[User]
Raske tellis kukub poole meetri kõrguselt jäigale lühikesele vedrule. Põrge
on elastne ja tellis lendab peaaegu algsele kõrgusele tagasi. Kui
kõrgele maast kerkib vedru pärast põrget?

Solution:
Hetkel, mil tellis vedrust eemalduma hakkab, liigub vedru ülemine ots koos tellisega kiirusega $v$ üles. Tellise kiirus on piisav, et kerkida tagasi esialgsele kõrgusele $H\propto v^2$. Eeldame, et vedru on pikisuunas ühtlane, sellisel juhul liigub vedru massikese ülemise ja alumise otsa keskmise kiirusega ülesse. Niisiis, $v\idx{vedru}=v/2$. Seega tõuseb vedru massikese kõrgusele $h\propto v\idx{vedru}^2=v^2/4$, ehk
\[h=\frac{H}{4}.\]

Vedru

Raske tellis kukub poole meetri kõrguselt jäigale lühikesele vedrule. Põrge
on elastne ja tellis lendab peaaegu algsele kõrgusele tagasi. Kui
kõrgele maast kerkib vedru pärast põrget?

Tellise eemaldumise hetkel on vedru alumine ots paigal, aga ülemine ots liigub tellisega sama kiirusega üles.


[User]
4. A student performed an experiment to determine the acceleration due to gravity (g) using a simple
   pendulum which has a spherical bob of diameter d hung with a long string. She varied the length

Page 4        Code 65                       Questions                     Last six digits of Roll No.:

  of the string l, and measured the period of oscillation T every time. She calculated the value of g
  from each measurement as shown in the table below.
  She noticed that not only was the average value of g smaller than the expected value, each one of
  the measurements had yielded a value smaller than the true value.
  Next, she plotted a graph between T 2 and l from the same data, and obtained the value of
  g = 981 cm/s2 from the slope of the best ﬁt line.

                                                                  5
                                                                          Experimental data
                                                                          Linear best ﬁt
                                                                  4
         l (in cm) T (in s)   g (in cm/s)
             20     0.93          912                             3

                                                  T 2 (s2 )
             40     1.29          948
             60     1.58          948                             2
             80     1.81          963
             100    2.02          967
                                                                  1
         Average g                947

                                                                  0
                                                                      0         40             80        120
                                                                                l (cm)

   (a) [3 marks] What do you think might be the main cause for the consistently low values of g
       that she obtained from each of her measurements?
   (b) [4 marks] Explain in detail why she still obtained a correct value of g from the slope of the
       graph plotted from the same data.
  Assume that the instruments of measuring time and length were accurate enough, and all the
  measurements of the stated quantities were correct within the accuracy of the instruments. It is
  veriﬁed that the graph and the linear best ﬁt were correctly plotted, and all numerical calculations
  in the above are correct. Note that you are not expected to plot any graph (no graph paper is
  provided to you).

5. [12 marks] A circuit consists of an emf source and ﬁve resistors with unknown resistances. When
   an ideal ammeter is connected between points 1 and 2, its reading is IA . If instead a resistor R
   is connected to the same two points, the current through that resistor is IR . If instead an ideal
   voltmeter is connected between points 1 and 2, its reading is V . Obtain V in terms of IA , R and
   IR only.

                                       R1                                 R3
                                                              1
                                       R2                     2           R4

                                       R0

                        **** END OF THE QUESTION PAPER ****

Solution:
4. A student performed an experiment to determine the acceleration due to gravity (g) using a simple
   pendulum which has a spherical bob of diameter d hung with a long string. She varied the length
   of the string l, and measured the period of oscillation T every time. She calculated the value of g
   from each measurement as shown in the table below.
  She noticed that not only was the average value of g smaller than the expected value, each one of
  the measurements had yielded a value smaller than the true value.
  Next, she plotted a graph between T 2 and l from the same data, and obtained the value of
  g = 981 cm/s2 from the slope of the best ﬁt line.

                                                              5
                                                                      Experimental data
                                                                      Linear best ﬁt
                                                              4
         l (in cm) T (in s)   g (in cm/s)
             20     0.93          912                         3

                                                  T 2 (s2 )
             40     1.29          948
             60     1.58          948                         2
             80     1.81          963
             100    2.02          967
                                                              1
         Average g                947

                                                              0
                                                                  0        40             80    120
                                                                                l (cm)

   (a) [3 marks] What do you think might be the main cause for the consistently low values of g
       that she obtained from each of her measurements?

           Solution: A low value of g from the formula g = 4π 2 l/T 2 can result either from an
           underestimation of l or an overestimation of T .
           The latter can happen in case of either a fast-running clock (stopwatch) or a consistent
           error in the measurement. We are told that the instruments were accurate and the
           measurements were properly made. So this possibility is ruled out.
           Under the given assumption that the measurements of l were accurate, the only way that
           the value of l can be underestimated systematically is by ignoring the size of the bob of
           the pendulum. Since the student recorded only the length of the string, and did not add
           the radius of the bob, this caused an underestimation of l, and consequently, of g.

   (b) [4 marks] Explain in detail why she still obtained a correct value of g from the slope of the
       graph plotted from the same data.
  Assume that the instruments of measuring time and length were accurate enough, and all the
  measurements of the stated quantities were correct within the accuracy of the instruments. It is
  veriﬁed that the graph and the linear best ﬁt were correctly plotted, and all numerical calculations
  in the above are correct. Note that you are not expected to plot any graph (no graph paper is
  provided to you).

    Solution: When T 2 is plotted against l, one should obtain a linear graph of the form

                                               T 2 = al + b

    where, in the ideal case, l represents the distance of the centre of mass of the bob from the
                                     4π 2
    suspension point, the slope a =       , and the intercept b = 0.
                                       g

Page 10     Code 65                      Questions            Last six digits of Roll No.:

    If the size of the bob is not included in the measurement of l, one should still get a linear
                                                          4π 2 r
    graph, with same value of a, but now with b =                , where r is the radius of the bob.
                                                            g
    Careful inspection of the graph indeed shows a positive intercept on the T 2 -axis, conﬁrming
    this scenario. However, since the slope of the line is unaﬀected by the underestimation of l,
    the value of g is still correctly determined from the slope.
    Note that an overestimation of T by ∆T , say, would have caused the slope of the graph, and
    hence the derived g, to change since T 2 would have been modiﬁed by a term proportional to
    T itself.

Page 11      Code 65                         Questions

5. [12 marks] A circuit consists of an emf source and ﬁve resistors with unknown resistances. When
   an ideal ammeter is connected between points 1 and 2, its reading is IA . If instead a resistor R
   is connected to the same two points, the current through that resistor is IR . If instead an ideal
   voltmeter is connected between points 1 and 2, its reading is V . Obtain V in terms of IA , R and
   IR only.

                                        R1                        R3
                                                        1
                                        R2              2         R4

                                        R0

    Solution:
                            R1                                              R3    I1 − IM
                                   I1
                                               RM
                            R2                                              R4
                                                       IM
                     I2                                           I2 + IM
                            R0
                                 I1 + I2
                                                                            ε

    Resistances, currents and emf are shown in the diagram. Imagine also there is a resistance
    RM between terminals 1 and 2. We can set later RM = 0 for an ideal ammeter and IM = 0
    for an ideal voltmeter placed between two terminals. Applying Krichoﬀ’s law
           ε = (I1 + I2 )R0 + I1 R1 + (I1 − IM )R3 = (R0 + R1 + R3 )I1 + I2 R0 − IM R3        (5.1)
                                                      = R013 I1 + R0 I2 − R3 IM               (5.2)
           ε = (I1 + I2 )R0 + I2 R2 + (I2 + IM )R4 = (R0 + R2 + R4 )I2 + I1 R0 + IM R4        (5.3)
                                                      = R024 I2 + R0 I1 + R4 IM               (5.4)

    Solving above equations
                     εR24 + IM (R0 R34 + R3 R24 )          εR13 − IM (R0 R34 + R4 R13 )
              I1 =                                and I2 =                                    (5.5)
                         R0 R1234 + R13 R24                    R0 R1234 + R13 R24
    Potential diﬀerence between terminal 1 and 2 is
                                   V12 = I2 R2 − I1 R1 = εA − IM B                            (5.6)
    where the coeﬃcients A and B depends only on resistances in the circuit. When an ideal
    ammeter is placed between terminals V12 = 0 and IM = IA .
                                                εA = IA B
    When a resistance R is placed between terminals, IM = IR and V12 = IR R. This yields
                                                                          IR R
                            IR R = εA − IR B = IA B − IR B ⇒ B =                              (5.7)
                                                                        IA − IR
    When an ideal voltmeter is placed between terminals, IM = 0. Hence V12 = εA − 0 = IA B or
                                                       IA IR
                                              V12 =           R
                                                      IA − IR

Page 12   Code 65          Questions   Last six digits of Roll No.:

                **** END OF THE QUESTION PAPER ****

INPhO 2022 Problem 4


[User]
Consider the triangular “plucked” shape of problem 2 again, but suppose that the
    string starts at rest, and its two outer corners are always held fixed.
      (a) Sketch what happens after the string is released. What is the period of the motion?
     (b) Confirm explicitly that the initial potential energy of the string is equal to the kinetic energy
         of the string when it is purely horizontal.
      (c) What would prevent a real string from achieving this ideal motion? What will the string look
          like after a few oscillations?
     (d) To further build your intuition, try to visualize what happens if the string begins with the
        “pluck” off-center. Also, consider what happens if the string starts horizontal and at rest, but
         it instantaneously receives an impulse at its center.

