IPhO Problems and Solutions (1967-2010)

International

Physics

Olympiad

(IPhO)

1967-2010

Problems and Solutions

Compiled & edited by

scroungehound

Visitors

Appendix to the Statutes of the International Physics Olympiads

General

a. The extensive use of the calculus (differentiation and integration) and the use of complex numbers or solving differential equations should not be required to solve the theoretical and practical problems.

b. Questions may contain concepts and phenomena not contained in the Syllabus but sufficient information must be given in the questions so that candidates without previous knowledge of these topics would not be at a disadvantage.

c. Sophisticated practical equipment likely to be unfamiliar to the candidates should not dominate a problem. If such devices are used then careful instructions must be given to the candidates.

d. The original texts of the problems have to be set in the SI units. A. Theoretical Part

The first column contains the main entries while the second column contains comments and remarks if necessary.

1. Mechanics

a) Foundation of kinematics of a point mass Vector description of the position of the point mass, velocity and acceleration as vectors b) Newton's laws, inertial systems Problems may be set on changing mass c) Closed and open systems, momentum and

energy, work, power

d) Conservation of energy, conservation of linear momentum, impulse

e) Elastic forces, frictional forces, the law of gravitation, potential energy and work in a gravitational field

Hooke's law, coefficient of friction (F/R = const), frictional forces, static and kinetic, choice of zero of potential energy f) Centripetal acceleration, Kepler's laws

2. Mechanics of Rigid Bodies

a) Statics, center of mass, torque Couples, conditions of equilibrium of bodies b) Motion of rigid bodies, translation, rotation,

angular velocity, angular acceleration, conservation of angular momentum

Conservation of angular momentum about fixed axis only

c) External and internal forces, equation of motion of a rigid body around the fixed axis, moment of inertia, kinetic energy of a rotating body

Parallel axes theorem (Steiner's theorem), additivity of the moment of inertia

d) Accelerated reference systems, inertial forces

Knowledge of the Coriolis force formula is not required

3. Hydromechanics

No specific questions will be set on this but students would be expected to know the elementary concepts of pressure, buoyancy and the continuity law.

4. Thermodynamics and Molecular Physics

a) Internal energy, work and heat, first and second laws of thermodynamics

Thermal equilibrium, quantities depending on state and quantities depending on process b) Model of a perfect gas, pressure and

molecular kinetic energy, Avogadro's number, equation of state of a perfect gas, absolute temperature

Also molecular approach to such simple phenomena in liquids and solids as boiling, melting etc.

c) Work done by an expanding gas limited to isothermal and adiabatic processes

Proof of the equation of the adiabatic process is not required

d) The Carnot cycle, thermodynamic efficiency, reversible and irreversible processes, entropy (statistical approach), Boltzmann factor

Entropy as a path independent function, entropy changes and reversibility, quasistatic processes

5. Oscillations and waves

a) Harmonic oscillations, equation of harmonic oscillation

Solution of the equation for harmonic motion, attenuation and resonance -qualitatively b) Harmonic waves, propagation of waves,

transverse and longitudinal waves, linear polarization, the classical Doppler effect, sound waves

Displacement in a progressive wave and understanding of graphical representation of the wave, measurements of velocity of sound and light, Doppler effect in one dimension only, propagation of waves in homogeneous

and isotropic media, reflection and refraction, Fermat's principle

c) Superposition of harmonic waves, coherent waves, interference, beats, standing waves

Realization that intensity of wave is proportional to the square of its amplitude. Fourier analysis is not required but candidates should have some understanding that

complex waves can be made from addition of simple sinusoidal waves of different

frequencies. Interference due to thin films and other simple systems (final formulae are not required), superposition of waves from secondary sources (diffraction)

6. Electric Charge and Electric Field

a) Conservation of charge, Coulomb's law

b) Electric field, potential, Gauss' law Gauss' law confined to simple symmetric systems like sphere, cylinder, plate etc., electric dipole moment

c) Capacitors, capacitance, dielectric constant, energy density of electric field

7. Current and Magnetic Field

a) Current, resistance, internal resistance of source, Ohm's law, Kirchhoff's laws, work and power of direct and alternating currents, Joule's law

Simple cases of circuits containing non-ohmic devices with known V-I characteristics

b) Magnetic field (B) of a current, current in a magnetic field, Lorentz force

Particles in a magnetic field, simple applications like cyclotron, magnetic dipole moment

c) Ampere's law Magnetic field of simple symmetric systems like straight wire, circular loop and long solenoid

d) Law of electromagnetic induction, magnetic flux, Lenz's law, self-induction, inductance, permeability, energy density of magnetic field e) Alternating current, resistors, inductors and capacitors in AC-circuits, voltage and current (parallel and series) resonances

Simple AC-circuits, time constants, final formulae for parameters of concrete resonance circuits are not required

a) Oscillatory circuit, frequency of oscillations, generation by feedback and resonance b) Wave optics, diffraction from one and two slits, diffraction grating,resolving power of a grating, Bragg reflection,

c) Dispersion and diffraction spectra, line spectra of gases

d) Electromagnetic waves as transverse waves, polarization by reflection, polarizers

Superposition of polarized waves

e) Resolving power of imaging systems

f) Black body, Stefan-Boltzmanns law Planck's formula is not required

9. Quantum Physics

a) Photoelectric effect, energy and impulse of the photon

Einstein's formula is required

b) De Broglie wavelength, Heisenberg's uncertainty principle

10. Relativity

a) Principle of relativity, addition of velocities, relativistic Doppler effect

b) Relativistic equation of motion, momentum, energy, relation between energy and mass, conservation of energy and momentum

11. Matter

a) Simple applications of the Bragg equation b) Energy levels of atoms and molecules (qualitatively), emission, absorption, spectrum of hydrogen like atoms

c) Energy levels of nuclei (qualitatively), alpha-, beta- and gamma-decaysalpha-, absorption of radiation, halflife and exponential decay, components of nuclei, mass defect, nuclear reactions

The Theoretical Part of the Syllabus provides the basis for all the experimental problems. The experimental problems given in the experimental contest should contain measurements.

Additional requirements:

1. Candidates must be aware that instruments affect measurements.

2. Knowledge of the most common experimental techniques for measuring physical quantities mentioned in Part A.

3. Knowledge of commonly used simple laboratory instruments and devices such as calipers, thermometers, simple volt-, ohm- and ammeters, potentiometers, diodes, transistors, simple optical devices and so on.

4. Ability to use, with the help of proper instruction, some sophisticated instruments and devices such as double-beam oscilloscope, counter, ratemeter, signal and function generators, analog-to-digital converter connected to a computer, amplifier, integrator, differentiator, power supply, universal (analog and digital) volt-, ohm- and ammeters. 5. Proper identification of error sources and estimation of their influence on the final result(s). 6. Absolute and relative errors, accuracy of measuring instruments, error of a single

measurement, error of a series of measurements, error of a quantity given as a function of measured quantities.

