[User]
EFPhO 2006, problem 1. (Note that the comma in the density of air in part 4
    denotes a decimal point.)

   3    General Phase Transitions
   In this section we’ll illustrate some of the ideas needed to analyze phase transitions in general.

       Idea 5
       A nonideal gas can be described by the van der Waals equation of state,
                                           an2

                                       P + 2 (V − nb) = nRT.
                                            V

  You derived the pressure correction in T1 assuming weak attractive intermolecular forces;
  the modification of the volume accounts for the fact that the molecules can’t overlap each
  other. Remarkably, this equation of state also contains a liquid-gas phase transition!

  To see this, consider a plot of the isotherms on a PV diagram.

  At low temperatures, the isotherms can have negative compressibility, meaning that the
  pressure decreases as the volume decreases. This is unphysical, and means that the fluid is
  unstable at these points: if you push on it, it’ll just keep shrinking, until it condenses into a
  dense liquid. Therefore, parts of these isotherms should be replaced with horizontal lines;
  along these horizontal parts liquid and gas coexist, in varying proportions.

  Specifically, everything underneath the dotted line should be replaced with horizontal lines.
  As described in more detail in section 26.1 of Blundell, this can be shown by demanding
  that the liquid and gas have equal Gibbs free energy. As a result, the total area on the PV
  diagram of the isotherm that goes underneath the horizontal line equals the area that goes
  above it; this is called Maxwell’s equal area rule.

  There is a critical isotherm marked in bold above. Above this temperature, there is no
  liquid-gas phase transition at all; instead we just have one phase, called a supercritical fluid.
  Specifically, this is the temperature of the first isotherm that no longer has a local minimum
  in pressure, which means                       2
                                        ∂P           ∂ P
                                               =             =0
                                        ∂V T         ∂V 2 T

       at the critical temperature T = Tc . As you will see in problem 14, this occurs at
                                                         8a                a
                                   Vc = 3nb,    Tc =         ,     Pc =        .
                                                        27Rb              27b2
       This point, marked above, is called the critical point.

       Remark
       The van der Waals equation of state is accurate for a sparse gas with weak attractive inter-
       actions; you shouldn’t expect it to be accurate for dense gases or the liquid state. However,
       it still is extremely important because it is one of the simplest equations of state that gives
       a liquid-gas phase transition. What’s more, if you zoom in near the critical point and write
       the pressure, volume, and temperature as multiples of the critical pressures, volumes, and
       temperatures, it turns out that all equations of state give the same results! This deep phe-
       nomenon is known as universality, but unfortunately I can’t explain the reason why without
       using statistical field theory.

Solution:
See the official solutions here.

   3      General Phase Transitions
   In this section we’ll illustrate some of the ideas needed to analyze phase transitions in general.

  Idea 5
  A nonideal gas can be described by the van der Waals equation of state,

                                      an2

                                  P + 2 (V − nb) = nRT.
                                       V

  You derived the pressure correction in T1 assuming weak attractive intermolecular forces;
  the modification of the volume accounts for the fact that the molecules can’t overlap each
  other. Remarkably, this equation of state also contains a liquid-gas phase transition!

  To see this, consider a plot of the isotherms on a PV diagram.

  At low temperatures, the isotherms can have negative compressibility, meaning that the
  pressure decreases as the volume decreases. This is unphysical, and means that the fluid is
  unstable at these points: if you push on it, it’ll just keep shrinking, until it condenses into a
  dense liquid. Therefore, parts of these isotherms should be replaced with horizontal lines;
  along these horizontal parts liquid and gas coexist, in varying proportions.

  Specifically, everything underneath the dotted line should be replaced with horizontal lines.
  As described in more detail in section 26.1 of Blundell, this can be shown by demanding
  that the liquid and gas have equal Gibbs free energy. As a result, the total area on the PV
  diagram of the isotherm that goes underneath the horizontal line equals the area that goes
  above it; this is called Maxwell’s equal area rule.

  There is a critical isotherm marked in bold above. Above this temperature, there is no
  liquid-gas phase transition at all; instead we just have one phase, called a supercritical fluid.
  Specifically, this is the temperature of the first isotherm that no longer has a local minimum

       in pressure, which means
                                                                      ∂2P

                                             ∂P
                                                              =                   =0
                                             ∂V       T               ∂V 2   T
       at the critical temperature T = Tc . As you will see in problem 14, this occurs at
                                                                   8a                    a
                                   Vc = 3nb,      Tc =                 ,         Pc =        .
                                                                  27Rb                  27b2
       This point, marked above, is called the critical point.

       Remark
       The van der Waals equation of state is accurate for a sparse gas with weak attractive inter-
       actions; you shouldn’t expect it to be accurate for dense gases or the liquid state. However,
       it still is extremely important because it is one of the simplest equations of state that gives
       a liquid-gas phase transition. What’s more, if you zoom in near the critical point and write
       the pressure, volume, and temperature as multiples of the critical pressures, volumes, and
       temperatures, it turns out that all equations of state give the same results! This deep phe-
       nomenon is known as universality, but unfortunately I can’t explain the reason why without
       using statistical field theory.


[User] Veekogu põhjas asub mullitaja --- õhuballoon väikese avausega, millest võrdsete ajavahemike $\Delta t = \SI{1}{s}$ järel väljuvad õhumullid raadiusega $R = \SI{0,3}{mm}$. Taolise mullikese liikumisel vees mõjub sellele takistusjõud $F = 6\pi \eta Rv$, kus $\eta$ on vedeliku voolamistakistust iseloomustav tegur ehk vedeliku viskoossus (vee korral on selle suuruse väärtuseks \SI{1e-3}{N.s/m^2} ) ja $v$ on mullikese kiirus. Võite lugeda, et mullikese liikumine toimub kogu aeg kiirusega, mis on määratud tingimusega, et kõigi talle mõjuvate jõudude resultant on null. Vee tihedus $\rho = \SI{1000}{kg/m^3}$, raskuskiirendus $g = \SI{9,8}{m/s^2}$, õhurõhk $p_0 = \SI{100}{kPa}$. Mitu korda muutub vahemaa naabermullikeste vahel tõusul põhjast pinnale, kui veekogu sügavus on $H = \SI{27}{m}$?

Solution:
Arvestades mullide arvu jäävust on ruumiline vahemaa nende vahel võrdeline mullikeste kiirusega. Viimase leiame võrdsustades takistus- ja üleslükkejõu ning arvestades mullikese ruumala muutust rõhu muutumisel sügavuse vähenemisel.

Mullikesele mõjuvad vee takistusjõud $F$ ja üleslükkejõud $F_A$. Nende võrdusest
\[
6\pi \eta Rv = g (\rho - \rho\idx{õhk}) V,
\]
kus $V$ on mullikese ruumala. Kuna ülesande tingimuste kohaselt $\rho \ll \rho\idx{õhk}$ ja tähistades indeksitega \enquote{H} ja \enquote{0} vastavalt situatsioone veekogu põhjas ja pinna lähedal saame
\begin{align}
6\pi \eta R_Hv_H &= g\rho V_H, \label{2005-v3g-07:eq1}\\
6\pi \eta R_0v_0 &= g\rho V_0. \label{2005-v3g-07:eq2}
\end{align}
Võrrandeid \ref{2005-v3g-07:eq1} ja \ref{2005-v3g-07:eq2} omavahel jagades ning kasutades seost $V_i = (4/3)\pi R_i^3$, saame mullikeste vahekauguse suhte
\[
\frac{L_{0}}{L_{H}}=\frac{v_{0}}{v_{H}}=\frac{V_{0} R_{H}}{V_{H} R_{0}}=\left(\frac{V_{0}}{V_{H}}\right)^{2 / 3}.
\]
Kuna $V_0p_0 = V_Hp_H$ ja $p_H = p_0 + g\rho H$, siis
\[
\frac{L_{0}}{L_{H}}=\left(1+\frac{\rho g H}{p_{0}}\right)^{2 / 3}=\num{2,4},
\]
st mullikeste vahemaa suureneb \num{2,4} korda.

Mullitaja

Veekogu põhjas asub mullitaja --- õhuballoon väikese avausega, millest võrdsete ajavahemike $\Delta t = \SI{1}{s}$ järel väljuvad õhumullid raadiusega $R = \SI{0,3}{mm}$. Taolise mullikese liikumisel vees mõjub sellele takistusjõud $F = 6\pi \eta Rv$, kus $\eta$ on vedeliku voolamistakistust iseloomustav tegur ehk vedeliku viskoossus (vee korral on selle suuruse väärtuseks \SI{1e-3}{N.s/m^2} ) ja $v$ on mullikese kiirus. Võite lugeda, et mullikese liikumine toimub kogu aeg kiirusega, mis on määratud tingimusega, et kõigi talle mõjuvate jõudude resultant on null. Vee tihedus $\rho = \SI{1000}{kg/m^3}$, raskuskiirendus $g = \SI{9,8}{m/s^2}$, õhurõhk $p_0 = \SI{100}{kPa}$. Mitu korda muutub vahemaa naabermullikeste vahel tõusul põhjast pinnale, kui veekogu sügavus on $H = \SI{27}{m}$?

Arvestades mullide arvu jäävust on ruumiline vahemaa nende vahel võrdeline mullikeste kiirusega. Viimane on leitav võrdsustades takistus- ja üleslükkejõu ning arvestades mullikese ruumala muutust rõhu tõttu.


[User]
IPhO 2006, problem 1. This is a neat problem which illustrates the effect of a
    gravitational field on quantum particles, as well as the basics of interferometry, a subject developed
    further in W2. Give this a try even if it looks tough; only the ideas introduced above are needed!

       Idea 3: Bohr Quantization

       In general, p dx may be replaced by any generalized momentum/position pair. For example,
                                             I
                                                L dθ = nh.

       When angular momentum is conserved, the left-hand side is simply 2πL, immediately giving
                                                   L = nℏ

       which is Bohr’s quantization condition. In a system of particles rotating together, L stands
       for the total angular momentum of the system.

       Compared to back-and-forth linear motion, covered in idea 2, rotation is different because
       it’s inherently periodic. For rotation, the integer n can be positive or negative, representing
       a particle going clockwise or counterclockwise. Also, there is no analogue of the α phase
       factor because the particle just rotates all the way around; it never gets reflected.

