       = 800 Hz                               (4.11)
                                                1 + vc
                                                     f0 ≈ 991 Hz                              (4.12)

    (d) Calculate the distance (D) between the detectors.                                                 [9]

         D=

         Solution: We have to find out Di , Do and angle θ between the detectors. From equations
         (4.10) and (4.11),                          0
                              C2                v = 78.6 m/s

                                                v = Rω = R
                                                R ≈ 300 m
                                                            2π
                                                            24

                                                           Ao
                                                                                (4.13)

                                                                                (4.14)
                                                                                (4.15)

         HB           A0i              Di
                                            O
                                             θ
                                             αi
                                                αo
                                                                     Do
                                                                                Dout

                                       Din
                                                                Ai

                                                     A0o

          By symmetry, Ao and A0o are equidistant from the detector 1. Similarly, Ai and A0i are
          equidistant from the detector 2. It takes 9 s from Ai to A0i (Ao and A0o ). Hence

                                                           3π
                                        2∠αo = 2∠αi =                                         (4.16)
                                                            4
                                             Di = R cos αi ≈ 115 m                            (4.17)
                                                    R
                                             Do =        ≈ 784 m                              (4.18)
                                                  cos αo
         There will be signal delay due to finite Do and Di .
         Let θ be the angular separation between the detectors. Time difference between the peak

INPhO 2020 Page 13          Questions & Summary Answers

          frequencies of detectors 1 and 2 is 4 seconds.
                                         θ     Ai Din Ao Dout
                                        4=  +         −                                        (4.19)
                                         ω        c       c
                                         θ     R sin αi Do sin αo
                                      4= +             −                                       (4.20)
                                         ω         c        c
                                         π ωR sin αi ωDo sin αo
                                     ⇒θ= −              +                                      (4.21)
                                         3          c         c
                                       = 1.4 rad                                               (4.22)
                                                   q
          The distance between the detectors =        OD2in + OD2out − 2ODin .ODout . cos(1.4) ≈ 773
          m

                            Detailed answers can be found on page numbers:

INPhO 2020 Problem 4


[User]
APHO II   2001    Theoretical Question 3                                                 p. 1 / 3

                                 Theoretical Question 3

                   Thermal Vibrations of Surface Atoms
     This question considers the thermal vibrations of surface atoms in an elemental
metallic crystal with a face-centered cubic (fcc) lattice structure. The unit cubic cell
of an fcc lattice consists of one atom at each corner and one atom at the center of
each face of the cubic cell, as shown in Fig. 3a. For the crystal under consideration,
we use (a, 0, 0), (0, a, 0) and (0, 0, a) to represent the locations of the three atoms on
the x, y and z axes of its cell. The lattice constant a is equal to 3.92 Å (i.e..the length
of each side of the cube is 3.92 Å).
                                 z

                                   G       (0,0, a)           F

                    A                         D

                                                              (0, a, 0)
                                                                            y
                                                              E

             ( a,0,0)
                        B                             C           Fig. 3a
              x
(1) The crystal is cut in such a way that the plane containing ABCD becomes a
   boundary surface and is chosen for doing low-energy electron diffraction
   experiments. A collimated beam of electrons with kinetic energy of 64.0 eV is
   incident on this surface plane at an incident angle            of 15.0o . Note that       is
   the angle between the incident electron beam and the normal of the surface plane.
   The plane containing             and the normal of the surface plane is the plane of
   incidence. For simplicity, we assume that all incident electrons are back scattered
   only by the surface atoms on the topmost layer.
(a) What is the wavelength of the matter waves of the incident electrons?
(b) If a detector is set up to detect electrons that do not leave the plane of incidence
    after being diffracted, at what angles with the normal of the surface will these
    diffracted electrons be observable?
(2) Assume that the thermal vibrational motions of the surface atoms are simple

                                                          1

APHO II   2001   Theoretical Question 3                                               p. 2 / 3

   harmonic. The amplitude of vibration increases as the temperature rises.
   Low-energy electron diffraction provides a way to measure the average
   amplitude of vibration. The intensity I of the diffracted beam is proportional to
   the number of scattered electrons per second. The relation between the intensity I
   and the displacement           (t) of the surface atoms is given by

                                         {                      }                      (1)

   In Eq.(1), I and I0 are the intensities at temperature T and absolute zero,
   respectively.           and         are wave vectors of incident electron and diffracted
   electron, respectively. The angle brackets < > is used to denote average over
   time. Note that the relation between the wave vector                and the momentum
   of a particle is        =2         / h, where h is the Planck constant.
      To measure vibration amplitudes of surface atoms of a metallic crystal, a
   collimated electron beam with kinetic energy of 64.0 eV is incident on a crystal
   surface at an incident angle of 15.0o. The detector is set up for measuring
   specularly reflected electrons. Only elastically scattered electrons are detected. A
   plot of ln ( I / I0 ) versus temperature T is shown in Fig. 3b.
      Assume the total energy of an atom vibrating in the direction of the surface
   normal        is given by kBT , where kB is the Boltzmann constant.
(a) Calculate the frequency of vibration in the direction of the surface normal for the
   surface atoms.
(b) Calculate the root-mean-square displacement, i. e. the value of ( < ux2> )1/2, in the
   direction of the surface normal for the surface atoms at 300 K.

                      01

                 -0.2
                   28

                 -0.4
                      88
                   0.6

                 -0.6
                   0.4

                 -0.8
                   0.2

                 -1.0
                           0     50     100 150 200 250 300 350 400                 T (K)

                                                      Fig. 3b
                                                      2221-2
                                                      2

APHO II   2001   Theoretical Question 3                  p. 3 / 3

The following data are given:
    Atomic weight of the metal M = 195.1
    Boltzmann constant kB = 1.38 10            J/K
    Mass of electron = 9.11 10-31 kg
    Charge of electron = 1.60 10-19 C
    Planck constant h = 6.63       10-34 J-s

                                                     3

Solution:
APHO II   2001    Theoretical Question 3                                      p. 1 / 4

[Solution]                          Theoretical Question 3

                      Thermal Vibration of Surface Atoms
(1) (a) The wavelength of the incident electron is

   (b) Consider the interference between the atomic rows on the surface as shown in
       Fig. 3c.
                                    e-           e-
                                                             1
                                                                         2

                                                      b          [011]

                                                                 Fig.3c

      The path difference between electron beam 1 and 2 is

     Given                 ,    =          and                      , two solutions

      are possible.

      (i) When n = 0,           o = 15.0                             (Answer 1)
      (ii) When n =1

                                                  1

 APHO II   2001   Theoretical Question 3                                        p. 2 / 4

 [Solution] (continued)              Theoretical Question 3

                            Vibration of Surface Atoms

                                                                          (Answer 2)

 For n = 2, no solution exists as                                   and        > 1.

(2)

                               Fig. 3d

                                                        surface

       For the specularly reflected beam, we have from Fig. 3d

       where      is the unit vector in the direction of the surface normal. Take the
       x-component of       , we then obtain

                                                                                (2)

       The vibration in the direction of the surface normal of the surface atoms is
       simple harmonic, take

       The total energy E is thus given by

                                               =              =

                                                   2

APHO II   2001   Theoretical Question 3                                         p. 3 / 4

[Solution] (continued)              Theoretical Question 3

                           Vibration of Surface Atoms
     Therefore, one obtains

                                                   = E/(        )

     where        is the mass of the atom. From either of the above two equations,
     one then has the following equality
                                                                                (3)

     From eq. (3) and eq. (2), one obtains

     where                       . Accordingly,

                                                                                (4)

     From the plot of            versus Tone obtains the slope

                                                                                (5)

     The slope of the curve can be estimated from Fig. 3b and leads to the result

                                                           .

