estonian finnish physics olympiad (2003-2014)

8/10/2019 Estonian Finnish Physics Olympiad (2003-2014)

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ESTONIANFINNISHPHYSICSOLYMPIAD

PROBLEMS & SOLUTIONS (2003-2014)


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PROBLEMS & SOLUTIONS (2003)


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r >100


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V1 = 20 3

t1 = 25

V2 = 10 3

t2 = 1

cp = 1005

V =V1+ V2

T

1,0

1,5

2,0

0 5 10 15 20 25

t(0C)

(g/m3)

r0 = 100 r

r > 100

m

0 = 1 189 3

q= 2500

m 1

e

E0 E0e gm g

q

H

a

h

R1= 1.5

R2 = 6.2

V I

V I

Uc

C

R

E

U=E U0et/RC

l


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L

l

= 0 p

g

c

T = x + x

P

T

a

T1

a p

p g

= bab+a b

l

b

a

c

T

T

P

x

d = 1

= 1

C

C

T

U

eU kT e

k

M

B l

U

e

r

r l

M M M+ M

M

r

eU kT

M

M

M + M

M

b

I0

b < r

b 2I0bb

I/ = 2I0bb// = 2I0b(d/db)1

(b)

b


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(a)


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m

k

v

L

v

Lv

P =P0cos(t)

t

T(t)

T0

L

S

1 C 1 m

C

heatertemperature sensor

Ti nano-disk thermalbridge

R

R

T(t) T =T0+ Tsin(t + )

T(t)

C

0

c c l c

a

b

v0 = 30km/h a

a

b

(a)

(b)

B

z

x2 +y2 < R2

v = RBe/m e m

y = 0

y = a

a < R


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R = 10k

I= 2 mA

I/I 0

r1 = R = 0.65 R

r2 = R = 2.3

v0 = 30km/h

R0 = 6400km

g = 9.8 m/s2 E = GmM/2a m

a

M

v

vr vt

ur ut

w

h = 100km

= 650nm

t= R/2v

O B


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v

v/2

=

2k/m

v

v

t

T = 2

m/2k

L

2L/v = T(n+ 12),

L= v

n +

1

2

m/2k.

L/S

R= L/4S

P

= T /R

Q = CT = CT

P0cos(t) =CT+ T/R.

T = A cos(t + )

= arcsin(C/

C22 + R2)

P0cos(t) =A

C

2

2

+ R

2

cos(t + ).

=

A= P0/C22 + R2 T =T0+

P0cos

t + arcsin(C/

C22 + R2)

C22 + R2

.

A = P0C22+R2

C

dA/dC dA/dC = P0(C22 +

R2)3/2C2

x= (C)2

x

ln

(dC/dA)2

= 3 ln(x + R2

) ln x + ln C.

3x= x + R2

x= R2/2

= 1/

2CR.

AcSL AS/(Lc) c /cL2.

B

C B

|BC| =ut

u

|AB| = v0t

|AB|

|BC

u= v0|BC||AC| =

18mm

42mm30km/h 13km/h.

C

|OC| =|AO| = v0t C

B

D

OD = ut

u= v0|OD||AO| =

27mm

39mm30km/h 21km/h.

R

DABE

AB

O

B

COBO

R

BO

OC

BO

= CO

B = AOB + AOC=

2 arcsin a

R .

B

O

C

B

CO

arctan

= tanBC O = tanOAB =|OB|/|AB| = R/|AB|

R 40mm

2 |AB|

B

I= 2 mA

u= 1 V

P= 2 mW

U

(t) = U0cos(t) u

U0cos(t) < U

(t) + u

I/I

E 1 i t


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U

(t) IR U0 = I R+ u= 21V

U = Q/C

Q= I t

t

t 1/ I/I= U/U = Q/CU = Q/CIR=1/CR

C 100/R= 200 F

Q= C IR

P1= Qu= CI Ru= 200mW.

pV =

(pV /T) =

p1/T =

( 1) dpp dT

T = 0

20K = T =T 1

p

p .

p= gh

= p/RT 1.2 kg/m3

=cp/cV = (cV +R)/cV

T = R

cV + R

gh

R ,

h=

1 +cV

R

T Rg

2040m.

x

y

mx= eE0 cos t,my= eE0sin t.

mx= eE0 1 sin t,my= eE0

1(1 cos t).

x= eE0m2

(cos t

1),

y = eE0m2

cos t +eE0m

t.

y

u = eE0m

R = eE02m

= 2u/= 2R

a = 12

(r1+ r2) = 12

(+)R

E

m = M

2a =

1

2v2 M

R .

v20 = MR

v20 +

=v20 12 v2.

v = v0

2[1 ( + )1] 34.5 km/h.

vr = vp

v

vp = v0

2

1

1

+

= v0

2

+ 37.5 km/h.

vt = v0

2 +

24.4 km/h.

vr =

v2 v2t 24.4 km/h.

ur = vr

ut = vt v0 5.6 km/h.

gR0

w=

u2t + u2r+ 2gR0 27.4 km/h

l = 2(n|CD | |AB|) = 2 (nd/ cos d sin sin ) = 2d(n/ cos sin2 /n);

sin = sin /n

l

d(l)d = a = L

L |AB| +|BC|

d(l)d = 2d(sin / cos

2 sin2/n) = 2d sin (cos2 2cos /n) L = 2ad sin (cos2 2cos /n) d = L/2a sin (cos2 2 cos /n)

n

n 1.4

B n= tan B

a

= 2/d

a = 2L/d

d = 2L/a

d 0.20mm


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. Pencil ( points)Equipment:pencil, paper, ruler.Determine the coefficient of friction of the pencils graphite

core against the paper. Estimate the uncertainty.

i) FindthecurrentthroughtheresistorsR1andR2 atthemomentof time t1= 5ms.ii) Find the current through the resistors R1 and R2 at the


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g p p y. Spring( points) Equipment: helical spring of known massm= 19 0.5 g, measuring tape, a load of unknown mass.

Determine the mass of the load. Estimate the uncertainty.. Soap film ( points) Lord Rayleigh had in a lecture abouttaking photos of physical processes. Among others, he showed aphoto of a soap film, which is falling apart (see Fig.). Instead of aflash, he used an electric spark (well, nowadays the flashes are also

based on electric sparks). Estimate, how precise must have beenthe timing, i.e. estimate the time for a soap film to fall apart. Letthe thickness of the soap film be h = 1 m, the ring diameterD= 10 cmand the surface tension= 0.025 N/m.

Hint: you may use a model, according to which the alreadybroken part of the soap film gathers into a single front and movesall together towards the still preserved part of the film.

. Magnetic pulse ( points) Consider an electric circuitconsisting of a coil of negligibly small inductance, consisting ofN = 10turns and with the surface area of a single loop S =10cm2), resistorsR1 = R2 = 3 , capacitorC = 0.2 F,and an inductanceL = 1H, connected as shown in Fig. At themoment of timet = 0, a magnetic field, parallel to the axis ofthe coil is switched on. e inductance of the magnetic field startsgrowing linearly, starting fromB = 0until the maximal valueB = 1 Tis achieved att = 10ms. Further, the inductance of

the magnetic field remains constant (and equal to T).

) g 1 2moment of time t2= 15ms.iii)What is the net charge passing through the resistorR2?. Stratostat ( points)i) Show that the pressure of an isothermal gas of molar massfollowsthelawp= p0e

z, wherep0is thepressure attheorigin,

andz is theheight. Find theconstant. e temperature isT, thefree fall acceleration is g.ii) Consider a stratostat, the envelope of which (a freelydeformable non-elastic sack) is filled at the Earths surface byhelium to the volume fraction of = 10%. At which heighthdoes the helium expand so that it fills the entire volume of thestratostat? e molar masses of the air and helium area =29 g/molandHe = 4 g/mol, respectively. You may neglect

the temperature variations of the atmosphere, and use the valueT= 250K.. Wedge( points)A wedge of massMis kept at rest on anhorizontal surface, and a block of massm is kept on the wedge atthe heighth from the surface. e angle of the wedge is, seeFig. ere is no friction neither between the block and wedge nor

between the surface and the wedge. e system is released intoa free motion. Find the timetneeded for the block to reach thesurface.


