Competition “Physics Cup – IPhO2012"

8/3/2019 Competition Physics Cup IPhO2012"

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Competition Physics Cup IPhO2012

Eligibility: age- and educational restrictions for the participants are the same as for IPhO-2012.

Registration: please send e-mail to [email protected], indicating:

1. Your given name;2. Family name;3. Date of birth;4. E-mail address;5. Full mailing address;6. Your school;7. Your physics teacher

It is recommended to register as soon as possible, prior to the publication of the first problem (18.

Sept. 2011). However, you can also register at any later stage.

Distribution of problems: At 2 pm (GMT) of the third Sunday of each month, between September

2011 and June 2012, a new problem is published at the homepage of 43rd IPhO (www.ipho2012.ee)

and also sent by e-mail to all the registered participants.

Submitting the solutions: The contestants are asked to submit the answer as fast as they can, but

not later than the publication time of the next problem. The formulae can be submitted using

the LaTeX symbols; these can be included directly into the e-mail text (e.g. m=m_0/\sqrt{1-

v^2/c^2}). Alternatively, these can be also submitted as formulae in OpenOffice or MSWordfiles,

or scanned(digitally photographed)images of a clear handwritten text. Also, the students are asked to

send their solutions (using one of the above mentioned formats), but this can be done witha delay

of up to 48 hours (from the moment of submitting the answer). These solutions should contain as

few text (in English) as possible (few words are typically enough), mostly using formulae and

figures. When sending solutions, please use the subject line "Problem No 1" ("Problem No 2",

etc), exactly as written here (but without quotation marks); in that case you will receive an auto-reply

confirming that your solution has been received. If you do not want to receive an auto-reply,

make sure the subject line does not contain (not necessarily begin with) the text "Problem No 1"

(for instance just drop "No" and write "Problem 1").

Grading: Each problem costs 1.0 pts; this serves as a base score, which will be multiplied with factors

corresponding to bonuses and penalties. Either a full or zero credit is given; on a weekly basis (for thefirst three weeks on Sunday, for the fourth week on Thursday), the competitors are notified if their

solutions are (a) complete, (b) incomplete (e.g. some missing motivations), (c) with minor mistakes,

or (d) incorrect. Cases (b)-(c) incur a penalty factor of 0.9 and the case (d) 0.8. The students can

send the corrected solutions either upon finding an error or receiving the notification of

incompleteness; each revision incurs a penalty factor of 0.9. The first 10 correct answers

(supplemented later with a complete solution) receive a bonus factor according to the


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formula k= 1.111-n, where n is the order number. The best solution will receive a bonus factor

ofe (=2.718) and will be published as the official solution at the web page. If there is no single

solution, which is better than the others, the factor e may be distributed over several solutions of

similar quality in such a way that the product of all the factors equals still to e. If, in addition to the

best solutions, there are other good solutions which (due to certain reasons, eg. the usage of a

significantly different approach) deserve publishing, these published alternative solutions will receive abonus factor of 1.1. There is also an additional rule for those who send many incorrect solutions

before finding a correct one: if the product of penalty factors with the speed bonus factors gives a

number which is smaller than e-2/3, the factor e-2/3 is used, instead. Example: a student submits an

answer, discovers an error and therefore submits a revised answer (k1 = 0.9), together with the

solution; upon being notified that the solution has still a minor mistake (k2 = 0.9), he submits a third

revision (k3 = 0.9) of the answer and the solution. This solution turns out to be the fifth correct

solution, so k4 = 1.16; later on, it is found that this is the best solution which will be published at the

web-page (k5 = e). Thus, the overall score will be 0.931.16e = 4.673

Publication of results: The names and results of half of the students with best scores are publishedat the web-page; the list is updated monthly.

Distribution of awards: The awards are announced and handed over at the closing ceremony of the

43rdIPhO or sent to the mailing address (if the recipient is not present at the ceremony).

In your solutions what you may assume well-known and what not: things which are not in

theformula sheet need to be motivated/derived. (You don't need to check your typical high-school

formulae are there.)

Questions regarding thecompetition: e-mail to [email protected].

See also: Frequently asked questions.

Problem No 0 you will not receive points for this, but you can submit your answer and let us

know how long time did it take for you to solve it.

Physics solvers mosaic

http://www.ipho2012.ee/physicscup/physics-solvers-mosaic/

What is needed to be able to solve problems so well that you could get a gold medal at IPhO? Is itenough to be justvery gifted?Of course not, there are other students, who have solved a lot of

problems while you are thinking hard trying to "invent a bicycle", they are already writing the

solution, because they had solved a similar problem earlier. Is itenough to solve a lotofproblems and

read a lotofproblem solutions?Most often, no. Just solving or reading solutions, of course, will

increase your technical skills, but you also need to think over, what were the main ideas which made

it possible to solve the problem, and take these ideas into your permanent arsenal; if you solve too


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many problems, you don't have time to think over. Is itpossible to learn "the artofproblem solving"

andifyes then how?Well, 99% of the Olympiad problems are solved using a rather limited set of

ideas (for mathematics, that set is somewhat larger). So, ifyou acquire those ideas well enough so

thatyou can recognize them even ifthey are carefully hidden then the IPhO gold will be yours! Do

not worry, no-one expects you to discover a solving technique which has been never seen before,

because that would be an achievement worth of a Nobel Prize!

Since we started the topic of Nobel Prize is itenough to be the absolute winner ofthe IPhOto get, at

a later stage, a NobelPrize?(Each year, there is one Nobel Prize in physics similarly to the absolute

winner of IPhO.) Of course, it is not; however, you'll have better chances than anyone else. Becoming

a great physicist requires several components, one of which is having brilliant problem solving skills

(tested at IPhO). Another one is ability to make solvable models - formulate problems which can be

solved and which reflect important aspects of reality. Third component is ability to distinguish, which

problems are important and which are not. You can be very skilful and smart, but if you study

problems of marginal interest, no-one will pay attention to your research results. Finally, you need a

considerable amount of luck. Indeed, that particular field of physics in which you start your studies,eg. start making your PhD thesis, depends on somewhat random decisions it is almost impossible to

foresee, where are the biggest scientific challenges after five or ten years. Also, in order to perfect

yourself in regard of the above-mentioned three components, you need excellent supervisors and

excellent lab; although you have some freedom in choosing your supervisor and lab, you still need to

be very lucky to find outstanding ones!

I coined to name this section as "mosaic", because we shall describe here a set of solving techniques,

fragments of the whole arsenal needed for a perfect problem solver. With a large number of pieces,

the picture would become recognizable, but we need to start making it piece by piece While some

"tiles" will be useful for solving a spectrum of problems, other tiles are aimed to give more insightinto certain physical concepts.

Jaan Kalda, Academic Committee of IPhO-2012

Tile 1: minimum or maximum?

It is well-known that a system is stable at the minimum of its potential energy. But why? Why is a

minimum different from a maximum? For Fermat' principle it is clear: there is no longest optical path

between two points the ray could just go "zig-zag" -, but there is definitely one which is theshortest!

The reason is simple at an equilibrium state, the kinetic energy has always minimum (as long as

masses are positive). What we actually do need for a stability is a conditional extremumof one

conserved quantity (such as the net energy), under the assumption that the other conserved

quantities are kept constant (unconditional extremum is OK, too). Consider the motion of a body


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alongx-axis and let us describe it on the phase plane, with coordinates x andp (the momentum). The

overall energy isE=U(x)+p2/2m.Now, if we depict this energy as a surface in 3-dimensional space with

coordinates x,pandE, the point describing the state of the system will move along the intersection

line of that surface with a horizontal plane E=Const. At the minimum ofU(x), with p=0, this intersection

line would be just a single point, because this is the lowest point of that surface. The near-by

trajectories will be obtained if we ascend the horizontal plane a little, E=Emin+e, so that it no longer justtouches the surface, but cuts a tiny ellips from it. All the points of that trajectory (the ellips) are close

to the equilibrium point, so the sate is, indeed, stable.