7. Transformation of a dependence to the linear form by appropriate choice of variables and fitting a straight line to experimental points.

8. Proper use of the graph paper with different scales (for example polar and logarithmic papers).

9. Correct rounding off and expressing the final result(s) and error(s) with correct number of significant digits.

10. Standard knowledge of safety in laboratory work. (Nevertheless, if the experimental set-up contains any safety hazards the appropriate warnings should be included into the text of the problem.)

Problems of the 1st International Physics Olympiad

(Warsaw, 1967)

Waldemar Gorzkowski

Institute of Physics, Polish Academy of Sciences, Warsaw, Poland2

Abstract

The article contains the competition problems given at he 1st International Physics Olympiad (Warsaw, 1967) and their solutions. Additionally it contains comments of historical character.

Introduction

One of the most important points when preparing the students to the International Physics Olympiads is solving and analysis of the competition problems given in the past. Unfortunately, it is very difficult to find appropriate materials. The proceedings of the subsequent Olympiads are published starting from the XV IPhO in Sigtuna (Sweden, 1984). It is true that some of very old problems were published (not always in English) in different books or articles, but they are practically unavailable. Moreover, sometimes they are more or less substantially changed.

The original English versions of the problems of the 1st IPhO have not been conserved. The permanent Secretariat of the IPhOs was created in 1983. Until this year the Olympic materials were collected by different persons in their private archives. These archives as a rule were of amateur character and practically no one of them was complete. This article is based on the books by R. Kunfalvi [1], Tadeusz Pniewski [2] and Waldemar Gorzkowski [3]. Tadeusz Pniewski was one of the members of the Organizing Committee of the Polish Physics Olympiad when the 1st IPhO took place, while R. Kunfalvi was one of the members of the International Board at the 1st IPhO. For that it seems that credibility of these materials is very high. The differences between versions presented by R. Kunfalvi and T. Pniewski are rather very small (although the book by Pniewski is richer, especially with respect to the solution to the experimental problem).

As regards the competition problems given in Sigtuna (1984) or later, they are available, in principle, in appropriate proceedings. “In principle” as the proceedings usually were published in a small number of copies, not enough to satisfy present needs of people interested in our competition. It is true that every year the organizers provide the permanent Secretariat with a number of copies of the proceedings for free dissemination. But the needs are continually growing up and we have disseminated practically all what we had.

The competition problems were commonly available (at least for some time) just only from the XXVI IPhO in Canberra (Australia) as from that time the organizers started putting the problems on their home pages. The Olympic home page www.jyu.fi/ipho contains the problems starting from the XXVIII IPhO in Sudbury (Canada). Unfortunately, the problems given in Canberra (XXVI IPhO) and in Oslo (XXVII IPhO) are not present there.

The net result is such that finding the competition problems of the Olympiads organized prior to Sudbury is very difficult. It seems that the best way of improving the situation is publishing the competition problems of the older Olympiads in our journal. The

1 _{This is somewhat extended version of the article sent for publication in Physics Competitions in July 2003.} 2 _{e-mail: [email protected]}

question arises, however, who should do it. According to the Statutes the problems are created by the local organizing committees. It is true that the texts are improved and accepted by the International Board, but always the organizers bear the main responsibility for the topics of the problems, their structure and quality. On the other hand, the glory resulting of high level problems goes to them. For the above it is absolutely clear to me that they should have an absolute priority with respect to any form of publication. So, the best way would be to publish the problems of the older Olympiads by representatives of the organizers from different countries.

Poland organized the IPhOs for thee times: I IPhO (1967), VII IPhO (1974) and XX IPhO (1989). So, I have decided to give a good example and present the competition problems of these Olympiads in three subsequent articles. At the same time I ask our Colleagues and Friends from other countries for doing the same with respect to the Olympiads organized in their countries prior to the XXVIII IPhO (Sudbury).

I IPhO (Warsaw 1967)

The problems were created by the Organizing Committee. At present we are not able to recover the names of the authors of the problems.

Theoretical problems Problem 1

A small ball with mass M = 0.2 kg rests on a vertical column with height h = 5m. A bullet with mass m = 0.01 kg, moving with velocity v0 = 500 m/s, passes horizontally through

the center of the ball (Fig. 1). The ball reaches the ground at a distance s = 20 m. Where does the bullet reach the ground? What part of the kinetic energy of the bullet was converted into heat when the bullet passed trough the ball? Neglect resistance of the air. Assume that g = 10 m/s2. Fig. 1 M s h m v0

Solution

Fig. 2

We will use notation shown in Fig. 2.

As no horizontal force acts on the system ball + bullet, the horizontal component of momentum of this system before collision and after collision must be the same:

. 0 mv MV mv = + So, V m M v v= ₀− .

From conditions described in the text of the problem it follows that

.

V v>

After collision both the ball and the bullet continue a free motion in the gravitational field with initial horizontal velocities v and V, respectively. Motion of the ball and motion of the bullet are continued for the same time:

. 2 g h t= d M s h

m v0 v – horizontal component of the velocity

of the bullet after collision

V – horizontal component of the velocity of the ball after collision

It is time of free fall from height h.

The distances passed by the ball and bullet during time t are:

Vt s= and d=vt, respectively. Thus . 2h g s V = Therefore h g s m M v v 2 0 − = . Finally: s m M g h v d = ₀ 2 − . Numerically: d = 100 m.

The total kinetic energy of the system was equal to the initial kinetic energy of the bullet: 2 2 0 0 mv E = .

Immediately after the collision the total kinetic energy of the system is equal to the sum of the kinetic energy of the bullet and the ball:

2 2 mv E_m = , 2 2 MV E_M = . Their difference, converted into heat, was

) ( 0 Em EM E E= − + ∆ .

It is the following part of the initial kinetic energy of the bullet:

. 1 0 0 E E E E E p=∆ = − m+ M

      + − = m m M g h s v h g v s m M p 2 2 2 0 2 0 2 . Numerically: p = 92,8%. Problem 2

Consider an infinite network consisting of resistors (resistance of each of them is r) shown in Fig. 3. Find the resultant resistance RAB between points A and B.

Fig. 3

Solution

It is easy to remark that after removing the left part of the network, shown in Fig. 4 with the dotted square, then we receive a network that is identical with the initial network (it is result of the fact that the network is infinite).

Fig. 4

Thus, we may use the equivalence shown graphically in Fig. 5.

Fig. 5 A r r r r r r A B r r r r r r A B RAB r RAB r

⇔

Algebraically this equivalence can be written as AB AB R r r R 1 1 1 + + = . Thus 0 2 2 _{− − =} r rR R_{AB AB} . This equation has two solutions:

r R_AB = ₂1(1± 5) _.