     Using the following data in Eq.(5),

                       = 195.1            /(6.02     ) = 3.24       kg/atom

                                                    3

APHO II    2001   Theoretical Question 3                                           p. 4 / 4

[Solution] (continued)               Theoretical Question 3

                            Vibration of Surface Atoms
     one finds

     The solution for frequency is then

                                              (new)                           Answer (a)

     From                                          ,   one finally obtains

                                                                          (new)

     and

                                                              (new)

                                                                                Ans

                                                4

APhO 2001 Theory T3


[User] Kitsas elektronkiir läbib vaakumis tasaparalleelsete plaatide vahelise pilu ja langeb seejärel fluorestseeruvale ekraanile, mis asub plaatide ekraanipoolsemast servast kaugusel $l = \SI{15}{cm}$. Kui plaatidele antakse pinge $U = \SI{50}{V}$, nihkub helendav punkt ekraanil endisest asukohast kaugusele $s = \SI{21}{mm}$. Plaatidevaheline kaugus $d = \SI{18}{mm}$, plaatide mõõtmed elektronide liikumise suunas on $b = \SI{6}{cm}$. Milline on elektronide algkiirus plaatide vahele sattumisel? Elektroni laengu ja massi suhe $e/m \approx \SI{1,76e11}{C/kg}$.

Solution:
Olgu $v_0$ elektronide algkiirus plaatide vahele sattumisel. Aeg, mille jooksul üks elektron viibib plaatide vahel, on $t = \frac{b}{v_o}$.

Plaatide vahel on elektriväli tugevusega $E = \frac{U}{d}$. Newtoni II seadusest $eE = ma$ leiame, et elektron liigub plaatide vahel kiirendusega $a = \frac{eU}{md}$. Läbides plaatide vahelise tee, kalduvad elektronid vahemaa $s_0$ esialgsest trajektoorist kõrvale, kus
\[
s_{0}=\frac{a t^{2}}{2}=\frac{U e b^{2}}{2 d m v_{0}^{2}}.
\]
Elektronide liikumise kiirus $v$ plaatide vahelisest ruumist väljudes koosneb kahest komponendist:\\
-- paralleelsest ekraaniga $v_{y}=a t=\frac{e U b}{m d v_{0}}$,\\
-- risti ekraaniga $v_x = v_0$.\\
Seega veedavad elektronid plaatidest ekraanini aja $t' = \frac{l}{v_x} = \frac{l}{v_0}$. Selle aja jooksul lisandub täiendav ekraaniga paralleelne nihe
\[
s^{\prime}=v_{y} t^{\prime}=\frac{e U b l}{m d v_{0}^{2}}.
\]
Kogu nihe on niisiis
\[
s=s_{0}+s^{\prime}=\frac{e U b}{m d v_{0}^{2}}\left(\frac{b}{2}+l\right).
\]
Seega
\[
v_{0}=\sqrt{\frac{e}{m} \frac{U b}{d s}\left(\frac{b}{2}+l\right)} \approx \SI{1,58e7}{m/s}.
\]
Sooritame kontrolli, kas elektroni nihe plaatide vahel on väiksem kui plaatide vaheline kaugus:
\[
s_{0}=\frac{U e b^{2}}{2 d m v_{0}^{2}}=\frac{bs}{2l + b} = \SI{3.5}{mm} < d.
\]

Elektronkiir

Kitsas elektronkiir läbib vaakumis tasaparalleelsete plaatide vahelise pilu ja langeb seejärel fluorestseeruvale ekraanile, mis asub plaatide ekraanipoolsemast servast kaugusel $l = \SI{15}{cm}$. Kui plaatidele antakse pinge $U = \SI{50}{V}$, nihkub helendav punkt ekraanil endisest asukohast kaugusele $s = \SI{21}{mm}$. Plaatidevaheline kaugus $d = \SI{18}{mm}$, plaatide mõõtmed elektronide liikumise suunas on $b = \SI{6}{cm}$. Milline on elektronide algkiirus plaatide vahele sattumisel? Elektroni laengu ja massi suhe $e/m \approx \SI{1,76e11}{C/kg}$.

Selleks, et siduda elektronide kiirust ekraani-sihilise nihkega, tasub ülesanne jagada kaheks eraldi osaks: elektronide viibimine plaatide vahel ning plaatide ja ekraani vahelises ruumis. Esimeses osas mõjub elektronile ühtlane kiirendus, teises osas liigub elektron sirgjooneliselt.


[User]
Let us observe the following simplified model of the Earth’s atmosphere where the atmosphere layer surrounding the Earth a) reflects $\mu=\SI{30}{\%}$ of the Sun’s radiation falling on it back to the cosmos and lets the rest of the radiation through without absorbing any of it; b) absorbs completely all the infrared radiation coming from the Earth’s surface. The irradiance coming from the Sun is $w_0=\SI{1400}{\frac{W}{m^2}}$. Find the average temperature of the Earth’s surface.\\
\emph{Hint.} The Stefan-Boltzmann law applies – the power radiated by a black body per unit of area is given as $w=\sigma T^4$ where $\sigma =5,67 \cdot 10^{-8} \SI{}{\frac{W}{m^2 \cdot K^4}}$. Assume that the Earth’s surface only radiates infrared radiation and that it can be looked at as an absolutely black body due to this. The surface also absorbs all of the Sun light reaching it.

Solution:
Päikselt jõuab maale koguvõimsus $P_p=w_0 \left(1-\mu\right)\pi R^2$, kus $R$ on maa raadius. Kuna atmosfäär on maalt tuleva kiirguse jaoks läbimatu, peab tasakaalu korral atmosfäär väljapoole kiirgama selle sama võimsuse: $P_p=P_a$, kus $P_a$ on atmosfääri poolt väljapoole kiiratav võimsus. Maalt kiiratav võimsus avaldub kui $P_m=4 \pi R^2 \sigma T_m^4$, kus $T_m$ on maapinna temperatuur. Tasakaalu korral on see võrdne Päikeselt ja atmosfäärist tagasi kiirgunud võimsuste summaga:
\[P_m=P_p+P_a=2P_p.\]

Maakera temperatuur avaldub kui:

\[T_m=\sqrt[4]{\frac{w_0\left(1-\mu\right)}{2\sigma}}=\SI{303}{K}.\]

Kasvuhooneefekt

Vaatleme järgnevat Maa atmosfääri lihtsustatud mudelit, kus Maad ümbritsev atmosfäärikiht a) peegeldab kosmosesse tagasi $\mu=\SI{30}{\%}$ pealelangevast päikesekiirgusest ning ülejäänu laseb läbi ilma kiirgust neelamata; b) neelab täielikult kogu maapinnalt tuleva infrapunakiirguse. Päikeselt tulev kiiritustihedus on $w_0=\SI{1400}{\frac{W}{m^2}}$. Leidke maapinna keskmine temperatuur.

\textit{Vihje.} kehtib Stefan-Boltzmanni seadus -- musta keha poolt kiiratav võimsus pindalaühiku kohta avaldub kui $w=\sigma T^4$, kus $\sigma=\SI{5.67e-8}{\frac{W}{m^2 \cdot K^4}}$. Eeldada, et maapind kiirgab ainult infrapunakiirgust, ning et teda saab selle jaoks lugeda absoluutselt mustaks kehaks. Samuti neelab maapind kogu temani jõudva päikesevalguse.

Lisaks läbi atmosfääri tulevale päikesekiirgusele tuleb arvestada atmosfääri poolt kiiratava võimsusega. Juhul, kui Maa pinnalt jõuab atmosfääri kiirgus võimsusega $P_m$, siis atmosfäär kiirgab Maast välja ja Maale sisse kiirgust võimsusega $\frac{P_m}{2}$.


[User]
1. CURLıNG (8 points)

— Oskar Vallhagen.
In the sport of curling, participants take
turns sliding near‐cylindrical stones across
an ice court towards a target, trying to get
their stones as close to the target as possible
after using a set of stones. A vertical cross
section of a stone is depicted below, showing
that the stone is in contact with the ice on a
thin ring of radius r. The full radius of the
stone is R, the mass of the stone is m and the
coefficient of friction with the ice is μ.
Consider the case when the stone is re‐
leased at a speed v0 with the aim of knocking
out an opponent’s stone at a distance s.
i) (1 point)Give an expression for the sliding
speed vs as a function of the time t since the
stonewasreleaseduntilthestonehitstheop‐
ponent’s stone.
ii) (1 point)What is the sliding speed vhit just
before the stone hits the opponent’s stone?
Now the stone is given a small rotation at
the initial angular speed ω0 (this can be done
toalterthedeflectionangleofthestonewhen
it hits the opponent’s stone). Assume that
the rotation speed ω remains small through‐
out the sliding motion: ωr ≪ vs. Keep only
the main non‐vanishing terms in your calcu‐
lations, i.e. among the terms with a factor
(ωr/vs)n
, keep only the term with the smal‐
lestn. Dependingonyourapproachyoumay
use the following approximations for x ≪ 1:
(1+x)α
≈ 1+αx+ 1
2α(α−1)x2
, sin(α + x) ≈
sinα+xcosα, cosx ≈ 1−x2
/2. Youmayalso
need the integral
∫
(at+b)−1
dt = a−1
ln|at+
b| + C.
iii) (2 points) By how much does the friction
force on the stone change due to its rotation?
Express your answer in terms of the current
angular speed ω and sliding speed vs.
iv)(2points)Giveanexpressionforthetorque
T exerted on the stone.
v) (2 points)What is the angular speed of the
stone just before it hits the opponent’s stone?