It appears that a system can be stable also because of a conditional maximum of the net energy:

while an unconditional extremum of the kinetic energy can only be a minimum, things are different for

conditional extrema. Perhaps the simplest example is the rotation of a rigid body. Let us consider a

rectangular brick with length a, widthb, and thickness c (a>b>c). LetIcbe its moment of inertia for the

axis passing its centre of mass and perpendicular to the (a,b)-plane; Ib and Ia are defined in a similar

way. For a generic case, the moment of inertia Iwill depend on the orientation of the rotation axis,

but it is quite clear that Ic>= I>= Ia(it can be shown easily once you learn how to use tensorcalculations). Now, let us throw the brick rotating into air and study the motion in a frame which

moves together with the centre of mass of the brick (in that frame, we can ignore gravity). There are

two conserved quantities: angular momentumL, and rotation energy K=L2/2I. We see that for a

fixed L, the system has minimal energy for I= Ic (axis is parallel to the shortest edge of the brick), and

maximal energy for I= Ia (axis is parallel to the longest edge of the brick). You can easily check

experimentally that both ways of rotation are, indeed, stable! Not so for the axis parallel to the third

edge This phenomenon is demonstrated in a video made by NASA on the International Space

Station.

W

ell, actually the rotation with the minimal energy is still a little bit more stable than that of with themaximal energy; the reason is in dissipation. If we try to represent the motion of the system in the

phase space (as described above), we would start with touching a top of an hill with a horizontal

plane E=Emax(so that the intersection is just a point), but due to dissipation, the energy will

decrease, E=Emin e, and the phase trajectory would be a slowly winding-out spiral. So, while you are

probably used to know that dissipation draws a system towards a stable state, here it is vice versa, it

draws the system away from the stable state!


2. Fast or slow?

What is an adiabatic process? Most of the readers would probably answer that this is a process with agas which is so fast that there is no heat exchange with the surroundings. However, this is only a halfof the truth, and actually the less important half. In fact, it is quite easy to understand that this is notentirely correct: consider a cylinder, which is divided by a thin wall into two halves; one half is filledwith a gas at a pressure p, and the other one is empty. Now, let us remove momentarily the wall: thegas from one half fills the entire cylinder. Since no external work is done (the wall can be removed


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without performing a work), the energy of the gas is preserved, hence, the temperature remains thesame as it was at the beginning. Meanwhile, for an adiabatic process we would expect a decrease oftemperature by a factor of 2-1: part of the internal energy is supposed to be spent on a mechanicalwork performed by the expanding gas. However, if the piston moves faster than the speed of sound,the gas will be unable to catch up and push the piston. So, the adiabatic law was not followed becausethe process was too fast!

It appears that the adiabatic law for thermodynamics has also a counterpart in classical mechanics theconservation of the adiabatic invariant. For mechanical systems (oscillators) performingperiodic motion, the adiabatic invariant is defined as the area of theclosed curve drawn by thesystem in phase space(which is a graph where the momentum p is plotted as a function of therespective coordinatex), and is (approximately) conserved when the parameters of the system arechanged adiabatically, ie. slowly as compared with the oscillation frequency. For typicalapplications, the accuracy of the conservation of the adiabatic invariant is exponentially good and canbe estimated as e-f, where f is the eigenfrequency of the oscillator, and is the characteristic periodof the variation of the system parameters.

How are relatedto each other (a) adiabaticinvariantand (b) adiabaticprocess with a gas? The easiestway to understand this is to consider a one-dimensional motion of a molecule between two walls,which depart slowly from each other (Figure 1). Let us use the system of reference where one of the

walls is at rest, and the other moves with a velocity u


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that p0 = mx00, where 0 is the circular eigenfrequency of the oscillator; therefore, the full energy ofthe oscillator (calculated as the maximal kinetic energy) is E=p02/2m = p0x00/2 =J 0/2 =Jf0.Hence, the adiabatic invariantJ= E/f0: during adiabatic processes, the oscillation energy isproportional to the frequency. According to the quantum mechanics, the stationary energy levels ofthe oscillator are given byEn=hf0(n +1/2), where n isan integer representing the order number of theenergy level. Comparing the classical and quantum-mechanical results leads us to the conclusionthat during adiabatic processes,n = Const: the system will remain at the stationary state of

the same order number where it was(Figure 2) . (While it is not always completely correct tocombine classical and quantum-mechanical results, classical mechanics is a macroscopic limit of thequantum mechanics and hence, the conservation laws of both theories need to be compatible.)

Now, suppose our bi-atomic molecule is forced by an electromagnetic field in the form of an adiabaticpulse. In terms of classical mechanics we say that such a forcing is unable to pump energy intooscillations of the molecule, because the adiabatic invariant is conserved and hence, the energy ofoscillations depends only on the current eigenfrequency. In terms of quantum mechanics well say

exactly the same, but the motivation will be different: the adiabatic pulse contains no photons whichare resonant with the oscillator.

Another important role of the adiabatic invariant is protecting us from the cosmic radiation (incollaboration with the magnetic field of the Earth). It appears that the motion of a charged particlein a magnetic field can be represented as an Hamiltonian motion (we skip here the definition of theHamiltonian motion as it would go too deeply into the subject of theoretical mechanics), with a re-defined momentum. It appears also that with this new momentum (the so-called generalizedmomentum), the adiabatic invariant of a gyrating (helicoidally moving) charged particle is its magneticdipole moment (which is proportional to the magnetic flux embraced by the trajectory, hence this fluxis also conserved). So, if a charged particle moves helicoidally along magnetic field lines towards astronger magnetic field, due to the conservation of its magnetic moment, the perpendicular (to themagnetic field) component of its velocity will increase. Owing to the conservation of its kinetic energy,the parallel component of the velocity will decrease, and at a certain point, it becomes equal to zero:

the particle is reflected back (Figure 3). This is exactly what happens with a majority of the chargedparticles approaching Earth along the field lines of its magnetic field.


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Adiabatic invariant has simple every-day applications, too. Suppose you try to carry a cup of coffee this will be quite simple even if the cup is completely full. Now try the same with a plate of soup atleast with full plate, this will be quite difficult! Finally, with a large full photographic tray, this will benearly impossible! The reason is that when you try to keep your hands motionless, they still moveslightly, but the feedback from your vision allows you to correct the mistakes. The characteristic time-scale of such a motion of hands is of the same order of magnitude as your reaction time, in the rangeof 0.2 0.4 s. This is to be compared with the reciprocal of the circular eigenfrequency 0

-1 of thewater level oscillations. (0-1 differs from the full period Tby 2; 0-1 serves as a better referencehere, because the corrective motion of hands represents no more than a quarter of a full period of anoscillatory motion.) For a plate of depth h and length L, the smallest eigenfrequency can be estimatedas the frequency of standing waves of wavelength 2L (see also problem No 2 of IPhO-1984). Thespeed of shallow water waves is (gh)1/2, so that the eigenfrequency will be f0 = (gh)

1/2/2L. For a cupof coffee, the diameter and depth can be estimated as 7cm, hence the characteristic time scale ofoscillations will be 0-1 0.03s; with respect to such oscillations, the hand motion is adiabatic evenif we apply our smallest estimate of 0.2s (note that counter-intuitively, here a slow reaction is betterthan a fast one). For a plate ofH= 3cm and L = 25cm we get 0

-1 0.15s the hand motion isalready not very adiabatic. Finally, for a photographic tray ofH= 3cm and L = 60cm, we obtain 0-1 0.35s, which is really difficult to handle.

Finally, in the context of adiabaticity, it is interesting to analyse the IPhO problem about tides, whichwas posed in 1996 in Oslo (as Problem No 3). The problem is, indeed, very interesting: you are givena simplified model of a complex and important phenomenon, which, regardless of simplicity, gives youreasonable estimate and teaches valuable physical concepts. Let us read its text and comment themodel assumptions.

In this problem we consider some gross features ofthe magnitude ofmid-ocean tides on earth. We

simplifythe problem by making the following assumptions:(i) The earth andthe moon are consideredto be an isolated system,

/a very reasonable assumption: even the effect of the Sun is small in the reference frame of Moon-Earth centre of mass, where the inertial force and Sun gravity cancel each other out/(ii) the distance between the moon andthe earth is assumedto be constant,

/also reasonable: there are small variations, but nothing to worry about/(iii) the earth is assumedto be completelycovered by an ocean,

/this is definitely not the case, but at least the Pacific Ocean is very large; as a model, why not /(iv) the dynamiceffects ofthe rotation ofthe earth aroundits axis are neglected, and

/Did you understand what they wanted to say? If not, you need to learn reading the problem texts!


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Well, it means that the forcing of the water by the Moon is to be assumed to be adiabatic (slow), sothat the water level will take a quasi-equilibrium position (ie. equilibrium, where the equilibrium statechanges slowly in time). The validity of this assumption will be discussed below./(v) the gravitational attraction ofthe earth can be determined as ifall mass were concentrated atthecentre ofthe earth.

Again, a perfectly reasonable assumption: the gravitational field of a sphere (assuming that the massdensity depends only on the distance from the centre) is outside the sphere the same as that of a

point mass. The departure of the Earth's shape from a sphere is small, indeed.