       Example 1

       Find the energy levels and orbit radii of the electron in the hydrogen atom using Bohr
       quantization.

       Solution
       We postulate a circular orbit, and quantize the angular momentum. We have

                                     mv 2     e2
                                          =         ,       L = mvr = nℏ.
                                      r     4πϵ0 r2
       Solving the second equation for v and plugging into the first gives

                                                4πϵ0 ℏ2 2
                                           r=          n = a0 n2
                                                 me2
       where a0 = 5.3 × 10−11 m is called the Bohr radius; these are the allowed orbit radii. To get
       the energies, we use the standard result for circular motion with an inverse square force that
       the total energy is half the potential energy, so

                                             e2          me4      1
                                     E=−           =−               .
                                            8πϵ0 r    2(4πϵ0 ) ℏ n2
                                                              2 2

       Evidently, they get more and more closely spaced together as n increases. The big constant
       in front is called the Rydberg, and is equal to 13.6 eV.

Solution:
IPhO 2006, problem 1. This is a neat problem which illustrates the effect of a
    gravitational field on quantum particles, as well as the basics of interferometry, a subject developed
    further in W2. Give this a try even if it looks tough; only the ideas introduced above are needed!

       Idea 3: Bohr Quantization

       In general, p dx may be replaced by any generalized momentum/position pair. For example,
                                             I
                                                L dθ = nh.

       When angular momentum is conserved, the left-hand side is simply 2πL, immediately giving

                                                  L = nℏ

       which is Bohr’s quantization condition. In a system of particles rotating together, L stands
       for the total angular momentum of the system.

       Compared to back-and-forth linear motion, covered in idea 2, rotation is different because
       it’s inherently periodic. For rotation, the integer n can be positive or negative, representing
       a particle going clockwise or counterclockwise. Also, there is no analogue of the α phase
       factor because the particle just rotates all the way around; it never gets reflected.

       Example 1

       Find the energy levels and orbit radii of the electron in the hydrogen atom using Bohr
       quantization.

       Solution
       We postulate a circular orbit, and quantize the angular momentum. We have

                                     mv 2     e2
                                          =         ,       L = mvr = nℏ.
                                      r     4πϵ0 r2
       Solving the second equation for v and plugging into the first gives

                                                4πϵ0 ℏ2 2
                                           r=          n = a0 n2
                                                 me2
       where a0 = 5.3 × 10−11 m is called the Bohr radius; these are the allowed orbit radii. To get
       the energies, we use the standard result for circular motion with an inverse square force that
       the total energy is half the potential energy, so

                                             e2          me4       1
                                     E=−           =−                 .
                                            8πϵ0 r    2(4πϵ0 )2 ℏ2 n2

       Evidently, they get more and more closely spaced together as n increases. The big constant
       in front is called the Rydberg, and is equal to 13.6 eV.

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


[User]
A mass, which is free to move on a horizontal frictionless plane, is attached to one end of a massless
string which wraps partially around a frictionless vertical pole of radius r, as shown (top view).

   1. At time t = 0, the mass has speed v0 in the tangential direction along the dotted circle of
      radius R shown. You pull on the string so that the mass keeps moving along the dotted circle,
      so that the string remains in contact with the pole at all times. Find the speed of the mass
      as a function of time.

   2. Now suppose that, at time t = 0, the mass has speed v0 opposite the direction shown. You
      hold the string, keeping your hand stationary. Using the approximation r ≪ R, find the time
      when the mass hits the pole.

Solution:
A mass, which is free to move on a horizontal frictionless plane, is attached to one end of a massless
string which wraps partially around a frictionless vertical pole of radius r, as shown (top view).

   1. At time t = 0, the mass has speed v0 in the tangential direction along the dotted circle of
      radius R shown. You pull on the string so that the mass keeps moving along the dotted circle,
      so that the string remains in contact with the pole at all times. Find the speed of the mass
      as a function of time.
   2. Now suppose that, at time t = 0, the mass has speed v0 opposite the direction shown. You
      hold the string, keeping your hand stationary. Using the approximation r ≪ R, find the time
      when the mass hits the pole.
Solution. This is an extension of BAUPC 2002, problem 2, whose official solution is here. The
answers are:
   1. The centripetal acceleration√is v 2 /R, so by similar triangles, the tangential acceleration is
      v 2 tan θ/R, where tan θ = r/ R2 − r2 . We thus have
                                               dv   v2
                                                  =    tan θ
                                               dt   R
      and separating and integrating gives
                                             t tan θ   1  1
                                                     =   − .
                                                R      v0 v
      Solving for v gives
                                                            −1
                                               1      tr
                                       v=        − √             .
                                              v0 R R 2 − r 2
      Of course, this expression eventually blows up, which indicates that it’ll break down before
      that point, e.g. because the string will snap.
   2. This is a new part. Here energy is conserved, so the speed remains v0 . Every time the string
      wraps around the pole, the length r′ of the free string decreases by 2πr. In addition, each
      wrapping takes an approximate time 2πr′ /v0 , where we approximated the trajectory for one
      revolution as circular. Therefore, we have
                                             dr′      r
                                                 ≈ − ′ v0 .
                                             dt       r

      Separating and integrating gives tf = R /(2v0 r).


[User]
1. SATELLITE (8 points)

— Taavet Kalda. A
solar powered satellite with starting velocity
v0 is launched from Earth onto an elliptical
heliocentric orbit with the intent of gather-
ing as much solar energy as possible. The
angle of departure can be freely changed.
i) (1 point) What is the minimal required
launching speed vm for the satellite to reach
any heliocentric orbit?
ii) (2 points)What is the speed of the satellite
just after exiting Earthʼs gravitational field?
iii) (2.5 points)Express the long-time-average
solar irradiance of the satellite in terms of
its semi-major axis a, angular momentum J,
orbital period T and mass m.
iv) (2.5 points) What is the maximal average
solar irradiance the satellite can gather and
what is the needed launch angle relative to
Earthʼs motion?
The Sunʼs mass is M ̄, Earthʼs orbital ra-
dius R⊕, free fall acceleration on Earth g,
Earthʼs radius r⊕ and Sunʼs luminosity L ̄.

Solution:
1. SATELLITE

— Taavet Kalda.
i) (1 point)From the energy conservation law,
mv2
m
2
−
GM⊕m
r⊕
= 0,
hence
vm =
√
2GM⊕
r⊕
.
ii) (2 points)Let the speed of the satellite just
after leaving Earthʼs gravitational field be v1.
From energy conservation,
mv2
0
2
−
GM⊕m
r⊕
=
mv2
1
2
,
where M⊕ is Earthʼs mass. Furthermore, we
have g = GM⊕
r2
⊕
. Thus,
v1 =
√
v2
0 −
2GM⊕
r⊕
=
√
v2
0 − 2gr⊕.
iii) (2.5 points) The average solar irradiance
can be expressed as
Iavg =
1
T
∫ T
0
I(t)dt,
where I(t) the solar irradiance at time t. We
can express the solar irradiance as I(t) =
L⊙
4πr(t)2 . Furthermore, it might be more con-
venient to integrate over the angle instead of
time so we can use dt = dα
ω(α), where ω is the
angular velocity of the satellite. This yields
Iavg =
1
T
∫ 2π
0
L⊙
4πr2
dα
ω
=
L⊙
4πT
∫ 2π
0
dα
ωr2
.
Notethatthedenominatorintheintegrandis
verysimilartotheangularmomentumofthe
satellite. Indeed, the angular momentum is
J = mvr = mωr2
= Const. Thus,
Iavg =
L⊙
4πT
∫ 2π
0
dαm
J
=
L⊙m
4πTJ
∫ 2π
0
dα =
L⊙m
2TJ
.
iv) (2.5 points) Since L⊙ and m are constant,
we need to minimise the quantity TJ. Note
that the minimal angular momentum cor-
responds to the case when the satellite is
launched directly opposite to the motion of
Earth. It turns out that this also corresponds
to the minimal orbital period. Let ⃗ v2 be the
satelliteʼs velocity in Sunʼs frame. Then ⃗ v2 =
⃗ v1 − ⃗ v⊕, where v⊕ =
√
GM⊙
R⊕
is Earthʼs ve-
locity. For convenience, letʼs write x = v2
v⊕
.
Consider the total energy of an elliptical or-
bit Etot = −GM⊙m
2a . On the other hand, the
total energy is Etot =
mv2
2
2 − GM⊙m
R⊕
. combin-
ingthetwoequationsandrearranging, R⊕
a =
2 −
v2
2R⊕
GM⊕
= 2 − x2
. From Keplerʼs III Law,
T2
a3 = 4π2
GM⊙
. Thus, T =
2πR3
⊕
√
GM⊙
(
2 − x2
)−3/2
.
As we can see, in order to minimise T, v2
needs to be minimal as well.
In conclusion, Iavg is maximal when the
satellite is launched directly against the mo-
tion of Earth. The corresponding value for
Iavg is
Iavg =
L⊙
4πR2
⊕
√
GM⊙
R⊕
1
v2
(
2 − x2
)3/2
=
L⊙
4πR2
⊕
(2 − x2
)3/2
x
,
where x =
√
R⊕
GM⊙
√
v2
0 − 2gr⊕ − 1.

SATELLITE


[User]
USAPhO 1997, problem B1. A nice problem on the dynamics of a plasma.
    (Note that the assumption made in part (e) is somewhat arbitrary, without much physical meaning.

                    01^
    It’s just made to make part (f) a bit simpler.)

Solution:
USAPhO 1997, problem B1. A nice problem on the dynamics of a plasma.
    (Note that the assumption made in part (e) is somewhat arbitrary, without much physical meaning.
    It’s just made to make part (f) a bit simpler.)


[User]
Estimating some integrals with sharp peaks. For all cases, you only need to find
    an estimate for the leading contribution in the limit ϵ → 0, and you can drop numeric prefactors.