Solution:
(a) The answer is shown in the early stages of this video. This generally shows how
         standing waves are produced. From an initial pluck, the reflections from the ends naturally
         create the oppositely-moving waves needed to form a standing wave. Since the wave becomes
         inverted after bouncing off the ends, it needs
                                                     p to travel a distance of 2L before it reaches its
         original state, giving a period of 2L/v = 2L µ/T .

     (b) Let the string have a height h initially, so that y(x, t) = f (x − vt) + f (x + vt) where f (x) a
         triangle of height h/2 and base L. Since f (x) was in the form of f (x) = x(h/2)/(L/2) = hx/L,
         then ẏ = hv/L. When the string is straight, each waveform has its peak at the end with
         the other half reflected, so the velocity adds up to give a velocity of 2hv/L at every point
         uniformly across the string, so K = 12 (µL)(2hv/L)2 . The potential energy can be found with
         U/L = 12 T y ′2 = 21 T (2h/L)2 , which gives U = 2T h2 /L = 2µv 2 h2 /L = K.

      (c) Of course, the amplitude decreases over time. But also, sharp features in the string are
          damped faster, due to both energy losses in the bending string, and drag with the air. As a
          result, after a while, the initially sharp features get smoothed out. You can start to see this
          happen in the latter part of the video linked in part (a).

     (d) You can check your answers here and here.

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


[User] Calculate I in the previous example by explicitly performing either Cartesian integral.

Solution:
Starting from the second expression in the example,
                             Z R     Z √R2 −y2                Z R
                          M                     2        M
                     I=           dy   √       x  dx =            2(R2 − y 2 )3/2 dy.
                         πR2 −R           2
                                      − R −y 2         3πR  2
                                                               −R

    Let y = R sin θ. Then we have
                                                         Z π/2
                                             2M R2
                                          I=                     cos4 θ dθ.
                                              3π          −π/2
    This integral can be done by repeatedly using the double angle formula,
                                 1 + cos(2θ) 2
      Z π/2              Z π/2                    Z π/2
                                                            1            1 1             3π
             cos4 θ dθ =                       dθ =           + cos(2θ) + + cos(4θ) dθ =    .
        −π/2              −π/2        2              −π/2   4            8  8             8
    Personally, I can never remember all the trigonometric formulas, and I usually just expand everything
    in complex exponentials. Here that method gives a slick solution, as
                                 Z π/2                 Z π/2

                                       cos4 θ dθ =           (eiθ + e−iθ )4 dθ.
                                  −π/2              16  −π/2

    Now note that expanding with the binomial theorem gives terms of the form e2inθ for integers n,
    which integrate to zero unless n = 0. So the only term that matters gives
                                 Z π/2                Z π/2
                                           4        1        4       3π
                                        cos θ dθ =              dθ =    .
                                   −π/2            16  −π/2  2        8

    Whichever method you used, we conclude the answer is I = M R2 /4, as expected.


[User]
01@  IPhO 1993, problem 3. (Hint: to do the final part of this problem, you should
    not try to compare the total path lengths traversed by the electrons. That would be very hard, and
    worse, it won’t give the right answer, because the potential from the wire also affects the electrons’
    phases. Instead, you should use the facts about wavefronts mentioned in W2. That is, waves always
    propagate perpendicular to wavefronts, and all points on a wavefront have the same phase.)

    3    The Uncertainty Principle
        Idea 5: Heisenberg Uncertainty

        So far we have treated a quantum particle as having a well-defined position and momentum,
        but in reality the uncertainties in the position and momentum obey
                                                            ℏ
                                                  ∆x ∆p ≥

        where, as in P2, the uncertainties may be interpreted as standard deviations. The “semi-
        classical limit” used in the rest of this problem set simply corresponds to the case where the
        required uncertainty is relatively small, which is reached for energy levels n ≫ 1. Occasionally,
        Olympiad questions will ask you to use the Heisenberg uncertainty principle to make a very
        rough estimate. In these cases, the constant factors will not matter.

        Idea 6: Energy-Time Uncertainty

        There are two commonly used versions of the energy-time uncertainty principle. If the energy
        of a system is only measured for a finite time ∆t, it must have a finite uncertainty ∆E in
        its energy. In addition, if a system significantly changes its state in time ∆t, then its energy
        must have been uncertain by a finite amount ∆E. In both cases, we have
                                                            ℏ
                                                 ∆E ∆t ≥      .

        A third common statement of the energy-time uncertainty principle is “for a short time ∆t, a
        system can violate energy conservation by an amount ∆E”. This is wrong, because quantum
        systems always conserve energy; systems that naively seem to violate energy conservation
        simply didn’t have a well-defined energy in their initial state to begin with. However, thinking
        this way will usually get you the right answers, essentially because of dimensional analysis.

  Example 3

  Consider once again a particle of mass m attached to a one-dimensional spring, with natural
  angular frequency ω. Use the uncertainty principle to estimate the minimum possible energy
  of the particle, and compare it with the result of problem 4.

  Solution
  Suppose the uncertainties in position and momentum are ∆x and ∆p. Then the potential
  energy is of order k(∆x)2 /2 and the kinetic energy is of order (∆p)2 /2m. Dropping constants,

                                          (∆p)2              ℏ2
                          E ∼ k(∆x)2 +          ≳ k(∆x)2 +
                                            m              (∆x)2 m

  where we applied the uncertainty
                            √      principle. The ground state minimizes√ the energy,
                                                                                p which is

  achieved when (∆x) ∼ ℏ/ km. In this case, the energy is of order kℏ/ km ∼ ℏ k/m ∼ ℏω,
  which is just what we found earlier. (A similar derivation can be used to derive the energy
  of the ground state of hydrogen, along with the Bohr radius; try it!)

  Remark
  We can also “solve” the above problem with the energy-time uncertainty principle incorrectly.
  The only timescale in the problem is 1/ω, so

                                               ℏ
                                        ∆E ≳      ∼ ℏω
                                               ∆t
  so E ≳ ℏω. However, in reality the ground state has no energy uncertainty; its energy is
  simply the ground state energy. Another way of saying this is that a particle can hang out
  in the ground state forever, so ∆t is infinite and hence ∆E is zero. This incorrect derivation
  gives the right answer just because it’s the only possible answer by dimensional analysis.
  Thus, a sloppy problem might ask you to do it.

  Example 4

  Consider a single slit diffraction experiment, where photons of wavelength λ pass through a
  slit of width a. If the screen is a large distance D away, roughly how wide is the resulting
  diffraction pattern on the screen?

  Solution
  The photon has a momentum px = ℏk = h/λ, and passing through the slit necessarily gives
  it a transverse momentum uncertainty of
                                               h
                                            ∆py ∼
                                                a
  where we dropped order one constants, which means an angle uncertainty of
                                               ∆py  λ
                                        ∆θ ∼       ∼ .
                                               px   a

       Therefore, using basic geometry, the size of the pattern on the screen is
                                                           Dλ
                                            ∆y ∼ D∆θ ∼        .
                                                            a
       This is the approximate width of the central maximum for single slit diffraction, as we found
       in W2. The reason the result is the same is that light acts like a wave both classically and
       quantum mechanically; the quantum version of the derivation is just the same as the classical
       version, but with “everything multiplied by h”. What’s new about this derivation is that it
       also applies for matter particles, which have λ = h/p.

       Example 5

       The Higgs boson has a mass of 125 GeV and a lifetime of about τ = 1.6 × 10−22 s. About
       what percentage uncertainty must a measurement of a Higgs boson’s mass have?

       Solution
       Decay is a significant change in the particle’s state, and this change happens over a time τ ,
       which means the energy uncertainty is
                                          ℏ
                                   ∆E ∼     = 7 × 10−13 J = 0.004 GeV.
                                          τ
       When we measure the Higgs boson’s mass, we really measure the E = mc2 energy released
       when it decays, so the unavoidable uncertainty of the mass is ∆E/E ∼ 0.003%. (But the
       actual measured uncertainties are much higher, due to a variety of other effects.)

[1] Problem 18 (Krane 4.39). An apparatus is used to prepare an atomic beam by heating a collection
    of atoms to a temperature T and allowing the beam to emerge through a hole of diameter d in
    one side of the oven. Show that the uncertainty principle causes the diameter of the beam, after
                                                                        √
    traveling a length L, to be larger than d by an amount of order Lℏ/d mkB T , where m is the mass
    of an atom.

[2] Problem 19 (Insight 8.26). When helium is cooled to extremely cold temperatures, it becomes
    a superfluid, an exotic type of liquid that can flow with zero dissipation. These strange properties
    occur because quantum mechanical effects are large, making the quantum uncertainty in the position
    of each helium atom on the same order as the separation between atoms.

      (a) Given that superfluid helium has density ρ and a helium atom has mass m, estimate the
          temperature T at which helium becomes a superfluid. This is closely related to, but not quite
          the same thing as Bose–Einstein condensation, a phase transition that bosons undergo at low
          temperatures.

     (b) Numerically evaluate T , given that ρ ∼ 100 kg/m3 and m ∼ 7 × 10−27 kg.

Solution:
The energy of the particles
    travel the length L is t ∼ L/v ∼ L m/kB T . Now, the uncertainty in the vertical direction is d, so
    the range of vertical momenta is ∼ ℏ/d, so the range of vertical speeds is ∼ ℏ/md. Thus, in the
    time t, we get a spread of order
                                               tℏ      Lℏ
                                                  = √         ,
                                              md     d m kB T
    as desired.