And so, is the tide forcing really adiabatic? We need to compare the period of forcing with theeigenfrequency, or, the speed of the "piston" with the speed of waves. The speed of the "piston" is theEarth perimeter divided by 24 h, ie. v = 460 m/s. The relevant wave is, in effect, a tsunami with theestimated speed of (gH)1/2 = 200 m/s (here, H = 4000 m is an estimate for the average ocean depth).So, the forcing is far from being adiabatic, we could say that the assumption (iv) is horribly wrong. Onthe other hand, if we solve the problem according to these assumptions, we obtain for the tideamplitude h = 27 cm, which has at least a correct order of magnitude; why? Well, because for atypical resonance response curve, the response amplitude at a double eigenfrequency (which wewould need as the "piston" speed is ca twice the wave speed) is of the same order of magnitude asthat of a zero frequency (which is obtained in this Problem). Further, since the tidal motion of thewater is by no means quasi-stationary, the ocean boundaries will play an important role. What willhappen is very similar to the motion of tea in a cup, when you push the tea by a spoon: basinboundaries reflect the moving water, creating vortices and complex pattern of tidal heights. Toconclude, we learned that the above tide model fails for watertides (providing a very rough estimateof the tidal height); perhaps it can be used somewhere else with a better accuracy? The answer is"yes,forthe tides ofthe Earth crust"! Indeed, the mantle thickness is of the order of few thousandskm, which corresponds to almost ten-fold tsunami speed and makes the Moon as a "piston"reasonably adiabatic. The relative crust deformation due to tidal movements is so small that theelastic response of the crust is also negligible: the result h = 27 cm is indeed very close to reality.


3. Force diagrams or generalized coordinates?

Typically you are taught in high school that in order to solve problems with interacting bodies you

need to draw force diagrams, and write down the force balance equations (based on Newton II law)

forxand ycomponents (for three-dimensional problems, also the z-component). However, for

problems which are more difficult than the elementary ones, this is typically far from being the

simplest approach. Meanwhile, there is a very powerful method based on generalized coordinates,

which provides in most cases the easiest route to the solution. The basic idea of the method is as

follows.

Suppose the state of a system can be described by a single parameter , which we call the

generalized coordinate (the method can be also applied with two or more parameters, but this will

complicate things, and in most cases, one parameter is perfectly enough). Then, what you need to do

is to express the potential energy of the system in terms of , , and the kinetic energy

in terms of , the time-derivative of : . Then, if there is no dissipation and external


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forces, the net energy is conserved: . Upon taking time-derivative of this

equality, we obtain , from where we can express the acceleration of the

generalized coordinate:

Note that most often, is constant, because the kinetic energy is proportional to , and

plays the role of an effective mass . In some cases, it may happen that depends also on

and/or depends also on ; then, the above formula will not work, but the technique itself remains

still applicable (cf. the example of rotating spring below).

In order to illustrate this method, let us start with a simple wedge problem.Consider a

system where a ball of mass lays on a wedge of mass , and is attached with a weightless rope

and pully to a wall as depicted in Figure; you are asked to find the acceleration of the wedge,

assuming that all the surfaces are frictionless, and there is a homogeneous gravity field .

When using the force diagram method, it would be a good idea to use the (non-inertial) reference

frame associated with the wedge (introducing thereby the inertial forces and ), because

otherwise, it would be difficult to write down equation describing the fact that the ball will remain on

the inclined surface of the wedge. Here, however, we leave this for the reader as an exercise, and

describe the state of the system via the displacement of the wedge. Then, the velocity of the wedge

is ; the velocity of the ball with respect to the wedge is also , implying that the vertical component

of the ball's velocity is , and the horizontal component is . Hence, we find that

Upon taking time derivative of this equation and cancelling out , we obtain an expression for the

wedge acceleration:


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As another example, let us consider an old IPhO problem (5th IPhO in Sofia, 1971, Problem No

1). The set-up is quite similar to the previous problem, but there is no wall, there are two bricks

instead of one ball, and the wedge has two inclined surfaces (see Figure); we ask again, what is the

acceleration of the wedge.

You might think that the method does not work here, because there are two degrees of freedom: the

wedge can slide on the table, and the bricks can slide with respect to the wedge. However, if we make

use of the conservation of the centre of mass (there are no external horizontal forces), we can express

the displacement of the bricks (with respect to the wedge) via the displacement of the wedge

:

What is left to do, is to write

substitute by , take time derivative of the full energy, and express . Well, there is some math do

be done, but that is actually just an algebra. If you do it correctly, you obtain

.

A really simpleexample is provided by water level oscillations in U-tube. Let the water occupy

length of the U-tube, and let us use the water level height (with respect to the equilibrium level)

as the generalized coordinate. For a state with , a water column of height from one arm has

been lifted by an height difference and moved into the other arm of the U-tube, which corresponds

to the potential energy ; meanwhile, . So, upon applying our technique we

obtain , which describes an harmonic oscillator of circular frequency .

Actually, when in hurry and oscillation frequency is needed, two steps of the scheme (taking time

derivative and writing the equation of motion) can be skipped. Indeed, for an harmonic oscillator,

both and need to be quadratic in and , respectively, ie. should have form

and , where and are constants; then, .


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Next, the techniquecan be used to analyse oscillations in simple rotating systems, such as,

for instance, a system of two balls of mass , connected with a spring of length and stiffness ,

rotating with angular momentum (which is perpendicular to the spring). Here, again, an additional

(to the energy) conservation law (of angular momentum) reduces the effective number of degrees of

freedom down to one. Let us use the deformation of the spring as the generalized coordinate. Then,

This case is different in that the kineticenergy depends not only on , but also on ; in

effect, the second term of the kinetic energy behaves as a potential one, and can be combined into an

effective potential energy in the expression for the full energy. Following our technique,

This equation of motion can be linearised around the state of equilibrium (such that for ,

the right-hand-side turns to zero), by introducing . Linearisation means approximating anon-linear function with a linear one, and is typically done by neglecting in the Taylor

expansion quadratic and higher terms, ie. by substituting with ; this is

legitimate if the argument varies in a narrow range, in this case for . As a result, we

obtain

which gives us immedieately the circular frequency of small oscillations, .

What we did here can be also called a linear stability analysis (which is a very popular technique in

physics). Indeed, it is easy to see that regardless of the parameter values, the circular frequency is

always a real number, ie. the circular trajectories of the balls are always stable (meanwhile, imaginary

circular frequency would mean that the solution includes a component which grows exponentially in

time, ie. the regular motion along the circular trajectory would be unstable).

Note that almost exactly the same analysis which was done here for the rotating spring, was used in

the"official" solution of the Problem 1 (subquestion 3) of IPhO-2011. However, it appears that

for the mentioned problem, this technique cannot be applied as easily: there is one mistake in the

solution, and another one among the assumptions of the problem; for more details, see the mosaic

tile "Are trojans stable?".

Up til now we have dealt with problems where the task was to find an acceleration. What to do, if

you are asked to find a force?For instance, a sphere and a wedge are placed on two facing ramps

as shown in Figure; all the surfaces are frictionless. Find the normal force between the wedge and the

sphere.


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Well, it would be very easy to find the acceleration of the ball (or that of the wedge) using the method

of generalized coordinates (ball displacement can be used as the coordinate). But once we know the

acceleration, it is also easy to find the normal force between the wedge and the ball from the

Newton II law! (The answer is .)

The method of generalized coordinates is designed to work for dissipation-less systems.. However, in

some

c

ase

s it is also possible

to take

into acc

ount the

fric

tion. To illustrate this, let us modifythe previous problem so that the right ramp remains frictionless, but the left ramp has high friction, so

that the ball will rotate along it, and the friction between the wedge and the ball is described by kinetic

friction coefficient .

The idea here is to "fix" theenergy conservation law by adding the work performed by the

friction force. Initally, such an equation will involve the normal force as a parameter, but it can be

determined later: we express the normal force in the same ways as for the previous problem, and this

will be the equation for finding . So, and ; the

contact point leaves "traces" both on the wedge (of length ) and on the ball (of length ),

corresponding to the net work of . So, the energy conservation law is written

as

from where

.

Now, assuming that we have heavy wedge, and the system moves leftwards, the Newton II law for

the wedge can be written as

,

and hence,


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.