Solution:
1. CURLıNG

— Solution by Jaan
Kalda, Oskar Vallhagen, grading schemes by Os
kar Vallhagen.
i) (1 point) As the stone is gliding over the
ice everywhere on the contact surface, the
friction force is Ffx = μmg, i.e. the ac‐
celeration is μg, directed in the negative x‐
direction (with the x‐direction pointing to‐
wards the target stone). Thus, we get
dvs
dt
= −μg ⇒ vs = v0 − μgt.
Grading:
• Correct friction force (0.5pts)
• Correct acceleration (0.2pts)
• correct vs (0.3pts)
ii) (1 point) The simplest way to obtain the
final velocity is to use energy conservation,
noting that the friction force does a work
Wf = μmgs. Thus, we get
mv2
0
2
=
mv2
hit
2
+ μmgs ⇒ vhit =
√
v2
0 − 2μgs.
Grading:
• Idea of using energy conservation (0.2pts)
• Calculating the work done by the friction
force (0.3pts)
• Correct total energy conservation equation
vs (0.3pts)
• Correct vhit (0.2pts)
Alternatively, one can first calculate the
time thit before the stone hits the opponents
stone. Integrating the expression for vs up to
this time gives
s = v0thit − μg
t2
hit
2
.
Solving for thit, we get
thit =
v0
μg
−
√
v2
0
μ2g2
−
2s
μg
(the solution with the plus sign gives a negat‐
ive velocity). Inserting back into the expres‐
sion for vs and simplifying gives
vhit =
√
v2
0 − 2μgs.
Grading:
• Integration of the expression for vs
(0.4pts)
• Solving for thit (0.4pts)
• Inserting into expression for vs and final
answer (0.2pts)
iii) (2 points) Consider the motion of a point
at the ring in contact with the ice at an angle
θfromthex‐axis,movingatavelocityv asde‐
picted in the figure below. The components
of this motion along the x and y axes can be
seentobevx = vs−ωrsinθandvy = ωrcosθ,
respectively.
x
y
R
θ
φ
θ − φ
r
vs
ωr v
vs
The friction force per unit length has the
magnitude
F′
f =
μmg
2πr
everywhere on the circle in contact with the
ice, asbefore, butisnowdirectedoppositeto
the local velocity v rather than the sliding ve‐
locityvs. Thecomponentoftheforcedensity
in the x‐direction thus becomes
F′
fx = −F′
f cosφ = −F′
f
vx
v
.
From this point on, there are two ways
to work out the mathematical calculations.
First, consider an infinitesimal arc of the
contact ring of central angle dθ; the friction
force exerted on it has magnitude dFf =
μmg
2π dθ. The total force is obviously directed
antiparallel to the velocity of the stone as y‐
directional components of the acting on arcs
at θ and π − θ cancel pairwise out. The pro‐
jection of the friction force acting on our arc
is given by
dFx =
μmg
2π
cosφdθ
Using the small parameter δ ≡ ωr/vs ≈
ωr/v ≪ 1, we conclude that φ ≪ 1, hence
we can use approximation
dFx ≈
μmg
2π
(
1 −
1
2
φ2
)
dθ.
The angle φ can be found from the sine
theorem assuming sinφ ≈ φ: φ ≈
sin
(π
2 − θ
)ωr
v ≈ cosθωr
vs
≡ δ cosθ. Alternat‐
ively, a similarly accurate expression cab be
found by noting that
φ ≈ tanφ =
vy
vx
≈
vy
vs
=
ωrcosθ
vs
.
Thus,
dFx =
μmg
2π
(
1 −
δ2
2
cos2
θ
)
dθ.
Recognising that
∫2π
0
cos2
θdθ = π, this can
be integrated to obtain
Fx = μmg
(
1 −
ω2
r2
4v2
s
)
so that the final answer is
∆Fx = −
μmgω2
r2
4v2
s
.
The second way is by continuing
F′
fx = F′
f
vx
v
= F′
f
vs − ωrsinθ
√
v2
s − 2vsωrsinθ + ω2r2
= F′
f
1 − δ sinθ
√
1 − 2δ sinθ + δ2
.
To second order in δ, we get, using the ap‐
proximation suggested in the problem text
for x = −2δ sinθ + δ2
and doing some ele‐
mentary algebra,
F′
fx ≈ F′
f(1 −
1
2
δ2
cos2
θ).
The total friction force can now be obtained
byintegratingoverthelengthelementofsize
rdθ along the ring in contact with the ice,
yielding, after simplifying and re‐inserting
the expression for F′
f,
Ffx =
∫ 2π
0
F′
fxrdθ = μmg
(
1 −
1
4
δ2
)
.
Grading:
• Understanding that the friction force is
everywhere antiparallel with the local ve‐
locity (0.4pts)
• Calculating the magnitude of the force per
unit length (0.3pts)
• Correct expression for the x component
(0.2pts)
First solution:
• ApproximateexpressionforF′
fx ordFfx for
small φ (0.3pts)
• Expressing φ in terms of δ and θ (0.5pts)
• Correct integral and final answer (0.3pts)
Second solution:
• Expressing F′
fx or dFfx in terms of δ and θ
(0.3pts)
• ApproximateexpressionforF′
fx ordFfx for
small δ (0.5pts)
• Correct integral and final answer (0.3pts)
iv) (2 points)For the torque τ′
per unit length
(around the central axis of the stone), we see
from the figure that the angle between the
local force per unit length and the radius is
θ − φ, yielding
τ′
= F′
frsin(θ − φ).
From here on, there are again two ways to
continue. First, we can use Taylor series
and keep only the linear term: sin(θ − φ) ≈
sinθ − φcosθ; by using the previously ob‐
tainedexpressionφ ≈ δ cosθ, wecanexpress
dτ = μmgr(sinθ − δ cos2
θ)
dθ
2π
integration of which yields
τ = −
μmgωr2
2vs
.
The second way is to write
τ′
= F′
frsin(θ − φ)
= F′
fr(sinθcosφ − cosθsinφ)
= F′
fr(sinθ
vx
v
− cosθ
vy
v
),
which, using the above expressions for vx
and vy, can be simplified to
τ′
= F′
fr
sinθ − δ
√
1 − 2δ sinθ + δ2
.
Applying the same approximation as in the
previous part and keeping only the first, now
leading, order terms in δ, we get
τ′
≈ F′
fr(sinθ − δ cos2
θ).
Integrating to find the total torque, we get
τ =
∫ 2π
0
τ′
rdθ = −
1
2
δμmgr.
Grading:
• Finding the angle between the local force
per unit length and the radius (0.5pts)
• Correct expression for τ′
(0.4pts)
First solution:
• Approximate expression for τ′
or dτ for
small φ (0.3pts)
• Expressing φ in terms of δ and θ (0.5pts)
• Correct integral and final answer (0.3pts)
Second solution:
• Expressing τ′
or dτ in terms of δ and θ
(0.3pts)
• Approximate expression for τ′
or dτ for
small δ (0.5pts)
• Correct integral and final answer (0.3pts)
v) (2 points) The equation describing the ro‐
tation is, now expressed in terms of ω and vs
rather than δ,
I
dω
dt
= τ = −
1
2
μmgr2 ω
vs
,
wherethemomentofinertiaisI ≈ 1
2mR2
.As
the correction to Ffx is only of second order
in δ, to the first, leading order approxima‐
tion, we may neglect the effect of rotation on
Ffx. Thus, the sliding speed vs has the same
time dependenca as in part i). The equation
above can then be separated as
∫ ωhit
ω0
dω
ω
=
∫ thit
0
−μg
r2
R2
dt
v0 − μgt
⇒
ln
ωhit
ω0
=
r2
R2
ln
v0 − μgthit
v0
,
which, recognising that v0 − μgthit = vhit,
yields
ωhit = ω0
(
vhit
v0
) r2
R2
.
Grading:
• Correct equation of motion for the rotation
(0.5pts)
• Realising that vs is unaffected by the rota‐
tion to leading order (0.5pts)
• Using separation of variabels (0.5pts)
• Final answer (0.5pts)