[2] Problem 19 (Insight 8.26). When helium is cooled to extremely cold temperatures, it becomes
    a superfluid, an exotic type of liquid that can flow with zero dissipation. These strange properties
    occur because quantum mechanical effects are large, making the quantum uncertainty in the position
    of each helium atom on the same order as the separation between atoms.

      (a) Given that superfluid helium has density ρ and a helium atom has mass m, estimate the
          temperature T at which helium becomes a superfluid. This is closely related to, but not quite

         the same thing as Bose–Einstein condensation, a phase transition that bosons undergo at low
         temperatures.

     (b) Numerically evaluate T , given that ρ ∼ 100 kg/m3 and m ∼ 7 × 10−27 kg.
                                                                                √           √
    Solution. (a) The energy is of order kB T , so the momentum is of order p ∼         2mE ∼       mkB T .
         This leads to a spread in position by the uncertainty principle of
                                                     ℏ     ℏ
                                              ∆x ∼     ∼√        .
                                                     p    m kB T

         The volume per helium atom is m/ρ, giving a typical separation of (m/ρ)1/3 . Setting this
         equal to ∆x and solving for T gives

                                                           ρ2/3 ℏ2
                                                 T ∼               .
                                                          kB m5/3

     (b) Plugging in the numbers gives T ∼ 0.7 K. The actual answer is 2.172 K, so this isn’t bad for
         such a rough estimate!


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

                                 Theoretical Question 1

      When will the Moon become a Synchronous Satellite?
      The period of rotation of the Moon about its axis is currently the same as its
period of revolution about the Earth so that the same side of the Moon always faces
the Earth. The equality of these two periods presumably came about because of
actions of tidal forces over the long history of the Earth-Moon system.
      However, the period of rotation of the Earth about its axis is currently shorter
than the period of revolution of the Moon. As a result, lunar tidal forces continue to
act in a way that tends to slow down the rotational speed of the Earth and drive the
Moon itself further away from the Earth.
      In this question, we are interested in obtaining an estimate of how much more
time it will take for the rotational period of the Earth to become equal to the period of
revolution of the Moon. The Moon will then become a synchronous satellite,
appearing as a fixed object in the sky and visible only to those observers on the side
of the Earth facing the Moon. We also want to find out how long it will take for the
Earth to complete one rotation when the said two periods are equal.

       Two right-handed rectangular coordinate systems are adopted as reference
frames. The third coordinate axes of these two systems are parallel to each other and
normal to the orbital plane of the Moon.
(I)The first frame, called the CM frame, is an inertial frame with its origin located at
    the center of mass C of the Earth-Moon system.
(II)The second frame, called the xyz frame, has its origin fixed at the center O of the
    Earth. Its z-axis coincides with the axis of rotation of the Earth. Its x-axis is along
    the line connecting the centers of the Moon and the Earth, and points in the
    direction of the unit vector as shown in Fig.1a. The Moon remains always on the
    negative x-axis in this frame.

      Note that distances in Fig.1a are not drawn to scale. The curved arrows show
the directions of the Earth's rotation and the Moon's revolution. The Earth-Moon
distance is denoted by r.

                                           Earth
                                                                                Moon
       Fig. 1a                        O
                   x                                                          M
                                          C

                                  y

                                              1

APHO II   2001   Theoretical Question 1                                                      p. 2 / 3

The following data are given:
(a) At present, the distance between the Moon and the Earth is                        and
     increases at a rate of 0.038 m per year.
(b) The period of revolution of the Moon is currently     = 27.322 days.
(c) The mass of the Moon is                        .
(d) The radius of the Moon is                        .
(e) The period of rotation of the Earth is currently   = 23.933 hours.
(f) The mass of the Earth is                          .
(g) The radius of the Earth is                    .
(h) The universal gravitational constant is                                  .

The following assumptions may be made when answering questions:
(i) The Earth-Moon system is isolated from the rest of the universe.
(ii) The orbit of the Moon about the Earth is circular.
(iii) The axis of rotation of the Earth is perpendicular to the orbital plane of the Moon.
(iv) If the Moon is absent and the Earth does not rotate, then the mass distribution of
      the Earth is spherically symmetric and the radius of the Earth is .
(v) For the Earth or the Moon, the moment of inertia I about any axis passing through
      its center is that of a uniform sphere of mass M and radius R, i.e.            .
(vi) The water around the Earth is stationary in the xyz frame.
Answer the following questions:
(1) With respect to the center of mass C, what is the current value of the total angular
   momentum L of the Earth-Moon system?
(2) When the period of rotation of the Earth and the period of revolution of the Moon
   become equal, what is the duration of one rotation of the Earth? Denote the answer
   as T and express it in units of the present day. Only an approximate solution is
   required so that iterative methods may be used.
(3) Consider the Earth to be a rotating solid sphere covered with a surface layer of
   water and assume that, as the Moon moves around the Earth, the water layer is
   stationary in the xyz-frame. In one model, frictional forces between the rotating
   solid sphere and the water layer are taken into account. The faster spinning solid
   Earth is assumed to drag lunar tides along so that the line connecting the tidal
   bulges is at an angle with the x-axis, as shown in Fig.1b. Consequently, lunar
   tidal forces acting on the Earth will exert a torque Γ about O to slow down the
   rotation of the Earth.
          The angle δ is assumed to be constant and independent of the Earth-Moon
   distance r until it vanishes when the Moon's revolution is synchronous with the
   Earth's rotation so that frictional forces no longer exist. The torque Γ therefore

                                              2

APHO II   2001    Theoretical Question 1                                        p. 3 / 3

  scales with the Earth-Moon distance and is proportional to   .

        According to this model, when will the rotation of the Earth and the
  revolution of the Moon have the same period? Denote the answer as      and
  express it in units of the present year.

                                                               Moon
      Fig.1b                               δ
                                     O
                                           Earth

  The following mathematical formulae may be useful when answering questions:
  (M1)      For          and            :

  (M2)       If           and                  , then

                                                 3

Solution:
APHO II   2001   Theoretical Question 1                                                 p. 1 / 5

[Solution]                       Theoretical Question 1

      When will the Moon become a Synchronous Satellite?
(1) The total orbital angular momentum         of the Earth-Moon system with
   respect to C can be calculated as follows.
         Since all angular momenta are along the z-direction, only the z-component of
   each angular momentum will be calculated. The distance between the center of
   mass C and the center of the Earth O is

  The angular speed of the Moon's revolution is
                                                                                (1a)

  The orbital angular momentum of the Moon about C is

  The angular speed of the Moon's spinning or rotational motion is

  The spin angular momentum of the Moon is

  This is much smaller than the Moon's orbital angular momentum and can therefore
  be neglected.
        The orbital angular momentum of the Earth about C is

  The angular speed of the Earth's spinning motion is

  The moment of inertia of the Earth about its axis of rotation is
                                                                                (1b)

  The spin angular momentum of the Earth is

                                           1

APHO II    2001   Theoretical Question 1                                                    p. 2 / 5

[Solution] (continued)               Theoretical Question 1

      When will the Moon become a Synchronous Satellite?
          Thus, the total angular momentum of the Earth-Moon system L is given by

                                                                                (2)

  Note that                          .

(2) According to Newton's form for Kepler's third law of planetary motions, the
   angular speed ω of the revolution of the Moon about the Earth is related to the
   Earth-Moon distance r by
                                                                                (3)
  Therefore, the orbital angular momentum of the Earth-Moon system with respect
  to C is

                                                                                (4)

  (Note:                                 ,                      so that              .)

  When the angular speed of the Earth's rotation is equal to the angular speed ω of the
  orbiting Moon, the total angular momentum of the Earth-Moon system is, with the
  spin angular momentum of the Moon neglected, given by

                                                                                (5a)

  The last equality follows from conservation of total angular momentum and Eq.(2).
  For an initial estimate of ω, the spin angular momentum of the Earth may be
  neglected in Eq.(5a) to give

                                                 rad /s   (first iteration)

  An improved estimate may be obtained by using the above estimated value             to
  compute the spin angular momentum of the Earth and use Eq.(5a) again to solve
  for ω. The result is

                                             2

APHO II   2001     Theoretical Question 1                                                          p. 3 / 5

[Solution] (continued)                Theoretical Question 1

     When will the Moon become a Synchronous Satellite?

                                                        rad /s   (second iteration)         (5b)

  Further iterations of the same procedure lead to the same value just given. Thus,
  the period of rotation of the Earth will be

  (3) Since the total torque Γ is proportional to            , we conclude
                                                                                (6)

          Fig.1e
                                                                                Moon
                                                δ
                                            O
                                                Earth

       Let the current values of r and Γ be, respectively,              and    .    From Eq.(6),
  we then have
                                                                                (7)

  The torque Γ is equal to the rate of change of spin angular momentum                    of the
  Earth so that
                                                                                (8)
  By Newton's law of action and reaction or by the law of conservation of the total
  angular momentum,         is equal to the rate of change of the total orbital angular
  momentum      of the Earth-Moon system so that
                                                                                (9)

  But according to Eq.(3), we have

  and Eq.(4) may be written as

                                                    3

APHO II   2001   Theoretical Question 1                                                     p. 4 / 5

                                                                                (10)

  This implies

                                                                                (11a)

  The value of         can be determined from Eq. (11a) as follows:

                                                                                (13)

  Starting with Eq.(11a) in the following form

  we may use Eq.(3) to express r in terms of ω and obtain the following equations:

[Solution] (continued)              Theoretical Question 1

     When will the Moon become a Synchronous Satellite?

  where the constant b stands for the expression in the square brackets. The last
  equation leads to the solution

  where      is the length of time needed for the angular speed of the rotation of the
  Earth to be equal to that of the Moon's revolution about the Earth.
        Using the values of      and obtained in Eqs.(1a) and (5b) and the value of
      in Eq.(13), we have

                                            4

APHO II   2001   Theoretical Question 1       p. 5 / 5

                                          5

APhO 2001 Theory T1


[User]
\begin{wrapfigure}[7]{r}{0.4\textwidth}
	\vspace{-10pt}
	\includegraphics[width = 0.4\textwidth]  {2017-lahg-10-delta.pdf}
\end{wrapfigure}
In the given figure a hinged construction is pictured, it consists of two rods with a length $2l$. One of the rod’s tip is fixed to an unmoving point $A$ and the other rod’s tip $C$ moves with a constant velocity $v$ along a direction which passes the point $A$ at the distance $l$. Find the acceleration of the connection point $B$ of the rods during the moment when the distance between the points $A$ and $C$ is $2l$.