The solution corresponding to “-“ in the above formula is negative, while resistance must be positive. So, we reject it. Finally we receive

r RAB 2(1 5)

1 +

= .

Problem 3

Consider two identical homogeneous balls, A and B, with the same initial temperatures. One of them is at rest on a horizontal plane, while the second one hangs on a thread (Fig. 6). The same quantities of heat have been supplied to both balls. Are the final temperatures of the balls the same or not? Justify your answer. (All kinds of heat losses are negligible.)

Fig. 6

Solution

Fig. 7

As regards the text of the problem, the sentence “The same quantities of heat have been supplied to both balls.” is not too clear. We will follow intuitive understanding of this

B A B A B A

sentence, i.e. we will assume that both systems (A – the hanging ball and B – the ball resting on the plane) received the same portion of energy from outside. One should realize, however, that it is not the only possible interpretation.

When the balls are warmed up, their mass centers are moving as the radii of the balls are changing. The mass center of the ball A goes down, while the mass center of the ball B goes up. It is shown in Fig. 7 (scale is not conserved).

Displacement of the mass center corresponds to a change of the potential energy of the ball in the gravitational field.

In case of the ball A the potential energy decreases. From the 1st principle of thermodynamics it corresponds to additional heating of the ball.

In case of the ball B the potential energy increases. From the 1st principle of thermodynamics it corresponds to some “losses of the heat provided” for performing a mechanical work necessary to rise the ball. The net result is that the final temperature of the ball B should be lower than the final temperature of the ball A.

The above effect is very small. For example, one may find (see later) that for balls made of lead, with radius 10 cm, and portion of heat equal to 50 kcal, the difference of the final temperatures of the balls is of order 10-5 K. For spatial and time fluctuations such small quantity practically cannot be measured.

Calculation of the difference of the final temperatures was not required from the participants. Nevertheless, we present it here as an element of discussion.

We may assume that the work against the atmospheric pressure can be neglected. It is obvious that this work is small. Moreover, it is almost the same for both balls. So, it should not affect the difference of the temperatures substantially. We will assume that such quantities as specific heat of lead and coefficient of thermal expansion of lead are constant (i.e. do not depend on temperature).

The heat used for changing the temperatures of balls may be written as B A i t mc Qi = ∆ i, where = or ,

Here: m denotes the mass of ball, c - the specific heat of lead and ∆ - the change of the t_i temperature of ball.

The changes of the potential energy of the balls are (neglecting signs): B A i t mgr Ei = ∆ i, where = or ∆ α .

Here: g denotes the gravitational acceleration, r - initial radius of the ball, α - coefficient of thermal expansion of lead. We assume here that the thread does not change its length.

Taking into account conditions described in the text of the problem and the interpretation mentioned at the beginning of the solution, we may write:

A E

A Q

Q= _A − ∆ _A, for theball , B E

A Q

Q= B + ∆ B, for theball .

A denotes the thermal equivalent of work:

J cal 24 . 0 ≈

A . In fact, A is only a conversion ratio between calories and joules. If you use a system of units in which calories are not present, you may omit A at all.

Thus

A t

Amgr mc

Q=( − α)∆ _A, for theball , B t

Amgr mc

Q=( + α)∆ _B, for theball and α Amgr mc Q t_A − = ∆ , α Amgr mc Q t_B + = ∆ . Finally we get 2 2 2 2 ) ( 2 mc AQgr m Q Agr c Agr t t t A B α α α ≈ − = ∆ − ∆ = ∆ .

(We neglected the term with α as the coefficient α is very small.) 2

Now we may put the numerical values: Q= 50 kcal, A≈0.24cal/J, g≈9.8m/s2, ≈

m 47 kg (mass of the lead ball with radius equal to 10 cm), r=0.1 m, c≈0.031cal/(g⋅K),

≈

α 29⋅10-6

K-1. After calculations we get ∆t≈1.5⋅10-5 K.

Problem 4

Comment: The Organizing Committee prepared three theoretical problems. Unfortunately, at the time of the 1st Olympiad the Romanian students from the last class had the entrance examinations at the universities. For that Romania sent a team consisting of students from younger classes. They were not familiar with electricity. To give them a chance the Organizers (under agreement of the International Board) added the fourth problem presented here. The students (not only from Romania) were allowed to chose three problems. The maximum possible scores for the problems were: 1st problem – 10 points, 2nd problem – 10 points, 3rd problem – 10 points and 4th problem – 6 points. The fourth problem was solved by 8 students. Only four of them solved the problem for 6 points.

A closed vessel with volume V0 = 10 l contains dry air in the normal conditions (t0 =

0°C, p0 = 1 atm). In some moment 3 g of water were added to the vessel and the system was

warmed up to t = 100°C. Find the pressure in the vessel. Discuss assumption you made to solve the problem.

Solution

The water added to the vessel evaporates. Assume that the whole portion of water evaporated. Then the density of water vapor in 100°C should be 0.300 g/l. It is less than the density of saturated vapor at 100°C equal to 0.597 g/l. (The students were allowed to use physical tables.) So, at 100°C the vessel contains air and unsaturated water vapor only (without any liquid phase).

Now we assume that both air and unsaturated water vapor behave as ideal gases. In view of Dalton law, the total pressure p in the vessel at 100°C is equal to the sum of partial pressures of the air pa and unsaturated water vapor pv:

v a p p

p= + .

As the volume of the vessel is constant, we may apply the Gay-Lussac law to the air. We obtain:       + = 273 273 0 t p pa .

The pressure of the water vapor may be found from the equation of state of the ideal gas: R m t V p_v µ = + 273 0 ,

where m denotes the mass of the vapor, µ - the molecular mass of the water and R – the universal gas constant. Thus,

0 273 V t R m pv + = µ and finally 0 0 273 273 273 V t R m t p p= + + + µ . Numerically: atm. 88 . 1 atm ) 516 . 0 366 . 1 ( + ≈ = p Experimental problem

The following devices and materials are given: 1. Balance (without weights)

2. Calorimeter 3. Thermometer 4. Source of voltage 5. Switches 6. Wires 7. Electric heater 8. Stop-watch 9. Beakers 10. Water 11. Petroleum

12. Sand (for balancing)

Determine specific heat of petroleum. The specific heat of water is 1 cal/(g⋅°C). The specific heat of the calorimeter is 0.092 cal/(g⋅°C).

Solution

The devices given to the students allowed using several methods. The students used the following three methods:

1. Comparison of velocity of warming up water and petroleum; 2. Comparison of cooling down water and petroleum;

3. Traditional heat balance.

As no weights were given, the students had to use the sand to find portions of petroleum and water with masses equal to the mass of calorimeter.