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Estonian-Finnish Olympiad -

. Charges in E( points)Two particles (the blue and the red)of massmare connected with a spring, the stress-free length of

is related to).i) ( pt) Knowing that the chicken leg is approximatelyl =15 cm tall, estimate the length of a tyrannosaur leg L. You may

th t th l th f l l th l th f th h l


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p g gwhich isL and stiffnessk; the blue carries chargeq(q > 0)and the red is chargeless. In the regionx > 0, there is anhomogeneous electric fieldE, antiparallel to thex-axis; In theregionx < 0, there is no electric field. Initially, the dumbbell

of charges moves in region x < 0with velocityv, parallel to thex-axis; the dumbbells axis is also parallel to thex-axis and thespring is stress-free. It is known that aer a while, the dumbbellmoves in the regionx < 0with velocityv, and that the redparticle never enters the regionx > 0. Also, the springs lengthachieves minimum only once.i)(. pt) How long timedoes the blue particle spend in theregion x >0?

e process takes place exactly as described, if one equality

and one inequality are satisfied for the quantities m,v,k,q,Eand L.ii) ( pt) Which equality must be satisfied?iii) (. pt) Which inequality must be satisfied?. ermos bole ( points) In order to study the thermalproperties of a thermos bole, let us model it as two concentricspherical vessels, with radiiR1 = 7 cmand R2 = 10cm. egap between the walls of the vessels contains vacuum (hence, the

heat conductivity can be neglected).i)(, pt) Find the radiative heat flux (i.e. transmied heat perunit time) between the walls of the bole, assuming that theambient temperature is T2 = 293Kand the inner sphere isfilled with liquid nitrogen at the boiling temperature T1 = 77 K.e emissivities of all the surfaces are equal to that of stainlesssteel: = 0.1. Remark: e emied heat flux per unit areais given Stefan-Boltzmanns law P = T4, where =5.67 108 W/m2K4 (assuming thatis independent of the

wavelength.ii) (, pt) Estimate, how long time does it take for a fullevaporizationof the liquid nitrogen (thevaporescapes through anover pressurevalve). Forthe liquid nitrogen, density= 810g/land latent heat for vaporization = 5.580 kJ/mol).NB! If youwere unable to findP( for question i), express the evaporization timesymbolically (i.e. using the symbolP).. Tyrannosaur (T. Rex) ( points) Paleontologists havediscovered tracks of a tyrannosaur where the footprints of thesame leg areA = 4.0meters apart. ey have also recovereda piece of a tyrannosaur leg bone that has bone cross-sectionaloarea N= 10000times that of a chicken (which the tyrannosaur

assume that the length of a leg scales as the length of the wholeanimal, and that the bone stress (force per area) is the same for

both animals. Is your result consistent with the step length A?ii) ( pt)Estimate thenatural walking speedof thetyrannosaur byapproximating thewalking motion of a legwith a freely oscillating

pendulum motion. State clearly all the assumptions you make.. Ball ( points) Massivesphericalballhasamass M= 100kg;an aempt is made to roll the ball upwards, along a vertical wall,

by applying a force F to some point Pon the ball. e coefficientof friction between the wall and the ball is = 0.7.i)( pt) What is the minimal force Fminrequired to achieve thisgoal?ii) ( pt) On a side view of the ball and the wall, constructgeometrically the pointP, where the force has to be applied to,together with the direction of the applied force.. Elastic thread( points)Equipment: ruler, tape, an elasticthread, a wooden rod, marker, a known weight.

e purpose of this problem is to study the elastic propertiesof an elastic thread for large relative deformations = (l l0)/l0, where l0and l are thelengths in initial andstretched states,respectively. If the Hooks law were valid, the ratioF /of theelastic force Fand would be constant: F/= SE, where Sis

the cross-section area of the thread and E the Young modulusof the thread material.i) Collect the data needed to plot the ratioF /as a function, up to 4. Plot the appropriate graph, and indicate theuncertainties.ii)By making assumption that the Young modulusE = F/Sremains constant, study, how does the volume of the threaddepend on . Plot the appropriate graph.

. Charges in B( points)ere is an homogeneous, parallel tothez-axis magnetic field of inductanceB in regionx > 0. Inregion x


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and charge q. Initially, the particles have coordinates y= z = 0,and respectivelyx = L0and x = 2L0(withL0 > 0).Initial velocity of both particles is v, along the x-axis, towards themagnetic field. Neglect the electrostatic repulsionforce of the twocharges.

i) (, pt) Sketch the trajectory of the first particle, and thedependance of its y-coordinate on time.ii) (, pt) Sketch the distance L between the particles as afunction of timet, assuming thatmv/Bq > L. What is theminimal distance Lmin?. Satellite ( points)i)( pt) A large ball of mass m1is kept at the height h from thefloor (so that the center of the ball is at the height h + d/2, where

d is its diameter). A small ball of massm2 is placed upon thelarge one, and the system is released (so that it starts falling). Towhich height (fromthe floor) will the small ball rise, assumingthatthe collision between the lower ball and ground, and the collision

between the balls are absolutely elastic, and m1 m2?ii) ( pt) Consider the followingsatellitelaunching project. ereareNabsolutely elastic balls of massesm1 m2 ... mN: the first ball (the heaviest) is the lowest; the second ball isplaced on top of the first; the third on top of the second etc.

e upmost ball is supposed to become a satellite, i.e. to obtainthe velocityvN = 7.8 km/s). e lowest ball is at the heighth= 1m from the floor, and the system is released. What should

be the number of balls N? What should be the mass of the lowestball, ifmi/mi+1 = 10, and the mass of the satelliteMN = 1 kg?. Sprinkler ( points)

A sprinkler has a shape of hemisphere, which has small holesdrilled into the spherical part of its surface. From these small

holes, water flows out with velocity v = 10 m/s. Nearthe sprinkler, the water flow is distributed evenly over all thedirections of the upper half-space. e sprinkler is installed at theground level so that its axis is vertical. In what follows, the airresistance can be neglected, and the dimensions of the sprinklercan be assumed to be very small.

i) (. pt) Find the surface area of the ground watered by the

. Power supply( points)

i) ( pt) Consider the cirquit given in Fig (a), where the diodecan be assumed to be ideal (i.e. having zero resistance for forwardcurrent and infinite resistance for reverse current. e key isswitched on for a time cand then switched off, again. e inputand output voltages are during the whole process constant andequal toUiand Uo, respectively (2Ui < Uo). Plot the graphsof input and output currents as functions of time.ii)( pt) Now, the key is switched on and off periodically; eachtime, the key is kept closed for time intervalcand open alsofor c. Find the average output current.

iii) ( pt)Now, cirquit (a) is substituted by cirquit (b); theswitchis switched on and off as in part ii. What will be the voltage onthe load R, when a stationary working regime has been reached?

You may assume that c RC, i.e. the voltage variation on theload (and capacitor) is negligible during thewhole period (i.e. thecharge on the capacitor has no time to change significantly).. Ice-rally( points)e car accelerates on a slippery groundso that the wheels are always at the limit of slipping (e.g. via using

an electronic traction control). Such an acceleration would resultin the velocity vs time graph as given in the Figure.

1 = 0.5 s, during which there is no driving force (so that thecar decelerates due to air friction). Except for that period, theacceleration follows the law given by the graph. As a result, theterminal velocityvt= 40 m/s is reached 2= 1.0 s later than it

would have been reached, if there were no delay caused by the gear

change. Upon reaching the terminal velocity, the car continuesmoving at constant speed. In your calculations, you can assumethat the air friction was constant during the gear change period.ii) (, pt) At which speed the gear was changed?iii)(, pt) How many meters shorter distance will be coveredduring thefirst seconds, as comparedto ideal acceleration (i.e.

without the delay due to the gear change)?. Black box( points)Equipment: a black box, multimeter,

baery, timer (on the screen).Determine the electrical scheme inside the black box, and the

values of all the resistors inside it. Estimate the characteristics ofother electrical components. Itis knownthat apart from thewires,the total number of components is three.


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Estonian-Finnish Olympiad -

. Spool ( points)A spool with inner radius r and outer radiusR lies on a horizontal table; the axis of the spool is horizontal. A

i htl i d d th i t h i th

plates equals still to E. What is the hydrostatic pressure betweenthe plates if the atmospheric pressure isp0and the pressureof the

water column can be neglected?. Charged cylinder( points)Dielectric cylinder of radius r


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weightless rope is wound around the inner part as shown in thepicture. e loose end of the rope makes an angle with the hori-zont (the angle can be also negative). e moment of inertia ofthe spool isJand mass M. In what follows you may assumethat the spool rolls on the table without slipping.i) ( pts ) We pull the loose end of the rope with velocityu (par-allel to the loose part of the rope; that loose part can be thoughtto be very long). What is the velocity of the spool?ii) (pts)Supposenowthatthespoolisatrest,andweapplyforceF to the loose end of the rope (parallel to the loose part of therope). What is the acceleration of axis of the spool?iii) ( pts) How large does the coefficient of friction need to be(asafunctionof) to ensure that there is no slipping between the

spool and the table?iv) (. pts) Now the spool rolls, again, with velocityu; howeverthere is no rope. e spool hits a threshold of height H(see Fig-ure); the impact is perfectly inelastic. What is the speed vof thespool immediately aer the impact?v) (. pts) What is the speed w of the spool aer rolling over thethreshold? Assume that u is such that the spool will roll over thethreshold without losing contact with its edge.vi) ( pts) If the speed u is too large, u > u0, the spool will jump

up and lose contact with the edge of the threshold. Determine u0.