As a final example illustrating this method, let us consider a somewhat more difficult problem, posed

by W.H. Besant in 1859, and solved by Lord Rayleighin 1917: in an infinite space filled with an

incompressible liquid of density at pressure , there is a spherical "bubble" of radius , which hasvacuum inside. Due to the pressure (far away, it is kept equal to ), the "bubble" starts collapsing;

find the collaps time of the "bubble". Here we use the radius of the "bubble" as the generalized

coordinate; there is no potential energy, but there is work done by the pressure,

. What is left to do, is to express the kinetic energy of the fluid in terms of . Due to the

incompressibility of the fluid, the volume flux of liquid through any spherical surface of radius

around the centre of the "bubble" is independent of : .

So, the kinetic energy can be found as

So, the energy balance can be written as

This equation could be used to find the acceleration ; however, we need to know the collapse time;

so we put , and express in terms of and :

Thus, we were able to obtain an answer, which contains a dimensionless

integral: substituting allowed us to get rid of the dimensional quantities under the

integral (if possible, always use this technique to convert integrals into dimensionless numbers). This

result could be left as is, since finding an integral is a task for mathematicians. The mathematicians,

however, have been up to the task: where denotes the gamma

function. So, we can write

Finally, to close the topic of the generalized coordinates, it should be mentioned that this

technique can be developed into generic theories Lagrangian and Hamiltonian formalisms,

which are typically taught as a main component of the course oftheoretical mechanics. In particular,

the Hamiltonian formalism makes it possible to prove the conservation of adiabatic invariant, as well


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as the KAM (Kolmogorov-Arnold-Mozer)theorem, as well as to derive conservation laws from the

symmetry properties of the Hamiltonian (orLagrangian) using the Noether's theorem. The Hamiltonian

approach differs from what is described here by using the generalized momentum , instead of the

generalized velocity . For the most typical cases when the kinetic energy is proportional to the

square of the generalized velocity, one can just use the effective mass (defined

above): . Then, the expression for the full energy is

considered as a function of and , , and is called the Hamiltonian; the equation of

motion is written in the form of a system of equations, , . However, for

the practical application ofproblem solving,the simplified approach to the generalizedcoordinates

provided above is justenough!


4. Are Trojans stable?

To begin with, what are Trojans? These are small celestial bodies which move together with two

heavy bodies (typically the sun and a planet) in such a way that (a) the relative position of the three

bodies does not change (they rotate as if forming a solid body); (b) the motion of these small bodies

is stable: small fluctuations in the relative position will not be amplified. It appears that for a two-body

system, eg, the Sun and the Jupiter, there are five points, where a small (third) body could move so

that the condition (a) will be satisfied the so called Lagrangian points, denoted by L1, L2, L3, L4

and L5. The first three of these lay at the same line with the Sun and Jupiter. In addition, as wasshown in IPhO problem 1989-2, the condition (a) will be also satisfied, if the three bodies form an

equilateral triangle; the respective points are denoted by L4 and L5. It appears that the Lagrangian

points L1, L2 and L3 are always unstable, but the points L4 and L5 can be stable. In particular, for the

Sun-Jupiter system, L4 and L5 are stable, and there are actually a considerable number of asteroids

"trapped" into the vicinity of these points. These asteroids are named after the figures of the Trojan

war, which is why the satellites in Lagrangian points L4 and L5 are called the Trojans (the term is not

limited to the Sun-Jupiter system).

And so, the Trojans are stable by definition, and the title here is somewhat inaccurate; the actual

question is, are

the

Lagrangian points L4 and L5 always stable?

The question is motivated bythe Problem 1 (subquestion iii) of IPhO 2010 which made an attempt of studying the stability of L4 for

a system consisting of two equal point masses (actually, small oscillations of a small body moving

around L4). The official solution, concluded that the small body will oscillate, ie. the position is stable.

However, a careful analysis shows that the stability of L4 and L5 is achieved only if the ratio of the two

large masses is large enough larger than , ie. for two equal masses the

equilibrium is unstable! So, what went wrong in the IPhO Problem 2010-1-iii?


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To begin with, let us mention that stability for such a system is actually quite a surprising thing.

Indeed, according to the Earnshaw's theorem, there are no stable equilibrium configurations for

particles with Coulomb potential (gravitational potential is identical to the Coulomb one). Indeed, if

there were a point Pwhich is a stable equilibrium for positive charges, in the immediate vicinity ofP,

all the field lines need to be directed towards P, because otherwise, a positive charge would escape

from P along the outgoing field lines. This, however, would be in contradiction with the Gauss law for asmall spherical neighbourhood ofP: the flux of the force field needs to be negative (there are only

incoming field lines), but equals strictly to zero for Coulomb potentials. Here we hope that L4 will be a

stable equilibrium in the system of reference co-rotating with the two heavy masses; in that system,

there is also the force field of the centrifugal force. Unfortunately, centrifugal force is of no help,

because it leads to the creation of field lines in vacuum, making the flux around P strictly positive

(recall that stability requires a negative flux). Now, let us recall that besides the gravitational and

centrifugal forces, we have also the Coriolis force, which acts, however, only on moving bodies.

Hence, the stability can be created only by the Coriolis force!

Unfortunately, the Coriolis force is not included into the Syllabus of IPhO. Quite often, the usage ofCoriolis force can be avoided, most typically by using non-rotating systems of reference (the origin

can move along a circle, though), or studying only potential energies (Coriolis force does not perform

work). Here, however, neither of these tricks can be used: the system of reference needs to rotate

(because the net gravitational field is stationary only in such a system), and as we saw, we cannot

work with the potentials only, because the Coriolis force is needed to achieve the stability. The authors

of the problem believed to have been found a work-around: assume that there is an approximate

conservation of the angular momentum (with respect to the centre of mass O of the whole system, cf.

Figure) of the small body, and apply the method of generalized coordinates: if the radial

displacement from the equilibrium point L4 (or L5; marked in Figure as P) is used as the

coordinate, the tangential velocity can be expressed via the radial one , allowing us towrite down the energy balance equation (recall that the Coriolis force cancels out from that equation

as it does not create any work). From that equation, one could immediately obtain the circular

frequency of small oscillations. However, we have made two mistakes here. First, the angular

momentum is not conserved, even not approximately. Indeed, angular momentum is conserved if

the force field is rotationally symmetric. However, a superposition of the gravitational fields of two

point masses has no such symmetry. Approximate conservation of would require that such a

symmetry is local: near L4, the curvature radius of the equipotential surface needs to be equal to the

distance from the origin O; regrettably, this is also not the case. Second, the gravitational energy

depends not only on the radial coordinate , but also on the tangential displacement from L4; note

that there is no way of expressing via , even the (non)conservation of the angular momentum isuseless.


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So, how to obtain a correct solution to this problem what is the frequency of small oscillationsaround the Lagrangian point L4, assuming that the two heavy masses are equal? Well, we just need to

follow the standard way of doing such things: first,we write down theequations ofmotion for

both coordinates, and , and second, use linear approximation (which is valid for small

displacements), ie. neglect the terms which involve second and higher powers of and ; when

working with the gravitational potential, this corresponds to neglecting the terms with third and higher

powers. In such a way, we obtain linear equations of motion. The third step is to find the

eigenfrequencies of that system of equations, ie. such values of that with a proper choice of ,

the equations will be satisfied with and . If there is at least one

eigenfrequency with a positive imaginary part then the system is unstable. On the other

hand, if for all the eigenfrequencies , the system is stable (unless there is aneigenfrequency , in which case the linear analysis is not sufficient for proving stability). As

afourth useful idea, let us note that with more than one point mass, it is much more convenient

tocalculate the gravitational potential, rather than the resultant gravitational force.

According to what has been said, we need an expression for the Coriolis force. Of course, we could

just take a ready formula, but it would be better to understand how it can be obtained (if you are not

interested, please skip this part). And so, consider a system of reference, which rotates around the

origin with an angular velocity (the vector defines the rotation axis according to the corkscrew

rule). Consider a point , which is motionless in the rotating system, and let us denote . In

the lab system of reference, the point moves with velocity , and when studyingthe direction of the velocity , one can see that . Now, if the point moves in the

rotating system of reference with velocity (let us use to measure the time in the rotating

system), then this additional velocity needs to be added to what would have been for a motionless

point:


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So, we can conclude that the time-derivatives of vectors in rotating and lab systems of reference are

related via equality

This is written in the form of an operator, which means that we can write any vector (eg or )

rightwards of all the three terms. In particular, we can apply this formula to the right- and left-hand-

sides of the equality :

Here we need to bear in mind that when taking derivatives of vectors and products of vectors, all the

well-known rules can be applied; in particular,

and . We also need the rule for the double cross

product, ; you can memorize this equality by keeping in mind thatthe double product is a linear combination of the vectors from the inner braces, and that the sign '+'

comes with the vector from the middle position. And so, bearing in mind that and ,

and assuming that , we obtain

Let us recall that is the acceleration of the point as seen in the lab system of reference, and

is the same as seen in the rotating system of reference. Now, if is a point mass , and there is

an external force acting on , then and hence,

ie. in the rotating system of reference, the body behaves as if there were additional forces: the

Coriolis force , and the centrifugal force .