First method: comparison of velocity of warming up

If the heater is inside water then both water and calorimeter are warming up. The heat taken by water and calorimeter is:

1 1

1 m c t m c t

Q = _{w w}∆ + _{c c}∆ ,

where: m denotes mass of water, _w m - mass of calorimeter, _c c - specific heat of water, _w c - _c specific heat of calorimeter, ∆ - change of temperature of the system water + calorimeter. t₁

On the other hand, the heat provided by the heater is equal:

1 2 2 τ R U A Q = ,

where: A – denotes the thermal equivalent of work, U – voltage, R – resistance of the heater, τ1 – time of work of the heater in the water.

Of course, 2 1 Q Q = . Thus 1 1 1 2 t c m t c m R U A τ = _{w w}∆ + _{c c}∆ . For petroleum in the calorimeter we get a similar formula:

2 2 2 2 t c m t c m R U A τ = p p∆ + c c∆ .

where: m_p denotes mass of petroleum, c_p - specific heat of petroleum, ∆ - change of t₂ temperature of the system water + petroleum, τ2 – time of work of the heater in the petroleum.

2 2 1 2 1 t c m t c m t c m t c m c c p p c c w w ∆ + ∆ ∆ + ∆ = τ τ .

It is convenient to perform the experiment by taking masses of water and petroleum equal to the mass of the calorimeter (for that we use the balance and the sand). For

c p

w m m

m = =

the last formula can be written in a very simple form:

2 2 1 1 2 1 t c t c t c t c c p c w ∆ + ∆ ∆ + ∆ = τ τ . Thus c w c c t t c t t c _      ∆ ∆ − − ∆ ∆ = 2 2 1 1 2 2 1 1 1 τ τ τ τ or c w c c k k c k k c _      − − = 2 1 2 1 1 , where 1 1 1 _τ t k =∆ and 2 2 2 _τ t k =∆

denote “velocities of heating” water and petroleum, respectively. These quantities can be determined experimentally by drawing graphs representing dependence ∆ and t₁ ∆ on time t₂ (τ). The experiment shows that these dependences are linear. Thus, it is enough to take slopes of appropriate straight lines. The experimental setup given to the students allowed measurements of the specific heat of petroleum, equal to 0.53 cal/(g°⋅C), with accuracy about 1%.

Some students used certain mutations of this method by performing measurements at

1

t

∆ = ∆ or at t₂ τ₁ = . Then, of course, the error of the final result is greater (it is additionally τ₂ affected by accuracy of establishing the conditions ∆ = t1 ∆ or at t2 τ1= ).τ2

Second method: comparison of velocity of cooling down

Some students initially heated the liquids in the calorimeter and later observed their cooling down. This method is based on the Newton’s law of cooling. It says that the heat Q transferred during cooling in time τ is given by the formula:

τ ϑ s t h

Q= ( − ) ,

where: t denotes the temperature of the body, ϑ - the temperature of surrounding, s – area of the body, and h – certain coefficient characterizing properties of the surface. This formula is

correct for small differences of temperatures t−ϑ only (small compared to t and ϑ in the

absolute scale).

This method, like the previous one, can be applied in different versions. We will consider only one of them.

Consider the situation when cooling of water and petroleum is observed in the same calorimeter (containing initially water and later petroleum). The heat lost by the system water + calorimeter is t c m c m Q = w w + c c ∆ ∆ 1 ( ) ,

where ∆t denotes a change of the temperature of the system during certain period τ₁. For the system petroleum + calorimeter, under assumption that the change in the temperature ∆t is the same, we have

t c m c m Q = _{p p} + _{c c} ∆ ∆ ₂ ( ) .

Of course, the time corresponding to ∆t in the second case will be different. Let it be τ₂. From the Newton's law we get

2 1 2 1 τ τ = ∆ ∆ Q Q . Thus c c p p c c w w c m c m c m c m + + = 2 1 τ τ .

If we conduct the experiment at

c p w m m m = = , then we get c w p c T T c T T c _      − − = 1 2 1 2 1 .

As cooling is rather a very slow process, this method gives the result with definitely greater error.

Third method: heat balance

This method is rather typical. The students heated the water in the calorimeter to certain temperature t and added the petroleum with the temperature ₁ t . After reaching the thermal 2

equilibrium the final temperature was t. From the thermal balance (neglecting the heat losses) we have

) ( )

)(

(m_wc_w +m_cc_c t₁ −t =m_pc_p t−t₂ . If, like previously, the experiment is conducted at

c p w m m m = = , then 2 1 ) ( t t t t c c c_{p w c} − − + = .

In this methods the heat losses (when adding the petroleum to the water) always played a substantial role.

The accuracy of the result equal or better than 5% can be reached by using any of the methods described above. However, one should remark that in the first method it was easiest. The most common mistake was neglecting the heat capacity of the calorimeter. This mistake increased the error additionally by about 8%.

Marks

No marking schemes are present in my archive materials. Only the mean scores are available. They are:

Problem # 1 7.6 points

Problem # 2 7.8 points (without the Romanian students) Problem # 3 5.9 points

Experimental problem 7.7 points

Thanks

The author would like to express deep thanks to Prof. Jan Mostowski and Dr. Yohanes Surya for reviewing the text and for valuable comments and remarks.

Literature

[1] R. Kunfalvi, Collection of Competition Tasks from the Ist trough XVth International Physics Olympiads, 1967 – 1984, Roland Eotvos Physical Society and UNESCO, Budapest 1985

[2] Tadeusz Pniewski, Olimpiady Fizyczne: XV i XVI, PZWS, Warszawa 1969

[3] Waldemar Gorzkowski, Zadania z fizyki z całego świata (z rozwiązaniami) - 20 lat Międzynarodowych Olimpiad Fizycznych, WNT, Warszawa 1994 [ISBN 83-204-1698-1]

Problems of the 2

International Physics Olympiads

(Budapest, Hungary, 1968)

Péter Vankó

Institute of Physics, Budapest University of Technical Engineering, Budapest, Hungary

Abstract

After a short introduction the problems of the 2nd and the 9th International Physics Olympiad, organized in Budapest, Hungary, 1968 and 1976, and their solutions are presented.

Introduction

Following the initiative of Dr. Waldemar Gorzkowski [1] I present the problems and solutions of the 2nd and the 9th International Physics Olympiad, organized by Hungary. I have used Prof. Rezső Kunfalvi’s problem collection [2], its Hungarian version [3] and in the case of the 9th Olympiad the original Hungarian problem sheet given to the students (my own copy). Besides the digitalization of the text, the equations and the figures it has been made only small corrections where it was needed (type mistakes, small grammatical changes). I omitted old units, where both old and SI units were given, and converted them into SI units, where it was necessary.

If we compare the problem sheets of the early Olympiads with the last ones, we can realize at once the difference in length. It is not so easy to judge the difficulty of the problems, but the solutions are surely much shorter.