     (a) To warm up, find a rough estimate for
                                              Z ∞
                                                                    dx
                                        I1 =                                   .
                                                     0        (ϵ + x)2 (1 + x)

     (b) Find a rough estimate for                   Z ∞
                                                                      dx
                                              I2 =            √                 .
                                                     0            ϵ + x (1 + x)
      (c) Find a rough estimate for                  Z ∞
                                                                    dx
                                              I3 =                           .
                                                      0       (ϵ + x)(1 + x)

       Remark
       You might be wondering how good you have to be at integration to do Olympiad physics.
       The answer is: not at all! You need to understand how to set up integrals, but you almost
       never have to perform a nontrivial integral. There will almost always be a way to solve the

  problem without doing explicit integration at all, or an approximation you can do to render
  the integral trivial, or the integral will be given to you in the problem statement. The Asian
  Physics Olympiad takes this really far: despite   R having   some of the hardest problems ever
  written, they often provide information like “ x dx = xn+1 /(n + 1) + C” as a hint! This
                                                        n

  is because physics competitions are generally written to make students think hard about
  physical systems, and the integrals are just viewed as baggage.

  In fact, plain old AP Calculus probably has harder integrals than Olympiad physics. For
  example, in those classes everybody has to learn the integral
                               Z
                                 sec x dx = log |sec x + tan x| + C

  which has a long history. When I was in high school, I was shocked by how the trick for doing
  this integral came out of nowhere; it seemed miles harder than anything else taught in the
  class. And it is! Historically, it arose in 1569 from Mercator’s projection, where it gives the
  vertical distance on the map from the equator to a given latitude. For decades, cartographers
  simply looked up the numeric value of the integral in tables, where the Riemann sums had
  been done by hand. (They had no chance of solving it analytically anyway, since Napier only
  invented logarithms in 1614.) Gradually, tabulated values of the logarithms of trigonometric
  functions became available, and in 1645, Bond conjectured the correct result by noticing the
  close agreement of tabulated values of each side of the equation. Finally, Gregory proved the
  result in 1668, using what Halley called “a long train of Consequences and Complications of
  Proportions.” So it took almost a hundred years for this integral to be sorted out! (Though to
  their credit, they had the handicap of not knowing about differentiation or the fundamental
  theorem of calculus; they were finding the area under the curve with just Euclidean geometry.)

  Even though Olympiad physics tries to avoid tough integrals, doing more advanced physics
  tends to produce them, so physicists often get quite good at integration. By contrast,
  Spivak’s calculus textbook for math majors only covers integration techniques in a single
  chapter towards the end of the book. He justifies the inclusion of this material by saying:

        Every once in a while you might actually need to evaluate an integral [...] For
        example, you might take a physics course [...] Even if you intend to forget how to
        integrate (and you probably will forget some details the first time through), you
        must never forget the basic methods.

  That attitude is why physics students frequently win the MIT Integration Bee.

Solution:
(a) There’s a sharp peak near x = 0, of width ϵ and height 1/ϵ2 , so we estimate
         I1 ∼ 1/ϵ.

                                                              √                              √
 (b) There’s a sharp peak near x = 0, of width ϵ and height 1/ ϵ, so the peak contributes ∼ ϵ.
     But wait! That’s actually a small quantity in the limit ϵ → 0. In this case, the peak
     simply isn’t very high, and the dominant contribution actually comes from the rest of the
     integral. A very rough estimate can be found by neglecting the smaller part of each term in
     the denominator in each part of the integration range,
                       Z ϵ      Z 1       Z ∞
                           dx       dx          dx       √
                  I2 ∼     √ +      √ +         3/2
                                                    = O( ϵ) + O(1) + O(1) = O(1).
                        0    ϵ   ϵ    x    1   x
     That is, the integral is an order-one number, independent of ϵ.

 (c) This is in between the previous two cases, so let’s apply a similar rough approximation,
                   Z ϵ      Z 1      Z ∞
                       dx       dx       dx
              I3 ∼        +        +         = O(1) + O(log(1/ϵ)) + O(1) = O(log(1/ϵ)).
                    0 ϵ      ϵ  x     1  x2
     That is, the integral is only logarithmically divergent in ϵ, and most of the contribution comes
     the region away from the peak but before the falloff at infinity, ϵ ≪ x ≪ 1. This kind of
     behavior is more common than you might think, especially in more advanced physics.
     Of course, if you’re not sure about any of these steps, you can explicitly split the integration
     range up (e.g. into 0 < x < δ1 , δ1 < x < δ2 , and x > δ2 , with ϵ ≲ δ1 ≪ δ2 ≲ 1) and
     approximate within each region, to get the same result.

  Remark
  You might be wondering how good you have to be at integration to do Olympiad physics.
  The answer is: not at all! You need to understand how to set up integrals, but you almost
  never have to perform a nontrivial integral. There will almost always be a way to solve the
  problem without doing explicit integration at all, or an approximation you can do to render
  the integral trivial, or the integral will be given to you in the problem statement. The Asian
  Physics Olympiad takes this really far: despite   R having   some of the hardest problems ever
  written, they often provide information like “ x dx = xn+1 /(n + 1) + C” as a hint! This
                                                        n

  is because physics competitions are generally written to make students think hard about
  physical systems, and the integrals are just viewed as baggage.

  In fact, plain old AP Calculus probably has harder integrals than Olympiad physics. For
  example, in those classes everybody has to learn the integral
                               Z
                                 sec x dx = log |sec x + tan x| + C

  which has a long history. When I was in high school, I was shocked by how the trick for doing
  this integral came out of nowhere; it seemed miles harder than anything else taught in the
  class. And it is! Historically, it arose in 1569 from Mercator’s projection, where it gives the
  vertical distance on the map from the equator to a given latitude. For decades, cartographers
  simply looked up the numeric value of the integral in tables, where the Riemann sums had
  been done by hand. (They had no chance of solving it analytically anyway, since Napier only
  invented logarithms in 1614.) Gradually, tabulated values of the logarithms of trigonometric
  functions became available, and in 1645, Bond conjectured the correct result by noticing the
  close agreement of tabulated values of each side of the equation. Finally, Gregory proved the

  result in 1668, using what Halley called “a long train of Consequences and Complications of
  Proportions.” So it took almost a hundred years for this integral to be sorted out! (Though to
  their credit, they had the handicap of not knowing about differentiation or the fundamental
  theorem of calculus; they were finding the area under the curve with just Euclidean geometry.)

  Even though Olympiad physics tries to avoid tough integrals, doing more advanced physics
  tends to produce them, so physicists often get quite good at integration. By contrast,
  Spivak’s calculus textbook for math majors only covers integration techniques in a single
  chapter towards the end of the book. He justifies the inclusion of this material by saying:

       Every once in a while you might actually need to evaluate an integral [...] For
       example, you might take a physics course [...] Even if you intend to forget how to
       integrate (and you probably will forget some details the first time through), you
       must never forget the basic methods.

  That attitude is why physics students frequently win the MIT Integration Bee.


[User] Siledal pinnal asub kerge ristkülikuline anum, mis on täidetud vedelikuga tihedusega $\rho_0$, vedeliku ruumala on $V_0$. Anuma põhja sattunud põrnikas ruumalaga $V$ ja tihedusega $\rho$ hakkab anuma põhja suhtes roomama kiirusega $u$. Millise kiirusega hakkab anum pinnal liikuma? Anuma mass on tühine, veetase jääb kogu aeg horisontaalseks. Eeldada, et pinna ja anuma vahel hõõre puudub.

Solution:
Vaatame algul liikuva anumaga seotud taustsüsteemi. Kui põrnikas (massiga $\rho V$) roomab mööda anuma põhja, siis selle peale liigub ka põrnikat ümbritsev vedelik. Põrnika liikumist võib mugavuse mõttes ette kujutada virtuaalse põrnika liikumisega, mille tihedus on $\rho - \rho_0$. Sellisel juhul liigub virtuaalne põrnikas vedelikku tõrjumata ning ülesanne taandub mugavamale dünaamika ülesandele.

Liigume nüüd põrandaga seotud taustsüsteemi. Vastavalt impulsi jäävuse seadusele on põrnika ja anuma koguimpulss kogu aeg konstantne (sest põrnika+anuma süsteemile ei mõju väliseid jõude). Kuna see oli alguses 0, on põrnika ja anuma impulsid vastassuunalised ning absoluutväärtuse poolest võrdsed. Olgu anuma kiirus põranda suhtes $v$. Siis põrnika kiirus põranda suhtes on $u - v$. Impulsi jäävuse seadusest tulenevalt
\[
\rho_0 (V_0 + V ) v = (\rho - \rho_0) V (u - v),
\]
ehk
\[
v=\frac{V\left(\rho_{0}-\rho\right)}{\rho V+\rho_{0} V_{0}}u.
\]

Anum

Siledal pinnal asub kerge ristkülikuline anum, mis on täidetud vedelikuga tihedusega $\rho_0$, vedeliku ruumala on $V_0$. Anuma põhja sattunud põrnikas ruumalaga $V$ ja tihedusega $\rho$ hakkab anuma põhja suhtes roomama kiirusega $u$. Millise kiirusega hakkab anum pinnal liikuma? Anuma mass on tühine, veetase jääb kogu aeg horisontaalseks. Eeldada, et pinna ja anuma vahel hõõre puudub.


[User] A house has a mirror symmetric gable roof: the vertical plane of symmetry is East-West directional and the North and South sides of the roof are perpendicular to each other. At both sides of the roof there is a rainwater gutter that collects the water falling on the roof and directs it into a barrel. It is raining and South wind is blowing $u= \SI{6,0}{m/s}$. The barrel at the South side fills 2,0 times faster than the barrel at the North side. You can assume that the direction of the falling rain drops does not change significantly. What is the average speed of the falling drops (meaning the velocity’s vertical component)?

