# Problems of the 2 nd and 9 th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)

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2 nd IPhO (Budapest 968) Theoetical poblems Poblem On an inclined plane of 30 a block mass m = 4 kg is joined by a light cod to a solid cylinde mass m = 8 kg adius = 5 cm (Fig ) Find the acceleation if the bodies ae eleased The coefficient of fiction between the block and the inclined plane µ = 0 Fiction at the beaing and olling fiction ae negligible m m F F m gsin µ m gcos m gsin S Figue Figue If the cod is stessed the cylinde and the block ae moving with the same acceleation a Let F be the tension in the cod S the fictional foce between the cylinde and the inclined plane (Fig ) The angula acceleation of the cylinde is a/ The net foce causing the acceleation of the block: m a = mg sin µ mg cos F and the net foce causing the acceleation of the cylinde: m a = m g sin S F The equation of motion fo the otation of the cylinde: S a = I (I is the moment of inetia of the cylinde S is the toque of the fictional foce) Solving the system of equations we get: a = g I S = ( m m ) g sin µ cos m m ( m m ) m I sin µ m I m m () cos ()

3 I I sin µ m cos F = mg (3) I m m The moment of inetia of a solid cylinde is m I = Using the given numeical values: ( m m ) sin µ m cos a = g = 0 337g = 35 m s 5m m m g ( m m ) sin µ m cos S = = 30 N 5m m ( 5µ cos 05sin ) m F = mg = 09 N 5m m Discussion (See Fig 3) The condition fo the system to stat moving is a > 0 Inseting a = 0 into () we obtain the limit fo angle : m tan = µ m m µ = 3 = = 3 8 Fo the cylinde sepaately = 0 and fo the block sepaately = tan = µ 3 If the cod is not stetched the bodies move sepaately We obtain the limit by inseting F = 0 into (3): m tan = µ = 3 µ = I The condition fo the cylinde to slip is that the value of S (calculated fom () taking the same coefficient of fiction) exceeds the value of µ m g cos This gives the same value fo 3 as we had fo The acceleation of the centes of the cylinde and the block is the same: g ( sin µ cos ) the fictional foce at the bottom of the cylinde is µ m g cos the peipheal acceleation of the cylinde is m µ g cos I Poblem 06 = Thee ae 300 cm 3 toluene of 0 C tempeatue in a glass and 0 cm 3 toluene of 00 C tempeatue in anothe glass (The sum of the volumes is 40 cm 3 ) Find the final volume afte the two liquids ae mixed The coefficient of volume expansion of toluene β = 000 C Neglect the loss of heat ( ) β a g = 3 F S a β F S (N) Figue

4 If the volume at tempeatue t is V then the volume at tempeatue 0 C is V0 = V ( β t ) In the same way if the volume at t tempeatue is V at 0 C we have V0 = V ( β t ) Futhemoe if the density of the liquid at 0 C is d then the masses ae m = V0d and m = V0d espectively Afte mixing the liquids the tempeatue is t m t m = m t m The volumes at this tempeatue ae V0 ( β t) and V ( β t) The sum of the volumes afte mixing: V 0 = V = V = V 0 ( β t) V ( β t) = V V β ( V V ) V V m m mt mt β = d m m mt mt β = V0 βv d d ( β t ) V0 ( β t ) = V V t V 0 t = βv The sum of the volumes is constant In ou case it is 40 cm 3 The esult is valid fo any numbe of quantities of toluene as the mixing can be done successively adding always one moe glass of liquid to the mixtue Poblem 3 Paallel light ays ae falling on the plane suface of a semi-cylinde made of glass at an angle of 45 in such a plane which is pependicula to the axis of the semi-cylinde (Fig 4) (Index of efaction is ) Whee ae the ays emeging out of the cylindical suface? t 0 = A O ϕ β D E B C Figue 4 Figue 5 Let us use angle ϕ to descibe the position of the ays in the glass (Fig 5) Accoding to the law of efaction sin 45 sin β = sin β = 0 5 β = 30 The efacted angle is 30 fo all of the incoming ays We have to investigate what happens if ϕ changes fom 0 to 80 4