Now we are finally ready to tackle the IPhO Problem 2010-1-iii. As mentioned above, the first

step is writing down the potential energy in the rotating system of reference (see Figure above):

note that the last term corresponds to the potential energy of the centrifugal force. When working with

this potential energy, we can forget about the constant part of it; additionally, we can also forget

about the linear part, because it gives us the force, which is exactly zero: our point L4 has been

chosen so as to provide an equilibrium. Owing to that equilibrium, we have also

equality . We approximate the potential using the formula


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(which includes the first two terms of the Taylor expansion); keeping in mind that

we obtain

This can be further simplified:

hence, the resultant force of the gravitational and centrifugal forces can be written

as and . Component-wise, the Coriolis force can be

written as , . Finally, theequations ofmotion can be written

as

Now wecan proceed with the final step, finding theeigenfrequencies.We look for the

solutions in the form , , upon substituting these expressions we obtain

This is a quadratic equation for , which results in This can be brought to

the form

So, we can conclude that due to the presence of an unstable solution

,

theequilibrium point L4 (and L5) is not stable in thecase of a binary gravitational system

with two equal masses.

Final notes. This mosaic tile is different from the others in that it is not motivated by a (more or less)

universal problem solving technique or an important physical concept; instead, it is mainly aimed to

clarify a single IPhO problem. The assumptions of physics contest problems don't need to be entirely

correct. However, for physicists, it is very important to be aware, how firm or loose are the

assumptions of their study, and to which degree can the the conclusions of their study be affected by

the mismatch between the assumptions and the real life. Studies based on wrong assumptions can be

useful, but the fact that the assumptions are not valid needs to be emphasized. The contest problems


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serve mostly educational purposes and are no different ifan invalid assumption is made,itshould

be clearlypointed out, and, if possible, explained why an incorrect assumption was made. Of course,

no-one is secured against accidental mistakes; in particular, the more interesting your newlyinvented

problem is, the higher are the chances thatthere are some mistakes. Meanwhile, the IPhO problems

serve as a well-tested pool of exercises, tested by the contestants and leaders of many countries, and

it is better to make sure that there are no unresolved issues in these problems. This is the reasoningwhich led to the current mosaic tile. Although we are not able to close here the list of all such

problems (for instance, there are problems1988-2-iv and 2000-3-iv,v; you can let me know if you

found out what is wrong there), more recent problems get typically more attention.


Problem 0

What is the diameter of the lens which was used to make the photo below?

Remark: although the photographic lenses are made of several optical components, for many practical

calculations including this problem they can be considered as idealthin lenses.


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Solution

Although this was not a competition problem, two correct solutions have been submitted: by Dinis

Cheian and Ivan Ivashkovskiy.

The Ivan's solution is very thorough with error analysis (as should be for such a semi-experimental problem) and is presented below on two images. Note that he defines n as the ratio

of two lengths, him and l, which he can measure from the photo and thus calculate n. On the other

hand, he finds expressions for him and l, and using these expressions shows that n =H/D. Knowing the

object lenth H, he can find now D.

In addition to that solution, I have two comments. First, the usage of the Newton formula would have

simplified slightly the mathematics: x1x2 =F2, where x1 =dFand x2 =f Fare the distances of the object

and of the image from the respective focal planes (using Ivan's notations, dand fare the respective

distances from the lens). So, the diameter of the circle of confusion l=Dx2/F=FD/x1=FD/ (d

F). Second,pay attention to the main result: the diameter of the lens equals to the diameter ofthecircle ofconfusion created by a far-away dot-source, ifmeasured by the image of an in-

focus ruler.So, if you have a subject for your photo (eg. friends face for a portrait) and you want to

blur the (far-away) background, the degree of the background blur is defined purely by the diameter

of the lens: a tele-lens 600mm/4 creates circles of confusion of the size of the face, leaving no visible

details at the background; a point-and-shoot camera with a small sensor and 8mm/4 lens creates

almost no background blur: the diameter of the circles of confusion are smaller than the pupils of the

portrait.

Finally, let us analyse the benefits of a large camera: you have a full-frame DSLR, and your friend has

a small point-and-shoot camera of a 4 times smaller sensor (in linear size). Your friend takes a photo

of something, zooming to the focal length of 13 mm and is using the full aperture of F/4 (ie. the focal-

length-to-diameter ratio equals to 4). You want to obtain exactly the same result: in order to have the

same angle of view (and perspective), you need to take the focal length equal to 13*4 = 52 (your

50mm/1.4 standard prime lens works well). In order to have the same blur, you need to shut down

the lens down to the aperture F/16 (using the diaphragm you decrease the effective diameter). On

the other hand, if you take a photo at the full aperture of F/1.4, your friend would need the aperture

of F/0.35, which is theoretically impossible (would violate the second law of thermodynamics).

Meanwhile, if you want to have both sharp foreground and sharp background, the point-and-shoot is

better: you can use F/22, which would correspond to F/88 for the DSLR. While theoretically this is

possible, the smallest aperture is typically only F/32 (starting from ca F/16, diffraction starts

degrading the image).

- Jaan Kalda Academic Committe -


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Problem No 1

A ballistic missile is launched from the north pole, the target is at the latitude \Phi (>0, if northern

hemisphere,


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Finally, what you may assume well-known and what not: things which are not in the formula

sheet need to be motivated/derived. (You don't need to check your typical high-school formulae are

there.)

Results after Problem 1

The list of the contest leaders after the first problem:

Points Name Country School

4.8156 Brahim Saadi AlgeriaPreparatory School for Science & Technology of

Annaba

2.5937 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau 2.5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye

2.1436 Ivan Tadeu Ferreira Antunes Filho Brazil Colgio Objetivo, Lins, So Paulo

1.7538 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau

1.6105 Jakub Supe Poland 14th School of Stanisaw Staszic, Warsaw

1.6105 Nikita Sopenko Russia Lyceum No.14, Tambov

1.1 Kohei Kawabata Japan Nada High School

1.1 Lars Dehlwes Germany Ohm-Gymnasium Erlangen

1.1 SZAB Attila Hungary Lewey Klra High School, Pcs

1 Ion Toloaca Moldova liceul "Mircea Eliade"

1 Jakub afin Slovak Pavol Horov Secondary, Michalovce 1 Jaan Toots Estonia Tallinn Secondary Science School

1 Lus Gustavo Lapinha Dalla Stella Brazil Colgio Integrado Objetivo, Barueri, Brazil

1 Lev Ginzburg Russia Advanced Educational Scientific Center, MSU, Mosco

1 Alexandra Vasileva Russia Lyceum "Second School", Moscow

1 Task Ohmori Japan Nada High School

1 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau

1 Sharad Mirani India Prakash Higher Secondary School

0.9801 Mekan Toyjanow TurkmenistanTurgut Ozal Turkmen Turkish High School

0.8733 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau 0.81 Meylis Malikov TurkmenistanTurgut Ozal Turkmen Turkish High School

0,81 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India

0.81 Liara Guinsberg Brazil Colgio Integrado Objetivo, So Paulo, Brazil

0.6561 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum


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The list of those who completed successfully the problem, ordered according to the arrival time of the

correct solution:

1. Dinis Cheian (Moldova),

2. Mikhail Shirkin (Russia),

3. Ivan Tadeu Ferreira Antunes Filho (Brazil),

4. Cristian Zanoci (Moldova),

5. Brahim Saadi (Algeria).

6. Nikita Sopenko (Russia)

7. Jakub Supe (Poland)

8. Papimeri Dumitru (Moldova)

9. Mekan Toyjanow (Turkemistan)

10. Kohei Kawabata (Japan)

11. Lars Dehlwes (Germany)

12. Meylis Malikov (Turkemistan)

13. Szab Attila (Hungary)

14. Jakub afin (Slovakia)

15. Ion Toloaca (Moldova)

16. Jaan Toots (Estonia)

17. Bharadwaj Rallabandi (India)

18. Lus Gustavo Lapinha Dalla Stella (Brazil)

19. Lev Ginzburg (Russia)

20. Liara Guinsberg (Brazil)

21. Alexandra Vasileva (Russia)


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22. Task Ohmori (Japan)

23. Ilie Popanu (Moldova)

24. Nadezhda Vartanian (Russia)

25. Sharad Mirani; (India)

The best solution: Brahim Saadi.