Solution:
Vaadeldaval hetkel on kiirusvektori ja $AC$ vahelise nurga siinus $\frac 12$, st nurk ise on $30^\circ$ ja seetõttu $AB$ on risti kiirusvektoriga. Järelikult on punkti $B$ kiirusvektori suund sama, mis $\vec v$. Varda $BC$ venimatuse tõttu peavad otspunktide kiirusvektorite projektsioonid varda sihile olema võrdsed. Et nurgad nende kiirusvektorite ja varda sihiga on võrdsed, siis peavad ka kiiruste moodulid olema võrdsed. Niisiis punkti $B$ kiirus on $\vec v$. Seetõttu on punkti $B$ kesktõmbekiirendus $v^2/2l$. Punktil $B$ võib olla veel mingi tangentsiaalkiirendus, hetkel me teame vaid kogukiirenduse projektsiooni $AB$ sihile. Läheme nüüd kiirusega $\vec v$ liikuvasse taustsüsteemi, kus punkt $C$ on kogu aeg paigal ja punkt $B$ on hetkeliselt paigal. Et $B$ on hetkeliselt paigal, siis selle kesktõmbekiirendus pöörlemisel ümber $C$ on antud hetkel null, st $B$ kiirenduse vektor on risti $BC$-ga. Seega me teame kiirenduse suunda ja projektsiooni $v^2/2l$ sihile $AB$ (mis moodustab vektori suunaga nurga $30^\circ$). Selle põhjal saame avaldada kiirenduse mooduli
\[
a=\frac{v^2}{2l\cos30^\circ}=\frac{v^2}{\sqrt 3l}.
\]

Vardad

\begin{wrapfigure}[7]{r}{0.4\textwidth}
	\vspace{-10pt}
	\includegraphics[width = 0.4\textwidth] {2017-lahg-10-delta.pdf}
\end{wrapfigure}

Juuresoleval joonisel on kujutatud kahest vardast pikkusega $2l$ koostatud \v sarniirne konstruktsioon. Ühe varda otspunkt on fikseeritud liikumatuna punkti $A$ ning teise varda otspunkt $C$ liigub konstantse kiirusega $v$ piki sihti, mis möödub punktist $A$ kaugusel $l$. Leidke varraste ühenduspunkti $B$ kiirendus hetkel, mil punktide $A$ ja $C$ vahekaugus on $2l$.

Punkti $B$ kiirust on võimalik leida, kasutades varraste venimatust ning asjaolu, et $A$ on fikseeritud. Punkti $B$ kiirenduse $AB$-sihilist komponenti saab leida kesktõmbekiirenduse kaudu ning kiirenduse suuna määramiseks on kasulik minna kiirusega $\vec{v}$ liikuvasse taustsüsteemi.


[User] A kettle with a heating power $N$ is filled with water. The area of the kettle’s spout is $S$. What is the biggest speed of the water vapor exiting the kettle? Water’s enthalpy of vaporization is $L$, the universal gas constant is $R$, the air pressure is $p$ and the molar mass of water is $\mu$. The efficiency factor of the kettle without accounting for the heat losses is $\gamma$.

Solution:
Suurima kiiruse saavutab veeaur siis, kui vesi on kuumutatud keemistemperatuurini $T=\SI{100}{\degreeCelsius}=\SI{373}{K}$. Veekeedukannu tehtud töö $A=\gamma Nt$ läheb siis vee aurustamiseks ehk $A=Q=Lm$. Seega aja $t$ jooksul aurustunud vee mass on
\[
m=\frac{\gamma Nt}{L}
\]
ning ruumala
\[
V=\frac{\gamma NtRT}{Lp\mu}.
\]
Kuna veeaur pääseb välja avast pindalaga $S$, siis peab aja $t$ jooksul eraldanud veeaur läbima vahemaa $s=V/S$ ning veeauru väljumise kiiruseks saame
\[
v=\frac{s}{t}=\frac{\gamma NtRT}{Lp\mu S}.
\]

Veekeedukann

Veekeedukann küttekeha võimsusega $N$ on täidetud veega. Kannu tila ava pindala on $S$. Milline on suurim joonkiirus, millega veeaur kannu avast väljub? Vee aurustumissoojus on $L$, ideaalse gaasi konstant on $R$, õhurõhk on $p$ ning vee molaarmass on $\mu$. Veekeedukannu kasutegur soojuskadusid arvestades on $\gamma$.

Suurima kiiruse saavutab veeaur siis, kui vesi on kuumutatud keemistemperatuurini. Sellisel juhul kulub kogu küttekeha võimsus vee aurustamiseks.


[User]
01W   USAPhO 1997, problem A3.

[3] Problem 10 (Purcell 4.18). Consider the two RC circuits below.

      (a) The circuit shown below contains two identical capacitors and two identical resistors, with
          initial charges as shown above at left. If the switch is closed at t = 0, find the charges on the
          capacitors as functions of time.

     (b) Now consider the same setup with an extra resistor, as shown above at right. Find the
         maximum charge that the right capacitor achieves. (Hint: the methods of M4 can be useful.)

Solution:
(a) Let the two loop currents be I1 and I2 , both counterclockwise. The loop equations
         are Q1 /C = I1 R and Q2 /C = I2 R. We also have Ik = −Q̇k . Thus, Qk + RC Q̇k = 0 for
         k = 1, 2. Based on the initial conditions, we see then that the solutions are Q1 (t) = Q0 e−t/RC
         and Q2 (t) = 0. (The simple reason Q2 (t) is zero is because the middle wire effectively shorts
         out the right half of the circuit.)
     (b) Again, with the same setup of variables, we get that
                                              Q1 /C − 2I1 R + I2 R = 0
                                              Q2 /C − 2I2 R + I1 R = 0.
         This is a system of two linear differential equations, which can be solved using the methods
         of M4. However, in this case we can just add and subtract the equations, giving
                         (Q1 + Q2 )/C − (I1 + I2 )R = 0,    (Q1 − Q2 )/C − 3(I1 − I2 )R = 0.
         That is, the sum of the two acts like an RC circuit with time constant RC, while the difference
         acts like one with time constant 3RC. (These are the “normal modes”.) By superposing these
         solutions and fitting the initial conditions, we get
                                    Q0 −t/RC                           Q0 −t/RC
                         Q1 (t) =     (e     + e−t/3RC ),   Q2 (t) =     (e     − e−t/3RC ).
                                    2                                  2

          We want to maximize |Q2 |, so setting the derivative to zero gives t = 23 RC log(3), so

                                                                Q0
                                                     |Q2 |max = √ .
                                                               3 3


[User]
\begin{wrapfigure}[10]{r}{0.4\textwidth}
\vspace{-20pt}
\begin{resizebox}{\linewidth}{!}{
\begin{circuitikz}
\draw
(1,0) to[battery1,l_=${U_0=\SI{14}{V},\,I_0=\SI{10}{A}}$] (8,0) to (1,0) to (1,2.5) to (2,2.5) to[short, *-] (2,4) to[resistor,l=${R_1}$, -*] (4.5,4) to[resistor,l=${R_4}$] (7,4) to[short, -*] (7,2.5) to (8,2.5) to (8,0)
(2,2.5) to (2,1) to[resistor,l=${R_2}$, -*] (4.5,1) to[resistor,l=${R_5}$] (7,1) to (7,2.5)
(4.5,1) to[resistor,l_=$R_3$] (4.5,4)
;
\draw[->,thick] (4,2) -- (4,3);
\end{circuitikz}}
\end{resizebox}
\end{wrapfigure}
Five resistors are connected to a current source. Three of them have a resistance of $\SI{1}{\ohm}$, the other two have a same but an unknown resistance. The voltage of the current source is $U_0=\SI{14}{V}$ and its current is $I_0=\SI{10}{A}$. The voltage and the current strength on the third resistor are respectively $U_3=\SI{2}{V}$ and $I_3=\SI{2}{A}$. The direction of the current in the resistor $R_3$ is shown in the drawing. Determine all the resistances of the resistors.