CURLıNG


[User]
Question A1
An object of mass m is sitting at the northernmost edge of a stationary merry-go-round of radius
R. The merry-go-round begins rotating clockwise (as seen from above) with constant angular
acceleration of α. The coefficient of static friction between the object and the merry-go-round is
µs .

  a. Derive an expression for the magnitude of the object’s velocity at the instant when it slides
     off the merry-go-round in terms of µs , R, α, and any necessary fundamental constants.

  b. For this problem assume that µs = 0.5, α = 0.2 rad/s2 , and R = 4 m. At what angle, as
     measured clockwise from north, is the direction of the object’s velocity at the instant when
     it slides off the merry-go-round? Report your answer to the nearest degree in the range 0 to
     360◦ .

Solution:
Question A1
An object of mass m is sitting at the northernmost edge of a stationary merry-go-round of radius
R. The merry-go-round begins rotating clockwise (as seen from above) with constant angular
acceleration of α. The coefficient of static friction between the object and the merry-go-round is
µs .

  a. Derive an expression for the magnitude of the object’s velocity at the instant when it slides
     off the merry-go-round in terms of µs , R, α, and any necessary fundamental constants.

  b. For this problem assume that µs = 0.5, α = 0.2 rad/s2 , and R = 4 m. At what angle, as
     measured clockwise from north, is the direction of the object’s velocity at the instant when
     it slides off the merry-go-round? Report your answer to the nearest degree in the range 0 to
     360◦ .

                                             Solution
  a. The maximum possible acceleration provided by friction is µs g. The object will begin to slide
     once the acceleration exceeds this. We have
                                    q
                                a = a2r + a2t , at = αR, ar = ω 2 R

     where at is the tangential acceleration and ar is the centripetal acceleration. Plugging in,

                                          α2 R2 + ω 4 R2 = µ2s g 2

     and using ω = αt and solving for t, we find
                                             s r
                                               1 µ2s g 2
                                         t=              − 1.
                                               α α2 R 2

     When the object slides off, it has a tangential velocity v = Rω, so
                                                q p
                                     v = Rαt = R µ2s g 2 − α2 R2 .

  b. The angular position of the object is θ = αt2 /2, so
                                              1 p 2 2
                                        θ=       µs g − α 2 R 2 .
                                             2αR
     Plugging in the numbers, we find
                               1
                                      (0.5)2 (9.8)2 − (0.2)2 (4)2 = 3.021 rad = 173◦ .
                                     p
                      θ=
                           2(0.2)(4)

     The velocity of the object is rotated 90◦ relative to the position, so the answer is 263◦ .

                  Copyright c 2010 American Association of Physics Teachers

2010 Semifinal Exam                              Part A                                              4

USAPhO 2010 Question A1


[User]
This is a rewrite of EFPhO 2007 problem 3, which has some typos and ambiguities.
    Suppose particles of mass m, charge e > 0, and kinetic energy eU0 are produced at point A, all
    traveling to the right. The particles are not produced at exactly the same time, but we would like
    them to arrive at point D at the same time. This is known as temporal focusing.

    To do this, we place a pair of parallel plates along the path, with width a. The plates have the
    same time-dependent voltage U (t), while the voltage outside the plates is held at zero. Thus, the
    electric field is only nonzero near point B, where the particles enter the plates, and point C, where
    the particles exit the plates. The particles then travel a distance b ≫ a to point D.

      (a) Suppose the first particle reaches point B at time t = 0, and that U (0) = U ′ (0) = 0. Find the
          U (t) such that all the next particles reach point D at the same time. Assume that |U (t)| ≪ U0 .

      (b) The voltage cannot become arbitrarily high, so every time T it resets to zero and the process
          begins again. As a result, particles are periodically focused into clumps. On average, what
          fraction of the particles do not make it into a clump? Assume that T is much larger than the
          time it takes a particle to cross the plates.

Solution:
The official solutions are here, but the answer for part 3 has a factor of 2 error, and
there are other places where it differs. Stefan Ivanov’s solution here is better. Here’s a solution
adapted for my version of the problem.

 (a) Note that the particle does not accelerate when it is between the plates, even when U (t)
     changes, since the electric field vanishes there. The situation is analogous to a ball rolling on
     a flat table while the entire table is being lifted up.
      Now consider the particle that enters the plates at time t and exits at time t′ . It loses a kinetic
      energy eU (t) when it enters, then gains a kinetic energy eU (t′ ) when it exits. Therefore, if U (t)
      is time-dependent, the particle can have a net change in speed, which allows later particles to
      move faster to D.
     To make this concrete, let ta = a/v0 and tb = b/v0 , where eU0 = mv02 /2. The approximations
     of the problem allow us to neglect the particles’ change in speed while between the plates,
     since it’s penalized by factors of both a/b and U (t)/U0 . Then a particle that enters the plates
     at time t exits at time t′ ≈ t + ta . The extra energy imparted must shorten the time it takes
     to go from C to D by an amount t + t0 , where t0 is an arbitrary constant. Then
                                            ∆v   tb ∆K   tb U (t + ta ) − U (t)
                              t + t0 ≈ tb      ≈       ≈                        .
                                             v   2 eU0   2          U0
      In other words, the finite difference of U (t) is a linear function of t, which means that U (t) is
      a quadratic polynomial. The given conditions U (0) = U ′ (0) = 0 fix U (t) ∝ t2 , and matching
      the coefficients of t on both sides gives
                                                       U0 2 2eU02 2
                                            U (t) =         t =     t .
                                                      ta tb     abm

 (b) When the voltage resets to zero, all the particles that were between the plates won’t get
     focused correctly. So the fraction that don’t get focused is approximately
                                                    r
                                             ta   a     m
                                                =           .
                                             T    T 2eU0

          Note that for this solution to make sense, we need ta ≪ T , but we also need T to be short
                                                                √
          enough so that |U (t)| ≪ U0 , which corresponds to T ≪ ta tb . Both conditions can be satisfied
          simultaneously, since a ≪ b.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User] Juku has an iron ball ($\varrho_\mathrm{Fe}=\SI{7,9}{\gram\per\centi\meter\cubed}$) of radius $r=\SI{10}{\centi\meter}$ and mass $m=\SI{30}{\kilo\gram}$. Juku knows that inside the ball there is a spherical hollowness and he wants to find the distance $d$ from the center of the ball to the center of the hollowness. To do that he hanged the ball to a thread two times, using two opposite points on the ball as the hanging points. At the first time, the axis that connected the two hanging points formed an angle $\alpha=\ang{60}$ with respect to the horizontal axis, the second time that angle was $\beta=\ang{45}$. Find $d$.