Solution:
Tähistame katuse vertikaallõikel harja tähega $A$, räästad tähtedega $B$ ja $C$ ning lõigu $BC$ keskpunti tähega $O$.
Tähistame sümbolitega $s_1$, $s_2$ ja $s_3$ selliste piiskade trajektoorid, mis tabavad vastavalt põhjaräästast punktis $B$, harja punktis $A$ ning lõunaräästast punktis $C$. Sirgete $s_1$ ja $s_2$ vahelisse ribasse jäävad piisad tabavad põhjakatust ning sirgete $s_2$ ja $s_3$ vahelisse ribasse jäävad
piisad tabavad lõunakatust. Seega on veehulkade suhe võrdne ribade laiuste suhtega, mis omakorda on võrdne lõikude $BD$ ja $CD$ pikkuste suhtega, kus $D$ on sirge $s_2$ lõikepunkt lõiguga $BC$. Seetõttu $|BD|=\frac 13 |BC|$ ning järelikult
\[
|DO|=|BO|-|BD|=\frac 16 |BC|.
\]
Paneme tähele, et vihmapiiskade
kiirusvektori horisontaal- ja vertikaalkomponentide suhe on võrdne lõikude $DO$ ja $OA=\frac 12 BC$ pikkuste suhtega; et $\frac {|AO|}{|DO|}=3$, siis piiskade
langemise kiirus on $3u=\SI{18}{m/s}$.

\begin{center}
\includegraphics[width=0.5\linewidth]{2012-v2g-07-katus}
\end{center}

Vihmasadu

Viilkatusega maja katus on peegelsümmeetriline: vertikaalne sümmeetriatasand on
ida-läänesuunaline ning katuse
põhja- ja lõunaküljed on omavahel risti. Mõlemal katusepoolel on vihmaveerenn,
mis kogub katusele langeva vee ning suunab selle tünni.
Sajab vihma ning puhub lõunatuul $u= \SI{6,0}{m/s}$; lõunaküljel paiknev tünn
täitub 2,0 korda kiiremini kui põhjaküljel paiknev tünn; võib
lugeda, et katuse läheduses piiskade langemissuund oluliselt ei muutu.
Milline on piiskade langemise keskmine kiirus (st kiiruse vertikaalkomponent)?

Horisontaalsuunalise tuule tõttu langevad vihmapiisad teatud nurga all vertikaali suhtes. Kuna lõunaküljel olev tünn täitub kaks korda kiiremini, siis lõunakatuse ristlõikepindala peab langevate piiskade risttasandis olema kaks korda suurem kui põhjaküljel.


[User]
Theoretical competition – Problem No.1
                       Measure the mass in the weightless state

    In the spacecraft orbiting the Earth, there is weightless state, so that one cannot use
ordinary instruments to measure the weight and then to deduce the mass of the astronaut.
Skylab 2 and some other spacecrafts are supplied with a Body Mass Measurement Device
which consists of a chair attached to one end of a spring. The other end of the spring is
attached to a fixed point of the spacecraft. The axis of the spring passes through the center of
mass of the craft. The force constant (the hardness) of the spring is k = 605.6 N/m.

1. When the craft is fixed on the pad, the chair (without person) oscillates with the period
T0 = 1.28195 s.
    Calculate the mass m0 of the chair [2 pts].

2. When the craft orbits the Earth the astronaut straps himself into the chair and measures the
period T’ of the chair oscillations. He obtains T’ = 2.33044 s, then calculates roughly his
mass. He feels some doubt and tries to find the true value of his mass. He measures again the
period of oscillation of the chair (without person), and find T0’ = 1.27395 s.

       What is the true value of the astronaut’s mass and the craft’s mass? [4pts]

Note: The mass of the spring is negligible and the astronaut is floating.

Solution:
Solution of Problem No. 1
       Measure the mass in the weightless state

1/ Formula for period of oscillation :
                      2π             m0
               T0 =       = 2π                                      (1)
                      ω0              k
One can deduce :
                           k        2
                  m0 =          T = 25,21 kg                       (2)
                          4π  2 0

2/ When the craft orbits the Earth, oscillating system is the spring with one end attached
to the chair of mass m0 and other end attached to he craft with mass M. This system
oscillates like an object with mass
                                          m0 M
                            m0’ =                                  (3)
                                        m0 + M
attached to an end of the spring, other end of the spring is fixed (m0’ is the reduced mass
of the system craft-chair). The period T0’ of the system is also given by (1) and (2). We
can deduce :
                  m0      T             1,28195 2
                       = ( 0 )2 = (               )                  (4)
                 m0 '     T0 '          1,27395
The mass M of the craft can be calculate from (3) :
                   m0           25,21               25,21
          M=              =                 =                 = 2001 kg      (5)
                 m0            T0 2            1,28195 2
                      −1 ( ) −1 (                        ) −1
                 m0 '          T0 '            1,27395
Let m be the mass of astronaut and chair, the coressponding reduce mass is m’:
                                mM
                       m’ =                                              (6)
                               m+M
the expression of m is then :
                                                    m'
                                            m=
                                                      m'
                                                 1−
                                                      M
The reduce mass m’ can be calculated from the oscillation period T’ by using formula
(2) :
                               605,6 2,33044 2
                       m’ =             .(          ) = 83,31 kg
                                 4         3,1416
the true value of the mass m is :
                               83,31
                      m=                   = 86,93kg
                                 83,31
                           1−
                                 2001
the true value of the mass of astronaut :
                         86,93 – 25,21 = 61,72 kg

APhO 2004 Theory T1


[User] Average day, usually just called a day, is known as the average period during which the Sun seems to make a full circle in the sky for an observer located on Earth. The length of an average day is 24 h or 86 400 s. It takes Earth 365,256 average days to make one circle around the Sun. The direction of the Earth’s rotation around its axis matches the direction of its orbiting around the Sun. Using this data, find the rotation period of the Earth with the accuracy of a second.

Solution:
Päikese näivat liikumist taevas põhjustavad nii Maa pöörlemine kui ka tiirlemine. Maa tiirlemise tõttu erineb Maa täispöörete arv aastas ühe võrra keskmiste päikeseööpäevade arvust. Kuna Maa tiirlemise suund ühtib Maa pöörlemise suunaga, teeb Maa ühe aasta jooksul ühe täispöörde rohkem. Seega on Maa pöörlemisperioodiks
\[
P=\frac{\num{365,256}}{\num{366,256}} \cdot \SI{86400}{\second}=\SI{86164}{\second}=\SI{23}{\hour} \SI{56}{\minute} \SI{4}{\second}.
\]

\vspace{0.5\baselineskip}
\emph{Alternatiivne lahendus}\\
Päike teeb täistiiru taevas sagedusega $f_k=\frac{1}{\SI{86400}{\second}}$. Maa tiirlemise sagedus on $f_t=\frac{1}{\num{365,256}\cdot86400\text{s}}$. Kuna Maa pöörlemis- ja tiirlemissuunad ühtivad, siis kehtib võrrand $f_k=f_p-f_t$, kus $f_p$ on Maa pöörlemise sagedus. Siit saame avaldada Maa pöörlemisperioodi
\[
P=\frac{1}{f_p}=\frac{1}{f_k+f_t}=\SI{86164}{\second}=\SI{23}{\hour} \SI{56}{\minute} \SI{4}{\second}.
\]

\emph{Märkused.}

\vspace{-5pt}
\begin{itemize}[noitemsep, leftmargin=*]
\item Nimetuse keskmine päikeseööpäev tingib asjaolu, et Maa elliptilise orbiidi tõttu on Päikese näiv nurkkiirus taevas veidi muutlik.
\item Maa tiirlemisperioodi nimetatakse ka sideeriliseks aastaks.
\item Enamasti mõistetakse aastana troopilist, mitte sideerilist aastat, mis on defineeritud pööripäevade kordumise põhjal. Troopilise ning sideerilise aasta erinemise põhjustab Maa telje pretsessioon. Igapäevaelus ei ole olulised mitte Maa pöörlemine ning tiirlemine vaid hoopis Päikese ööpäevane liikumine taevas ning aastaaegade kordumine, mistõttu laialdaselt kasutatavad ööpäeva ning aasta mõisted erinevadki Maa pöörlemis- ning tiirlemisperioodidest.
\end{itemize}

Maa pöörlemisperiood

Keskmiseks päikeseööpäevaks ehk tavatähenduses ööpäevaks nimetatakse keskmist perioodi, mille jooksul Päike näib Maaga seotud vaatleja jaoks tegevat taevas täisringi. Keskmise päikeseööpäeva pikkuseks on \SI{24}{\hour} ehk \SI{86400}{\second}. Maal kulub ühe tiiru tegemiseks ümber Päikese \SI{365.256}{} keskmist päikeseööpäeva. Maa pöörlemissuund ümber oma telje ühtib selle tiirlemissuunaga Päikese ümber. Leidke nende andmete põhjal Maa pöörlemisperiood sekundi täpsusega.

Maa tiirlemise tõttu erineb Maa täispöörete arv aastas keskmiste päikeseööpäevade arvust ühe võrra.


[User] Ühtlase lumekihiga kaetud katus on horisondi suhtes kaldu $\alpha = \ang{40}$ nurga all. Katus on ristküliku kujuline ja laius harjast räästani mööda katuse pinda on $L$. Katuse ja lume vaheline hõõrdetegur on $\mu = \num{1}$. Katuse harjast hakkab lumekihi ja katuse vahele voolama vesi, mis muudab märja katuse ja lumekihi vahelise hõõrdeteguri nulliks. Kui vesi jõuab katuseharjast kaugusele $l$, hakkab lumekiht alla libisema. Leidke suhe $l/L$.

Solution:
Vaatleme lund katusel kui kahte vastasmõjus olevat keha: üheks kehaks on lumi, mille all on vesi ning millele hõõrdejõud ei mõju, ja teiseks kehaks kuival katusel olev lumi. Nende kahe osa vahel mõjuva jõu $F$ võime lugeda katuse sihiliseks (selle sihi valime $x$-teljeks, $y$-telg on katuse sihiga risti). Arvestame, et lumi on ühtlase paksusega ja seega osade massid on võrdelised nende pikkustega:
\[
\frac{m_1}{m_2} = \frac{l}{L-l}.
\]
$y$-telje sihiline tasakaaluvõrrand kuiva osa jaoks:
\[
N_2 = m_2g \cos \alpha,
\]
kus $N_2$ on kuivale osale mõjuva katuse rõhumisjõud. $x$-telje sihiline tasakaaluvõrrand vesise ja kuiva osa jaoks:
\[
F = m_1g \sin \alpha,
\]
\[
\mu N_2 = F + m_2g \sin \alpha.
\]
Elimineerides kahest viimasest võrrandist $F$-i leiame
\[
\mu N_2 = (m_1 + m_2) g \sin \alpha.
\]
Asendades siia $N_2$ leiame
\[
m_2\mu g \cos \alpha = (m_1 + m_2) g \sin \alpha.
\]
Jagades läbi $m_2g$-ga ja asendades esimesest võrrandist suhte $m_1/m_2$, saame
\[
\frac{L}{L-l}=\mu \cot \alpha,
\]
millest
\[
\frac{l}{L}=1-\mu^{-1} \tan \alpha \approx \num{0,16}.
\]