Besides, there was one almostcorrectsolutions and 23 wrong solutions. The overall number of

registered participants: 186.

-

NB! Don't forget to try solving Problem No 2, the faster ones will get more points!

Jaan Kalda Academic Committee of IPhO-2012

Solution

I hope you found the first problem to be interesting. I am fond of it myself, because (a) it extends

nicely the widely-known fact that 45 degrees is the optimal throwing angle; (b) is exactly solvable

regardless of seeming difficulty; (c) optimal solution is technically quite simple; (d) the answer has a

real practical relevance. According to the results, it was not a simple one; still the difficulty wasperhaps close to optimal it kept the most of you busy the whole month, and still 13% of you were

able to solve it. The second problem is probably at least as difficult (maybe the third one will be

simpler).

Regardless of the relative difficulty, there was one of you, for whom it was not difficult enough, and

who solved a more generic star-war-problem: the launching site and target are at different distances

from the centre of Earth. This is not a cosmetic change, because the problem becomes non-

symmetric. Solution-wise, it adds one more step, which makes use of the geometric property of

hyperbolas. For a professional physicist, it is really important to be able to figure out if the problem

(or model) under study can be made more generic while maintaining solvability, so for this problem,we have a clear winner of the best solution. (However, I am inclined to think that for the major part of

the contest problems, it will be impossible to make such nice, non-cosmetic generalizations, which

would justify giving the award of the best solution.)

And so, the award for the best solution, a bonus factor e, goes to Brahim Saadi.


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The second-best solution, the one made by Mikhail Shirkin, is actually the one I had in mind when

giving you the problem. Mikhail has written it down in a very laconic way: for a research paper, this

writing style is definitely not appropriate, but for the solution of an Olympiad problem, it is just

perfect! So, I am giving him a bonus factor of1.1.

Another factor of1.1 goes to the solution ofSzab Attila, which represents a good style of researchpapers: the model assumptions are clearly formulated, formulae are put into correct sentences, text is

understandable even for people who are not very well-prepared.

There are two more solutions which I found useful to show you (both will also get a factor of1.1). The

first one is ofLars Dehlwes, who had made a useful graph ofhowthe minimal velocity depends on

the latitude ofthe target(note that small differences in the required velocity imply actually a large

economy, because the fuel mass of the missile depends exponentially on the terminal velocity, cf. Eq

IV-21 of the latest formula sheet).

The second one is ofJakub Supe, who was not the only one to derive the formula E=-GMm/2a, butwas among the first ones to do so, and did it nicely in LaTeX. While actually you did not need to derive

this formula as it is in my formula sheet, it is useful for you to know, how it is done.

Many of you (majority, in fact) used the "brute force" approach: express the launch velocity (or the

square of it) via some parameter (eg. launch angle or ellipticity of the orbit) and find the minimum

from the condition that the derivative is zero. This was definitely a difficult way of doing it, but I was

quite amazed by your technical skills! There was not a single mistake in very long mathematical

manipulations, ending up in perfectly correct results! (Though, some final answers were left non-

simplified.)

Last but not least, what is the lesson of this problem?First, quite often, problems on extrema can

be solved withouttaking derivatives, geometrically, which is typically a much simpler

way. Second, when solving the problems put on the Kepler's laws,the geometrical and optical

properties ofellips (see Eq XII-7 on the formula sheet; by the way, these properties are connected

with each other via the Fermat's principle) are always very useful. Third, expression for the full

energy, E=-GMm/2ais extremely handy(I'd like to call it the Kepler's fourth law but it was not

derived by Kepler.)

And one more thing: one of you, Comoglio Lorenzo, pointed out that there is a satellite simulation

written in Java, if you want to play with trajectories, have a look.


Best solutions.


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Brahim Saadi:


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Mikhail Shirkin:

Szab Attila:


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Lars Dehlwes:

J

akub Supe

:


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Problem No 2

Consider a brick-shaped ferromagnetic of relative magnetic permeability >> 1, which has

dimensions 2a 2a a and a narrow slit of width dand depth a sawed into it as shown in Figure. Youmay assume that d>> a >> d. A circular loop of diameter a and inductance L, made of a

superconducting material, is put into that slit; the loop carries electric current I.What is the

mechanical workA needed to be done in order to pull the loop out of the slit and move it to a large

distance from the ferromagnetic?

Remarks: (a) the inductance of the loop equals to L when it is far away from the ferromagnetic. (b)

The current in the loop equals to Iwhen the loop is inside the slit. (c) You may assume that the

hysteresis of the ferromagnetic is negligible, and is constant (independent ofB).


The list of the contest leaders after the second problem:

Points Name Country School

5.6695 Brahim Saadi Algeria

Preparatory School for Science & Technology of

Annaba

4.9694 SZAB Attila Hungary Lewey Klra High School, Pcs

4.5041 Nikita Sopenko Russia Lyceum No.14, Tambov

4.4353 Ivan Tadeu Ferreira Antunes Filho Brazil Colgio Objetivo, Lins, So Paulo

2.7363 Jakub afin Slovak Pavol Horov Secondary, Michalovce

2.6785 Lars Dehlwes Germany Ohm-Gymnasium Erlangen

2.5937 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau


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2.5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye

2.435 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau

1.9801 Lus Gustavo Lapinha Dalla Stella Brazil Colgio Integrado Objetivo, Barueri, Brazil

1.9703 Alexandra Vasileva Russia Lyceum "Second School", Moscow

1.81 Ion Toloaca Moldova liceul "Mircea Eliade"

1.7643 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau

1.7538 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau

1.6105 Jakub Supe Poland 14th School of Stanisaw Staszic, Warsaw

1.1 Kohei Kawabata Japan Nada High School

1 Jaan Toots Estonia Tallinn Secondary Science School

1 Lev Ginzburg Russia Advanced Educational Scientific Center, MSU, Mosco

1 Sharad Mirani India Prakash Higher Secondary School

1 Task Ohmori Japan Nada High School

0.9801 Mekan Toyjanow TurkmenistanTurgut Ozal Turkmen Turkish High School

0.81 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India

0.81 Liara Guinsberg Brazil Colgio Integrado Objetivo, So Paulo, Brazil

0.81 Meylis Malikov TurkmenistanTurgut Ozal Turkmen Turkish High School

0.6561 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum

Points for Problem No 2:

Pts Name Country

3,8694 SZAB Attila Hungary

2,8935 Nikita Sopenko Russia

2,2917 Ivan Tadeu Ferreira Antunes Filho Brazil

1,7363 Jakub afin Slovak

1,5785 Lars Dehlwes Germany

1,435 Ilie Popanu Moldova

0,9801 Lus Gustavo Lapinha Dalla Stella Brazil

0,9703 Alexandra Vasileva Russia

0,891 Papimeri Dumitru Moldova

0,8539 Brahim Saadi Algeria0,81 Ion Toloaca Moldova

Correct solutions (ordered according to the arrival time):

1. Szab Attila (Hungary).


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2. Ivan Tadeu Ferreira Antunes Filho (Brazil)


4. Jakub afin (Slovak)



7. Brahim Saadi (Algeria)





Overall number of registered particpants: 204 (from 38 countries).

During the first week, only one correct solution has been submitted (by Szab Attila). So, the

problem was judged to be very difficult, and after the first week, few hints were published:

"You needto understand howto calculate fields using the circulation theorem andGauss law; relevant

formulas (from the formula sheet) are IX-2,IX-3 andIX-6; I also recommend studying the formulae

VIII-8, VIII-9, VIII-13,IX-27,IX-28, andIX-29. Andfinally this is a common mistake the

inductance ofa coilis notalways multiplied by a factor of when you supplyitwith a ferromagnetic

core (in fact,itis multiplied by onlyfor very specificcases); the inductance will depend on the

geometry ofthe ferromagnetic!"

During the second week, two more correct solutions were received: by Ivan Tadeu Ferreira Antunes

Filho and Nikita Sopenko. So, another hint was added:

"In particular,itwould be helpfulto studythe magneticfieldcreated by electrictransformers with

closed (forinstance,toroidal) ferromagneticcores; Wikipedia is nottoo useful (there is no calculation

ofB), exceptforthe figure for leakage flux (which is small/negligible,ifis large)."