Solution:
\begin{wrapfigure}[9]{r}{0.4\linewidth}
 \vspace{-8pt}
	\begin{resizebox}{\linewidth}{!}{
		\begin{circuitikz}
			\draw
				(8,0) to[battery1] (1,0) to (1,2.5) to (2,2.5) to[short, *-] (2,4) to[resistor,l=${R_1=\SI{2}{\ohm}}$, -*] (4.5,4) to[resistor,l=${R_4=\SI{1}{\ohm}}$] (7,4) to[short, -*] (7,2.5) to (8,2.5) to (8,0)
				(2,2.5) to (2,1) to[resistor,l=${R_2=\SI{1}{\ohm}}$, -*] (4.5,1) to[resistor,l=${R_5=\SI{2}{\ohm}}$] (7,1) to (7,2.5)
				(4.5,1) to[resistor,l_=${R_3=\SI{1}{\ohm}}$] (4.5,4)
				;
		\end{circuitikz}}
	\end{resizebox}
\end{wrapfigure}
Elektrivoolul on võimalik ahelat läbida kolmel moel: läbi ülemise haru (takistid $R_1$ ja $R_4$), läbi keskmise haru (takistid $R_2$, $R_3$ ja $R_4$) ning läbi alumise haru (takistid $R_2$ ja $R_5$). Pingelang igal takistil on võrdne selle takistuse ning seda läbiva voolutugevuse korrutisega ning pingelangude summa kõigil kolmel teekonnal peab olema võrdne pingega vooluallikal. Niisiis $I_1R_1+I_4R_4=I_2R_2+I_3R_3+I_4R_4=I_2R_2+I_5R_5=U_0$. Neist võrranditest saab tuletada avaldised $I_1R_1=I_2R_2+I_3R_3$ ning $I_3R_3+I_4R_4=I_5R_5$, mis kirjeldavad pingelange skeemi vasakul ning paremal poolel. Lisaks saame panna kirja võrrandid voolutugevuste jaoks: $I_1=I_0-I_2$, $I_4=I_1+I_3$ ning $I_5=I_2-I_3$.

Paneme tähele, et kui $R_1=R_2$ ja $R_4=R_5$, siis sümmeetriakaalutlustel peaks kehtima $R_3=0$.

On ilmne, et $R_3=\SI{1}{\ohm}$. Oletame esmalt, et $R_1=R_4=R_3$. Sellisel juhul avaldub pingelang ülemises harus kujul $(2I_1+I_3)R_3=U_0$, millest järeldub $I_1=\SI{6}{A}$ ning $I_2=\SI{4}{A}$. Vaadeldes voolu teekonda läbi keskmise haru, saame kirjutada $I_2R_2+U_3+(I_1+I_3)R_3=U_0$, millest saab järeldada $R_2=R_5=\SI{1}{\ohm}$. Selline skeem rahuldab eespool mainitud sümmeetriat, seega tehtud eeldus pole tõene.

Oletame nüüd, et $R_2=R_5=R_3$. Skeemi alumise haru jaoks saame kirjutada $(2I_2-I_3)R_3=U_0$ ehk $I_2=\SI{8}{A}$ ning $I_1=\SI{2}{A}$. Pingelangud skeemi vasakus pooles peavad rahuldama võrrandit $I_1R_1=(I_2+I_3)R_3$, millest järeldub $R_1=\SI{5}{\ohm}$. Sellisel juhul on aga pingelang skeemi ülemisel harul $(2I_1+I_3)R_1=\SI{30}{V}\not=U_0$.

Oletame nüüd, et $R_1=R_5=R_3$. Pingelang skeemi keskmisel harul on $I_2R_2+U_3+(I_0-I_2+I_3)R_2=U_3+(I_0+I_3)R_2=U_0$ ehk $R_2=R_4=\SI{1}{\ohm}$. Tegemist on juba vaadeldud juhuga.

Ainus järelejäänud võimalus on $R_2=R_4=R_3$. Skeemi vasakust poolest saame $(I_0-I_2)R_1=I_2R_3+U_3$, paremast $(I_2-I_3)R_1=U_3+(I_0-I_2+I_3)R_3$. Nende põhjal $(I_0-I_3)R_1=2U_3+(I_0+I_3)R_3$ ehk $R_1=\SI{2}{\ohm}$. On lihtne veenduda, et leitud väärtused rahuldavad kõiki võrrandeid pingelangude jaoks.

Takistuste tuvastamine

\begin{wrapfigure}[10]{r}{0.4\textwidth}
\vspace{-20pt}
\begin{resizebox}{\linewidth}{!}{
\begin{circuitikz}
\draw
(1,0) to[battery1,l_=${U_0=\SI{14}{V},\,I_0=\SI{10}{A}}$] (8,0) to (1,0) to (1,2.5) to (2,2.5) to[short, *-] (2,4) to[resistor,l=${R_1}$, -*] (4.5,4) to[resistor,l=${R_4}$] (7,4) to[short, -*] (7,2.5) to (8,2.5) to (8,0)
(2,2.5) to (2,1) to[resistor,l=${R_2}$, -*] (4.5,1) to[resistor,l=${R_5}$] (7,1) to (7,2.5)
(4.5,1) to[resistor,l_=$R_3$] (4.5,4)
;
\draw[->,thick] (4,2) -- (4,3);
\end{circuitikz}}
\end{resizebox}
\end{wrapfigure}

Vooluallikaga on ühendatud viis takistit. Neist kolme takistus on \SI{1}{\ohm}, ülejäänud kaks on tundmatu, kuid ühesuguse takistusega. Vooluallika pinge $U_0=\SI{14}{V}$ ning voolutugevus selles $I_0=\SI{10}{A}$. Pinge ja voolutugevus kolmandal takistil on vastavalt $U_3=\SI{2}{V}$ ning $I_3=\SI{2}{A}$. Joonisel on märgitud elektrivoolu suund takistis $R_3$. Määrake kõigi takistite takistused.

Takistuste leidmiseks tasub kõik \SI{1}{\ohm} takistuste kombinatsioonid läbi vaadata ning süstemaatiliselt valed konfiguratsioonid elimineerida.


[User]
01T
                       IPhO 2012, problem 2. A challenging electrostatics and fluids problem; some
    prior exposure to surface tension is helpful. (For more about the kinds of bubbles encountered in
    this problem, see section 5.9 of Physics of Continuous Matter by Lautrup.)

    2    Charges in Fields
[3] Problem 7 (BAUPC). A particle with charge q and mass m is initially at the origin in a region with
    constant magnetic field Bẑ, and velocity v0 ŷ. The particle experiences a frictional force F = −αv.
    Find the final position of the particle.

[3] Problem 8 (APhO 2006). Two large, identical conducting plates α and β with charges −Q and
    +q (where Q > q > 0) are parallel to each other and fixed in place. Another identical plate γ with
    mass m and charge +Q is parallel to the original plates at distance d, as shown.

    The plates have surface area A. The plate γ is released from rest and bounces elastically off the
    plate β. Assume that charges have sufficient time to redistribute between the plates during the

                  01W
    collision. When the plate γ returns to its original position, what is its speed?

Solution:
Newton’s second law is ma = qv × B − αv, and its components are
                                  max = qBvy − αvx ,      may = −qBvx − αvy .
    This is a set of two linear differential equations, so we can find the solution by guessing exponentials,
    as discussed in M4. But in this case, since we know the initial and final velocities, and only want
    the final position, there’s an easier way. (I thank Stefan Ivanov for pointing this out.) We integrate
    both sides of the above equations with respect to time, from zero to infinity, to get
                              m∆vx = qB∆y − α∆x,          m∆vy = −qB∆x − α∆y.
    The initial velocity is v0 ŷ, and the final velocity is zero, so
                                             ∆vx = 0,     ∆vy = −v0 .
    This yields a system of two equations, which can be solved to yield
                                             mv0 qB                    mv0 α
                                    ∆x =              ,       ∆y =              .
                                           (qB)2 + α2                (qB)2 + α2

[3] Problem 8 (APhO 2006). Two large, identical conducting plates α and β with charges −Q and
    +q (where Q > q > 0) are parallel to each other and fixed in place. Another identical plate γ with
    mass m and charge +Q is parallel to the original plates at distance d, as shown.

    The plates have surface area A. The plate γ is released from rest and bounces elastically off the
    plate β. Assume that charges have sufficient time to redistribute between the plates during the
    collision. When the plate γ returns to its original position, what is its speed?

    Solution. Initially, the electric field due to the plates α and β at γ is
                                                           Q−q
                                                    E=
                                                           2ϵ0 A
    so the force is
                                                               Q(Q − q)
                                               F = QE =
                                                                 2ϵ0 A
    towards plate β. During the collision, β and γ effectively become one plate with total charge Q + q.
    In order to shield the field of plate α, the difference of the charges on the left end of β and the right
    end of γ must be Q, which means that the right end of γ gets a charge q/2, while the left end of β
    gets a charge Q + q/2. After the collision, the electric field due to α and β is
                                                              q
                                                    E′ =
                                                            4ϵ0 A
    so the magnitude of the force is
                                                      q      q2
                                                 F ′ = E′ =
                                                      2     8ϵ0 A
    away from plate β. The total work done is
                                                                 d
                                          W = (F + F ′ )d =          (2Q − q)2
                                                               8ϵ0 A

    and setting this equal to mv 2 /2 gives
                                                   r
                                                          d
                                              v=             (Q − q/2).
                                                       ϵ0 mA


[User]
You are given a small light bulb, a filter which lets only red light pass through, and
    a light meter which measures the total light intensity on it. How can you accurately measure the

                  01^
    intensity of the red light given off by the bulb in a given direction, at a given distance?

Solution:
Obviously, you have to fix the locations of the bulb and light meter, put the filter between
    them, and point the light meter directly at the bulb. But some other things impact the accuracy of
    the result.

     • Of course, you should first turn off all of the other lights that you can, and keep them off.