The problems of the 2nd Olympiad followed the more than hundred years tradition of physics competitions in Hungary. The tasks of the most important Hungarian theoretical physics competition (Eötvös Competition), for example, are always very short. Sometimes the solution is only a few lines, too, but to find the idea for this solution is rather difficult.

Of the 9th Olympiad I have personal memories; I was the youngest member of the Hungarian team. The problems of this Olympiad were collected and partly invented by Miklós Vermes, a legendary and famous Hungarian secondary school physics teacher. In the first problem only the detailed investigation of the stability was unusual, in the second problem one could forget to subtract the work of the atmospheric pressure, but the fully “open” third problem was really unexpected for us.

The experimental problem was difficult in the same way: in contrast to the Olympiads of today we got no instructions how to measure. (In the last years the only similarly open experimental problem was the investigation of “The magnetic puck” in Leicester, 2000, a really nice problem by Cyril Isenberg.) The challenge was not to perform many-many measurements in a short time, but to find out what to measure and how to do it.

Of course, the evaluating of such open problems is very difficult, especially for several hundred students. But in the 9th Olympiad, for example, only ten countries participated and the same person could read, compare, grade and mark all of the solutions.

2

2

IPhO (Budapest, 1968)

Theoretical problems Problem 1

On an inclined plane of 30° a block, mass m2 = 4 kg, is joined by a light cord to a solid

cylinder, mass m1 = 8 kg, radius r = 5 cm (Fig. 1). Find the acceleration if the bodies are

released. The coefficient of friction between the block and the inclined plane µ = 0.2. Friction at the bearing and rolling friction are negligible.

Solution

If the cord is stressed the cylinder and the block are moving with the same acceleration a. Let F be the tension in the cord, S the frictional force between the cylinder and the inclined plane (Fig. 2). The angular acceleration of the cylinder is a/r. The net force causing the acceleration of the block:

F g m g m a m2 = 2 sinα −µ 2 cosα+ ,

and the net force causing the acceleration of the cylinder: F S g m a m₁ = ₁ sinα − − .

The equation of motion for the rotation of the cylinder:

I r a r

S = ⋅ .

(I is the moment of inertia of the cylinder, S⋅r is the torque of the frictional force.) Solving the system of equations we get:

(

)

(

)

3 2 2 1 2 2 1 2 sin cos r I m m r I r I m g m F + + −       ₊ ⋅ = α α µ . (3)

The moment of inertia of a solid cylinder is

2

2 1r

m

I = . Using the given numerical values:

(

)

(

)

(

)

Discussion (See Fig. 3.)

The condition for the system to start moving is a > 0. Inserting a = 0 into (1) we obtain the limit for angle α1:

0667 . 0 3 tan 2 1 2 1 = = + ⋅ =µ µ α m m m , α₁ = 813. °.

For the cylinder separately α₁=0, and for the block separately α₁=tan−1µ=11.31°.

If the cord is not stretched the bodies move separately. We obtain the limit by inserting F = 0 into (3): 6 . 0 3 1 tan 2 1 2 = =      + ⋅ =µ µ α I r m , α2 =30.96°. The condition for the cylinder to

slip is that the value of S (calculated from (2) taking the same coefficient of friction) exceeds the value of µm₁gcosα. This gives the same value for α3 as we had for α2. The

acceleration of the centers of the cylinder and the block is the same:

(

)

g , the frictional force at the bottom of the cylinder is µm1gcosα , the peripheral acceleration of the cylinder is

α µ cos 2 1 _g I r m ⋅ ⋅ . Problem 2

There are 300 cm3 toluene of 0°C temperature in a glass and 110 cm3 toluene of

C

100° temperature in another glass. (The sum of the volumes is 410 cm3.) Find the final volume after the two liquids are mixed. The coefficient of volume expansion of toluene

( )

β .Neglect the loss of heat.

β r, a g α 0° 30° 60° 90° F, S (N) α1 α2=α3 10 20 F S β r a Figure 3

4

Solution

If the volume at temperature t1 is V1, then the volume at temperature 0°C is

(

)

1

10 V 1 t

V = +β . In the same way if the volume at t2 temperature is V2, at 0°C we have

(

)

2

20 V 1 t

V = +β . Furthermore if the density of the liquid at 0°C is d, then the masses are d

V

m₁ = ₁₀ and m₂ =V₂₀d, respectively. After mixing the liquids the temperature is

2 1 2 2 1 1 m m t m t m t + + = .

The volumes at this temperature are V₁₀

(

)

(

)

(

)

(

)

(

)

(

)

(

)

The sum of the volumes is constant. In our case it is 410 cm3. The result is valid for any number of quantities of toluene, as the mixing can be done successively adding always one more glass of liquid to the mixture.

Problem 3

Parallel light rays are falling on the plane surface of a semi-cylinder made of glass, at an angle of 45°, in such a plane which is perpendicular to the axis of the semi-cylinder (Fig. 4). (Index of refraction is 2 .) Where are the rays emerging out of the cylindrical surface?

Solution

Let us use angle ϕ to describe the position of the rays in the glass (Fig. 5). According to the law of refraction sin45° sinβ = 2, sinβ =0.5, β = 30°. The refracted angle is 30° for all of the incoming rays. We have to investigate what happens if ϕ changes from 0° to 180°. Figure 4 Figure 5 ϕ α β A C D O B E

5

It is easy to see that ϕ can not be less than 60° (AOB∠ 60= °). The critical angle is given by sinβcrit = n1 = 2 2 ; hence βcrit = 45°. In the case of total internal reflection

° = ∠ 45

ACO , hence ϕ=180°−60°−45°=75°. If ϕ is more than 75° the rays can emerge

the cylinder. Increasing the angle we reach the critical angle again if OED∠ 45= °. Thus the rays are leaving the glass cylinder if:

° < <

° 165

75 ϕ ,

CE, arc of the emerging rays, subtends a central angle of 90°.

Experimental problem

Three closed boxes (black boxes) with two plug sockets on each are present for investigation. The participants have to find out, without opening the boxes, what kind of elements are in them and measure their characteristic properties. AC and DC meters (their internal resistance and accuracy are given) and AC (5O Hz) and DC sources are put at the participants’ disposal.

Solution

No voltage is observed at any of the plug sockets therefore none of the boxes contains a source.

Measuring the resistances using first AC then DC, one of the boxes gives the same result. Conclusion: the box contains a simple resistor. Its resistance is determined by measurement.

One of the boxes has a very great resistance for DC but conducts AC well. It contains a capacitor, the value can be computed as

C X C ω 1 = .

The third box conducts both AC and DC, its resistance for AC is greater. It contains a resistor and an inductor connected in series. The values of the resistance and the inductance can be computed from the measurements.