Solution:
\begin{center}
\includegraphics[width=\textwidth]{2013-v2g-09-kera.png}
\end{center}

Kui tegemist poleks õõnsa keraga, oleks selle mass $M=\frac{4}{3}\pi r^3 = \SI{33}{kg}$. Mudeldame õõnsusega kera kui tervet kera, mille sees on negatiivse massiga $m'=\SI{3}{kg}$ kera. Kera pöörab ennast nii, et tema masskese jääks kinnituspunkti alla. Olge õõnsuse keskpunkti (Õ) asukoha koordinaadid kera keskpunkti suhtes $x$ ja $y$. Nende leidmiseks kasutame jõumomentide tasakaalu nööri kinnituspunkti suhtes. Saame võrrandisüsteemi:
\[Mr \cos \alpha = m'(r-y+x \tan \alpha) \cos \alpha,\]
\[Mr \cos \beta = m'(r+y+x \tan \beta) \cos \beta.\]
Õõnsuse ja kera keskpunktide vaheline kaugus on $d=\sqrt{x^2+y^2}$,
\[ d=\frac{2r(\frac{M}{m'}-1)}{\tan\alpha+\tan\beta}\sqrt{1+\frac{1}{4}(\tan\alpha-\tan\beta)^2} = \SI{0,8}{cm}. \]

Õõnes kera

Jukul on rauast kera
($\varrho_\mathrm{Fe}=\SI{7,9}{\gram\per\centi\meter\cubed}$) raadiusega
$r=\SI{10}{\centi\meter}$ ja massiga $m=\SI{30}{\kilo\gram}$. Juku teab, et kera
sees on sfääriline õõnsus, mille keskpunkti kaugust $d$ kera keskpunktist ta üritab
leida. Selleks riputas ta kuuli kaks korda nööri otsa rippuma, kasutades
riputuskohtadeks kera vastaspunkte. Ühel korral moodustas neid kinnituspunkte
ühendav telg horisondiga nurga $\alpha=\ang{60}$, teisel korral aga nurga
$\beta=\ang{45}$. Leidke $d$.

Õõnsusega kera saab vaadelda positiivse tihedusega täidetud kera ja õõnsuse suuruse negatiivse tihedusega kera superpositsioonina. Sellisel juhul saab mugavalt kirja panna jõumomentide tasakaalu mõlema stsenaariumi jaoks.


[User] Kas basseini kohal rippuv punktvalgusallikas, mida vaadeldakse basseini põhjast, on heledam siis, kui bassein on veest tühi, või siis, kui ta on veega täidetud ja kaugus silmast veepinnani võrdub valgusallika kõrgusega veepinna kohal? Mitu korda? Veepinnalt peegeldub tagasi $r = \SI{2}{\%}$ valgust, vee murdumisnäitaja on $n = \num{1,33}$ ja neeldumine vees on tühine. Allika heledus on võrdeline silmaavasse sattuva valguse energiaga, silmaava läbimõõdu loeme samaks kõigis vaatlustingimustes ja väikeseks võrreldes vaatleja sügavusega.

Solution:
\begin{wrapfigure}{r}{0.5\textwidth}
	\begin{center}
		\includegraphics[width=0.95\linewidth]{2008-v3g-05-lah}
	\end{center}
\end{wrapfigure}
Allika heledus on ligikaudselt võrdeline vaatleja silmaava nurkläbimõõdu ruuduga allika asukohast vaadatuna. Olgu $\alpha$ ja $\alpha_0$ nurkläbimõõdud vastavalt veega täidetud ja veeta basseinis.

Jooniselt
\[
2L\tan (\alpha_0 /2) = L\tan (\alpha /2) + L\tan (\beta /2),
\]
kus langemis- ja murdumisnurkade $\alpha /2$ ja $\beta /2$ vahel kehtib seos
\[
\frac{\sin(\alpha/2)}{\sin(\beta/2)} = n.
\]
Kasutades väikeste nurkade lähendust $\tan\alpha\approx\sin\alpha\approx\alpha$, saame
\[
\left(\frac{\alpha}{\alpha_0}\right)^2 = \frac{4n^2}{(n+1)^2}.
\]
Arvestades ka peegeldumist veepinnalt, saame heleduste suhteks
\[
\frac{I}{I_{0}}=\left(\frac{\alpha}{\alpha_0}\right)^2(1-r) = \frac{4(1-r) n^{2}}{(n+1)^{2}}=\num{1,28}.
\]

Veealune valgus

Kas basseini kohal rippuv punktvalgusallikas, mida vaadeldakse basseini põhjast, on heledam siis, kui bassein on veest tühi, või siis, kui ta on veega täidetud ja kaugus silmast veepinnani võrdub valgusallika kõrgusega veepinna kohal? Mitu korda? Veepinnalt peegeldub tagasi $r = \SI{2}{\%}$ valgust, vee murdumisnäitaja on $n = \num{1,33}$ ja neeldumine vees on tühine. Allika heledus on võrdeline silmaavasse sattuva valguse energiaga, silmaava läbimõõdu loeme samaks kõigis vaatlustingimustes ja väikeseks võrreldes vaatleja sügavusega.


[User]
Pictured below is a so-called vacuum cannon. To load it, a ball meant for ammunition is put into the left end of the vacuum cannon. Next, both ends of the tube are covered with an easily rupturing airproof membrane, for example foil, and then the air is pumped out of the tube. The vacuum cannon is now ready for firing. For firing, the membrane on the left is intentionally ruptured. As a result the ball starts to charge towards the right end of the tube. If the tube is long enough the ball ruptures the membrane on the right end and is fired out of the tube. Let us say that the diameter of the ball is equal to the inner diameter of the tube $d=\SI{4,0}{cm}$, the ball’s mass is $m=\SI{24}{g}$ and before the firing the ball is positioned at a distance $l=\SI{150}{cm}$ from the right end of the tube. What is the speed of the ball directly before rupturing the membrane on the right? Air pressure is $P_0=\SI{100}{kPa}$. Do not account for friction.
\begin{center}
  \includegraphics[width=0.75\textwidth]{2014-lahg-02-vaakumkahur_ing}
\end{center}

Solution:
Pall jaotab kahuritoru kaheks kambriks. Enne tulistamist on mõlemas kambris rõhk võrdne nulliga. Vasakpoolse membraani purustamisel täitub vasak pool torust välisõhuga ning palli poolte vahel tekib rõhkude vahe $\Delta P=P_0-0=P_0$. Pallile hakkab mõjuma jõud, mis on võrdne palli ristlõikepindala ja rõhkude vahe korrutisega: $F=P_0\pi d^2/4$. Newtoni 2. seaduse abil saame leida palli kiirenduse $a=F/m$. Ühtlaselt kiireneva liikumise korral kehtib seos
\[ l=\frac{v^2-v_0^2}{2a}, \]
kus $l$ on läbitud vahemaa ning $v_0$ ja $v$ on vastavalt alg- ja lõppkiirus. Kuna palli algkiirus on võrdne nulliga ning tahame leida lõppkiirust, siis avaldame
\[ v=\sqrt{2la}=\sqrt{\frac{2lP_0\pi d^2}{4m}}=d\sqrt{\frac{lP_0\pi}{2m}}.\]
Kasutades ülesandes antud arvväärtusi, saame palli kiiruseks vahetult enne parempoolse membraani läbimist $v=\SI{130}{m/s}$. Membraani purustamiseks kulub energiat ning seetõttu on palli väljumiskiirus sellest veidi väiksem, kuid võite isegi ette kujutada, et õhuke fooliumileht ei takista \SI{470}{km/h} kihutava palli lendu just märkimisväärselt.

Vaakumkahur

Joonisel on kujutatud niinimetatud vaakumkahur. Laadimiseks pistetakse laskemoonaks olev pall vaakumkahuri toru vasakpoolsest otsast sisse. Seejärel kaetakse toru mõlemad otsad kergestirebeneva õhukindla membraaniga, näiteks fooliumiga, ning pumbatakse torust õhk välja. Nüüd on vaakumkahur laskevalmis. Tulistamiseks purustatakse vasakpoolne membraan, mille tagajärjel hakkab pall toru parempoolse otsa poole sööstma. Piisavalt pika toru korral purustab pall parempoolse membraani ning lendab torust välja. Olgu palli läbimõõt võrdne toru siseläbimõõduga $d=\SI{4,0}{cm}$, palli mass $m=\SI{24}{g}$ ja asugu pall enne tulistamist $l=\SI{150}{cm}$ kaugusel toru parempoolsest otsast. Kui suur on palli kiirus vahetult enne parempoolse membraani läbimist? Õhurõhk on $P_0=\SI{100}{kPa}$. Hõõrdumisega pole tarvis arvestada.
\begin{center}
 \includegraphics[width=0.75\textwidth]{2014-lahg-02-vaakumkahur.pdf}
\end{center}

Pall jaotab kahuritoru kaheks kambriks. Pärast vasakpoolse membraani purustamist hakkab vasakpoolses kambris olev gaas palli paremale poole suruma.