     • You should ensure the result isn’t affected by reflections. The best way to do this would be to
       put the bulb and light meter underneath a black box. If one isn’t available, the next best thing
       is to remove highly reflective objects, such as pieces of paper lying on your desk.

     • You should measure what the meter reads when the light bulb is turned off. Your answer is the
       difference of what you measure with the light bulb on, and this background value.

     • You should place the filter next to the light meter, not next to the bulb. If the filter is next to
       the bulb, it might partially scatter the bulb’s light, throwing off the angular distribution. Also,
       if the filter is covering the light meter, then it will also filter out non-red background light. This
       significantly lowers the background, allowing a more precise result.


[User]
01h
                       IPhO 2012, problem 3. This elegant and tricky problem covers the early stages

                            01r
    of star formation, and serves as a review of T1.

Solution:
IPhO 2012, problem 3. This elegant and tricky problem covers the early stages

                         01r
    of star formation, and serves as a review of T1.


[User]
INPhO 2014, problem 7. This is an instructive general application of the liquid drop
    model. The official solutions are already on the page, so you can check your work as you go. As

                     01T
    Stefan Ivanov notes, there’s a minor typo in part E.

Solution:
INPhO 2014, problem 7. This is an instructive general application of the liquid drop
    model. The official solutions are already on the page, so you can check your work as you go. As

                    01T
    Stefan Ivanov notes, there’s a minor typo in part E.


[User]
1. Volleyball (8 points)

Considerasimplifiedmodelofavolleyball: athin
spherical envelope filled with air. The envelope
material is non-stretchable, easily foldable. The
excess pressure inside the ball ∆p = 20kPa, the
ball radius R = 10cm and mass m = 400g (the
air mass inside the ball is negligible). You can
neglectthedependenceoftheexcesspressureon
the deformation of the ball.
1) The ball is pressed between two parallel rigid
plates,thedistance betweenofwhich is2R−2h
(so that the height of the deformed segments is
h = 1cm). Find the force between the ball and a
plate N.
2) Ball moves with velocity v0 = 2m/s and hits
a rigid wall. Find the maximal height of the de-
formed segment hm ...
3) ...andthe collision time τ.
4) For small excess pressures, the ball can lose
its spherical shape even in those points, which
are not in touch with the wall. Which condition
between the quantities ∆p, R, m and h has to
be satisfied in order to ensure that such a loss of
sphericityis negligible?

Solution:
1. Volleyball

1) F = ∆pS, where S = πr2
is the segment ba-
se surface. It is easyto see that r2
= (2R − h)h,
hence F = ∆pπh(2R − h) ≈ 120N.
2) During the collision the ball is deformed as
shown in Figure: the envelope is not stretchable,
henceit retainsthe sphericalshape (except whe-
re in touch with the wall). Using the approxima-
tion h  R we can neglect the term h2
in the
expression for the force. Then, the force is pro-
portional to h, ie. the ball behaves as a spring of
stiffness k = 2πR∆p. According to the ener-
gy conservation law mv2
= 2πR∆ph2
, hence
h = v

m/2πR∆p ≈ 11mm.
3) This is the half of the harmonic oscillations
period, τ = π

m/2πR∆p =

πm/2R∆p ≈
18ms.
4) Let us use the ball’s system of reference. The
envelope surface element dS is exerted by the
force of inertia dFi = amdS/4πR2
, where a =
∆pπh(2R − h)/m. Thus, dFi = ∆ph(2R −
h)dS/4R2
. In order to keep the spherical sha-
pe,this forcehas to be compensatedbythe force
due to the excess pressure dFr = ∆pdS, hence
h(2R−h) < 4R2
. Thisconditionisalwayssatis-
fied, no additional constraint is needed. Notice
that we consideredonly the worst case requiring
thelargestcompensatingforcewhen theforceof
inertia is normal to the surface. Remark: The ca-
seofstretchableenvelopeiscompletelydifferent,
sphericity disappears over all the surface (try to
press a balloon against a glass!).

Volleyball


[User]
41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010            1/4

2. Chimney physics
Introduction
Gaseous products of burning are released into the atmosphere of temperature TAir through a high
chimney of cross-section A and height h (see Fig. 1). The solid matter is burned in the furnace which
is at temperature TSmoke. The volume of gases produced per unit time in the furnace is B.

Assume that:

         the velocity of the gases in the furnace is negligibly small
         the density of the gases (smoke) does not differ from that of the air at the same temperature
          and pressure; while in furnace, the gases can be treated as ideal
         the pressure of the air changes with height in accordance with the hydrostatic law; the
          change of the density of the air with height is negligible
         the flow of gases fulfills the Bernoulli equation which states that the following quantity is
          conserved in all points of the flow:
                                             ,
          where ρ is the density of the gas, v(z) is its velocity, p(z) is pressure, and z is the height
         the change of the density of the gas is negligible throughout the chimney

Fig 1. Sketch of a chimney of height h with a furnace at temperature TSmoke .

Task 1
    a) What is the minimal height of the chimney needed in order that the chimney functions
       efficiently, so that it can release all of the produced gas into the atmosphere? Express your

     41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010      2/4

       result in terms of B, A, TAir, g=9.81m/s2, ΔT=TSmoke-TAir. Important: in all subsequent tasks
       assume that this minimal height is the height of the chimney. (3.5 points)
    b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections
       are identical, but are designed to work in different parts of the world: one in cold regions,
       designed to work at an average atmospheric temperature of -30 oC and the other in warm
       regions, designed to work at an average atmospheric temperature of 30 oC. The temperature
       of the furnace is 400 oC. It was calculated that the height of the chimney designed to work in
       cold regions is 100 m. How high is the other chimney? (0.5 points)
    c) How does the velocity of the gases vary along the height of the chimney? Make a
       sketch/diagram assuming that the chimney cross-section does not change along the height.
       Indicate the point where the gases enter the chimney. (0.6 points)
    d) How does the pressure of the gases vary along the height of the chimney? (0.5 points)

Solar power plant
         The flow of gases in a chimney can be used to construct a particular kind of solar power plant
(solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of
area S with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air
rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding
(thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air
through the chimney powers a turbine, resulting in the production of electrical energy. The energy of
solar radiation per unit time per unit of horizontal area of the collector is G. Assume that all that
energy can be used to heat the air in the collector (the mass heat capacity of the air is c, and one can
neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the
ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to
its entry into the chimney.

      41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010     3/4

Fig 2. Sketch of a solar power plant.

Task 2
    a) What is the efficiency of the solar chimney power plant? (2.0 points)
    b) Make a diagram showing how the efficiency of the chimney changes with its height. (0.4
       points)

Manzanares prototype
        The prototype chimney built in Manzanares, Spain, had a height of 195 m, and a radius 5 m.
The collector is circular with diameter of 244 m. The specific heat of the air under typical operational
conditions of the prototype solar chimney is 1012 J/kg K, the density of the hot air is about 0.9 kg/m3,
and the typical temperature of the atmosphere TAir = 295 K. In Manzanares, the solar power per unit
of horizontal surface is typically 150 W/m2 during a sunny day.

Task 3
    a) What is the efficiency of the prototype power plant? Write down the numerical estimate.
       (0.3 points)
    b) How much power could be produced in the prototype power plant? (0.4 points)
    c) How much energy could the power plant produce during a typical sunny day? (0.3 points)

   41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010   4/4

Task 4
  a) How large is the rise in the air temperature as it enters the chimney (warm air) from the
     surrounding (cold air)? Write the general formula and evaluate it for the prototype chimney.
     (1.0 points)
  b) What is the mass flow rate of air through the system? (0.5 points)

Solution:
41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010   1/4

Solution - Chimney physics
This problem was inspired and posed by using the following two references:

         W.W. Christie, Chimney design and theory, D. Van Nostrand Company, New York, 1902.
         J. Schlaich, R. Bergermann, W. Schiel, G. Weinrebe, Design of Commercial Solar Updraft
          Tower Systems — Utilization of Solar Induced Convective Flows for Power Generation, Journal
          of Solar Energy Engineering 127, 117 (2005).

Solution of Task 1
    a) What is the minimal height of the chimney needed in order that the chimney functions
       efficiently, so that it can release all of the produced gas in the atmosphere?

Let p(z ) denote the pressure of air at height z; then, according to one of the assumptions
p( z )  p(0)   Air gz , where p(0) is the atmospheric pressure at zero altitude.

Throughout the chimney the Bernoulli law applies, that is, we can write

1                                                       (1)
   Smoke ( z ) 2   Smokegz  pSmoke( z )  const. ,
2
where pSmoke(z ) is the pressure of smoke at height z,  Smoke is its density, and v(z ) denotes the
velocity of smoke; here we have used the assumption that the density of smoke does not vary
throughout the chimney. Now we apply this equation at two points, (i) in the furnace, that is at point
 z   , where  is a negligibly small positive number, and (ii) at the top of the chimney where z  h
to obtain:

1                                                          (2)
   Smoke (h) 2   Smokegh  pSmoke(h)  pSmoke( )
2
On the right hand side we have used the assumption that the velocity of gases in the furnace is
negligible (and also   Smokeg  0 ).