3

International Physics Olympiad

1969, Brno, Czechoslovakia

Problem 1. Figure 1 shows a mechanical system consisting of three carts A,

B and C of masses m1 = 0.3 kg, m2 = 0.2 kg and m3 = 1.5 kg respectively.

Carts B and A are connected by a light taut inelastic string which passes over a light smooth pulley attaches to the cart C as shown. For this problem, all resistive and frictional forces may be ignored as may the moments of inertia of the pulley and of the wheels of all three carts. Take the acceleration due to gravity g to be 9.81 m s−2_. µ´ ¶³ µ´ ¶³ i ¡ e e e e - _C B A ~ F Figure 1:

1. A horizontal force ~F is now applied to cart C as shown. The size of ~F

is such that carts A and B remain at rest relative to cart C. a) Find the tension in the string connecting carts A and B. b) Determine the magnitude of ~F .

2. Later cart C is held stationary, while carts A and B are released from rest.

a) Determine the accelerations of carts A and B. b) Calculate also the tension in the string.

Solution:

Case 1. The force ~F has so big magnitude that the carts A and B remain

at the rest with respect to the cart C, i.e. they are moving with the same acceleration as the cart C is. Let ~G1, ~T1 and ~T2 denote forces acting on particular carts as shown in the Figure 2 and let us write the equations of motion for the carts A and B and also for whole mechanical system. Note that certain internal forces (viz. normal reactions) are not shown.

-6 x y 0 µ´ ¶³ µ´ ¶³ i ¡ e e e e -6 ? - _C B A ~ F ~ T2 ~ T1 ~ G1 Figure 2:

The cart B is moving in the coordinate system Oxy with an acceleration

ax. The only force acting on the cart B is the force ~T2, thus

T2 = m2ax. (1)

Since ~T1 and ~T2 denote tensions in the same cord, their magnitudes satisfy

T1 = T2.

The forces ~T1 and ~G1 act on the cart A in the direction of the y-axis. Since, according to condition 1, the carts A and B are at rest with respect to the cart C, the acceleration in the direction of the y-axis equals to zero,

ay = 0, which yields

T1− m1g = 0 .

Consequently

T2 = m1g . (2)

So the motion of the whole mechanical system is described by the equation

F = (m1 + m2+ m3) ax, (3)

because forces between the carts A and C and also between the carts B and C are internal forces with respect to the system of all three bodies. Let us remark here that also the tension ~T2 is the internal force with respect to the system of all bodies, as can be easily seen from the analysis of forces acting on the pulley. From equations (1) and (2) we obtain

ax =

m1

m2

g .

Substituting the last result to (3) we arrive at

F = (m1 + m2+ m3) m1 m2 g . Numerical solution: T2 = T1 = 0.3 · 9.81 N = 2.94 N , F = 2 · 3 2· 9.81 N = 29.4 N .

Case 2. If the cart C is immovable then the cart A moves with an accelera-tion ay and the cart B with an acceleration ax. Since the cord is inextensible (i.e. it cannot lengthen), the equality

ax = −ay = a

holds true. Then the equations of motion for the carts A, respectively B, can be written in following form

T1 = G1− m1a , (4)

T2 = m2a . (5)

The magnitudes of the tensions in the cord again satisfy

T1 = T2. (6)

The equalities (4), (5) and (6) immediately yield (m1+ m2) a = m1g .

Using the last result we can calculate a = ax = −ay = m1 m1+ m2 g , T2 = T1 = m2m1 m1+ m2 g . Numerical results: a = ax = 3 5· 9.81 m s −2 _{= 5.89 m s}−2_, T1 = T2 = 1.18 N .

Problem 2. Water of mass m2 is contained in a copper calorimeter of mass m1. Their common temperature is t2. A piece of ice of mass m3 and temperature t3 < 0oC is dropped into the calorimeter.

a) Determine the temperature and masses of water and ice in the equilib-rium state for general values of m1, m2, m3, t2and t3. Write equilibrium equations for all possible processes which have to be considered. b) Find the final temperature and final masses of water and ice for m1 =

1.00 kg, m2 = 1.00 kg, m3 = 2.00 kg, t2 = 10oC, t3 = −20oC.

Neglect the energy losses, assume the normal barometric pressure. Specific heat of copper is c1 = 0.1 kcal/kg·oC, specific heat of water c2 = 1 kcal/kg·oC, specific heat of ice c3 = 0.492 kcal/kg·oC, latent heat of fusion of ice l = 78, 7 kcal/kg. Take 1 cal = 4.2 J.

Solution:

We use the following notation:

t temperature of the final equilibrium state,

t0 = 0oC the melting point of ice under normal pressure conditions,

M2 final mass of water,

M3 final mass of ice,

m0

2 ≤ m2 mass of water, which freezes to ice,

m0

3 ≤ m3 mass of ice, which melts to water.

a) Generally, four possible processes and corresponding equilibrium states can occur:

1. t0 < t < t2, m02 = 0, m03 = m3, M2 = m2+ m3, M3 = 0.

Unknown final temperature t can be determined from the equation (m1c1+ m2c2)(t2− t) = m3c3(t0− t3) + m3l + m3c2(t − t0) . (7) However, only the solution satisfying the condition t0 < t < t2 does make physical sense.

2. t3 < t < t0, m02 = m2, m03 = 0, M2 = 0, M3 = m2+ m3.

Unknown final temperature t can be determined from the equation

m1c1(t2− t) + m2c2(t2− t0) + m2l + m2c3(t0− t) = m3c3(t − t3) . (8) However, only the solution satisfying the condition t3 < t < t0 does make physical sense.

3. t = t0, m02 = 0, 0 ≤ m03 ≤ m3, M2 = m2+ m03, M3 = m3− m03. Unknown mass m0

3 can be calculated from the equation

(m1c1+ m2c2)(t2− t0) = m3c3(t − t3) + m03l . (9) However, only the solution satisfying the condition 0 ≤ m0

3 ≤ m3 does

make physical sense.

4. t = t0, 0 ≤ m02 ≤ m2, m03 = 0, M2 = m2− m02, M3 = m3 + m02. Unknown mass m0

2 can be calculated from the equation

(m1c1+ m2c2)(t2− t0) + m02l = m3c3(t0− t3) . (10) However, only the solution satisfying the condition 0 ≤ m0

2 ≤ m2 does

make physical sense.

b) Substituting the particular values of m1, m2, m3, t2and t3to equations (7), (8) and (9) one obtains solutions not making the physical sense (not satisfying the above conditions for t, respectively m0

3). The real physical process under given conditions is given by the equation (10) which yields

m0₂ = m3c3(t0− t3) − (m1c1+ m2c2)(t2− t0)

l .