[User] The mass of a basketball meeting the standards of NBA is $m=\SI{600}{g}$, circumference $C=\SI{76}{cm}$ and gauge pressure in the ball $p_1=\SI{55}{kPa}$. How deep should the basketball be pushed into water so that it would start to sink down by itself? The density of water is $\rho=\SI{1000}{kg/m^3}$, gravitational acceleration $g=\SI{9,8}{m/s^2}$ and the air pressure on the water's surface is $p_0=\SI{100}{kPa}$. You can assume that during the sinking the air temperature in the ball does not change and that the mass of the ball’s shell is negligible.

Solution:
Sügavale vee alla sukeldamisel surub veesamba rõhk palli kokku. Pall hakkab ise põhja vajuma, kui tema ruumala väheneb nii palju, et pallile mõjuv üleslükkejõud saab väiksemaks raskusjõust ehk palli keskmine tihedus muutub väiksemaks vee tihedusest. Kriitilisel sügavusel $\rho=m/V$, kus $V$ on kokkusurutud korvpalli ruumala. Lõpptulemusena otsime vedelikusamba kõrgust, mis tekitab piisava rõhu, et palli ruumala väheneks väärtuseni $V=m/\rho$. Seepärast leiame esmalt pallis oleva õhu rõhu ja ruumala vahelise seose. Sukeldamata pallis on rõhk võrdne õhurõhu ja palli kestas tekkivast elastsusjõust tingitud rõhu summaga: $p_0+p_1$. Kriitilisel sügavusel moodustab palli ruumala alla 10 \% oma esialgsest väärtusest, mis tähendab, et pall on lömmi vajutatud ja elastsusjõud enam rolli ei mängi. Rõhk pallis on siis võrdne vedelikusamba rõhu ja õhurõhu summaga: $p_v+p_0$. Boyle'i-Mariotte'i seadusest on teada, et konstantse temperatuuri juures on ideaalse gaasi rõhk ja ruumala pöördvõrdelises sõltuvuses ehk
\[
\frac{p_0+p_1}{p_0+p_v}=\frac{V_0}{V}.
\]
Sellest valemist saame avaldada vedelikusamba kriitilise rõhu:
\[
p_v = (p_0+p_1) \frac{V_0}{V} - p_0.
\]
Teades, et vedelikusammas kõrgusega $h$ tekitab rõhu $p_v=\rho g h$, leiame samaväärse avaldise kriitilise sügavuse jaoks:
\[
h = \frac{(p_0+p_1) \frac{V_0}{V} - p_0}{\rho g}.
\]
Nüüd tuleb veel palli alg- ja lõppruumala avaldada ülesandes etteantud suuruste kaudu. Lõppruumala on juba lahenduse alguses leitud ($V=m/\rho$). Algruumala jaoks kasutame ringi ümbermõõdu ja kera ruumala valemeid
\[
C = 2\pi r \; \text{ja} \; V_0 = \frac{4}{3}\pi r^3,
\]
millest saame, et $V_0=C^3/6\pi^2$ ning lõpuks
\[
h = \frac{(p_0+p_1) \frac{C^3 \rho}{6\pi^2 m} - p_0}{\rho g}.
\]
Kriitilise sügavuse arvuline väärtus on $h=190$ m.

Korvpall

NBA standarditele vastava korvpalli mass on $m=\SI{600}{g}$, ümbermõõt
$C=\SI{76}{cm}$ ning ülerõhk palli sees $p_1=\SI{55}{kPa}$. Kui sügavale vee
alla tuleks korvpall suruda, et see iseenesest põhja hakkaks vajuma? Vee tihedus
$\rho=\SI{1000}{kg/m^3}$, raskuskiirendus $g=\SI{9,8}{m/s^2}$ ja õhurõhk
veepinnal $p_0=\SI{100}{kPa}$. Võib eeldada, et sukeldamise jooksul palli sees
õhutemperatuur ei muutu ja palli kesta ruumala on tühine.


[User] Find the Taylor series for 1/ cos x up to and including the fourth order (O(x4 )) term.

Solution:
The derivatives of cos(x) at x = 0 are 0, −1, 0, 1, so

                                                     x2 x4
                                       cos x = 1 −     +    + O(x6 ).
                                                     2   4!
    To expand the inverse, note that

                                              = 1 + u + u2 + O(u3 )
                                          1−u

    where in our case, u = x2 /2 − x4 /24. Plugging this in gives

                    1                                                             x2 5x4
                        = 1 + (x2 /2 − x4 /24) + (x2 /2 − x4 /24)2 + O(x6 ) = 1 +   +    .
                  cos x                                                           2   24


[User]
Devise and perform an experiment to find the coefficients of static friction between (1) paper and
paper, and (2) paper and graphite mixture, the residue left behind when one draws on paper with
a pencil, as accurately as possible. You may use a ruler, a movable hard surface, a mass, and the
pencil and paper you brought to this exam.
    To receive credit, you should describe the method you used and justify why you have chosen it.
You do not have to do any graphing for this problem. State your results for the two coefficients of
friction along with uncertainties. (Most of the credit for this question is for producing a sensible
experimental method, not your final numeric results.)

Part B

Solution:
Devise and perform an experiment to find the coefficients of static friction between (1) paper and
paper, and (2) paper and graphite mixture, the residue left behind when one draws on paper with
a pencil, as accurately as possible. You may use a ruler, a movable hard surface, a mass, and the
pencil and paper you brought to this exam.
    To receive credit, you should describe the method you used and justify why you have chosen it.
You do not have to do any graphing for this problem. State your results for the two coefficients of
friction along with uncertainties. (Most of the credit for this question is for producing a sensible
experimental method, not your final numeric results.)

Solution. This is AuPhO 2013, problem 14, whose official solution is here. I bet you weren’t
expecting an experimental physics problem! This is why I like the AuPhO, despite its occasional
typos and ambiguities.
    As you might expect, there are many possible solutions. The most crucial point is that you
shouldn’t use any method that depends on the weight of the paper itself (e.g. just putting a scrap
of paper on another piece of paper and tilting), because it’s far too small to get reliable results.
The normal and friction forces will be overwhelmed by other, irrelevant forces, like air resistance or
even electrostatic forces.
    One possible method is to fold a little box out of paper and put other stuff in it to weigh it down.
Then, for the first part, put the box on another sheet of paper on a surface, tilt the surface until
sliding begins, measure the angle, and use µ = tan θ. For the second part, shade in the bottom of
the box with your pencil and repeat.
    Another possible method is to directly apply an extra vertical force by pushing downward with
your pencil, taking care to keep your pencil as vertical as possible. For the first part, you push
the pencil on a scrap of paper which is on top of another sheet of paper on a tilted surface. For
the second part, you can directly press the pencil onto the sheet of paper, and see when the pencil
slips. For an effective measurement, you want to avoid the problem with normal forces at corners
mentioned in M2. That is, since a sharp tip might deform the paper, you should make sure the tip
of your pencil is blunt.

Part B


[User]
Laetud osake laengu ja massi suhtega $q/m = \SI{1}{C/kg}$ seisab algselt paigal. Seejärel hakkab ta liikuma $x$- ja $y$-telje sihis
toimivate elektrivälja impulsside mõjul. Elektrivälja vastavate komponentide $E_x$ ja $E_y$ sõltuvus ajast on toodud graafikul (graafiku mastaap
ei ole korrektne, juhenduda tuleb graafikul näidatud numbritest impulsi
kestvuse $\tau = \SI{1}{ms}$ ja amplituudi $E_0 = \SI{1}{kV/m}$ ning perioodi $T = \SI{2}{s}$
jaoks). Visandage osakese trajektoor ja leidke keskmine kiirus (visandi tegemisel ja arvutustes võib lugeda ajavahemiku $\tau = \SI{1}{ms}$ jooksul
toimuvad muutused hetkelisteks).

\begin{center}
	\includegraphics[width=0.6\linewidth]{2009-v2g-06-yl}
\end{center}

Solution:
Esimene impulss annab alguses laengule $x$-suunalise impulsi $mv_x = qE_x\tau$ , millest
\[
v_{x}=\frac{q}{m} E_{x} \tau= \SI{1}{m/s}.
\]
Ajavahemiku $t_1 = T /4$ jooksul kuni järgmise impulsini jõuab osake liikuda sirgjooneliselt piki $x$-telge kaugusele
$s_x = v_xT /4 = \SI{0,5}{m}$. Seejärel saab ta impulsi y-telje sihis, mistõttu omandab ka kiiruse $y$-komponent
samasuguse väärtuse: $v_y = v_x = \SI{1}{m/s}$, mistõttu ta liigub \num{45}-kraadise nurga all,
sooritades kuni järgmise impulsini nii $x$- kui $y$-telje sihis nihke $s_x = s_y = \SI{0,5}{m}$. Järgmine impulss peatab $x$-telje sihilise (kuid muutmata $y$-sihilist) liikumise, nii et
osake nihkub nüüd piki $y$-telge kaugusele $s_y = \SI{0,5}{m}$. Järgmine impulss peatab ka $x$-suunalise liikumise, nii et osake jääb paigale. Edasi kordub protsess otsast
peale. Eelpooltoodud tulemuste põhjal saame juuresoleva trajektoori.

\begin{center}
	\includegraphics[width=0.6\linewidth]{2009-v2g-06-lah}
\end{center}

Keskmise kiiruse leiame perioodi jooksul sooritatud nihke $s = \sqrt{1+1}\si{m}$ perioodi $T = \SI{2}{s}$ suhtena, $v \approx \SI{0,7}{m/s}$ .