We are interested in the minimal height at which the chimney will operate. The pressure of smoke at
the top of the chimney has to be equal or larger than the pressure of air at altitude h ; for minimal
height of the chimney we have pSmoke(h)  p(h) . In the furnace we can use pSmoke( )  p(0) . The
Bernoulli law applied in the furnace and at the top of the chimney [Eq. (2)] now reads

1                                                        (3)
   Smoke (h) 2   Smoke gh  p(h)  p(0) .
2
From this we get

                  Air                                 (4)
 (h)  2 gh            1 .
                  Smoke 
The chimney will be efficient if all of its products are released in the atmosphere, i.e.,

      41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010          2/4

           B                                                  (5)
 ( h)      ,
           A
from which we have

     B2 1      1                                              (6)
h                              .
     A 2 g  Air
      2
                           1
                  Smoke
We can treat the smoke in the furnace as an ideal gas (which is at atmospheric pressure p(0) and
temperature TSmoke ). If the air was at the same temperature and pressure it would have the same
density according to our assumptions. We can use this to relate the ratio  Air /  Smoke to TSmoke / TAir
that is,

  Air   T                                                    (7)
         Smoke , and finally
 Smoke   TAir
  B2 1      TAir       B 2 1 TAir                             (8)
h 2                             .
  A 2 g TSmoke  TAir A 2 2 g T
For minimal height of the chimney we use the equality sign.

     b) How high should the chimney in warm regions be?

             T (30)                                           (9)
 h(30)   T       T (30)
        Smoke           ; h(30)  145m .
h(30)      T (30)
        TSmoke  T (30)

     c) How does the velocity of the gases vary along the height of the chimney?

The velocity is constant,

                         T                                                  (10)
                                                 T
  2 gh Air  1  2 gh Smoke  1  2 gh      .
           Smoke          TAir              TAir
This can be seen from the equation of continuity Av  const. (  Smoke is constant). It has a sudden
jump from approximately zero velocity to this constant value when the gases enter the chimney from
the furnace. In fact, since the chimney operates at minimal height this constant is equal to B , that is
v  B/ A.

     d) At some height z, from the Bernoulli equation one gets

psmoke ( z )  p(0)  (  Air   Smoke ) gh   Smoke gz .                      (11)
Thus the pressure of smoke suddenly changes as it enters the chimney from the furnace and acquires
velocity.

      41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010   3/4

Solution of Task 2
    a) The kinetic energy of the hot air released in a time interval t is

          1                                 T       (12)
E kin      ( Avt Hot )v 2  Avt Hot gh      ,
          2                                 TAtm
Where the index “Hot” refer to the hot air heated by the Sun. If we denote the mass of the air that
exits the chimney in unit time with w  Av Hot , then the power which corresponds to kinetic
energy above is

              T                                     (13)
Pkin  wgh         .
              TAir
This is the maximal power that can be obtained from the kinetic energy of the gas flow.

The Sun power used to heat the air is

PSun  GS  wcT .                                   (14)
The efficiency is evidently

     Pkin   gh                                       (15)
              .
     PSun cT Atm

    b) The change is apparently linear.

Solution of Task 3
    a) The efficiency is

      gh                                             (16)
           0.0064  0.64% .
     cT Atm
    b) The power is

P  GS  G( D / 2) 2   45 kW.                    (17)
    c) If there are 8 sunny hours per day we get 360kWh.

Solution of Task 4
The result can be obtained by expressing the mass flow of air w as

                           T                        (18)
w  Av Hot  A 2 gh             Hot
                           TAir
     GS                                              (19)
w
     cT
which yields

       G 2 S 2TAtm 1 / 3                             (20)
T  ( 2 2 2         )  9.1 K.
      A c  Hot 2 gh

     41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010   4/4

From this we get

w  760 kg/s.                                      (21)

2. Chimney physics


[User]
8. Elastic collision (7 points)

Consider a perfectly elastic collision of two balls, one of
which has mass M and is moving with velocity v. The other
has mass m ≤ M and stays initially at rest. The collision is
not necessarily central. The surface of the balls is slippery, so
the balls will not rotate.
i. (1 pt) What are the momenta of the balls before the colli-
sion in the frame of reference where the centre of mass of the
whole system stays at rest?
ii. (3 pts) What are the momenta of the balls (by moduli)
after the collision in the mass centre’s frame of reference?
iii. (3 pts) What is the maximal angle α by which the tra-
jectory of the initially moving ball can be inclined as a result
of the collision?
Problem

Solution:
8. Elastic collision

i. (1 pt) The centre of mass moves with the velocity ~ u =
M
M+m~ v, and that will be the speed of the small ball in new
reference frame, hence its momentum ~ p = −m~ u = − Mm
M+m~ v.
Since in this frame, the centre of mass is at rest, the large ball
needs to have equal by modulus and opposite momentum.
ii. (3 pts) Let the balls change a momentum ~ q. The small
ball will have momentum ~ p′
= ~ p+~ q, and as the centre of mass
remains at rest, the large ball will have momentum −~ p′
. The
energy conservation law can be written now as follows:
~ p2
2m
+
~ p2
2M
=
~ p′2
2m
+
~ p′2
2M
⇒ |~ p| = |~ p′
|,
ie. the moduli of the momenta will remain unchanged.
iii. (3 pts) In the laboratory frame, the momentum of the
large ball will be
~ p′′
= M~ u − ~ p′
;
since |~ p′
| remains constant, the angle α between ~ p′′
and ~ u will
be maximal when ~ p′′
⊥ ~ p′
, with
α = arcsin
|~ p′
|
|M~ u|
= arcsin
m
M
.
Problem

Elastic collision


[User]
AuPhO 2012, problem 14. Another question about practical measurements.
    Now for some brief advice for more subtle experiments.

    Idea 2
    In trickier experiments, it might not be obvious how to perform a measurement. Sometimes,
    it’s because the experiment contains strong noise sources, so that an accurate result can only
    be extracted by a clever choice of procedure. Other times, it’s because the experimental
    apparatus itself might behave in an unpredictable way; for instance, it might contain a
    nonlinear circuit element whose general behavior you won’t be told. (As an extreme case,
    some experiments are “black boxes” where you have to figure out what’s inside.) Finally,
    in some cases there are several adjustable input variables which are related to the output
    variables in a complicated way, so that you won’t be able to solve for the parameter if you
    just take arbitrary data points.

    Doing these experiments requires two stages. The first is exploration. Before settling on
    a strategy, you should figure out everything the experiment can do. Figure out all the
    ways that the pieces of the experimental apparatus can be combined. For every adjustable
    parameter, quickly sweep over its full range, and observe what happens to the outputs. (For
    example, the output might just change linearly, or it might not change at all, or peak, or
    qualitatively change at some point.) Figure out why each part of the experiment kit is included.

    Second, you have to settle on a strategy. Usually, this involves setting up the experiment in a
    particular configuration, varying one parameter over a well-chosen range, and fixing all other
    parameters to appropriate values (e.g. minimum, maximum, or intermediate ones, depending
    on what works best). It’s hard to say more about this process without getting into specific
    examples, but doing simulation labs is a fantastic way to practice it, since it’s essentially
    those labs’ only source of difficulty. A variety of good simulation labs are listed at the end of
    this problem set.

4    Analyzing Data
    Idea 3
    The zeroth step of data analysis is to neatly organize your work.

      • Always make an organized data table, with a row for each data point. For each column,
        specify the physical quantity calculated, and give its units. If your data points have
        uncertainties, indicate them on the table.

      • Draw neat graphs using a ruler and graph paper. Always label the axes with the corre-
        sponding quantities, and their units.

      • Set the axis scale so that the graphed data points cover almost the entire page. To
        achieve this, it can help to start the x and/or y axes at a nonzero value, though you
        should let the x-axis include x = 0 if you need to find a y-intercept.

      • It’s usually not allowed to use a calculator to find best-fit slopes and intercepts. Instead,
        you should eyeball the best-fit line, draw it with a ruler, compute the slope using two

      distant points on it, and show this computation explicitly. If theory implies that y = 0
      when x = 0, you should force the best fit line to pass through the origin.

    • Often, you’ll have to transform your data points before plotting, e.g. by raising them to a
      power, or applying a trigonometric function. You should learn how to do this as quickly
      and efficiently as possible with your calculator. It also helps to avoid keeping too many
      insignificant figures; your data points should never have more than 2 uncertain digits.

  These are all small things, but they really do help get reliable results, and many rubrics
  include explicit penalties if they aren’t done.

  Idea 4
  How much data is required for full credit on a given task? It’s tempting to give a rule of
  thumb, like “always take X data points for a line”, but the real answer is that it depends on
  the context. Generally, rubrics expect the student to spend the entire 5 hours making their
  results as accurate as possible. This means that you should stop taking data once the result
  won’t get much better, or when it’s not reasonable in the time given.

  To gauge how much time is reasonable, note that a 5 hour experimental exam contains 20
  points, so 1 point should take 15 minutes. Usually, you’ll spend a bit less than half your time
  collecting data. (But in some experiments, you’ll have to collect data “passively”, waiting
  several minutes for each data point. In these cases, you should collect data for most of the
  time, doing your thinking and data analysis while waiting.)

  To plot a completely generic line, it’s reasonable to take 6 to 8 data points. But there are
  plenty of exceptions, depending on the context.

    • Sometimes, it takes an extremely long time to get each data point, or the equipment
      might only support a limited number of types of measurement. In these cases, fewer than
      6 data points might be acceptable.

    • Sometimes, the data has a lot of noise, and/or is very fast to collect. (For example, you
      might be trying to measure a very small displacement or voltage, or the lab might be a
      simulation that instantly gives results.) In this case, you should take more measurements,
      to reduce the uncertainty. This can be done by repeating the same measurement several
      times and averaging the results, which helps save time making adjustments.