Substituting given numerical values one gets m0

2 = 0.11 kg. Hence, t = 0oC,

M2 = m2− m02 = 0.89 kg, M3 = m3+ m02 = 2.11 kg. 5

Problem 3. A small charged ball of mass m and charge q is suspended from the highest point of a ring of radius R by means of an insulating cord of negligible mass. The ring is made of a rigid wire of negligible cross section and lies in a vertical plane. On the ring there is uniformly distributed charge Q of the same sign as q. Determine the length l of the cord so as the equilibrium position of the ball lies on the symmetry axis perpendicular to the plane of the ring.

Find first the general solution a then for particular values Q = q = 9.0 · 10−8 _{C, R = 5 cm, m = 1.0 g, ε}

0 = 8.9 · 10−12 F/m.

Solution:

In equilibrium, the cord is stretched in the direction of resultant force of ~G =

m~g and ~F = q ~E, where ~E stands for the electric field strength of the ring

on the axis in distance x from the plane of the ring, see Figure 3. Using the triangle similarity, one can write

x R = Eq mg. (11) @ @ @ @ @ -? @ @ @ R R x l ~ F ~ G Figure 3:

For the calculation of the electric field strength let us divide the ring to

n identical parts, so as every part carries the charge Q/n. The electric field

strength magnitude of one part of the ring is given by

∆E = Q

4πε0l2n

.

@ @ @ @ @ -? @ @ @ R R x l ∆Ex ∆E⊥ ∆E α Figure 4:

This electric field strength can be decomposed into the component in the direction of the x-axis and the one perpendicular to the x-axis, see Figure 4. Magnitudes of both components obey

∆Ex= ∆E cos α = ∆E x

l ,

∆E⊥= ∆E sin α .

It follows from the symmetry, that for every part of the ring there exists another one having the component ∆ ~E⊥of the same magnitude, but however oppositely oriented. Hence, components perpendicular to the axis cancel each other and resultant electric field strength has the magnitude

E = Ex= n∆Ex =

Q x

4πε0l3

. (12)

Substituting (12) into (11) we obtain for the cord length

l = 3 s Q q R 4πε0m g . Numerically l = 3 r 9.0 · 10−8_{· 9.0 · 10}−8_{· 5.0 · 10}−2 4π · 8.9 · 10−12_{· 10}−3_{· 9.8} m = 7.2 · 10 −2 _{m .}

Problem 4. A glass plate is placed above a glass cube of 2 cm edges in such a way that there remains a thin air layer between them, see Figure 5.

Electromagnetic radiation of wavelength between 400 nm and 1150 nm (for which the plate is penetrable) incident perpendicular to the plate from above is reflected from both air surfaces and interferes. In this range only two wavelengths give maximum reinforcements, one of them is λ = 400 nm. Find the second wavelength. Determine how it is necessary to warm up the cube so as it would touch the plate. The coefficient of linear thermal expansion is

α = 8.0 · 10−6 o_C−1_{, the refractive index of the air n = 1. The distance of the}

bottom of the cube from the plate does not change during warming up.

6 ? ???????? d h Figure 5: Solution:

Condition for the maximum reinforcement can be written as 2dn − λk

2 = kλk, for k = 0, 1, 2, . . . ,

i.e.

2dn = (2k + 1)λk

2 , (13)

with d being thickness of the layer, n the refractive index and k maximum order. Let us denote λ0 _{= 1150 nm. Since for λ = 400 nm the condition for} maximum is satisfied by the assumption, let us denote λp = 400 nm, where p is an unknown integer identifying the maximum order, for which

λp(2p + 1) = 4dn (14)

holds true. The equation (13) yields that for fixed d the wavelength λk increases with decreasing maximum order k and vise versa. According to the

assumption, λp−1< λ0 < λp−2, i.e. 4dn 2(p − 1) + 1 < λ 0 _< 4dn 2(p − 2) + 1 .

Substituting to the last inequalities for 4dn using (14) one gets

λp(2p + 1) 2(p − 1) + 1 < λ

0 _< λp(2p + 1) 2(p − 2) + 1 .

Let us first investigate the first inequality, straightforward calculations give us gradually λp(2p + 1) < λ0(2p − 1) , 2p(λ0− λp) > λ0+ λp, i.e. p > 1 2 λ0_{+ λ} p λ0_{− λ} p = 1 2 1150 + 400 1150 − 400 = 1. . . . (15)

Similarly, from the second inequality we have

λp(2p + 1) > λ0(2p − 3) , 2p(λ0− λp) < 3λ0+ λp, i.e. p < 1 2 3λ0_{+ λ} p λ0_{− λ} p = 1 2 3 · 1150 + 400 1150 − 400 = 2. . . . (16)

The only integer p satisfying both (15) and (16) is p = 2. Let us now find the thickness d of the air layer:

d = λp

4 (2p + 1) = 400

4 (2 · 2 + 1) nm = 500 nm . Substituting d to the equation (13) we can calculate λp−1, i.e. λ1:

λ1 = 4dn

2(p − 1) + 1 = 4dn 2p − 1. Introducing the particular values we obtain

λ1 = 4 · 500 · 1

2 · 2 − 1 nm = 666.7 nm .

Finally, let us determine temperature growth ∆t. Generally, ∆l = αl∆t holds true. Denoting the cube edge by h we arrive at d = αh∆t. Hence

∆t = d αh = 5 · 10−7 8 · 10−6_{· 2 · 10}−2 o_{C = 3.1}o_{C .} 9

1

Problems of the IV International Olympiad, Moscow, 1970 The publication is prepared by Prof. S. Kozel & Prof. V.Orlov

(Moscow Institute of Physics and Technology)

The IV International Olympiad in Physics for schoolchildren took place in Moscow (USSR) in July 1970 on the basis of Moscow State University. Teams from 8 countries participated in the

competition, namely Bulgaria, Hungary, Poland, Romania, Czechoslovakia, the DDR, the SFR Yugoslavia, the USSR. The problems for the theoretical competition have been prepared by the group from Moscow University stuff headed by professor V.Zubov. The problem for the

experimental competition has been worked out by B. Zvorikin from the Academy of Pedagogical Sciences.

It is pity that marking schemes were not preserved.

Theoreti ca l Probl ems Problem 1.

A long bar with the mass M = 1 kg is placed on a smooth horizontal surface of a table where it can move frictionless. A carriage equipped with a motor can slide along the upper horizontal panel of the bar, the mass of the carriage is m = 0.1 kg. The friction coefficient of the carriage is μ = 0.02. The motor is winding a thread around a shaft at a constant speed v0 = 0.1 m/s. The other end of the

thread is tied up to a rather distant stationary support in one case (Fig.1, a), whereas in the other case it is attached to a picket at the edge of the bar (Fig.1, b). While holding the bar fixed one allows the carriage to start moving at the velocity V0 then the bar is let loose.