. Capacitor( points)An ideal plate capacitor has plates withareaAand separationd and is charged so that the electric field

between the plates equals to E.i)( pts) Find the energy density of the electric field inside thecapacitor and the total energy of the field.ii)(. pts) What is the force required to keep the plates separ-ated?iii) (. pts) Now, this capacitor is submerged into distilled waterof dielectric permiivity = 80; the electric field between the

C g y ( p ) D e ect c cy de o ad uscarries a charge of surface density on its cylindrical surface androtates with angular velocity.i)( pts) Determine the magnetic inductionB inside the cylin-der. Remark: if you wish, you can use the expression for the in-

ductanceL = 0N2S/lof a solenoidal coil of radius r, lengthl r, area of cross-section Sand number of loops N.ii) ( pts) A radial conducting wire connects the axis of the cylin-der with the cylindrical surface (it rotates together with the cylin-der). Find the electromotive force (voltage) Ebetween the endsof the wire.iii)( pts) Suppose that the wire connecting the axis of the cyl-inder with the cylindrical surface is not radial and has an arbitaryshape (still, there are no segments protruding outside the cylin-der). Show that Edoes not depend on the shape of the wire.. Black box( points)Equipment: a black box with three ter-minals, voltmeter, timer. Inside the black box, there are two capa-citors and a baery, connected as shown in Figure. e capacit-anceC1 = (3400 400)F; you are asked to determine thecapacitance C2and estimate the ucertainty.Remark:the terminal+ is a wire, long enough to be conected to either terminal A orterminal B.

. Plutonium decay(points) Plutonium is an unstableelement,a Pu239 atom decays with a half-life of1/2 = 24000 years

by creating smaller nuclei, including an-particle. Find the-particle flux density (i.e the number of passing nuclei per unit

. Vacuum bulb( points)Let us study how a vacuum can becreated inside a bulb by pumping. Let the volume of the bulb beV, and the pump consist in a piston moving inside a cylinder of

volume V, where 1. epumping cyclesstartswith piston

i)( pts) e electronic component has been dissipating energywith a constant power ofP = 35Wfor a long time, and the av-erage plate temperature has stabilized at the valueT0 = 49

C.Now, the component is switched off, and the average plate tem-


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p y ( p g ptime and per unit cross-sectional area) near the surface of a plateof Pu239. e plate has thicknessd = 1 m m; its width andlength are much larger than that. e density of plutonium =19800 kg/m3. Remark: half-life is the period of time it takes

for a substance undergoing decay to decrease in size (in the num-ber of particles) by half. e mass of an atom of Pu239 ism0 =3.84 1025 kg.. Violin string( points) e motion of a bow puts a violinstring into a periodic motion. Let us make a simplified model ofthis process. e string has elasticity and inertia, so we substituteit by a block of massm, fixed via a spring of stiffness kto a mo-tionless wall and laying on a frictionless horizontal surface. e

bow is substituted by a horizontal plate, which is pressed withconstant force Ndownwards, and which moves with a constant

velocityu, parallel to the axis of the spring, see Figure. e staticcoefficient of friction between the plate and the block is 1, andthe kinetic coefficient of friction is 2 < 1. So, as long as theplate does not slide with respect to the block, the coefficient offriction equals to1; as soon as there is some slip, it decreasesdown to 2.

i)( pts) For questions (i) and (ii), let us assume that the speedof the plate u is very small as comparedto the maximal velocity ofthe block. What is the maximal velocity of the blockvmax(max-

imized over time)?ii) (pts) Sketchqualitativelythe graphof thedisplacement of the

block as a function of time and indicate on the graph the durationsof the prominent stages of the block motion (graph segments).iii) (, pts) Now, letus abandon theassumption aboutthe small-nessofu. Sketch qualitatively thegraph of thevelocityof theblockas a function of time.iv)(, pts) Determine the amplitudeA of the blocks oscilla-tions.v) ( pt) Which condition (strong inequality, or) must besatisfied for u inordertoensurethattheoscillationswillbealmostharmonic?

, p p g y pbeing pulled up; when the pressure inside the cylinder becomessmaller than inside the bulb, a valve VA(connecting the cylinderand the bulb) opens and remains openas longas thepiston movesup. When piston is released, it starts moving down, at that mo-

ment, the valveVAcloses. As long as the valveVAis open, thepressures of the bulb and the cylinder can be considered as equalto each other. When the piston moves down, the pressure in thecylinder increases adiabatically until becoming equal to the out-side pressurep0 = 10

5 Pa; at that moment, another valve VBopens leting the gas out of the cylinder. When the piston reachesthe boommost position, there is no residual air le inside thecylinder. Now, the piston is ready for being li up: the valve VBcloses and VAopens, marking the beginning of the next pumpingcycle. e air inside the bulb can be considered isothermal, withthe temperature being equal to the surrondin temperature T0 . eadiabatic exponent of air = cp/cV = 1.4.

i)( pts) How many pumping cyclesNneeds to be done to re-duce the pressure inside bulb fromp = p0down top = p0,

where 1?

ii)( pts) What is the net mechanical work done during such apumping (covering all the Ncycles)?

iii)( pts) What is the temperature of the air released from thecylinder to the surroundings at the end of the pumping process(when the pressure inside the bulb has become equal to p0)?

iv)( pts) According to the above described pumping scheme,there is a considerable loss of mechanical work during the period

when the piston is released and moves down. Such a loss canbe avoided if there is another pump, which moves in an oppos-ite phase: the force due to outside air pressure pushing the piston

down can be transmied to the other pump for liing the pistonup. What is the net mechanical work done when such a pumpingscheme is used?

. Heat sink( points) Consider a heat sink in the form of a cop-per plate of a constant thickness (much smaller than the diameterd of theplate). Anelectronic component is fixed totheplate, andatemperature sensor is fixed to the plate at some distance from thatcompnent. You may assume that the heat flux (i.e. power per unitarea) from the plate to the surrounding air is proportional to thedifference of the plate temperature at the given point (the coeffi-cient of proportionality is constant over the entire plate, includingthe site of the electronic component).

, p , g pperature starts dropping; it takes = 1 0 sto reach the valueT1= 48

C. Determine the heat capacityC(unitsJ/C)oftheplate. e capacities of the electronic component andthe temper-ature sensor are negligible.

ii)( pts) Now, the electronic component has been switched offfor a long time; at the moment t = 0, a certain amount of heatQis dissipated at it during a very short time. In the Figure andTable, the temperature is given as a function of time, as recorded

by the sensor. Determine the dissipated heat amount Q.

t (s) T(C) . . . . . .t (s) T(C) . . . . . .

. Coefficient of refraction (points)Equipment:A thick glassplate having the shape of a half-cylinder, a glass prism, a container

with an unknown liquid, a laser pointer, graph paper, ruler.

i) ( pts) Determine the coefficient of refraction of the half-cylindrical glass plate and estimate the uncertainty of the result.ii) ( pts) Determine thecoefficient of refraction of theliquid andestimate the uncertainty of the result.


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to repeat the procedure with the other capacitor and measure thetime t2and calculateC2 = C1t2/t1; theuncertainty is estimated

as C1= C1t1t1

+ t2t2

+ C1C1

.

Itis recommendedto check thenegligibilityof theleak current

straight line is tangent to thesinusoid. e length of a straight seg-ment can be calculated as

T1= 20/u= 2(1 2)N/ku;the sinusoidal segmentcorresponds to a half-period and therefore

iii)Due to adiabatic law,pV = Const; when combined withthe gas lawpV Twe obtainp1 T. During the lastdownwards motion of the piston, the pressure inside the cylinder

is increased by a factor of1/; thus, T =T011.


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Itis recommendedto check thenegligibilityof theleak currentacross the plates of the capacitor. To this end, one cancharge a ca-pacitor, measure the voltage, remove the voltmeter and wait forsome time (of the order t1and t2), and check again the voltage.