During the first half of the third week, the correct solution of Jakub afin was received. After

that, on 3rd Nov, another amendment to the hints was made, so that the final wording was as

follows:


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"you needto understand howto calculate fields using the circulation theorem andGauss law; relevant

formulae (from the formula sheet) are IX-2,IX-3 andIX-6; Ialso recommend studying the formulae

VIII-8, VIII-9, VIII-13,IX-27,IX-28, andIX-29.In particular,itwould be helpfulto studythe

magneticfieldcreated by electrictransformers with closed (forinstance,toroidal) ferromagnetic

cores; suggested reading from Wikipedia: figure for leakage flux (which is small,ifis large); how to

deal with magnetic circuits.Please bearin mindthatyou are notsupposedto copy directlyformulae

from the latter article, because the shape ofourferromagneticbrickdiffers from a simplified model of

a closed-core transformer; instead,itshould be considered as a reading which helps you understand,

whatis going on with the B-fieldin ourcase, and howto correctly applythe circulation theorem.


Solution

I expected this problem to be of the same difficulty level as the Problem No 1. However, it turned out

to be more difficult probably because typically in high schools, magnetism is not taught as well as

mechanics. On the other hand, the problem, indeed, tests the knowledge of several things: (a) the

property of magnetic materials to "attract" the magnetic field lines; (b) Ampere's law; (c) Gauss law;

(d) the property of superconducting loops to conserve the magnetic flux; (e) the energy of magnetic

fields.

Regarding the distribution of the award for the best solution: all the first three solutions are very

good, with different strong points (which will be commented below). I decided to distribute the award

between these three evenly, giving a small bias to Szab Attila, who was the only one to solve the

problem without any hints. So, the bonus factors are e0.4, e0.3 and e0.3. The next three solutions are

also partially published due to different reasons and receive bonus factors of 1.1.

Let us start with the solution by Szab Attila, which is almost perfect, including all the required

components: (a) noting that the dominant part of the magnetic flux is kept inside the ferromagnetic

(either using energy-based arguments, or applying Ampere's law as is done here); (b) showing that

B has the same order of magnitude both inside the slit and in the ferromagnetic using the Gauss

law; (c) recognizing that inside the ferromagnetic, B is not homogeneous and hence, the contribution

of the segment residing inside the ferromagnetic brick to the circulation integral of the Ampere's lawcan only be estimated (and not calculated precisely); (d) applying the Ampere's law to show that

inside the slit region surrounded by the current loop, B is homogeneous, and calculating the value of

that B; (e) calculating the initial energy as is done here, or via magnetic field energy; (f) applying

the flux conservation law for the superconducting loop to calculate the final energy; (g) calculating

work as a difference of energies. If there is anything to be desired then it would be a motivation that

in the slit, B is perpendicular to the plane of the loop (it is only stated as a fact, without motivation).


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The next solution is that of Nikita Sopenko. As compared with the first solution, it includes a

proof of the formula (which is not mandatory as it is covered by the formula sheet).

Further, his solution does not require a proof that in the slit, B is perpendicular to the current loop,

because what is used here is only the perpendicular component of B (which enters both into the

Ampere's law and the expression for the magnetic flux). Finally, he has nicely and explicitly shown the

continuity of B at the slit boundary using the Gauss law (though, he does miss explicit proof thatmajority of the flux resides inside the ferromagnetic).


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The solution of Ivan Tadeu Ferreira Antunes Filho is provided here as a .pdf file (it is too long to

present page-by-page); it differs from the first two solutions in that the approach is based on the

concept of reluctance. This is not as clear physically as the approach based on the direct application of

the Ampere's law (in particular leaving open the question of why B is homogeneous in the slit);

however, Ivan does manage to keep things correct (providing first a theoretical motivation of the

method, and then calculating the reluctances in a correct way). The reason why he does receive a

bonus is not motivated by his method, but by the fact that he does study, whatwill happen ifL

becomes so large (when made ofa verythin wire) thatthe expression in the braces would become

negative. In particular, he shows that then, the solution needs to be modified, and the work would bestill positive. Also, he applies the formula for the loop inductance to estimate if it is realistic to have

such large values ofL which would be comparable to the initial inductance of the loop (surrounded by

the ferromagnetic); the answer is "not really". Note that intuitively, all the other solutions just imply

that the ferromagnetic makes the initial inductance much larger than the inductance L of the stand-

alone loop.


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Next, the solution ofJakub afin; what is worth highlighting, is his way of motivating, why in the

slit, B is perpendicular to the plane of the current loop (while mathematically not as clear and correct

as the magnetic field line refraction law described by Ilie Popanu, see below, intuitively and

qualitatively these are very useful arguments):

"Now, we only needto fi


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ndthe initialinductance M. For large ,fi


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eldlines ofB are attractedto the ferromagnetic; inside the slit,therefore,these


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fieldlines tryto escape from the slit, so they'll be perpendicularto the loop (we can thinkofitas a

deformation ofmagneticfield ofthe loopin vacuum). (This won'tholdperfectlyforfieldlines close to

the edge ofthe loop,though, because they're curved, butfor large ,this is negligible.)"

As mentioned above, the initial energy can be also calculated via theenergy density of the

magnetic field (for this method, it is important to understand that themagnetic field fills only thecircular sub-region of the slit surrounded by the current loop). The first one to do so was Lars

Dehlwes:

Finally, Ilie Popanu proved that in the slit, B is perpendicular to the slit using the refraction law for

the magnetic field lines:


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Problem No 3

Determine or estimate the net heat flux density between two parallel plates at distance from

each other, which are at temperatures and , respectively. The space between the plates is filled

with a monoatomic gas of molar density and of molar mass . You may use the following

approximations:

1. The gas density is so low that the mean free path ;

2. .

3. When gas molecules bounce from the plates, they obtain the temperature of the respective plates

(for instance, this will happen if they are absorbed/bound for a short time by the molecules of the

plate, and then released back into the space between the plates).

4.Y

ou may neglect the black body radiation.

5. "Estimate" means that the numeric prefactor of your expression does not need to be accurate.


The list of thecontest leaders after the third problem:

Points No 1 No 2 No 3 Name Country School

7,5632 1,1 3,8694 2,5937 SZAB Attila Hungary Lewey Klra High School, Pcs

5,6695 4,8156 0,8539 Brahim Saadi AlgeriaPreparatory School for Science &

of Annaba

5,5041 1,6105 2,8935 1 Nikita Sopenko Russia Lyceum No.14, Tambov

5,4546 1 1,7363 2,7183 Jakub afin Slovak Pavol Horov Secondary, Michalov

5,4353 2,1436 2,2917 1Ivan Tadeu Ferreira

Antunes FilhoBrazil Colgio Objetivo, Lins, So Paulo

5,2722 1,1 1,5785 2,5937 Lars Dehlwes Germany Ohm-Gymnasium Erlangen

4,1679 1 0,81 2,3579 Ion Toloaca Moldova liceul "Mircea Eliade"

3,919 1 0,9703 1,9487 Alexandra Vasileva Russia Lyceum "Second School", Moscow

3,7517 1 0,9801 1,7716Lus Gustavo Lapinha Dalla

StellaBrazil Colgio Integrado Objetivo, Baru

3,3137 2,5937 0,72 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau

3,3062 1 1,435 0,8712 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau

3,0746 1,6105 1,4641 Jakub Supe Poland 14th School of Stanisaw Staszic,


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2,8424 0,8733 0,891 1,0781 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau

2,7105 1,1 1,6105 Kohei Kawabata Japan Nada High School

2,5937 2,5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye

2,4738 1,7538 0,72 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau

2 1 1 Jaan Toots Estonia Tallinn Secondary Science Schoo

1,81 0,81 1 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Ba

1,6561 0,6561 1 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum

1,1 1,1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw

1 1 Lev Ginzburg RussiaAdvanced Educational Scientific C

Moscow

1 1 Sharad Mirani India Prakash Higher Secondary Schoo

1 1 Task Ohmori Japan Nada High School

0,9801 0,9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish Hig

0,8712 0,8712 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksi

0,81 0,81 Liara Guinsberg Brazil Colgio Integrado Objetivo, So

0,81 0,81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish Hig

0,72 0,72 Rajat Sharma India Pragati Vidya Peeth,Gwalior

0,5648 0,5648 Lorenzo Comoglio Italy Liceo Scientifico del Cossatese e


2,7183 Jakub afin

2,5937 SZAB Attila

2,5937 Lars Dehlwes

2,3579 Ion Toloaca1,9487 Alexandra Vasileva

1,7716Lus Gustavo Lapinha

Dalla Stella

1,6105 Kohei Kawabata

1,4641 Jakub Supe

1,1 Krzysztof Markiewicz

1,0781 Papimeri Dumitru

1 Nikita Sopenko

1Ivan Tadeu Ferreira

Antunes Filho1 Jaan Toots

1 Bharadwaj Rallabandi

1 Nadezhda Vartanian

0,8712 Ilie Popanu

0,8712 Petar Tadic

0,72 Dinis Cheian


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0,72 Cristian Zanoci

0,72 Rajat Sharma

0,5648 Lorenzo Comoglio

Correct solutions (ordered according to the arrival time; best solutions in bold):



3. Szab Attila (Hungary).





8.Papimeri Dumitru (Moldova)


10. Petar Tadic (Montenegro)

11. Cristian Zanoci (Moldova)

12. Dinis Cheian (Moldova)




16.