Fig. 1 Fig. 2

By the moment the bar is released the front edge of the carriage is at the distance l = 0.5 m from the front edge of the bar. For both cases find the laws of movement of both the bar and the carriage and the time during which the carriage will reach the front edge of the bar.

2

Problem 2.

A unit cell of a crystal of natrium chloride (common salt- NaCl) is a cube with the edge length a = 5.6ּ10-10 m (Fig.2). The black circles in the figure stand for the position of natrium atoms whereas the white ones are chlorine atoms. The entire crystal of common salt turns out to be a repetition of such unit cells. The relative atomic mass of natrium is 23 and that of chlorine is 35,5. The density of the common salt ρ = 2.22ּ103 kg/m3 . Find the mass of a hydrogen atom.

Problem 3.

Inside a thin-walled metal sphere with radius R=20 cm there is a metal ball with the radius r = 10 cm which has a common centre with the sphere. The ball is connected with a very long wire to the Earth via an opening in the sphere (Fig. 3). A charge Q = 10-8 C is placed onto the outside sphere. Calculate the potential of this sphere, electrical capacity of the obtained system of conducting bodies and draw out an equivalent electric scheme.

Fig. 3 Fig. 4

Problem 4.

A spherical mirror is installed into a telescope. Its lateral diameter is D=0,5 m and the radius of the curvature R=2 m. In the main focus of the mirror there is an emission receiver in the form of a round disk. The disk is placed perpendicular to the optical axis of the mirror (Fig.7). What should the radius r of the receiver be so that it could receive the entire flux of the emission reflected by the mirror? How would the received flux of the emission decrease if the detector’s dimensions decreased by 8 times?

Directio ns: 1) When calculating small values α (α<<1) one may perform a substitution

2 1

1−α ≈ −α ; 2) diffraction should not be taken into account.

3

Experi men tal Probl em

Determine the focal distances of lenses.

List of ins tru ments: three different lenses installed on posts, a screen bearing an image of a

geometric figure, some vertical wiring also fixed on the posts and a ruler.

Solution s of th e p ro blems of th e I V In ternationa l Oly mpi ad, Mos cow, 197 0 Theoreti ca l Co mp etition

Problem 1.

a) By the moment of releasing the bar the carriage has a velocity v0 relative to the table and continues

to move at the same velocity.

The bar, influenced by the friction force Ffr = μmg from the carriage, gets an acceleration

a = Ffr/ M = μmg/M ; a = 0.02 m/s , while the velocity of the bar changes with time according to the

law vb = at. .

Since the bar can not move faster than the carriage then at a moment of time t = t0 its sliding will stop, that is vb = v0. Let us determine this moment of time:

s 5 0 0 0 = = = mg M v a v t µ

By that moment the displacement of the Sb bar and the carriage Sc relative to the table will be equal to

mg M v t v S µ 2 0 0 0 c = = , mg M v at Sb _µ 2 2 2 0 2 0 = = .

The displacement of the carriage relative to the bar is equal to m 25 . 0 2 2 0 c− = = = mg M v S S S b _µ

Since S<l, the carriage will not reach the edge of the bar until the bar is stopped by an immovable support. The distance to the support is not indicated in the problem condition so we can not calculate this time. Thus, the carriage is moving evenly at the velocity v0 = 0.1 m/s, whereas the

bar is moving for the first 5 sec uniformly accelerated with an acceleration a = 0.02 m/s and then the bar is moving with constant velocity together with the carriage.

b) Since there is no friction between the bar and the table surface the system of the bodies “bar-carriage” is a closed one. For this system one can apply the law of conservation of momentum:

4

where v and u are projections of velocities of the carriage and the bar relative to the table onto the horizontal axis directed along the vector of the velocity v0. The velocity of the thread winding v0 is

equal to the velocity of the carriage relative to the bar (v-u), that is v0 = v – u (2)

Solving the system of equations (1) and (2) we obtain: u = 0 , v = v0 .

Thus, being released the bar remains fixed relative to the table, whereas the carriage will be moving with the same velocity v0 and will reach the edge of the bar within the time t equal to

t = l/v0 = 5 s.

Problem 2.

Let’s calculate the quantities of natrium atoms (n1) and chlorine atoms (n2) embedded in a single

NaCl unit crystal cell (Fig.2).

One atom of natrium occupies the middle of the cell and it entirely belongs to the cell. 12 atoms of natrium hold the edges of a large cube and they belong to three more cells so as 1/4 part of each belongs to the first cell. Thus we have

n1 = 1+12⋅1/4 = 4 atoms of natrium per unit cell.

In one cell there are 6 atoms of chlorine placed on the side of the cube and 8 placed in the vertices. Each atom from a side belongs to another cell and the atom in the vertex - to seven others. Then for one cell we have

n2= 6⋅1/2 + 8⋅ 1/8 = 4 atoms of chlorine.

Thus 4 atoms of natriun and 4 atoms of chlorine belong to one unit cell of NaCl crystal. The mass m of such a cell is equal

m = 4(mrNa + mrCl) (amu),

where mrNa and mrCl are relative atomic masses of natrium and clorine. Since the mass of hydrogen

atom mH is approximately equal to one atomic mass unit: mH = 1.008 amu ≈ 1 amu then the mass of

an unit cell of NaCl is

m = 4(mrNa + mrCl)mH .

On t he ot her hand, it is equal m = ρa3 , h ence

(

)

Having no charge on the ball the sphere has the potential

V 450 4 1 0 0s = = R Q πε ϕ .

5

When connected with the Earth the ball inside the sphere has the potential equal to zero so there is an electric field between the ball and the sphere. This field moves a certain charge q from the Earth to the ball. Charge Q`, uniformly distributed on the sphere, doesn’t create any field inside thus the electric field inside the sphere is defined by the ball’s charge q. The potential difference between the balls and the sphere is equal

, 4 1 0 s b       − = − = ∆ R q r q πε ϕ ϕ ϕ (1)

Outside the sphere the field is the same as in the case when all the charges were placed in its center. When the ball was connected with the Earth the potential of the sphere φs is equal

. 4 1 0 s R Q q+ = πε ϕ (2)

Then the potential of the ball

0 4 1 4 1 0 0 s b =      + =       + _{+ −} = ∆ + = r q R Q R q r q R Q q πε πε ϕ ϕ ϕ (3) Which leads to R r Q q=− . (4)

Substituting (4) into (2) we obtain for potential of the sphere to be found:

(

)

The electric capacity of whole system of conductors is

44pF F 10 4 . 4 4 11 2 0 s = ⋅ = − = = − r R R Q C πε ϕ

The equivalent electric scheme consists of two parallel capacitors: 1) a spherical one with charges +q and –q at the plates and 2) a capacitor “sphere – Earth” with charges +(Q-q) and

–(Q–q) at the plates (Fig.5).