Another way is to discharge completely one capacitor byshort-circuiting its terminals and charge the other capacitor up tothe voltage of the baery. Further, we connect the terminalsAandBso that the capacitors re-distribute the charge Q =EC1and take the same voltage:Q1/C1 = (Q Q1)/C2 Q1 =QC1/(C1+ C2) =EC21/(C1+ C2). Consequently, the new

voltage(which we measure) is U=Q1/C1= EC1/(C1+C2),from where C2= (

EU 1)C1.

. Plutonium decay( points)

Let the number of Pu239-atoms be reduced during time intervalt = 1 sby a factor of1 (with 1). en, during thetime period of1/2, it is reduced by a factor of(1 )1/2/t e1/2/t = 1

2 = t ln 2/1/2. erefore, the number of

atom decay events is Nd = N t ln 2/1/2, where N =dS/m0is the number of atoms, i.e. the-particle flux is = Nd/2St(wherethefactoraccountsforthefactthattheparticlesareemit-ted towards the both sides of the plate). Upon bringing all the ex-

pressions together, we obtain =

d ln 2

21/2m0 2.36 1013 m2 s1.

. Violin string ( points)i) When the plate slides, there is a constant friction force2Nacting upon the block, which means that the equilibrium deform-ation of the spring is x0= 2N/k; the net force acting upon the

block (due to spring and friction) is given byF =k, wherewe have defined = x x0. erefore, while sliding, the blockoscillates harmonically around the point = 0. Slipping startswhen the static friction will be unable to keep equilibrium, i.e.atkx = 1N, which corresponds to0 = (1 2)N/k.If the plate moves slowly, the block is released with essentiallymissing kinetic energy, and the energy conservation law yields12

k20 = 12

mv2max vmax= 0

k/m.ii)As mentioned, when the plate slides, the motion of the block isharmonic, i.e. the graph ofx(t)is a segment of a sinusoid; when

there is no sliding, the block moves togetherwith theplate, i.e. thegraphofx(t) is a straight line. At themoment when slippingstartsor stops, theoscillatory speed is equal to thespeed of plate, i.e. the

has a length ofT2 =

m/k.

iii)e speed v(t) = dxdt

is the derivative ofx(t); therefore, thesinusoidal segment ofx(t)will correspond to a sinusoidal seg-ment ofv(t), and a straight segment ofx(t) to a horizontalsegment ofv(t). e resulting graph is depicted below.

iv)Let the amplitude of the oscillations be A, i.e. the sinusoidalsegments follow the law(t) =A cos(t), where =

k/m.

Correspondingly, v(t) = A sin(t) A sin(t) =v(t)/; hence, for any point at a sinusoidal segment, 2 +v2/2 = A2. At a point, where a sinusoid and a straight linemeet, the straight line and sinusoid have equal values for =0 = (1 2)N/kand v= u. Consequently,

(1 2)2N2/k2 + u2/2 =A2 A=

1

k

(1 2)2N2 + u2mk.

v) e oscillations will be almost harmonic when the straight seg-ments are very short, i.e. whenu/ (1 2)N/k u (1 2)N/mk.. Vacuum bulb ( points)i) Each pumping cycle reduces the number of molecules insidethe bulb by a factor of(1 ); therefore, aerNcycles, thenumber of molecules (and hence, the pressure) by a factor of= (1 )N eN

N = ln

.

ii) Majority of the pumping cycles are done when the pressure in-

side the bulb is negligible as compared to the outside pressure.During such a cycle, a work equal top0V is done. erefore,A N p0V = p0V| ln |.

/iv)According to the modified pumping scheme, the work/energyloss is only due to the release of the hot air. Note that if we hada cylinder of volume V, we could be able to create vacuum insidethere using only one pumping motion, i.e. by performing work

A = p0V and without any energy loss. Now, we perform an ex-cess work, which is converted into internal energy of the releasedhot air, which needs to be calculated. Let = p

p0be an inter-

mediate rarefaction factor; then, we can apply the previous resultto calculate the internal energy of released air, if its quantity is d

moles:dU = T0(11 1)cVd. Let us note that the num-

ber of moles inside the bulb is = p0VRT0

d = p0VRT0

d. So,

U =p0V cV

R10(11 1)d= ( 1)p0V

cV

R

. Now, recall

that = cp/cV = 1 + RcV, hence cVR

= 11

andU = p0V.is gives us the energy loss due to heating the released air; an-otherp0V is required for loss-free creationof the vacuum. Hence,the total required work is A= 2p0V.. Heat sink( points)i)When the average temperature is stable at T0, all the power dis-sipated at the electronic component is eventually given to the air:the air is being heated with power P. As the heat flux depends lin-

early on the temperature difference between a point on the plateand the air, the average heat flux and therefore the net power dis-sipated into the air depends linearly on the average temperature ofthe plate. e average temperature determines the radiated power.

Now consider the situation aer the heating has ended. eaverage temperature is initially the same, so the radiated heatpower is initially still P. By the definition of heat capacity, aninfinitesimal heat amount given to the surroundings is dQ =

C dTavgwith the minus sign encoding the direction of the heat

flow. us, at the first moment,P = dQdt

=CdTavgdt

. Assum-ing that during the average temperature depends approximatelylinearly on time (because T0 T1 = 1 Cis much less than theusual ambient temperature), dTavg

dt T1T0

and C P

T0T1=

350J/C. Actually the graph ofTavg(t) is slightly curved down-wards (as it is an exponential eventually stabilizing at the ambienttemperature) and initially somewhat steeper, so Cis a bit smaller.ii) e average temperature of the heat sink fallsoff exponentially,

therefore, if the tail of the given graph turns out to be expo-nential, we can presume the tail depicts the situation where thesensor issensing the average temperature and the initial bump

in the temperature distribution has evened out. Extrapolating theexponential to t= 0we get the initial average temperature Tavg,0(immediately aer theQhas been dissipated into the sink) and,

byQ = C(Tavg,0 Tamb), the heatQ. e ambient temper-T b d ff f h b i i f h i h

but not too much: Tcis exponentially sensitive to the T-interceptof the straight line fied to the tail (its crossing point with theT-axis) on the logarithmic plot. e bump has still not yet disap-peared completely enough. C ffi i f f i ( i t )


45/57

ature Tambcan be read off from the beginning of the given graphwhere the sensors surroundings have not yet heated up. is isfurthermore a check for the assumption T0 T1 Tambmadein the first part of the solution. From the table, Tamb= 20.0

C.

Letus analyse the(yet hypothetical) exponential TavgTambought to obey, so that eventually we expect T Tavg =Tamb+ Tce

ttc where Tcand tcare, respectively, a characteristictemperature and a characteristic time. (e meansisasymp-totical to or approaches.) We plot ln(T Tamb) using thedatafrom the table. en approximate the tail linearly (valuing theend of it most) to getln[(T Tamb)/C] 4.89 t300 s .erefore Tc e4.89C 133 C. On the other hand,plugging t = 0 into our exponential function shows thatTavg,0 Tamb= Tcand, finally, Q= C Tc 46700J.

Actually, quite a good result can be obtained without re-

ploing anything, by just considering the last three datapointsof the table. DenoteTi Ti Tamb. If the timest3t2 = t2 t1, then with an exponential we should observethat T3/T2 = T2/T1. e last three timepoints aregood indeed, so we checkT1 = 4.4

C, T2 = 2.3C

andT3 = 1.2C. eir ratios areT3/T2 0.522and

T2/T1 0.523, a splendid match. is confirms the ex-ponential tail. As in every equal time interval the T is mul-tiplied by the same number (that is the essence of exponentials),

Tc = Tavg,0 = T3T2T3

t3t3t2 114 C. From this,

Q39 900 J. is is discrepant from our previous calculation,

. Coefficient of refraction ( points)i) We direct the laser beam radially into the semi-cylinder: per-pendicularly through its cylindrical surface. e beam enters theplate without refraction and reaches the opposing flat face at the

axis of thecylinder. Dependingontheanglebetween that face andthe beam, there may or may not be a refracting beam, but there isalways a reflecting (from the flat face) beam . We rotate the semi-cylinder around its axis to find the position, when the refracting

beam appears/disappears; the angle between the flat face andthe incident beam correspond to the angle of complete internalreflection, i.e. n = 1/ cos . We can measurecos using thegraph paper: we draw the beam as a segment AO and the flat faceof the semi-cylinder as a lineBOso that ABO = /2; then,n =|AO|/|BO|. e uncertainty can be found using the for-mulan = n(|AO|

|AO| + |BO|

|BO| )and by estimating the uncer-

tainties of the direct length measurements |AO| and |BO|.ii)Wedrop the liquid on the prism and press it againstthe flat faceof the semi-cylindrical plate. Further we study the complete in-ternal reflection at the boundary between the semi-cylinder andprism (which is filled with the liquid) by repeating the abovedescribed experiment. ereby we measure new lengthsAO

andB O; the condition of complete internal reflection is nown/nl = |AO|/|BO| nl = n|BO|/|AO|, wherenlstands for the coefficient of refraction of the liquid. e uncer-

tainty is now calculated as nl= nl(|AO||AO|

+ |BO|

|BO| + n

n ).