J

akub afin (Slovak)

17. Krzysztof Markiewicz (Poland)

18. Ivan Tadeu Ferreira Antunes Filho (Brazil)

19. Jaan Toots (Estonia)


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20. Comoglio Lorenzo (Italy)

21. Rajat Sharma (India)

Overall number of registered participants: 214 (from 38 countries).

Before the beginning of the final week, the following hints were given:

Few hints: you'll be ready to tackle the problem as soon as you understand how the basic formulae of

the kinetic theory (formula sheet X-9) are derived;Wikipedia has good enough coverage. And, of

course you need the definition of the net heat flux density: it is the difference between the incoming

and outgoing thermal energies per unit time and unit surface area. Also, please pay attention that for

a molecule, the round trip time (between the plates) is dominated by the period when its velocity is

small. This is similar to what happens in a road reconstruction region: the distance between the cars

in seconds (the number of cars per minute) remains the same as it was in high-speed regions, hence,

the distance in meters will be much smaller (and the density of the cars will be respectively higher).

-


Solution

In the problem text, it was stated that all numeric prefactors are considered to be acceptable, and

thus, the problem was graded generously. Actually, there were only two completely correct answers(by Szab Attila and Jakub afin), and two answers which were also flawless except that instead of

the correct , approximation was used (Petar Tadic and

Krzysztof Markiewicz).

The most common mistake was not noticing that unlike in the case of a normal gas, both for the hot

"faction" and cold "faction", the molecules move only in one direction. Hence, the ready formulae,

such as are two times smaller than needed, and the Maxwell velocity distribution

function should be also multiplied by two. Other typical mistakes were that instead of the

projection or , the modulus of the vector ( or ) was used. Meanwhile, themodulus should be used when calculating the transferred energy, but in some solutions, there

was , instead.

The best solutions were judged to be those ofSzab Attila and Jakub afin. However, Szab

Attila sent first an approximate solution, which he later corrected late enough to lose his bonus

points due to speed. If the best solution bonus would have been divided between these two, Szab


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Attila would have got less points than when taking into account his speed bonus. Therefore, he was

given his speed bonus, and additionally a double 1.1-factor-bonus for using both his originally

submitted and the revised solutions on this web page. And so, the best solution bonus goes entirely to

Jakub afin; Petar Tadic and KrzysztofMarkiewicz both receive a 1.1-factor-bonus.

Finally, Lorenzo Comoglio recieves also a bonus of 1.1: he made a very nice visualization of the

process.

There are two ways of calculating the frequency of collisions: (a) using the round-trip time, and (b)

calculating first the densities of both "factions" of molecules (hot and cold). The solution of Szab

Attila follows method (a).


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The solution ofJakub afin is based on calculating the densities of "factions". Also, he makes a

very useful analysis of the results.


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The solution ofKrzysztofMarkiewicz:


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The solution of Petar Tadic:


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Finally, the initial solution of Szab Attila: while incorrect, the idea itself is very nice, and the

mistake is well hidden; so I judged it to be useful to display the first page, and analyse, why the

prefactor will be wrong, if calculated in such a way.


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Notice the nice trick of introducing and arranging the molecules according to the values of .

Unfortunately, the trick does not work here: re-arranging the order of the molecule speeds does

introduce false correlations. In such a way, we create molecules which are always faster than average,

and the ones which are slower than average; the amount of transported heat is defined by after the

hot wall, and the round-trip time is defined by after the cold wall; so, relatively large amountofheat

would be transportedin relatively shortertime, and therefore, the average of the product of thetransported heat with the collision frequency would not be equal to the product of the respective

averages; however, itwould be equalif these two quantities were uncorrelated, as is actually the

case!

Problem No 4

There are three point masses m, 2m and 3m, each of which is fixed to a weightless rod; the three

rods are of equal length L and are fixed to each other via a connector, which allows a free rotation(frictionless, torque-less) of the rods with respect to each other (so that the angles between the rods

will change). Initially the angle between the rods is 120o and the system is motionless; all the rods

and point masses lay on the same plane. The heaviest mass (3m) is hit so that it obtains

instantaneously a velocity v0, perpendicular to the rod to which it is fixed to and coplanar to the other

rods. Determine the accelerations of all three point masses immediately after the point mass 3m was

hit. Remark: there is no gravity field, the system can be thought to be in weightlessness, or on

frictionless horizontal surface.


The list of thecontest leaders after the fourth problem:


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Points Name Country School Physics teac

10,157 SZAB Attila Hungary Lewey Klra High School, Pcs Simon Pter

7,6476 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladi

6,8666 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Martin Perlet

6,7301 Jakub afin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek

6,0833Ivan Tadeu Ferreira

Antunes FilhoBrazil Colgio Objetivo, Lins, So Paulo

6,0244 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodie

5,6695 Brahim Saadi AlgeriaPreparatory School for Science & Technology

of AnnabaDerradji Nas

5,1968 Jakub Supe Poland 14th School of Stanisaw Staszic, Warsaw Wodzimierz

5,0589 Ion Toloaca Moldova liceul "Mircea Eliade"Igor Iurevici

Simboteanu

4,919 Alexandra Vasileva Russia Lyceum "Second School", MoscowA.R. Zilberm

Arabuly

4,1057 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodie

3,9424 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodie

3,8105 Kohei Kawabata Japan Nada High School

3,7997 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko

3,7517Lus Gustavo Lapinha Dalla

StellaBrazil Colgio Integrado Objetivo, Barueri, Brazil Ronaldo Fog

3,4738 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodie

3,4205 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekha

2,5937 Mikhail Shirkin Russia Gymnasium of Ramenskoye Petrova Elen

2,1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw Robert Stasi

2 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reim

1,4641 Hideki Yukawa Japan Nada high school

1,3896 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksic Ana Vujacic

1,21 Midhul Varma India Vidyadham Junior, Hyderabad Manikanta K

1 Task Ohmori Japan Nada High School T.Hamaguch

1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadan

1 Lev Ginzburg RussiaAdvanced Educational Scientific Center, MSU,

MoscowI.V. Lukjano

0,9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshku

0,81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshku0,81 Liara Guinsberg Brazil Colgio Integrado Objetivo, So Paulo, Brazil Ronaldo Fog

0,792 Ulysse Lojkine France Lyce Henri IV, Paris M. Lacas

0,72 Rajat Sharma India Pragati Vidya Peeth,Gwalior Mr. Rakesh R

0,5648 Ng Fei Chong Malaysia SMJK Chung Ling, Penang

0,5648 Lorenzo Comoglio Italy Liceo Scientifico del Cossatese e Valle Strona Chiara Band



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2,7183 Ilie Popanu

2,5937 SZAB Attila

2,1436 Nikita Sopenko

2,1436 Nadezhda Vartanian

2,1222 Jakub Supe

1,6105 Bharadwaj Rallabandi

1,5944 Lars Dehlwes

1,4641 Hideki Yukawa

1,2755 Jakub afin

1,21 Midhul Varma

1,1 Papimeri Dumitru

1,1 Kohei Kawabata

1 Alexandra Vasileva

1 Cristian Zanoci

1 Krzysztof Markiewicz0,891 Ion Toloaca

0,792 Dinis Cheian

0,792 Ulysse Lojkine

0,648Ivan Tadeu Ferreira

Antunes Filho

0,5648 Ng Fei Chong

0,5184 Petar Tadic

Correct solutions (ordered according to the arrival time; best solutions in bold):

1. Szab Attila (Hungary)






7. Jakub afin (Slovak)


9. Hideki Yukawa (Japan)


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10. Midhul Varma (India)

11. Dinis Cheian (Moldova)



14. Cristian Zanoci (Moldova)

15. Ng Fei Chong (Malaysia)



Documents

Competition “Physics Cup – IPhO2012"