Liikuv laeng

Laetud osake laengu ja massi suhtega $q/m = \SI{1}{C/kg}$ seisab algselt paigal. Seejärel hakkab ta liikuma $x$- ja $y$-telje sihis
toimivate elektrivälja impulsside mõjul. Elektrivälja vastavate komponentide $E_x$ ja $E_y$ sõltuvus ajast on toodud graafikul (graafiku mastaap
ei ole korrektne, juhenduda tuleb graafikul näidatud numbritest impulsi
kestvuse $\tau = \SI{1}{ms}$ ja amplituudi $E_0 = \SI{1}{kV/m}$ ning perioodi $T = \SI{2}{s}$
jaoks). Visandage osakese trajektoor ja leidke keskmine kiirus (visandi tegemisel ja arvutustes võib lugeda ajavahemiku $\tau = \SI{1}{ms}$ jooksul
toimuvad muutused hetkelisteks).

\begin{center}
	\includegraphics[width=0.6\linewidth]{2009-v2g-06-yl}
\end{center}

Osakese asukohta on mugavam leida vaadeldes eraldi osakese $x$- ja $y$-koordinaate. Laengu keskmine kiirus on leitav perioodi jooksul sooritatud nihke ja perioodi suhtena.


[User]
EFPhO 2012, problem 7. An elegant Lorentz force problem with wires. (If you enjoy
    this problem, consider looking at IdPhO 2020, problem 1B, which has a similar setup but requires
    three-dimensional reasoning. The official solutions are here.)

[4] Problem 10 (Purcell 6.35/INPhO 2008.6). Consider the arrangement shown below.

    The force between capacitor plates is balanced against the force between parallel wires carrying
    current in the same direction. A voltage alternating sinusoidally with angular frequency ω is applied
    to the parallel-plate capacitor C1 and also to the capacitor C2 , and the current is equal to the
    current through the rings. Assume that s ≪ a and h ≪ b.
        Suppose the weights of both sides are adjusted to balance without any applied voltage, and C2
    is adjusted so that the time-averaged downward forces on both sides are equal. Show that
                                                            r
                                            1      √          b C2
                                         √       = 2π aω            .
                                           µ0 ϵ0              h C1

    The left-hand side is equal to c, as we’ll show in E7, so this setup measures the speed of light.

Solution:
See the official solutions here.

[4] Problem 10 (Purcell 6.35/INPhO 2008.6). Consider the arrangement shown below.

The force between capacitor plates is balanced against the force between parallel wires carrying
current in the same direction. A voltage alternating sinusoidally with angular frequency ω is applied
to the parallel-plate capacitor C1 and also to the capacitor C2 , and the current is equal to the
current through the rings. Assume that s ≪ a and h ≪ b.
    Suppose the weights of both sides are adjusted to balance without any applied voltage, and C2
is adjusted so that the time-averaged downward forces on both sides are equal. Show that
                                                        r
                                        1      √          b C2
                                     √       = 2π aω            .
                                       µ0 ϵ0              h C1

The left-hand side is equal to c, as we’ll show in E7, so this setup measures the speed of light.

Solution. It can be a little tricky to read the diagram. The key point is that the triangles are
conductors. They represent the fulcrum of a see-saw, but they also allow the voltage to be applied
across the capacitors on the left and right. The charge buildup on the capacitors on the left causes
them to attract, while the current flowing through the circular wires on the right causes them to
attract as well.
   Let a current I flow on in the right-hand side. Since h ≪ b, the magnetic field created by the
bottom circular loop at a point on the top circular loop is approximately the same as that created
by an infinite wire, B = µ0 I/2πh. Thus, the force between the wires is

                                                     µ0 I   µ0 bI 2
                                      F = (2πbI)          =         .
                                                     2πh      h
This force oscillates over time. The charge on the capacitor C2 is

                                         Q2 (t) = C2 E0 cos(ωt)

so that
                                                                      C22 E02 ω 2
                              ⟨I 2 (t)⟩ = C22 E02 ω 2 ⟨sin2 (ωt)⟩ =               .

    Thus, the average force on the right is
                                                   µ0 C22 E02 ω 2 b
                                              ⟨F ⟩ =                .
                                                        2h
    On the left-hand side, the force between the plates is, by a result in E1,

                                               F = ϵ0 E 2 (πa2 )

    where E is the electric field inside the plates, and we have
                                                  1 2          E02
                                         ⟨E 2 (t)⟩ = ⟨E (t)⟩ =     .
                                                  s2           2s2
    Combining these results and eliminating s, since it doesn’t appear in the final result,
                                                           C12 E02
                                                 ⟨F ⟩ =            .
                                                          4πa2 ϵ0
    Equating the averaged forces gives
                                              µ0 C22 ω 2 b     C12
                                                           =         ,
                                                  h          2πa2 ϵ0
    which is equivalent to the desired result.


[User]
7. Vacuum bulb (8 points)

Let us study how a vacuum can be
created inside a bulb by pumping. Let the volume of the bulb be
V , and the pump consist in a piston moving inside a cylinder of
volumeαV ,whereα ≪ 1. Thepumpingcyclesstartswithpiston
being pulled up; when the pressure inside the cylinder becomes
smaller than inside the bulb, a valve VA (connecting the cylinder
andthebulb)opensandremainsopenaslongasthepistonmoves
up. When piston is released, it starts moving down, at that mo-
ment, the valve VA closes. As long as the valve VA is open, the
pressuresof the bulb and the cylinder can be considered as equal
to each other. When the piston moves down, the pressure in the
cylinder increases adiabatically until becoming equal to the out-
side pressure p0 = 105
Pa; at that moment, another valve VB
opens leting the gas out ofthe cylinder. When the piston reaches
the bottommost position, there is no residual air left inside the
cylinder. Now, the piston is ready for being lift up: the valve VB
closesandVA opens, markingthe beginning ofthenext pumping
cycle. The air inside the bulb can be considered isothermal, with
thetemperaturebeingequaltothesurrondintemperatureT0. The
adiabatic exponent ofair γ = cp/cV = 1.4.
i) (2 pts) How many pumping cycles N needs to be done to re-
duce the pressure inside bulb from p = p0 down to p = βp0,
where β ≪ 1?
ii) (2 pts) What is the net mechanical work done during such a
pumping (covering all the N cycles)?
iii) (2 pts) What is the temperature of the air released from the
cylinder to the surroundings at the end of the pumping process
(when the pressureinside the bulb hasbecomeequalto βp0)?
iv) (2 pts) According to the above described pumping scheme,
there is a considerable loss of mechanical work during the period
when the piston is released and moves down. Such a loss can
be avoided if there is another pump, which moves in an oppos-
ite phase: the forcedue tooutside airpressurepushing thepiston
down can be transmitted to the other pump for lifting the piston
up. What is the net mechanical work done when such a pumping
scheme isused?