46/57



Estonian-Finnish Olympiad 2013

1. PRISM(8 points)

The focal length of its camera f = 4.3mm

and the diameter of the lensD = 1.8mm. The

sensor is w= 4.6mmwide corresponding to

N 3264 i l

been bent into the shape of a square with

side length a. At some point on the wire is

a small ideal current source that keeps cur-

I fl i i h i i i ll i i


47/57

i) (4 points) A right prism that has an

equilateral triangular base with length a is

placed in a horizontal slit between two tables,

so that one of the side faces is vertical. Howsmall can the width d of the slit be made be-

fore the prism falls out of the slit? There is no

friction between the prism and the tables and

the prism is made of a homogeneous material.

The edges of the slit are parallel.

ii)(4 points)Now the prism is placed in the

slit so that one of its side faces is horizontal.

How small can the width l of the slit be made

before that position becomes unstable?

i)

a

d

ii)

l

2. CELLPHONE CAMERA(6 points)A pho-tographer focussed his camera to distance L

and took a photo. On the photo, all farther

objects (up to infinity) turned also out to be

sharp. Additionally, all closer objects down

to distance s were sharp.

i) (4 points) What is the minimum pos-

sibleL?

ii)(2 points)Find the correspondings.

Background. We consider the image of

a pointlike object to be sharp if its image is

smaller than one pixel on the sensor. Oth-

erwise the image is blurry. The lens of the

camera may be viewed as a convex lens. The

camera is focussed by changing the distance

between the sensor and the lens.

Parameters. Calculate the answer for a

cellphone made by a well-known company.

N=3264 pixels.

3. MISSION TO MARS (7 points) A crew ofastronauts is going to be sent to explore the

polar region of Mars and search for buried

water ice. Their spaceship will travel from

Earth to Mars along an elliptic transfer orbit

tangential to the orbits of both planets. Des-

pite its shortcomings, this orbit is commonly

used in space travel due to its relatively good

fuel economy. Future manned missions to

Mars are very likely based on this kind of

transfer. In this problem you will examine

some aspects of this orbit.The mean orbital radius of Mars is Ra =

1.52AU. The mean orbital radius of Earth

is Rg = 1 AU = 149600000km. Mars has a

mean radius of ra = 3397km and surface

gravity ga =3.71m/s2. Earth has a mean

radius ofrg = 6371km.

i) (1 point) Find the orbital period Ta of

Mars, i.e. find the length of Martian yearin Earth years.

ii)(1.5 points)How long (tt) does a one-way

trip to Mars take?

iii) (1.5 points)The spaceship is put into this

orbit by using a powerful rocket. It is more

efficient to burn fuel as a short burst when

the spaceship is still near Earth. How much

additional speed (v1) does the booster haveto be able to give to the spaceship to enter the

transfer orbit, starting from the north pole?

Neglect the air resistance.

iv)(1.5 points) Estimate the v2 needed to

enter a circular orbit close to Mars.

v)(1.5 points)What is the minimal duration

of the trip to Mars and back?

4. MAGNETIC DIPOLES (7 points) Let usconsider the following model for a magnetic

dipole. Some wire with no resistance has

rentIflowing in the circuit in all situations.

The magnetic momentm of a planar circuit

is given by the relation m =IA, with vector

mpointing in the normal direction of the cir-

cuit according to the right hand rule (Ais thearea bounded by the circuit).

i) (3 points) The dipole is placed inside a

homogeneous magnetic field B, so that the

angle between mand Bis . Find the angles

s andu that correspond to stable and un-

stable equilibria, respectively. Calculate the

amount of work (w) needed to rotate the di-

pole from s to u. Give your answer in termsofm and B.

We can use this model to calculate the

magnetic properties of materials containing

unpaired electrons that have negligibly weak

interactions with one another. Let us con-

sider a sample of material with n such un-

paired electrons per unit volume, placed in-

side a homogeneous magnetic field B. Due

to spin, each unpaired electron acts as a

small magnetic dipole. However, owing to the

quantum nature the electron, the projection

of its magnetic moment alongBcan only be

B or B (B is called the Bohr magneton).

ii)(4 points) Calculate M, the magnetic mo-

ment per unit volume of the sample, if the

temperature of the material is Tand the ex-

ternal magnetic field is B.

5. FRICTION OF A STRING (8 points)Measure the dynamic coefficient of friction

1 between the ballpoint pen and the string.

Estimate the uncertainty. It might help that

the dynamic coefficient of friction between

the pencil and the same string was measured

beforehand and 2 = 0.200.01 was obtained.

Equipment: dynamometer, string, ball-point pen, pencil and weight.


6. SPHERE AND CYLINDER (7 points) Asphere and a cylinder are lying on an inclined

8. ZENER DIODE (7 points) An inductanceLand a capacitor C are connected in series

with a switch. Initially the switch is open

d th it i i h N

come to a halt.

iv) (2 points) Find the decrease q in the

maximum positive value of the capacitors

10. RESISTIVE HEATING (8 points) Meas-ure the resistor. You are not asked to estim-

ate the uncertainty.


48/57

sphere and a cylinder are lying on an inclined

surface with inclination angle. Both have

massm and radiusr. The bodies are released

from equal initial heightsH. The moments of

inertia of the sphere and the cylinder are, re-spectively, Isph =

25mr2 and Icyl =

12mr2. The

coefficient of friction between the surface and

the bodies is.

i)(2 points)Which of the bodies comes down

faster? What was the relative lag of the

slower body= (t2 t1)/t1? The times t1 and

t2, respectively, denote the traveling times of

the faster or the slower body. Assume that

the rolling occurs without slipping.

ii) (2.5 points) Find the minimal angle of

inclination0 for which the cylinder starts

to slide in addition to rolling.

iii) (2.5 points) If 90, the bodies obvi-

ously lose contact with the surface and fall

down in free fall with equal times. What

is the minimal angle of inclination m

, for

which both the sphere and the cylinder come

down with equal times?

7. BURNING WITH A LENS (7 points) Sun-rays are focused with a lens of diameter

d = 10cm and focal length of f = 7 cm to a

black thin plate. Behind the plate is a mirror.

Angular diameter of the Sun is = 32 and

its intensity on the surface of the Earth is

I= 1000W/m2, Stefan-Boltzmann constant= 5.670108 W/(m2K4).

i) (4 points) Find the temperature of the

heated point of the plate.

ii) (3 points) Using thermodynamic argu-

ments, estimate the maximal diameter of the

lens for which this model can be used.

and the capacitor is given a charge q0. Now

the switch is closed.

i) (1 point) What are the charge q on the

capacitor and the current Iin the circuit as

functions of time? Draw the phase diagram

of the system the evolution of the system

on a Iqgraph and note the curves para-

meters. Note the direction of the systems

evolution with arrow(s).

AZener diodeis a non-linear circuit ele-

ment that acts as a bi-directional diode: it

allows the current to flow in the positive dir-

ection when a forward voltage on it exceedsa certain threshold value, but it also allows

a current to flow in the opposite direction

when exposed to sufficiently large negative

voltage. Normally the two voltage scales are

quite different, but for our purposes we will

take a Zener diode with the following volt-

ampere characteristics: for forward currents,

the voltage on the diode is Vd , for reverse

currents, the voltage on the diode is Vd ,

for zero current the voltage on the diode is

Vd


49/57

1. PRISM(8 points)

i)(4 points)The prism is acted on by three

forces: reaction force Rl from the left-hand

table, directed perpendicularly to the prisms

face; reaction force Rr from the right-hand

table, directed horizontally (with its point

of action to be determined yet); and gravit-

ational forcemg, directed vertically and ap-

plied at the triangles centre. (Considering a

planar triangular cross-section of the prism

is enough.)

On the verge of falling out, the force Rr

is applied at the lower corner of the triangle.If a body in equilibrium is acted on by three

forces, then their lines of action must inter-

sect at one point. This is because otherwise

the torque of one of those forces would not be

zero with respect to the intersection of the

lines of action of the two other forces.

As the distance between the triangles

centre and its side is

3

6 a, the distancebetween the points of action ofRl and Rr is

36 acos30 = 1

4a. Thus, d = 1

4acos30 =

3

8 a.

a

d

mg Rl

Rr

3

6 a

1

4a

sider a small rotation of the prism (assum-

ing it remains in contact with the corners).