\vspace{0.5\baselineskip}

\emph{Alternatiivne lahendus}

Vaatleme lund katusel tervikliku süsteemina. Valime $x$-teljeks katuse sihi, $y$-telg olgu katuse sihiga risti. Lumekihile mõjuvad järgnevad jõud: raskusjõud $mg$, katusepinna toereaktsioonijõud $N$ ning hõõrdejõud
\[
F_h = \frac{L-l}{L} \mu N.
\]
Kordaja $(L-l)/L$ tuleb sellest, et hõõrdejud mõjub vaid $(L-l)$-pikkusel katuseosal.
Tasakaaluvõrrand $y$-telje jaoks on
\[
N = mg \cos \alpha,
\]
$x$-telje jaoks aga
\[
m g \sin \alpha=F_{h}=\frac{L-l}{L} \mu N.
\]
Elimineerides kahest viimasest võrrandist $N$-i leiame
\[
m g \sin \alpha=\mu m g \cos \alpha \frac{L-l}{L}.
\]
Saame, et
\[
\frac{L-l}{L} = \frac{\tan\alpha}{\mu}.
\]
Seega
\[
\frac{l}{L}=1-\frac{\tan \alpha}{\mu} \approx \num{0,16}.
\]

Katus

Ühtlase lumekihiga kaetud katus on horisondi suhtes kaldu $\alpha = \ang{40}$ nurga all. Katus on ristküliku kujuline ja laius harjast räästani mööda katuse pinda on $L$. Katuse ja lume vaheline hõõrdetegur on $\mu = \num{1}$. Katuse harjast hakkab lumekihi ja katuse vahele voolama vesi, mis muudab märja katuse ja lumekihi vahelise hõõrdeteguri nulliks. Kui vesi jõuab katuseharjast kaugusele $l$, hakkab lumekiht alla libisema. Leidke suhe $l/L$.

Libisemise piiril tasakaalustab lund katuselt alla lükkavat raskusjõudu hõõrdejõud, kusjuures hõõrdejõud mõjub vaid lume kuivale osale.


[User]
A few useful facts about energy and momentum, for future reference.

     (a) Recalling the definition of the four-velocity from R1, show that

                                                   (E/c, p) = muµ

           where uµ is the four-velocity. Setting c = 1, this establishes pµ = (E, p) is a four-vector.

     (b) Suppose a particle is at rest in frame S ′ . Confirm explicitly that the components of the
         four-momentum pµ transform as expected when going to frame S.

        (c) Setting c = 1 for all future parts, show that the norm of the four-momentum is

                                              pµ pµ = E 2 − |p|2 = m2 .

           This is a very useful result that can simplify the solutions to many problems below, especially
           ones that simply ask for a final mass m. In this case one can often compute a single four-
           momentum and find its norm to get the answer.

     (d) The expressions in idea 1 for E and p don’t work for photons, since γ is infinite and m is zero.
         Instead, show that for a photon we have pµ = ℏk µ .

        (e) A system’s center of mass frame is the one where its momentum is zero. For a system with
            total energy E and momentum p, show that the center of mass has velocity v = p/E.

        (f) In Newtonian mechanics, the kinetic energy K of an object with fixed mass m satisfies
            dK = v · dp. Show that this also holds in relativity, assuming the rest mass m is fixed.

     (g) As we’ll discuss in more detail below, the force three-vector is defined as F = dp/dt in
         relativistic mechanics. Show that dK = F · dx, continuing to assume that m is fixed.

       Remark
       The result of part (e) is equivalent to saying that momentum p is always associated with the
       motion of energy Ev. This is a very general statement, which also holds at the differential
       level: momentum density is equal to energy flux density. One example of this was given
       in E7, where it was noted that the electromagnetic momentum density p was equal to the
       Poynting vector S, in units where c = 1.

       Idea 2
       In relativistic dynamics problems, it is almost always better to work with energy and mo-
       mentum than velocity; one typically shouldn’t even mention velocities unless the problem
       asks for or gives them.

    We’ll start with some very simple problems to warm up, setting c = 1 throughout.

       Example 1: KK 13.5

       A particle of mass m and speed v collides and sticks to a stationary particle of mass M . Find
       the final speed of the composite particle.

       Solution
       The total four momentum is (E, p) = (γm + M, γmv), so the final speed is
                                       p    γmv            v
                                vf =     =        =          √        .
                                       E   γm + M   1 + (M/m) 1 − v 2

       Example 2: Morin 12.2

       Two photons of energy E collide at an angle θ and create a particle of mass M . What is M ?

       Solution
       The total four-momentum is

                                       pµ = (2E, E(1 + cos θ), E sin θ).

       The mass is just the norm of the four-momentum, so
                      q                                     √
                M = 4E 2 − E 2 (1 + cos θ)2 − E 2 sin2 θ = E 2 − 2 cos θ = 2E sin(θ/2).

[1] Problem 2 (Morin 12.4). A stationary mass MA decays into masses MB and MC . What are the
    energies of these two masses?

Solution:
(a) We saw in R1 that uµ = (γc, γv). Multiplying by m gives

                                      muµ = (γmc, γmv) = (E/c, p)

     as desired.
                                                            u+v
 (b) Suppose we boost by velocity u. Then, the new speed is 1+uv , so the new value of γ is
                                                    −1/2
                                           (u + v)2

                               γ′ =    1−                 = (1 + uv)γu γ.
                                          (1 + uv)2
     Thus, the boosted values of E and p are

                          E ′ = γ ′ m = γu (E + up),   p′ = γ ′ mv = γu (p + uE).

     These are exactly the expected Lorentz transformation properties.

 (c) The norm is
                           E 2 − p2 = γ 2 m2 − γ 2 m2 v 2 = γ 2 m2 (1 − v)2 = m2
     as desired.

 (d) This follows directly from the de Broglie relations E = ℏω and p = ℏk.

 (e) In this frame, p′ = 0. Then using the result of part (b), we have p − vE = 0 where v is the
     velocity of the center of mass in the original frame. Therefore, v = p/E.

 (f) Starting with E 2 = p2 + m2 and taking the differential of both sides,

                                             2E dE = 2p · dp.

     Solving for dE, we have
                                               p
                                          dE =   · dp = v · dp
                                               E
     where we used the result of problem 1. Since K and E are the same up to a constant anyway,
     we conclude dK = v · dp as desired.

 (g) We have F · dx = (F dt) · (dx/dt) = v · dp = dK using the result of part (f), as desired.
   Remark
   The result of part (e) is equivalent to saying that momentum p is always associated with the
   motion of energy Ev. This is a very general statement, which also holds at the differential
   level: momentum density is equal to energy flux density. One example of this was given
   in E7, where it was noted that the electromagnetic momentum density p was equal to the
   Poynting vector S, in units where c = 1.

   Idea 2
   In relativistic dynamics problems, it is almost always better to work with energy and mo-
   mentum than velocity; one typically shouldn’t even mention velocities unless the problem
   asks for or gives them.

We’ll start with some very simple problems to warm up, setting c = 1 throughout.

       Example 1: KK 13.5

       A particle of mass m and speed v collides and sticks to a stationary particle of mass M . Find
       the final speed of the composite particle.

       Solution
       The total four momentum is (E, p) = (γm + M, γmv), so the final speed is
                                      p    γmv            v
                               vf =     =        =          √        .
                                      E   γm + M   1 + (M/m) 1 − v 2

       Example 2: Morin 12.2

       Two photons of energy E collide at an angle θ and create a particle of mass M . What is M ?

       Solution
       The total four-momentum is

                                       pµ = (2E, E(1 + cos θ), E sin θ).

       The mass is just the norm of the four-momentum, so
                      q                                     √
                M = 4E 2 − E 2 (1 + cos θ)2 − E 2 sin2 θ = E 2 − 2 cos θ = 2E sin(θ/2).

[1] Problem 2 (Morin 12.4). A stationary mass MA decays into masses MB and MC . What are the
    energies of these two masses?
    Solution. In the lab frame, the momenta of the masses B and C adds to zero, so p2B = p2C , so

                                           EB − MB2 = EC2 − MC2 .

    We also know that EB + EC = MA , so simplifying gives
                                                            MB2 − MC2
                                           EB − EC =                  .
                                                               MA
    Therefore, we conclude
                                    MA2 + MB2 − MC2                MA2 − MB2 + MC2
                             EB =                   ,       EC =                   .
                                         2MA                            2MA


[User] Two point light sources are positioned on different points of a convex lens’ optical axis. The images of these light sources formed by the lens coincide. It is known that one light source is at a distance $a=\SI{18}{cm}$ from the center of the lens. How far is the other light source distanced from this light source? The focal length of the lens is $f=\SI{9}{cm}$.

Solution:
\emph{Esimene lahendus}\\
Kui optilisel peateljel paiknev valgusallikas asub läätsest \SI{18}{cm} kaugusel, mis on võrdne kahekordse fookuskaugusega, siis selle valgusallika kujutis asub teisel pool läätse läätsest samuti kahe fookuskaugusel ehk \SI{18}{cm} kaugusel. Et kahe valgusallika kujutised kattuksid, peab teine kujutis olema näilik.
Konstrueerime valgusallika asukoha, kui kujutise asukoht on teada.
 \vspace{-10pt}
 \begin{center}
 \includegraphics[width=0.7\linewidth]{2018-v2g-02-valgusallikaslah}
 \end{center}
 \vspace{-10pt}

Sarnastest kolmnurkadest $\triangle KAO$ ja $\triangle KBF$ saame, et
\[ \frac{AO}{BF} = \frac{KO}{KF} = \frac{2f}{3f} = \frac{2}{3}. \]
Teisest sarnaste kolmnurkade paarist $\triangle SAO$ ja $\triangle OBF$ saame, et
\[ \frac{AO}{BF} = \frac{OS}{FO} = \frac{b}{f}, \]
kus b on teise valgusallika kaugus läätsest. Ühendades kaks seost, saame
\[ \frac{KO}{KF}=\frac{OS}{FO} \quad\Rightarrow\quad \frac{2}{3} = \frac{b}{f} \quad\Rightarrow\quad b = \frac{2f}{3} = \SI{6}{cm}. \]
Teine valgusallikas peab asuma läätsest \SI{6}{cm} kaugusel ning kahe valgusallika kaugus teineteisest on \SI{6}{cm} + \SI{18}{cm} = \SI{24}{cm}.