    • Sometimes, the trend isn’t a single line, but has features like peaks or transitions between
      different regimes. In these cases, you should take enough data to cleanly identify each
      feature, e.g. at least a few around each peak or transition region. In fact, this is the
      fundamental reason that an ordinary linear relation usually needs 5 to 7 data points:
      that’s how you show it’s linear in the first place.

       Idea 5
       If you’re plotting a line to extract a slope and intercept, there are a few important rules of
       thumb for collecting the data.

         • Ensure your data encompasses the widest possible valid range of parameters. The wider
           the range, the longer your “lever” is for getting an accurate result.

         • If you’re plotting a transformed quantity (such as T 2 , where you measure T ), then you
           should arrange your measurements so that the plotted data points are relatively evenly
           spaced. Otherwise, they may pile up in one end of the graph, making them less useful.

         • If the measurements in the experiment are tricky to make, then it’s quite likely that
           you’ll make a few mistakes, leading to data points that are far, far off the best fit line.
           So, when you see such an outlier, you should measure it again to see if it’s a real feature,
           or if you simply made a mistake.

       It should be noted that not all data analysis requires drawing lines. Sometimes it suffices to
       take the same data point several times and average the results. In addition, sometimes you
       may have to sketch more general curves through data points, such as resonance curves, and
       eyeball minima, maxima, or transition points.

       Remark
       If you’re unusually scrupulous, you might object to the point about outliers above:
       isn’t throwing out data a kind of scientific fraud? Well, in social science it’s a huge
       problem, because it’s one of the methods used for “p-hacking”, endlessly tweaking
       the data analysis to argue that a nonexistent effect is “significant”. That’s why there’s
       a movement in social science to preregister data analyses, fixing all the procedures in advance.

       By contrast, physics is inherently reproducible. In an Olympiad experiment, a measure-
       ment will give the same result (within uncertainty) every time you perform it correctly.
       And the relevant physical laws are quantitatively true, making reliable and concrete
       predictions. As a result, there’s just no way you can fake a result by retaking and replac-
       ing mistaken data points. Doing this will get you closer to the truth, rather than further away.

       Now, if you’re unusually unscrupulous, you might be wondering if you can just make up a
       bunch of data points. Students try this occasionally, but it’s an awful idea for many reasons.
       First, this actually is scientific fraud, because it can be used to justify false results as well as
       true ones. Second, it’s against the rules, often easily detectable, and grounds for immediate
       disqualification. Third, it doesn’t actually help you get good results. As mentioned above,
       rubrics generally demand you take the number of data points that gets you the most accurate
       result in the time available. Real data points make your answer more accurate, while fake
       data points don’t. So if the rubric wants at least 7, and you take 2 and fake 10 more, then
       the most likely result is that your final answer will be outside the window that gets credit.

Solution:
See the official solutions here.

   Now for some brief advice for more subtle experiments.

      Idea 2
      In trickier experiments, it might not be obvious how to perform a measurement. Sometimes,
      it’s because the experiment contains strong noise sources, so that an accurate result can only
      be extracted by a clever choice of procedure. Other times, it’s because the experimental
      apparatus itself might behave in an unpredictable way; for instance, it might contain a
      nonlinear circuit element whose general behavior you won’t be told. (As an extreme case,
      some experiments are “black boxes” where you have to figure out what’s inside.) Finally,
      in some cases there are several adjustable input variables which are related to the output
      variables in a complicated way, so that you won’t be able to solve for the parameter if you
      just take arbitrary data points.

      Doing these experiments requires two stages. The first is exploration. Before settling on
      a strategy, you should figure out everything the experiment can do. Figure out all the
      ways that the pieces of the experimental apparatus can be combined. For every adjustable
      parameter, quickly sweep over its full range, and observe what happens to the outputs. (For
      example, the output might just change linearly, or it might not change at all, or peak, or
      qualitatively change at some point.) Figure out why each part of the experiment kit is included.

      Second, you have to settle on a strategy. Usually, this involves setting up the experiment in a
      particular configuration, varying one parameter over a well-chosen range, and fixing all other
      parameters to appropriate values (e.g. minimum, maximum, or intermediate ones, depending
      on what works best). It’s hard to say more about this process without getting into specific
      examples, but doing simulation labs is a fantastic way to practice it, since it’s essentially
      those labs’ only source of difficulty. A variety of good simulation labs are listed at the end of
      this problem set.

4    Analyzing Data
    Idea 3
    The zeroth step of data analysis is to neatly organize your work.

      • Always make an organized data table, with a row for each data point. For each column,
        specify the physical quantity calculated, and give its units. If your data points have
        uncertainties, indicate them on the table.

      • Draw neat graphs using a ruler and graph paper. Always label the axes with the corre-
        sponding quantities, and their units.

      • Set the axis scale so that the graphed data points cover almost the entire page. To
        achieve this, it can help to start the x and/or y axes at a nonzero value, though you
        should let the x-axis include x = 0 if you need to find a y-intercept.

      • It’s usually not allowed to use a calculator to find best-fit slopes and intercepts. Instead,
        you should eyeball the best-fit line, draw it with a ruler, compute the slope using two
        distant points on it, and show this computation explicitly. If theory implies that y = 0
        when x = 0, you should force the best fit line to pass through the origin.

      • Often, you’ll have to transform your data points before plotting, e.g. by raising them to a
        power, or applying a trigonometric function. You should learn how to do this as quickly
        and efficiently as possible with your calculator. It also helps to avoid keeping too many
        insignificant figures; your data points should never have more than 2 uncertain digits.

    These are all small things, but they really do help get reliable results, and many rubrics
    include explicit penalties if they aren’t done.

    Idea 4
    How much data is required for full credit on a given task? It’s tempting to give a rule of
    thumb, like “always take X data points for a line”, but the real answer is that it depends on
    the context. Generally, rubrics expect the student to spend the entire 5 hours making their
    results as accurate as possible. This means that you should stop taking data once the result
    won’t get much better, or when it’s not reasonable in the time given.

    To gauge how much time is reasonable, note that a 5 hour experimental exam contains 20
    points, so 1 point should take 15 minutes. Usually, you’ll spend a bit less than half your time
    collecting data. (But in some experiments, you’ll have to collect data “passively”, waiting
    several minutes for each data point. In these cases, you should collect data for most of the
    time, doing your thinking and data analysis while waiting.)

    To plot a completely generic line, it’s reasonable to take 6 to 8 data points. But there are
    plenty of exceptions, depending on the context.
      • Sometimes, it takes an extremely long time to get each data point, or the equipment
        might only support a limited number of types of measurement. In these cases, fewer than

      6 data points might be acceptable.

    • Sometimes, the data has a lot of noise, and/or is very fast to collect. (For example, you
      might be trying to measure a very small displacement or voltage, or the lab might be a
      simulation that instantly gives results.) In this case, you should take more measurements,
      to reduce the uncertainty. This can be done by repeating the same measurement several
      times and averaging the results, which helps save time making adjustments.

    • Sometimes, the trend isn’t a single line, but has features like peaks or transitions between
      different regimes. In these cases, you should take enough data to cleanly identify each
      feature, e.g. at least a few around each peak or transition region. In fact, this is the
      fundamental reason that an ordinary linear relation usually needs 5 to 7 data points:
      that’s how you show it’s linear in the first place.

  Idea 5
  If you’re plotting a line to extract a slope and intercept, there are a few important rules of
  thumb for collecting the data.

    • Ensure your data encompasses the widest possible valid range of parameters. The wider
      the range, the longer your “lever” is for getting an accurate result.

    • If you’re plotting a transformed quantity (such as T 2 , where you measure T ), then you
      should arrange your measurements so that the plotted data points are relatively evenly
      spaced. Otherwise, they may pile up in one end of the graph, making them less useful.

    • If the measurements in the experiment are tricky to make, then it’s quite likely that
      you’ll make a few mistakes, leading to data points that are far, far off the best fit line.
      So, when you see such an outlier, you should measure it again to see if it’s a real feature,
      or if you simply made a mistake.

  It should be noted that not all data analysis requires drawing lines. Sometimes it suffices to
  take the same data point several times and average the results. In addition, sometimes you
  may have to sketch more general curves through data points, such as resonance curves, and
  eyeball minima, maxima, or transition points.

  Remark
  If you’re unusually scrupulous, you might object to the point about outliers above:
  isn’t throwing out data a kind of scientific fraud? Well, in social science it’s a huge
  problem, because it’s one of the methods used for “p-hacking”, endlessly tweaking
  the data analysis to argue that a nonexistent effect is “significant”. That’s why there’s
  a movement in social science to preregister data analyses, fixing all the procedures in advance.

  By contrast, physics is inherently reproducible. In an Olympiad experiment, a measure-
  ment will give the same result (within uncertainty) every time you perform it correctly.
  And the relevant physical laws are quantitatively true, making reliable and concrete

       predictions. As a result, there’s just no way you can fake a result by retaking and replac-
       ing mistaken data points. Doing this will get you closer to the truth, rather than further away.

       Now, if you’re unusually unscrupulous, you might be wondering if you can just make up a
       bunch of data points. Students try this occasionally, but it’s an awful idea for many reasons.
       First, this actually is scientific fraud, because it can be used to justify false results as well as
       true ones. Second, it’s against the rules, often easily detectable, and grounds for immediate
       disqualification. Third, it doesn’t actually help you get good results. As mentioned above,
       rubrics generally demand you take the number of data points that gets you the most accurate
       result in the time available. Real data points make your answer more accurate, while fake
       data points don’t. So if the rubric wants at least 7, and you take 2 and fake 10 more, then
       the most likely result is that your final answer will be outside the window that gets credit.