The trajectory of the tip is a circle ascribed

around the triangle ABC (it follows from the

property of the inscribed angles because theACBremains equal to 60). The radius ofthat circle r = l/3; its centre will be denotedbyO. Once the prism rotates by angle, so

that the new position of the tip will be D , the

central angleCOD = 2. Hence, the tip israised by r r cos(2) 2r2. The height ofthe centre of mass P of the prism is raised

because the tip is raised, and lowered be-

cause the vertical projection of the segment

CP is reduced by|CP |(1 cos) |CP |2/2.Here,|CP | = a/

3. So, the original position

is stable if a32/2 < 2 l

32, hencel > 1

4a.

a

l

A B

CD

O

P

2

P

2. CELLPHONE CAMERA(6 points) ThedistanceL is often called hyperfocal distance

in photography and it was calculated more

than one hundred years ago by Louis Derr

(the figure is taken from his book Photo-

graphy for students of physics and chemistry,

published in 1906).

Lets consider that the camera is focused

to distanceL and the image is formed exactly

on the sensors plane. The objects distance L

and its images distance a (corresponds to p

on the figure) are related by the lens formula1L

+ 1a= 1

f, thus

a= LfLf=

Lf

L(1f/L)L f

L

1 + f

L

=f+f2

L ,

where the approximation (1 +x)1 1 x(forsmall x) was used. Images distance exceeds

the focal length by a= af= f2/L.i) (4 points) The light coming from an in-

finitely far away object will pass the focalpoint F and form a cone which is cut by the

sensors plane. The diameter d of the cut on

the sensors plane can be found from similar

trianglesd/D =a/f, thusd =D f/L. Takinginto account the sharpness conditiond ,where =w/Nis the size of a single elementof the sensor, we find that the limiting value

ofL is L

=D f/

=D f N/w

5.5 m.

ii)(2 points)Well now find the shortest dis-

tance s satisfying the sharpness condition.

Object at distance s will have an image at

distance b=f+f2/s and the light passing thelens will converge behind the sensors plane

forming a cone. The diameter d2 of the cones

cut with the sensors plane can be calculated

from similar triangles: d2/D = (ba)/b. Ac-

counting for sharpness conditiond2 = , wecan express b = a/(1/D), and substituting

fs 1/D f 1 /D f

D

f

1+ 2D

.

Finally, f2/s = 2f/D, or s = 12D f/ = L/2

2.75 m.

3. MISSION TO MARS(7 points)

i) (1 point) We can find the orbital period

of Mars from Keplers third law Ra3/Rg

3 =Ta

2/Tg2, givingt t 1.87yr.

ii) (1.5 points) Again, we can use Keplers

third law to calculate half of the orbitalperiod.

tt =Tg

2

(Ra +Rg)3/2(2Rg)

3/2 0.707yr.

iii) (1.5 points) Background. v is import-

ant, because the sum of all v determines

how much fuel is needed for a given mission.

The fuel needed is exponential of total v andis described by Tsiolkovsky rocket equation.

Kinetic energy per unit mass of such a

transfer orbit where it intersects the Earths

orbit is GMsRg+Ra +

GMsRg

. Using the orbital

angular speed of Earth we can substitute

GMs = 42R3gTg

. The speed at the beginning of

the transfer orbit becomes

vt0=

2GMs

1

Rg 1Rg +Ra

32.7km/s

. The speed in Earths inertial frame is

vt0=vt0 vg 2.94km/h. To achive that, wefirst need to escape Earths gravity, so

v1 =2vt0

2

2 +GMg

rg

.

1

Using the surface gravity of Earth we can

substitute GMgrg

= ggrg so v1 11.2km/s.iv) (1.5 points) We can calculate the speed

u, we can again keep two sides perpendicu-

lar to B - the answer cannot depend on the

path, so we choose the simplest one. Integ-

rating = Bm sin from = 0 to gives us

dial component ofTd2

(where is measured

in radians). Therefore, we get a differen-

tial equation: dT= T d or d lnT= d,whence T =T0e

stant accelerationa directed parallel to thesurface. Lets express a from the equationv2 = 2ax:


50/57

of the transfer orbit where it intersects the

orbit of Mars from Keplers second lawv t1=vt0/1.52 21.5km/s. The speed of the space-craft relative to Mars isvt1 3.25km/s. Thespeed of the spacecraft once near Mars sur-

face is

vtm=

2vt1

2

2 + raga

5.98km/s

Since the speed of low Mars orbit is vea=raga 3.55km/s, we need to brake for

v2 2.43km/s.

v)(1.5 points) The EarthSunMars angle

at the launch of the mission needs to

be = watt 0.77 for the spacecraft toreach Mars. Likewise for the return trip

= wg tt 1.301. If we go to the coro-tating frame of referense with earth, we can

see that the minimal time between those two

angles is 2+wa

wg

1.96yr. The minimal dura-

tion of the trip is therefore longer by tt, giving

2.67yr.

4. MAGNETIC DIPOLES(7 points)

i) (3 points)There is no torque on the square

if =0 or =, so one of them is stableand the other unstable. If we start from

= 0 and turn the square to some , butkeep two sides of the square perpendicular

to B, Lorentz forces on these two sides give a

torque = BI a asin = Bm sin towardsdecreasing. By symmetry, we get the same

result if we keep the other keep two sides

of the square perpendicular to B. It is pos-

sible to conclude that the torque depends

only on (at least near =0), not on theexact orientation of the square. Since torque

acts to restore = 0, we find thats = 0 andu= . To find the work to get from s to

rating =Bm sin from = 0 to gives usw= 2Bm.ii) (4 points) Let us denote the number

of electrons (per unit volume) with mag-

netic moment projection +B as n+ and theones withB as n. Their sum is alwaysthe same, n+ + n= n. Also, in thermalequilibrium, their ratio is given by

nn+ =

exp 2BB

kBT

, where kB is Boltzmanns con-

stant. Solving the equations, we can find

n+ and n. The total magnetic moment perunit volume (in the direction ofB) is given by

M=B(n+ n). After substituting,

M=BN1exp

2BBkBT

1+exp

2BB

kBT

=BntanhBB

kBT

.

Additional comments. We see B and M

always have the same sign, therefore M is

parallel with B. This makes sense, as we

saw that = 0 (m parallel to B) orientationhad lowest energy. The graph ofMvs B goes

to Bn for very large B or toBn for verysmallB (all spins aligned with B). AtB = 0,M= 0 as well, since both spin orientationshave the same energy. Around zero, the curve

is linear, as tanhx x for small x gives usM

2BnB

kBT .

5. FRICTION OF A STRING (8 points)Lets first calculate the difference of tension

forceTbetween two ends of a sliding string

arced over a cylinder by an angle . Further-

more, lets look at a short piece of the arc

that subtends an angle d. On one hand,

dT=dR is the friction force acting on thepiece, wheredR is the reaction force. On the

other hand, dR Td, because both endsof the piece are pulled by a force with a ra-

whenceT=T0e .As a solution to the problem, we can meas-

ure the change of the tension force for differ-

ent angles (for example,

2, , 3

2 , 2, 5

2etc. for several turns; however, keeping the

strings vertical offers better precision) and

plotlnTwith respect to . The slope of the

graph is the to be measured.

Extra solution (not as exact). Those who

cannot derive the necessary formula can still

do the experiment by doing the same meas-

urements and noting from the plot that the

relationship between and T looks expo-nential. Thus, we can make an ansatz that

T= T0X: as = 0 must correspond toT= T0, we cannot reasonably write the anywhere else without over-complicating the

formula. Now, we can re-measure the given

pencil (it may be reasonably enough approx-

imated with a cylinder here; more exact ap-

proaches exist) and conclude that X

2.7.

From there on, the calculation is the same.

6. SPHERE AND CYLINDER (7 points)

i) (2 points) Since no energy is lost due to

friction on sliding, the change in potential

energy Ep =mgHis transformed to kineticenergy consisting of both translational and

rotational motion. Taking into account the

rolling conditionv

=r, we have

Ep =Ek =1

2mv2 + 1

2I2

= 12mv2 + 1

2kmv2 =1

2(1+k)mv2,

where general expression I= kmr2 for mo-ment of inertia is used. Therefore, v2 =2gH/(1+k).

On the other hand, the bodies travel dis-tancex =H/sinalong the slope with a con-

a = gsin/(1+k).The times are now easy to calculate as t =v/a

, giving

t=

1+k

2H

gsin2

12

.