\vspace{\baselineskip}
\emph{Teine lahendus}\\
Sarnaste kolmnurkade asemel saame kasutada läätse valemit. Teades, et valgusallikas asub läätsest $a = \SI{18}{cm}$ kaugusel ning läätse fookuskaugus on $f=\SI{9}{cm}$, saame leida kujutise kauguse $k$ läätsest:
\[ \frac{1}{a} + \frac{1}{k} = \frac{1}{f} \quad\Rightarrow\quad k = \SI{18}{cm}. \]
Teise valgusallika kujutis peab olema samuti samas punktis, kus esimese valgusallika oma, seega peab teise valgusallikaga tekitatud kujutis olema näiline ning valgusallikas peab asuma kujutisega samal pool läätse. Kasutades läätse valemit, leiame teise valgusallika asukoha
\[ \frac{1}{a} - \frac{1}{k} = \frac{1}{f} \quad\Rightarrow\quad a = \SI{6}{cm}. \]
Teine valgusallikas peab asuma läätsest \SI{6}{cm} kaugusel ning kahe valgusallika kaugus teineteisest on \SI{6}{cm} + \SI{18}{cm} = \SI{24}{cm}.

Kaks valgusallikat

Kaks punktikujulist valgusallikat asuvad kumerläätse optilisel peateljel erinevates punktides. Nendest valgusallikatest läätse abil tekitatud kujutised kattuvad. On teada, et üks valgusallikas asub läätse keskpunktist $a=\SI{18}{cm}$ kaugusel. Kui kaugel sellest valgusallikast asub teine valgusallikas? Läätse fookuskaugus $f=\SI{9}{cm}$.

Selleks, et kaks valgusallikat sama kujutise annaksid, peab üks olema näiline ja teine tegelik.


[User]
Find how much force does the blade of the bolt cutter apply on the bolt (see figure) if the force applied on the handle is $F = \SI{90}{N}$.
\begin{center}
    \includegraphics[width=0.7\textwidth]{2015-v2g-07-poldiloikur}
  \end{center}

Solution:
Jõu ülekandmine toimub kangi põhimõttel, kus jõumoment pöördtelje suhtes summaarselt on võrdne nulliga. Käepideme korral
\[ \SI{600}{mm}\cdot\SI{90}{N}-\SI{100}{mm}\cdot F_k = 0 \quad\Rightarrow\quad F_k = \SI{540}{N}, \]
kus $F_k$ on käepidemetelt lõiketeradele mõjuv jõud. Analoogselt lõiketerade korral
\[ \SI{160}{mm}\cdot F_k - \SI{80}{mm}\cdot F_l = 0 \quad\Rightarrow\quad F_l = 2F_k, \]
kus $F_l$ on lõikuri poolt poldile avaldatud jõud.\\
Nendest võrranditest saame leida lõiketeradele mõjuva jõu $F_l$
\[ F_l = \SI{1080}{N}.\]

Poldilõikur

Leida, kui suurt jõudu avaldab poldilõikuri tera poldile (vt joonis), kui käepidemetele avaldatud jõud on $F = \SI{90}{N}$.
 \begin{center}
 \includegraphics[width=0.7\textwidth]{2015-v2g-07-poldiloikur}
 \end{center}

Jõu ülekanne toimub kangi põhimõttel, kus summaarne jõumoment pöördtelje suhtes on võrdne nulliga. Vastavad jõumomentide tasakaalu saab kirja panna nii käepideme kui ka lõiketerade jaoks.


[User]
EuPhO 2017, problem 1. (Hint: don’t try to use fancy math here. EuPhO
    problems are designed to be solved with only elementary math and graph reading.)

       Idea 5
       Standing wave solutions also exist for waves in more than one spatial dimension. In the special
       case where the wave medium is uniform, and shaped like a rectangle (in two dimensions) or
       a rectangular prism (in three dimensions), all the standing wave solutions can be found by
       separation of variables. That is, they can all be written as

                                    ψ(x, y, z, t) = f (x)g(y)h(z) cos(ωt)

       where f , g, and h are sinusoids.

Solution:
See the official solutions here.

       Idea 5
       Standing wave solutions also exist for waves in more than one spatial dimension. In the special
       case where the wave medium is uniform, and shaped like a rectangle (in two dimensions) or
       a rectangular prism (in three dimensions), all the standing wave solutions can be found by
       separation of variables. That is, they can all be written as

                                     ψ(x, y, z, t) = f (x)g(y)h(z) cos(ωt)

       where f , g, and h are sinusoids.


[User]
Mootorpaat sõidab jõe ühelt kaldalt punktist A teisele kaldale punkti B. Paadi kiirus on $u=\SI{7}{m/s}$.\\
\osa Joonisel on näidatud paadi tekitatud veelained. Milline on jõe voolukiirus?\\
\osa On teada, et kui vee sügavus on $h$, siis lained levivad kiirusega $w=\sqrt{gh}$, kus $g$ on vabalangemise kiirendus.
Kui sügav on jõgi?

\begin{center}
\includegraphics[width=0.55\linewidth]{2009-v3g-08-paat.pdf}
\end{center}

Solution:
\osa
Veega seotud taustsüsteemis liiguvad lained paadi trajektoori suhtes sümmeetriliselt. Seega, veega seotud taustsüsteemis on
paadi trajektoor lainetest moodustatud nurga poolitaja. Paadi kiirusest $\vec u$, jõe voolukiirusest $\vec v$ ja paadi kiirusest maa suhtes moodustub kiiruste kolmnurk, vt joonis.
Jooniselt mõõdame selle kolmnurga teravama nurga siinuse, $\sin \alpha =v/u=\num{0.26}$, millest $v=\SI{1.8}{m/s}$.

\begin{center}
	\includegraphics[width=0.8\linewidth]{2009-v3g-08-paatlah.pdf}
\end{center}

\osa
Kui paat tekitas teatud punktis häirituse, siis levis see ajaga $t$ kaugusele $wt$ (nähtavaks paadilaineks on selliste ringide mähisjoon),
paat aga liikus kaugusele $ut$. Seega leiame jooniselt pikkuste suhte abil $w/u=|OP|/|OQ|=\num{0.64}$, millest $w=\SI{4.5}{m/s}$. Järelikult on vee sügavus
$h=w^2/g=2\,$m.

Kaater

Mootorpaat sõidab jõe ühelt kaldalt punktist A teisele kaldale punkti B. Paadi kiirus on $u=\SI{7}{m/s}$.\\
\osa Joonisel on näidatud paadi tekitatud veelained. Milline on jõe voolukiirus?\\
\osa On teada, et kui vee sügavus on $h$, siis lained levivad kiirusega $w=\sqrt{gh}$, kus $g$ on vabalangemise kiirendus.
Kui sügav on jõgi?

\begin{center}
\includegraphics[width=0.55\linewidth]{2009-v3g-08-paat.pdf}
\end{center}

Olukorda on mugavam vaadelda veega seotud taustsüsteemis, sest siis liiguvad lained paadi trajektoori suhtes sümmeetriliselt (lainefrondid on paadi varasematest asukohtadest eemalduvad ringid).


[User]
USAPhO 2006, problem A2.
[2] Problem 3 (Kalda). Not all circuits are made of only series and parallel combinations. The Y-∆
    transform is the next simplest tool you can use. Consider the two sets of resistors shown below.

    The two are equivalent provided that
                                    RAB RBC            1         1/RB RC
                         RA =                   ,         =
                                RAB + RAC + RBC       RBC   1/RA + 1/RB + 1/RC
    along with cyclic permutations. As an application, consider the circuit below.

    Find the current through the battery using a Y-∆ transform.

Solution:
It’s most convenient to apply the Y-∆ transform to the top vertex, turning it from a Y
    into a ∆ of three 9 Ω resistors. At this point the circuit can be simplified using the usual series and
    parallel rules, giving Req = 19/7 Ω and thus I = 21/19 A.


[User]
For a wave on a string, there are two contributions to the energy: potential energy
    from stretching, and kinetic energy from transverse motion.

      (a) Find the kinetic and potential energy density (i.e. energy per unit length) of the string in
          terms of T , µ, y, and its derivatives.

      (b) Evaluate the above quantities for y = A cos(kx − ωt). Is the total energy density uniform?

      (c) Show that for a general traveling wave of the form y = f (x−vt), the total kinetic and potential
          energy are equal.

      (d) Show that for any wave function y, total energy is conserved. This will require some integration
          by parts, as well as the wave equation itself; you should assume y goes to zero at infinity.

      (e) Compute the energy of the static configuration in problem 2(b), assuming the triangle has
          height h and base L, where h ≪ L.

    One warning: as we saw in E6, energy is quadratic, so it does not obey the superposition principle.
    Locally, the amount of energy can be more or less than the sum of the energies of the superposed
    waves, due to interference.

[2] Problem 7 (French 7.23). One end of a stretched string is moved transversely at constant velocity
    u for a time τ , and is moved back to its starting point with velocity −u during the next interval τ .
    As a result, a triangular pulse is set up on the string and moves along it with speed v. Show that
    the total energy of the pulse is equal to the work done on the string, working to lowest order in u/v.

       Remark
       How can we account for damping in the wave equation? The simplest thing would be to add
       a force proportional to vy , which e.g. could be due to air drag. Then

                                            ∂t2 y = v 2 ∂x2 y + A ∂t y.

        But what if the string is in a vacuum? Then the simplest kind of damping would be due to
        the energy lost in bending and unbending of the string, which takes the form

                                           ∂t2 y = v 2 ∂x2 y + A ∂t ∂x2 y

        because ∂x2 y describes the bending. This is called Kelvin–Voigt damping.

        In both cases, it’s straightforward to handle the damping since the wave equation remains
        linear; we just plug in a solution of the form ei(kx−ωt) and find the new relation between ω
        and k. If we pick k to be a real number, we will generally find ω to be complex, with its
        imaginary part corresponding to exponential decay of the wave over time.