Solution:
7. Vacuumbulb

i) Each pumping cycle reduces the number of molecules inside
the bulb by a factor of (1 − α); therefore, after N cycles, the
number of molecules (and hence, the pressure) by a factor of
β = (1 − α)N
≈ e−Nα
⇒
N = −
lnβ
α
.
ii)Majority ofthepumpingcyclesaredonewhenthepressurein-
side the bulb is negligible as compared to the outside pressure.
During such a cycle, a work equal to p0V α is done. Therefore,
A ≈ Np0V α = p0V |lnβ|.
iii) Due to adiabatic law, pV γ
= Const; when combined with
the gas law pV ∝ T we obtain pγ−1
∝ Tγ
. During the last
downwardsmotion ofthe piston,the pressureinside the cylinder
isincreasedby a factorof1/β; thus,T = T0β
1
γ −1
.
iv)Accordingtothemodifiedpumpingscheme,thework/energy
loss is only due to the release of the hot air. Note that if we had
acylinder ofvolumeV , we couldbe able to create vacuuminside
there using only one pumping motion, i.e. by performing work
A = p0V and without any energy loss. Now, we perform an ex-
cesswork, which is converted into internal energy ofthe released
hot air, which needs to be calculated. Let ξ = p
p0
be an inter-
mediate rarefaction factor; then, we can apply the previous result
tocalculate the internal energy ofreleasedair, ifits quantity is dν
moles: dU = T0(ξ
1
γ −1
− 1)cV dν. Let us note that the num-
ber of moles inside the bulb is ν = p0ξV
RT0
⇒ dν = p0V
RT0
dξ. So,
U = p0V cV
R
R 1
0 (ξ
1
γ −1
− 1)dξ = (γ − 1)p0V cV
R
. Now, recall
that γ = cp/cV = 1 + R
cV
, hence cV
R
= 1
γ−1
and U = p0V .
This gives us the energy loss due to heating the released air; an-
otherp0V isrequiredforloss-freecreationofthevacuum. Hence,
thetotal requiredworkisA = 2p0V .

Vacuum bulb


[User]
4. GRAVITATIONAL WAVES (7 points)

—
Artūrs Bērziņš. The power radiated in gravit-
ational waves by an orbiting binary system is
given by P(r,m1,m2) = 32
5
G4
c5
(m1m2)2(m1+m2)
r5 ,
where r is the distance between the centers of
thetwoorbitingmasses m1 and m2. Itisknown
thatthemostcompactobjectisablackhole. The
size of a black hole is defined by its Schwarz-
schild radius rs = 2Gm
c2 , where m is the mass of
the black hole.
i) (2 points) Estimate the upper limit of the
power that can ever be emitted in gravitational
waves by an orbiting binary system.
The gravitational wave detectors on Earth
functionbymeasuringthesocalledgravitywave
strain ε(t) over time, which characterizes the de-
formation of spacetime. Data processing then
yields the maximum strain ε and its correspond-
ing wave frequency f. With the help of a theor-
etical spacetime model, the energy density u as-
sociated with the wave can then be determined.
We will use the analogy of linear elasticity to ex-
amine this model.
ii) (1.5 points) Derive the energy density u =
u(ε,E) in a uniformly stretched elastic band in
terms of the strain ε and the Elastic (Young’s)
modulus E.
iii) (1.5 points) Use dimensional analysis to es-
timate the frequency-dependent elastic modu-
lus of spacetime E(f) in terms of the universal
gravitational constant G, speed of light c and
gravitational wave frequency f.
iv) (2 points) Estimate the maximum distance
z = z(ε,f) from the Earth to the gravitational
wave source as a function of the strain ε and fre-
quency f. Use the models derived previously in
this problem.

Solution:
4. GRAVITATIONAL WAVES

i) Black holes will merge at this distance.
rmin = rs1 +rs2 = 2G(m1+m2)
c2
Substituting for r in the power equation, we get
the maximum power for a system with masses
m1 and m2.
Pmax(m1,m2)= 1
5
c5
G
(m1m2)2
(m1+m2)4
Nowletusconsidertheeffectofmasses. Let
us define a := (m1m2)2
(m1+m2)4 . The intuition tells us,
that more power should be radiated off a bigger
system, thus the masses should tend to infinity
for maximum power.
However the limit lim
m1=m2→∞
a = 1 suggests
otherwise. So does the dimensional analysis:
the dimension of [a] = (M·M)2
M4 = 1 means that
the absolute values of the masses don’t contrib-
ute to the equation. This is intuitively explained
by the proportionally increasing Schwarzschild
radii.
However the mass distribution does matter,
since a isn’t the same for all m1, m2 either.
Let us determine the maximum value it can
take. Let us redefine it in terms of the total
mass M := m1 +m2 and the mass distribution
k := m1
M . Then
m1 = kM
m2 =(1−k)M
⇒ a= (m1m2)2
(m1+m2)4 = (kM(k−1)M)2
M4 =[k(k−1)]2
d
dk[k(k−1)]2
=0⇔ k = 1
2 ⇔ m1 = m2
⇒ amax =[1
2(1− 1
2)]2
= 1
16
So that finally
Pmax = 1
5
c5
G
1
16 = 1
80
c5
G =4.5×1050
W
Interestingly enough, the peak power emit-
ted in gravitational radiation by any merging
binary system will be the same. c5
G is also called
luminosity of the universe.
ii) U = 1
2 k∆l2
, energy stored in a uniformly de-
formed body
k = F
∆l
F =σS
∆l=εl
σ=εE
u = U
V , where V = Sl is the volume
⇒ u = U
V0
= 1
2Eε2
iii) [G]= L3
M−1
T−2
, from e.g. F = GMm
r2
[c]= LT−1
[f]= T−
1
[E]= ML−1
T−2
, from e.g. σ=εE
let E = Gα
cβ
fγ
. Solving for each dimension
yields
M :1=−α ⇒α=−1
L :−1=3α+β ⇒β=2
T :−2=−2α−β−γ ⇒γ=2
⇒ E = c2
G f2
iv) The gravitational wave spreads as a spherical
wave centered at the source. It’s wavefront occu-
pies a thin spherical shell with the surface area
S =4πz2
anditsassociatedinfinitesimalvolume
is dV = Sdz, where dz = cdt for a wave trav-
eling with the speed of light c. The power P
and energy density u at a distance z is coupled
through dE = P dt= udV ⇔ z =
√
P
4πuc.
Using the models from
P = 1
80
c5
G
u = U
V0
= 1
2Eε2
E = c2
G f2
we finally get
z = 1 p
160π
c
εf

GRAVITATIONAL WAVES


[User]
Two fences of heights h1 and h2 are erected on a horizontal plain, so that the tops
    of the fences are separated byp a distance d. Show that the minimum speed needed to throw a
    projectile over both fences is g(h1 + h2 + d).

Solution:
It’s very confusing to think about how to throw the projectile starting from the ground,
    because you need to figure out where to launch and at what angle, under the condition that the
    trajectory just touches the tops of both fences. A much better way is to imagine the projectile
    starts at the top of the higher fence; the goal is then to throw it with minimal energy so that it
    just touches the top of the lower fence. At some point, this projectile will then reach the ground,
    though we don’t have to worry about where. Since mechanics is time-reversible, its speed at this
    point (which is found easily by energy conservation) will be the minimal possible speed.
       Now there are many ways to do this problem. A very slick solution, which requires no computation
    at all, is presented in problem 34. However, we’ll present a more direct attack for completeness.
    Note that if you want to hit the top of the lower fence with the minimum velocity, it’s equivalent
    to maximizing your throwing range down an inclined plane, namely the plane that connects the
    tops of the two fences. Then the optimal launch angle is along the angle bisector, as we found in
    problem 31. Using the same starting point as the solution to that problem, we have
                                                              √
                                           h                g d2 − h2
                                     −√          = tan θ −
                                         d2 − h2             2v 2 cos2 θ
    where we let h = h2 − h1 > 0. That solution gives a simple expression for tan 2θ, so we massage
    this equation to
                                 g     sin 2θ        h

                                    =√          + 2       (1 + cos 2θ).
                                v        2
                                       d −h   2   d − h2
    We then plug in our previous results, which are
                                             √
                                               d2 − h2                   h
                                    sin 2θ =           ,      cos 2θ =
                                                 d                       d
    to get the result
                                       v 2 = (d − h)g = (d + h1 − h2 )g.
    By energy conservation, the speed at the ground is

                                       v02 = v 2 + 2h2 g = (d + h1 + h2 )g

    as desired.