Replacingks= 25 for sphere and kc= 12 forcylinder, we find that the sphere is faster by

a relative factor

=

1+kc1

+ks

1 =15

14 1 0.035.

ii)(2.5 points)As found in previous subpart,

the accelerations parallel component to the

slopea is smaller than the contribution bygravity gsin. The difference is contributed

by the friction force Ff=mgsinma. Slid-ing starts, if the necessary friction reaches

the maximal value Fmax= N= mgcos.Equating the two expressions gives

mgsinmgsin/(1+k) =mgcos,

tan=1+kk

.

For the cylinder the limiting angle is 0=arctan(3).

iii)(2.5 points)When the maximal friction

force is reached, the motion goes into rolling

and sliding mode, where the total force com-ponent along the surface is given by the dif-

ference of gravity and friction:

F =mgsinFmax =mgsinmgcos.We note that the acceleration in this mode

does not depend on the moment of inertia any

more.

Calculating the limiting angle of slippingmode also for the sphere sph = arctan( 72) >

2

0 shows that for all angles larger than

m= sph both bodies are in the slippingmode and thus have equal accelerations and

arrival times.

the frequency = 1LC

and we can immedi-

ately write q(t) = q0 cost, while I(t) = q(t) =q0 sint.

N t th t

summarize the equations as follows:

Lq+ qC

=Vd ifq< 0q

(for I= 0). If a trajectory reaches any ofthe points in that segment, it will stay there

forever. The extent of that region is 2CVd.

i ) (2 i t ) L t th h di g


51/57

arrival times.

7. BURNING WITH A LENS(7 points) Thesolar energy flux which is focused by the lens

to the image of the Sun can be calculated

as P=4d2I; the image of the Sun radiates

according to the Stefan-Boltzmann law with

the total power P= 4

(f)2T4. From the

heat balance we obtain 4d2I=

4(f)2T4,

hence

T=

df

I

4500K.

Due to the second law of thermodynamics,it is impossible to direct heat energy from a

lower temperature body to a higher temper-

ature body. Hence, the image temperature

cannot exceed the temperature of the Sun.

Now we can use the known temperature of

Sun T0= 6000K, but it is better to use theStefan-Boltzmann law for solar radiation flux

density: near the Suns surface, I0=T40 ,

with the total flux of Pt= 4R2sI0. Nearthe Earth, the total flux Pt= 4L2I; here,Rs is the Suns radius, and L the orbital

radius of the Earth. From here we obtain

I=I0R2s/L2 =T40R2s/L2; using the previousresult,

T= T0

d

f

Rs

L .

Let us note that L= 2Rs, hence

T= T0d

2f T0,

which means that d 2f.8. ZENER DIODE(7 points)

i) (1 point) Kirchoffs 2nd law gives L I+q/C

=0 or q

+ 1

LCq=

0. This is the equa-

tion of a simple harmonic oscillator with

Note that

q2 + 12

I2 = q20(sin2t+ cos2t) = q20,

and therefore the phase diagram of the sys-

tem is an ellipse centred at the origin, with

semi-axes q0 andq0. Alternatively, this re-

lation comes directly from the conservation

of energy:

LI2

2 + q

2

2C=E0 =

q20

2C.

By looking at q and I a quarter-period

later fromt = 0, say, its not hard to see thatthe system must evolve in a clockwise sense

on the phase diagram. Note that in this in-

stance, only q= 0 is an equilibrium point:for all non-zeroq there will be never-ending

oscillations in the circuit.

ii)(2 points)Now the sign of the voltage on

the diode depends on the direction of the cur-

rent, giving either ofL q+ qC

Vd = 0. We can

Lq+ qC

= Vd ifq> 0

Let us introduce the new variables q1,2

such that q1= qCVd and q2= q+CVd .Then we can rewrite the two equations above

in a more familiar form:

Lq1 +q1

C= 0 ifq< 0

Lq2 +q2

C= 0 ifq> 0

Thus the introduction of the diode only serves

to shift the equilibrium points for the oth-erwise simple harmonic orbits. For q>0,the equilibrium point is q2 = 0 or q= CVd ,while for q< 0 it is q =CVd . So the orbit willconsist of half-ellipses in the upper and the

lower parts of the I qdiagram, centred atq= CVd for the upper half and at q=CVdfor the lower half. As the evolution is continu-

ous, these half-ellipses will join up at I

=0.

iii) (2 points) We can see on the diagram

that there is a dead zone betweenCVd

iv) (2 points) Lets use the phase diagram

to figure this out. Suppose the capacitor

initially has the charge q0 CVd. Thenthe charge will first swing to the other way

ofCVd and will become qT/2=CVd (q0 CVd) = 2CVd q0. Then it will perform theother half-oscillation aroundCVd and thecharge at the end of that is qT= CVd +(CVd (2CVd q0)) = q0 4CVd , and there-fore q= 4CVd.

Note that we have the right to talk about

half- and full periods because the oscilla-

tions still happen at the immutable frequency= 1

LC. Therefore the time between the

two maxima is just a full period of oscillation,

T= 2

.

Once q(t) has a zero derivative inside the

region bounded byCVd, it will remain atthat particular value forever. For a large ini-

tial q0, we expect there to be approximately q0q = |q0|4CVd total oscillations.More exactly, the distance from the dead

zone is initially |q0|CVd and decreases dur-ing each half-oscillation by 2CVd. The total

number of half-oscillations is N= |q0|CVd

2CVd

and the total time t =NT

2= N

=N

LC.

9. GLASS CYLINDER (7 points) The axisof the half-cylinder is where the stripe and

its image coincide (form a straight line). The

front edge of the half-cylinder is at the 28th

line, counting from the axis, hence the radius

of the cylinder R 28. Let us consider therefracting ray s which is very close to a total

internal reflection. One can see the images

of 20 lines (ca 20.2, to be more precise), when

counting from the central line upwards; the

upper edge of the half-cylinder coincides with

3

the 42nd line at the background.

So, the ray s arrives at the camera at

the angle = arcsin(28/42) 41.8 with re-spect to the plane of the paper The pro

the perpendicular of the paper surface. The

incidence angle of the ray s is + 45.9,hence n = 1/sin(+) 1.39.10 RESISTIVE HEATING (8 points) After

Measure the temperature at the end, after

waiting a bit or stirring the calorimeter. We

want to get maximum temperature difference

for precise measurement. SinceP =RI2,

The resistance used was R = 0.475%.In the described circuit the batteries were

depleted in 10 to 15 minutes and the tem-

perature of the calorimeter rose by 7 to 10


52/57

spect to the plane of the paper. The pro-

jection of the refraction point to the paper

surface lies at the distance a= 28 sin=282/42

18.7 lines from the axis. There-

fore, before refraction, ray s forms an angle

= arcsin[(20.2 18.7)/(28 cos)] 4.1 with

10. RESISTIVE HEATING(8 points) Afternoting the temperature of the calorimeter,

connect the batteries, resistor (in the calor-

imeter) and ammeter in series. Choose a

convenient time interval t and note the am-

meter reading until batteries are depleted.

p ,

Q

Rn

t(In/2+In+1/2)2,

R = (cama+ cwmw)Tnt(In/2+In+1/2)2

.

p y

degrees.

4


53/57




1. DC-DC CONVERTER(8 points) In orderto obtain high voltage supply using a battery,

iv) (2 points) Assuming that Vmax V0,what is the average power dissipation on the

diode?

v) (2 points) Now, let the key K2 be closed,

Sun?

iii) (1.5 points) What is the speed of the

spaceship in the Earths frame of reference

when the distance from the Earth is much

that ball fly compared to the initial drop

heighth? Take f= 0.5 andn = 10.It maybe useful that that sequencea0 = 1,

ak+1 = ak+ has a general term an =n+


54/57

g g pp y g y,

the following circuit is used.

An electromagnetic switch K1 connects a bat-tery of electromotive force Eto an inductor

of inductance L: it is closed if there is no cur-

rent in the inductor (a spring keeps it closed),

but if the inductor current reaches a critical

value I0, magnetic field created by the in-

ductor pulls it open. Due to inertia, once the

key is open, it takes a certain time Kto close

again even if the current falls to zero.

For the diode D you may assume that

its current is zero for any reverse voltage

(VD


55/57

Assume as a simplification that the reaction

force between the boy and the ice stays con-

stant (in reality it varies with every push,

but the assumption is justified by the factthat the value averaged over one step stays

constant).

i)(2 points)What is the minimum time ne-

cessary for him to change his moving direc-

tion to point towards the east so that the final

speed is also v = 5m/s?ii) (2 points)What is the shape of the optimal

trajectory called?