Solution:
(a) We will assume that the displacements of the string is small, and take the lowest
     order terms. Using 12 mv 2 for kinetic energy of a piece moving in the transverse direction gets
                                         1         1      p
                                   ∆K = ∆mẏ 2 = (µ∆x 1 + y ′2 )ẏ 2 ,
                                         2         2
                                      dK    1    p            1
                                          = µẏ 2 1 + y ′2 ≈ µẏ 2 .
                                      dx    2                 2
     For
     p   the  potential energy,  the work done on stretching the string is ∆U = T ∆ℓ where ∆ℓ =
             ′2             1 ′2
       1 + y ∆x − ∆x ≈ 2 y ∆x since the displacements are small. Thus
                                                 dU  1
                                                    = T y ′2 .
                                                 dx  2
 (b) We have
                      dK      1                      dU    1
                           = µA2 ω 2 sin2 (kx − ωt),     = T A2 k 2 sin2 (kx − ωt).
                      dx      2                      dx    2
     Here, we can see that the total energy density is not uniform, but rather comes in “lumps”.
     This is also true for electromagnetic waves.

 (c) The densities are
                                     dK      1          dU     1
                                          = µv 2 f ′2 ,     = T f ′2
                                      dx     2          dx     2
     and for a wave traveling in one direction, these densities are exactly equal because v 2 = T /µ,
     so the total kinetic and potential energy are equal. This is also true for ideal simple harmonic
     oscillations.

 (d) The total energy is                     Z ∞
                                                    1 2 1 ′2
                                       E=             µẏ + T y   dx.
                                              −∞    2      2
     Taking the time derivative and applying the wave equation,
                                  Z ∞                      Z ∞
                           dE                      ′ ′
                                =     µẏ ÿ + T ẏ y dx ∝     ẏy ′′ + ẏ ′ y ′ dx
                            dt     −∞                       −∞

     where we used the wave equation in the second equality. Integrating the first term by parts,
                               Z ∞                    ∞   Z ∞
                                    ẏy ′′ dx = ẏy ′   −     ẏ ′ y ′ dx
                                      −∞                 −∞      −∞

     and the boundary term vanishes by our assumptions. The remaining term is just the opposite
     of the other term in dE/dt, so dE/dt = 0 as desired.

 (e) Since the string was initially held steady, there is only potential energy. The amount of
     potential energy is just T times the total length the string is stretched, so
                                             p                2T h2
                                      U = T ( 4h2 + L2 − L) ≈
                                                               L
     where we used h ≪ L in the last step.

[2] Problem 7 (French 7.23). One end of a stretched string is moved transversely at constant velocity
    u for a time τ , and is moved back to its starting point with velocity −u during the next interval τ .
    As a result, a triangular pulse is set up on the string and moves along it with speed v. Show that
    the total energy of the pulse is equal to the work done on the string, working to lowest order in u/v.
    Solution. First let’s compute the energy in the pulse. Recall that the energy density is
                                                                           2 !
                               µ ∂y 2 T ∂y 2 T               1 ∂y 2

                          dE                                                ∂y
                    u=       =           +            =       2
                                                                        +           .
                          dx   2 ∂t        2 ∂x          2 v      ∂t        ∂x

    The triangular pulse has height uτ , and the two halves of it have length vτ . Thus, |∂y/∂t| = u and
    |∂y/∂x| = u/v across the pulse, so the total energy is
                                                               2T τ u2
                                           U = (2vτ )u =               .
                                                                 v
    When lifting the string to create the pulse, the string was at an angle of u/v to first order in u/v,
    so the transverse force that needed to be applied was T u/v. The distance over which this force was
    applied was uτ , for a total work of T τ u2 /v. The same work was done when bringing it down with
    constant velocity, so the total work done was 2T τ u2 /v, as expected.
       Remark
       How can we account for damping in the wave equation? The simplest thing would be to add
       a force proportional to vy , which e.g. could be due to air drag. Then

                                            ∂t2 y = v 2 ∂x2 y + A ∂t y.

       But what if the string is in a vacuum? Then the simplest kind of damping would be due to
       the energy lost in bending and unbending of the string, which takes the form

                                           ∂t2 y = v 2 ∂x2 y + A ∂t ∂x2 y

       because ∂x2 y describes the bending. This is called Kelvin–Voigt damping.

       In both cases, it’s straightforward to handle the damping since the wave equation remains
       linear; we just plug in a solution of the form ei(kx−ωt) and find the new relation between ω
       and k. If we pick k to be a real number, we will generally find ω to be complex, with its
       imaginary part corresponding to exponential decay of the wave over time.


[User]
Consider an exponentially distributed quantity,
                                           (
                                             ae−ax x ≥ 0,
                                  p(x) =
                                             0       otherwise.

    Verify that the total probability is 1, and compute the mean and standard deviation. To perform
    the integrals, you will have to integrate by parts.

Solution:
First, to check normalization,
                          Z ∞           Z ∞          Z ∞
                                               −ax
                              p(x)dx =       ae dx =     e−u du = 1 − 0 = 1.
                           −∞               0                 0

    Now, the mean can be evaluated using integration by parts,
                      Z ∞                            Z ∞
                                                ∞                     1    ∞ 1
                ⟨x⟩ =      xae−ax dx = −xe−ax +          e−ax dx = 0 − e−ax = .
                       0                        0     0               a    0 a
    To calculate the standard deviation, we must evaluate
                               Z ∞                  Z ∞
                                    2 −ax                            2      2

                        ⟨x ⟩ =     x ae dx = 0 +        (2x)e−ax dx = ⟨x⟩ = 2 .
                                0                    0               a     a
   We thus conclude                                           r
                                        p                         2     1  1
                                 σX =       ⟨X 2 ⟩ − ⟨X⟩2 =         2
                                                                      − 2 = .
                                                                  a    a   a


[User]
Blackbody radiation is an electromagnetic phenomenon, so the radiation intensity
    depends on the speed of light c. It is also a thermal phenomenon, so it depends on the thermal
    energy kB T , where T is the object’s temperature and kB is Boltzmann’s constant. And it is a
    quantum phenomenon, so it depends on Planck’s constant h.

     (a) Using the relation E = hf , find the dimensions of h.

     (b) Using dimensional analysis, show that the power emitted by the blackbody per unit area,
         called the radiation intensity I, obeys I ∝ T 4 , and find the constant of proportionality up to
         a dimensionless constant.

        (c) How would the result change in a world with d spatial dimensions?

     (d) The result you derived in part (b) is known as the Stefan–Boltzmann law. But if it can be
         derived with pure dimensional analysis, without needing any detailed calculations or experi-
         mental data at all, then why is it considered a law at all? Isn’t it obvious?

    3      Electromagnetism
    These problems can be solved using the material in chapters 25 through 38 of Halliday and Resnick.

Solution:
Blackbody radiation is an electromagnetic phenomenon, so the radiation intensity
    depends on the speed of light c. It is also a thermal phenomenon, so it depends on the thermal
    energy kB T , where T is the object’s temperature and kB is Boltzmann’s constant. And it is a
    quantum phenomenon, so it depends on Planck’s constant h.

     (a) Using the relation E = hf , find the dimensions of h.

     (b) Using dimensional analysis, show that the power emitted by the blackbody per unit area,
         called the radiation intensity I, obeys I ∝ T 4 , and find the constant of proportionality up to
         a dimensionless constant.

        (c) How would the result change in a world with d spatial dimensions?

     (d) The result you derived in part (b) is known as the Stefan–Boltzmann law. But if it can be
         derived with pure dimensional analysis, without needing any detailed calculations or experi-
         mental data at all, then why is it considered a law at all? Isn’t it obvious?

    Answer. (c) I ∝ T d+1

    3      Electromagnetism
    These problems can be solved using the material in chapters 25 through 38 of Halliday and Resnick.


[User]
INPhO 2013, problem 2. A short problem on X-ray diffraction.

    To finish up, here are three neat questions that each cover a broad range of topics.

Solution:
See the official solutions here.

                   01h
   To finish up, here are three neat questions that each cover a broad range of topics.


[User]
Question B2
The nature of magnetic dipoles.

  a. A “Gilbert” dipole consists of a pair of magnetic monopoles each with a magnitude qm but
     opposite magnetic charges separated by a distance d, where d is small. In this case, assume
     that −qm is located at z = 0 and +qm is located at z = d.

                                         −qm      qm
                                                                  z
                                         z=0 z=d

     Assume that magnetic monopoles behave like electric monopoles according to a coulomb-like
     force
                                             µ0 qm1 qm2
                                       F =
                                             4π r2
     and the magnetic field obeys
                                          B = F/qm .

        i. What are the dimensions of the quantity qm ?
       ii. Write an exact expression for the magnetic field strength B(z) along the z axis as a
           function of z for z > d. Write your answer in terms of qm , d, z, and any necessary
           fundamental constants.
      iii. Evaluate this expression in the limit as d → 0, assuming that the product qm d = pm
           is kept constant, keeping only the lowest non-zero term. Write your answer in terms of
           pm , z, and any necessary fundamental constants.

  b. An “Ampère” dipole is a magnetic dipole produced by a current loop I around a circle of
     radius r, where r is small. Assume the that the z axis is the axis of rotational symmetry for
     the circular loop, and the loop lies in the xy plane at z = 0.

                                                      I

                                                                  z

        i. Write an exact expression for the magnetic field strength B(z) along the z axis as a
           function of z for z > 0. Write your answer in terms of I, r, z, and any necessary
           fundamental constants.
       ii. Let kIrγ have dimensions equal to that of the quantity pm defined above in Part aiii,
           where k and γ are dimensionless constants. Determine the value of γ.
      iii. Evaluate the expression in Part bi in the limit as r → 0, assuming that the product
           kIrγ = p0m is kept constant, keeping only the lowest non-zero term. Write your answer
           in terms of k, p0m , z, and any necessary fundamental constants.
      iv. Assuming that the two approaches are equivalent, pm = p0m . Determine the constant k
          in Part bii.
                  Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam               Part B                                              11

  c. Now we try to compare the two approaches if we model a physical magnet as being composed
     of densely packed microscopic dipoles.

                                                   L

                                                            R      z

    A cylinder of this uniform magnetic material has a radius R and a length L. It is composed
    of N magnetic dipoles that could be either all Ampère type or all Gilbert type. N is a very
    large number. The axis of rotation of the cylinder and all of the dipoles are all aligned with
    the z axis and all point in the same direction as defined above so that the magnetic field
    outside the cylinder is the same in either dipole case as you previously determined. Below is
    a picture of the two dipole models; they are cubes of side d << R and d << L with volume
    vm = d3 .

                                  d
                   d                                       current sheet I

                                            d
                                                                                z
               −qm
                                      +qm
                       Gilbert Dipole                     Ampère Dipole

       i. Assume that R  L and only Gilbert type dipoles, determine the magnitude and direc-
          tion of B at the center of the cylinder in terms of any or all of pm , R, L, vm , and any
          necessary fundamental constants.
      ii. Assume that R  L and only Ampère type dipoles, determine the magnitude and
          direction of B at the center of the cylinder in terms of any or all of pm , R, L, vm , and
          any necessary fundamental constants.

                 Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam      Answer Sheets                          12

                         Answer Sheets

 Following are answer sheets for some of the graphical portions of the test.

               Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                              Answer Sheets                      13

                                        240

                                        200

                                        160

                       Pressure (kPa)
                                        120

                                         80

                                         40

                                         0
                                              280   300      320   340     360     380   400
                                                              Temperature (K)

                                   240

                                   200

                                   160

                 Pressure (kPa)
                                   120

                                        80

                                        40

                                         0
                                              0     1        10    102     103     104   105
                                                          Volume (in terms of V0 )

               Copyright c 2015 American Association of Physics Teachers

Solution:
Question B2
The nature of magnetic dipoles.
  a. A “Gilbert” dipole consists of a pair of magnetic monopoles each with a magnitude qm but
     opposite magnetic charges separated by a distance d, where d is small. In this case, assume
     that −qm is located at z = 0 and +qm is located at z = d.

                                         −qm      qm
                                                                        z
                                        z=0 z=d

     Assume that magnetic monopoles behave like electric monopoles according to a coulomb-like
     force
                                             µ0 qm1 qm2
                                       F =
                                             4π r2
     and the magnetic field obeys
                                          B = F/qm .

        i. What are the dimensions of the quantity qm ?

                                                 Solution
          By the second expression, qm must be measured in Newtons per Tesla. But since Tesla
          are also Newtons per Ampere per meter, then qm is also measured in Ampere meters.

       ii. Write an exact expression for the magnetic field strength B(z) along the z axis as a
           function of z for z > d. Write your answer in terms of qm , d, z, and any necessary
           fundamental constants.

                                                 Solution
          Adding the two terms,
                                                 µ 0 qm   µ 0 qm
                                      B(z) = −        2
                                                        +             .
                                                 4π z     4π (z + d)2

      iii. Evaluate this expression in the limit as d → 0, assuming that the product qm d = pm
           is kept constant, keeping only the lowest non-zero term. Write your answer in terms of
           pm , z, and any necessary fundamental constants.

                                                 Solution
          Simplifying our previous expression,

                                              µ0             2 + d/z
                                       B(z) =    qm d                       .
                                              4π            z(z + d)2
          Thus in the limit d → 0 we have
                                                  µ 0 qm d   µ0 p m
                                        B(z) =          3
                                                           =        .
                                                  2π z       2π z 3

                  Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam               Part B                                             19

  b. An “Ampère” dipole is a magnetic dipole produced by a current loop I around a circle of
     radius r, where r is small. Assume the that the z axis is the axis of rotational symmetry for
     the circular loop, and the loop lies in the xy plane at z = 0.

                                                      I

                                                                   z

       i. Write an exact expression for the magnetic field strength B(z) along the z axis as a
          function of z for z > 0. Write your answer in terms of I, r, z, and any necessary
          fundamental constants.

                                                Solution
          Applying the Biot-Savart law, with s the vector from the point on the loop to the point
          on the z axis,
                                                dl × s
                                              I
                                         µ0 I            µ0 I 2πr
                                B(z) =                 =              sin θ
                                         4π       s3      4π r2 + z 2
          where θ is the angle between the point on the loop and the center of the loop as measured
          by the point on the z axis, so
                                                           r
                                             sin θ = √          .
                                                       r + z2
                                                         2

          Then we have
                                                  µ0 I     2πr2
                                          B(z) =                     .
                                                   4π (r2 + z 2 )3/2

       ii. Let kIrγ have dimensions equal to that of the quantity pm defined above in Part aiii,
           where k and γ are dimensionless constants. Determine the value of γ.

                                                Solution
          We know pm must have dimensions of Amperes times meters squared, so γ = 2.

      iii. Evaluate the expression in Part bi in the limit as r → 0, assuming that the product
           kIrγ = p0m is kept constant, keeping only the lowest non-zero term. Write your answer
           in terms of k, p0m , z, and any necessary fundamental constants.

                                                Solution
          Using our previous result,

                                       µ0 I   2πr2         µ0 I πr2   µ0 π p0m
                              B(z) =                     ≈          =
                                       4π (r2 + z 2 )3/2   2π z 3     2π k z 3

                  Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam               Part B                                             20

      iv. Assuming that the two approaches are equivalent, pm = p0m . Determine the constant k
          in Part bii.

                                                Solution
         By inspection, k = π.

  c. Now we try to compare the two approaches if we model a physical magnet as being composed
     of densely packed microscopic dipoles.

                                                  L

                                                            R      z

    A cylinder of this uniform magnetic material has a radius R and a length L. It is composed
    of N magnetic dipoles that could be either all Ampère type or all Gilbert type. N is a very
    large number. The axis of rotation of the cylinder and all of the dipoles are all aligned with
    the z axis and all point in the same direction as defined above so that the magnetic field
    outside the cylinder is the same in either dipole case as you previously determined. Below is
    a picture of the two dipole models; they are cubes of side d << R and d << L with volume
    vm = d3 .

                                  d
                   d                                       current sheet I

                                            d
                                                                                z
               −qm
                                      +qm
                       Gilbert Dipole                     Ampère Dipole

       i. Assume that R  L and only Gilbert type dipoles, determine the magnitude and direc-
          tion of B at the center of the cylinder in terms of any or all of pm , R, L, vm , and any
          necessary fundamental constants.

                                                Solution

                 Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                Part B                                             21

         The monopoles that make up the dipoles cancel out except on the flat surfaces. Then
         the cylinder acts like a parallel plate capacitor.
         If the size of a dipole is d, then the surface density of monopole charge is

                                                 σm = qm /d2 .

         Using the analogy with a parallel place capacitor, the magnitude of B is
                                                                 pm
                                              B = µ0 σ m = µ0
                                                                 d3
         and the direction is to the left.

      ii. Assume that R  L and only Ampère type dipoles, determine the magnitude and
          direction of B at the center of the cylinder in terms of any or all of pm , R, L, vm , and
          any necessary fundamental constants.

                                                 Solution
         The currents that make up the dipoles all cancel out except on the cylindrical surfaces.
         Then the cylinder acts like a solenoid, with
                                                        µ0 I
                                                   B=
                                                         d
         where I/d is the surface current density. The magnitude of B is

                                                   µ0 I     pm
                                              B=        = µ0 3
                                                    d       d
         and the direction is to the right.

                 Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam      Answer Sheets                          22

                         Answer Sheets

 Following are answer sheets for some of the graphical portions of the test.

               Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                              Answer Sheets                      14

                                        240

                                        200

                                        160

                       Pressure (kPa)
                                        120

                                         80

                                         40

                                         0
                                              280   300      320   340     360     380   400
                                                              Temperature (K)

                                   240

                                   200

                                   160

                 Pressure (kPa)
                                   120

                                        80

                                        40

                                         0
                                              0     1        10    102     103     104   105
                                                          Volume (in terms of V0 )

               Copyright c 2015 American Association of Physics Teachers

USAPhO 2015 Question B2


[User]
Anumasse, milles oli $V=\SI{1}{l}$ vett temperatuuril $t_1=\SI{20}{\degreeCelsius}$,
visati rauatükk massiga $m=\SI{100}{g}$ temperatuuril $t_0=\SI{500}{\degreeCelsius}$. Osa
veest aurustus. Mõne aja pärast mõõdeti vee temperatuuriks
$t_2=\SI{24}{\degreeCelsius}$. Kui palju vett aurustus? Vee erisoojus
$c_1=\SI{4200}{J/(kg\cdot\degreeCelsius)}$, aurustumissoojus $L=\SI{2,26 e6}{J/kg}$ ja
tihedus $\rho=\SI{1000}{kg/m^3}$; raua
erisoojus $c_2=\SI{460}{J/(kg\cdot\degreeCelsius)}$. Anum on tühise soojusmahtuvusega ning väliskeskkonnast
hästi isoleeritud.

Solution:
Algne vee mass on $M=\rho V=\SI{1000}{g}$. Olgu väljaaurustunud vee mass $m_0$. Selle soojendamiseks keemistemperatuurini $t=\SI{100}{\degreeCelsius}$ ning aurustamiseks läheb vaja energiat
\[
Q_1=m_0(c_1(t-t_1)+L).
\]
Ülejäänud vee soojendamiseks temperatuurini $t_2$ läheb energiat
\[
Q_2=(M-m_0)c_1(t_2-t_1).
\]
Raua jahtumisel eraldub energia $Q_3=mc_2(t_0-t_2)$. Energia jäävuse seaduse kohaselt $Q_1+Q_2=Q_3$, ehk
\[
m_0(c_1(t-t_1)+L)+(M-m_0)c_1(t_2-t_1)=mc_2(t_0-t_2).
\]
Siit
\[
m_0=\frac{mc_2(t_0-t_2)-Mc_1(t_2-t_1)}{c_1(t-t_2)+L}\approx \SI{2}{g}.
\]

Rauatükk

Anumasse, milles oli $V=\SI{1}{l}$ vett temperatuuril $t_1=\SI{20}{\degreeCelsius}$,
visati rauatükk massiga $m=\SI{100}{g}$ temperatuuril $t_0=\SI{500}{\degreeCelsius}$. Osa
veest aurustus. Mõne aja pärast mõõdeti vee temperatuuriks
$t_2=\SI{24}{\degreeCelsius}$. Kui palju vett aurustus? Vee erisoojus
$c_1=\SI{4200}{J/(kg\cdot\degreeCelsius)}$, aurustumissoojus $L=\SI{2,26 e6}{J/kg}$ ja
tihedus $\rho=\SI{1000}{kg/m^3}$; raua
erisoojus $c_2=\SI{460}{J/(kg\cdot\degreeCelsius)}$. Anum on tühise soojusmahtuvusega ning väliskeskkonnast
hästi isoleeritud.

Energia jäävuse seaduse kohaselt peab rauatükist eraldunud soojusenergia minema vee soojendamiseks ja aurustumiseks.