7. SPIN SYSTEM (8 points) Let us considera system ofN independent magnetic dipoles

(spins) in a magnetic fieldB and temperature

T. Our goal is to determine some properties

of this system by using statistical physics. It

is known that the energy of a single spin is

E= m, wherem =12

and =B.

i) (2 points)What is the probability p fora spin to be in exited state, i.e. have positive

energy?

ii) (2 points) What is the average value of

the total energy Es of the spin system as a

function ofB and T?

iii) (2 points) Using high temperature ap-

proximationT Bmk

, simplify the expres-

sion ofEs.

iv) (2 points) Using high temperature ap-

proximationT Bmk

, find the heat capacity

Cof the spin system.

8. MIRROR INTERFERENCE (5 points)A point source S emits coherent light of

wavelength isotropically in all directions;

thus, the wavefronts are concentric spheres.

The waves reflect from a dielectric surface

point source (see figure).

In what follows we use thex, y, and zco-

ordinates as defined in the figure. The screen

is parallel to the mirror and lies in the yz-

plane.

i) (2 points) At which values of the y-

coordinate (for z= 0) are the interference

maxima observed on the screen? You mayassume that yL.ii) (1 point)Sketch the shape of few smallest-

sized interference maxima on the screen (in

yz-plane).iii) (2 points)Now the flat screen is replaced

with a spherical screen of radius L, centred

around the point source. How many interfer-

ence maxima can be observed?9. THERMAL ACCELERATION (9 points)Consider a cube of side lengtha = 1 cm, madeof aluminium (density = 2.7g/cm3, molarmass MA=23g/mol). The heat capacitanceof one mole of aluminium is given as a func-

tion of temperature in the graph below. The

speed of light c = 3108 m/s, universal gasconstant R

=8.31J/(kgK). The initial tem-

perature of the cube is T0 = 300K.

i)(1 point)What is the total heat energy of

such a cube at the initial temperature T0?

ii) (3 points) Now, the cube has 5

faces painted in white (reflects all relevant

wavelengths) and one face painted in black

(absorbs all these waves). The cube is sur-rounded by vacuum at a very low temperat-

ure (near absolute zero); there is no gravity

field. Initially, the cube is at rest; as it cools

down due to heat radiation, it starts slowly

moving. Estimate its terminal speed v1.

iii) (2 points) At very low temperatures,

the heat capacitance of aluminium is pro-

portional to T3, where T is its temperat-

ure. Which functional dependance f(t) de-

scribes the temperature as a function of time

[T= A f(Bt), where Aand B are constants]

for such very low temperatures under the

assumptions of the previous question?

iv) (3 points) Now, the cube has 5 faces

covered with a thermal insulation layer (you

may neglect heat transfer through these

faces). One face is left uncovered. The cubeis surrounded by hydrogen atmosphere at

a very low temperature (molar mass of hy-

drogen molecules MH= 2g/mol). The cubestarts cooling down due to heat transfer to

the surrounding gas; you may neglect the

heat radiation. Initially, the cube is at rest;

as it cools down, it starts slowly moving. Es-

timate the order of magnitude of its terminal

speedv2. Assume that the surrounding gas

10. YOUNGS MODULUS OF RUBBER(12 points) The linear Hookes law for a

rope made from an elastic material is sup-

posed to held for small relative deformations

=x/L(which is also called strain), whereLis the undeformed length of the rope, and x

is the deformation. Once becomes too large,

the force-deformation relationshipF=kx isno longer linear; what is too large depends

on the material. For very elastic materials

which can reach relative deformations con-

siderably large than one, it may happen thatthe linear Hookes law with a constant stiff-

ness k fails, but if we take into account the

change of the cross-sectional area S of the

rope with k =ES/L, where E is the Youngsmodulus of the elastic material, such a non-

linear Hookes law remains valid. In that

case we can say that there is still a linear

stress-strain relationship =E , where thestress=F/S.i) (7 points) Measure the relationship

between the stress and strain in a rubber

string and plot it.

ii)(5 points)From your plot determine the

Youngs modulus E with its uncertainty, and

the maximum strain m until which it ap-

plies.

Note: the diameter of the thread is to bemeasured using the diffraction of laser light.

Equipment: rubber thread, stand, meas-

uring tape, 15 hex nuts with a known mass,

a plastic bag for hanging a set of nuts to the

thread, a green laser (=532nm), a screen.

WARNING: AVOID LOOKING INTO

A LASER BEAM, THIS MAY DAMAGE

YOUR EYES!

L

E

E =L dIdt I= Et/L I0 = EL/L

L = LI0/E.I0

V0 Vmax

i

LC

i = I0

i= 0

LC

1LI2

RE/2

23/2

T t t= 25/2T 64.6 days

F

=F

(d)d= kn

dn+1x,

d= x

F =

kn

dn+1 mg

l

x.


56/57

I0

L

R L R

L/R K

I0 Vmax= RI0

Vmax V0

L dIdt = RI = R dqdt q

I0

0

LI0= Rq

q= I0L/R

V0 A = V0q

P = A

L=

V0I0L

RL=

V0ER

.

RC

K1 LC

TLC= 2LC

RC TLC

2LI0

L

Q= V2avL/R

V2avLR

= 1

2LI20 Vav =I0

LR

2L=

EI0R

2 .

qC = LVav/R RC

LC

V = qC/C = LVav/(RC)

U0= V

2 =

LVav2RC

= I0L

2C

I02RE.

E = GMm2a

M

m

a

a

2a= RE+ rS RE rS

v0= 29.8 km/s vS

GMm2a

= GM mRE+ rS

= GMmRe

+mv2S

2 ,

vS =

GM

RE

2rSRE+ rS

.

v2

0 = GM

RE

vS =v0

2rS

RE+ rS=v0

2sin

2

v0

.

vS 2.8 km/s

vE = v0vS27.0 km/s

=

GMEm

R = gmR

gR + v2E2

= u2

2

ME u

u=

v2E+ 2gR 29.2 km/s.

mg

T

F

T

mg

(x/l)mg

l

x= 1 cm

F =F

(x/l)mg

F = 0

x

F = F x

l mg

F

= kdn

k

F = 0

n

k

k

dn xmg

l = 0,

kn

dn+1 mg

l = 0;

k

dn =

xmg

l ,

kn

dn+1 =

mg

l .

d/n= x

n= d/x = 4.

v0 = v k

vk

(k + 1)

v

u = vkfv1+f

vk+1 =

v+ 2vkfv

1+f = 2

1+fvk+ 1f

1+fv

v0 = v

v1 = 3f1+f

v = 4

1+f 1

v

n vn =

2 2

1+f

n 1 v

v2 = 2gh0 v2n= 2ghn

hn/h0= vn/v0 =

2 2

1 + fn

1.

f= 0.5

n= 10

940 nm

620nm

590 nm

525nm

470nm

450nm

I

R

Vd = E IR

Ep Ep =hc/(e)

IR =E Vd

IR

1/

E N 2/4kT

C= dEdT =N 2/4kT2

p Ec


57/57

IR= E 1

hc

e .

A = hc/e h =eA/c

A

A

A= 12 (Amax Amin ) h= hA/A

vx vy

A

(0, v)

B

(v, 0)

vx vy

g

v

2

t= v

2/g 7.2 s

p =

A em

A

Ae/2 + A e/2 = 1

A= 1

e/2kT + e/2kT =

1

2 cosh(/2kT).

p = e/2kT

e/2kT + e/2kT.

E= N

2

e/2kT e/2kTe/2kT + e/2kT

= N 2

tanh(/2kT).

y

= y/L

= 2l cos 2N N 2

= 2/ = 4N (2N 2 1)

2(2N n) n

=

n + 0.5N yn = Ln + 0.5N ,

n= 0, 1, . . . N

n

x

n

1 = 1

3 1.73

5 2.23

max = 4N+ min =

m= (max min )/2= 2N.

dq= CvdT T = 0

q =

T0 CvdT

q R560J/K

= a3/MA 0.117mol

Q= q 546J

E= h

p= h/ = h/c= E /c

p = E

c cos

cos

p = E

c

1

2 cos2 d

d = 2 sin d p =

Ec

/2

0sin cos2 d = Ec

/2

0cos

2 d cos = E3c

1/c

Q/c

a3v Q/c

v Qa3c

0.67mm/s.

1

3

v

0.22mm/s

AT3dT =ST4dt

A

S

dT

T = Bt T =A eBt.

1/vT

vT =

RT/MH

a3v Q/vT

v Qa3

MHRT

180m/s.

T4

a

n

n= 10

n/d = a/L L d

d= nL/a.

b

= b b0b0

,

b0

= 4N mg

d2 ,

N

m

estonian finnish physics olympiad (2003-2014)

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