5 It is easy to see that ϕ can not be less than 60 ( AOB = 60 ) The citical angle is given by sin β = n = ; hence β = 45 In the case of total intenal eflection cit cit ACO = 45 hence ϕ = = 75 If ϕ is moe than 75 the ays can emege the cylinde Inceasing the angle we each the citical angle again if OED = 45 Thus the ays ae leaving the glass cylinde if: 75 < ϕ < 65 CE ac of the emeging ays subtends a cental angle of 90 Expeimental poblem Thee closed boxes (black boxes) with two plug sockets on each ae pesent fo investigation The paticipants have to find out without opening the boxes what kind of elements ae in them and measue thei chaacteistic popeties AC and DC metes (thei intenal esistance and accuacy ae given) and AC (5O Hz) and DC souces ae put at the paticipants disposal No voltage is obseved at any of the plug sockets theefoe none of the boxes contains a souce Measuing the esistances using fist AC then DC one of the boxes gives the same esult Conclusion: the box contains a simple esisto Its esistance is detemined by measuement One of the boxes has a vey geat esistance fo DC but conducts AC well It contains a capacito the value can be computed as C = ω X C The thid box conducts both AC and DC its esistance fo AC is geate It contains a esisto and an inducto connected in seies The values of the esistance and the inductance can be computed fom the measuements 5

6 9 th IPhO (Budapest 976) Theoetical poblems Poblem A hollow sphee of adius R = 05 m otates about a vetical axis though its cente with an angula velocity of ω = 5 s - Inside the sphee a small block is moving togethe with the sphee at the height of R/ (Fig 6) (g = 0 m/s ) a) What should be at least the coefficient of fiction to fulfill this condition? b) Find the minimal coefficient of fiction also fo the case of ω = 8 s - c) Investigate the poblem of stability in both cases ) fo a small change of the position of the block β) fo a small change of the angula velocity of the sphee R S N mg mω Rsin R/ Figue 6 Figue 7 a) The block moves along a hoizontal cicle of adius R sin The net foce acting on the block is pointed to the cente of this cicle (Fig 7) The vecto sum of the nomal foce exeted by the wall N the fictional foce S and the weight mg is equal to the esultant: m ω Rsin The connections between the hoizontal and vetical components: mω Rsin = N sin S cos mg = N cos S sin The solution of the system of equations: ω Rcos S = mg sin g ω Rsin N = mg cos g 6

7 The block does not slip down if ω Rcos S g 3 3 µ a = sin = = 059 N ω Rsin 3 cos g In this case thee must be at least this fiction to pevent slipping ie sliding down ω Rcos b) If on the othe hand > some g fiction is necessay to pevent the block to slip upwads m ω Rsin must be equal to the esultant of foces S N and mg Condition fo the minimal coefficient of fiction is (Fig 8): S N ω Rcos g = sin = ω Rsin cos g µ b 3 3 = 9 = 079 S N mg mω Rsin Figue 8 c) We have to investigate µ a and µ b as functions of and ω in the cases a) and b) (see Fig 9/a and 9/b): 05 µ a ω = 5/s 05 µ b ω = 8/s ω < 5/s ω > 5/s ω < 8/s 90 Figue 9/a Figue 9/b ω > 8/s 90 In case a): if the block slips upwads it comes back; if it slips down it does not etun If ω inceases the block emains in equilibium if ω deceases it slips downwads In case b): if the block slips upwads it stays thee; if the block slips downwads it etuns If ω inceases the block climbs upwads - if ω deceases the block emains in equilibium Poblem The walls of a cylinde of base dm the piston and the inne dividing wall ae pefect heat insulatos (Fig 0) The valve in the dividing wall opens if the pessue on the ight side is geate than on the left side Initially thee is g helium in the left side and g helium in the ight side The lengths of both sides ae dm each and the tempeatue is 7

8 0 C Outside we have a pessue of 00 kpa The specific heat at constant volume is c v = 35 J/gK at constant pessue it is c p = 55 J/gK The piston is pushed slowly towads the dividing wall When the valve opens we stop then continue pushing slowly until the wall is eached Find the wok done on the piston by us dm dm dm Figue 0 The volume of 4 g helium at 0 C tempeatue and a pessue of 00 kpa is 4 dm 3 (mola volume) It follows that initially the pessue on the left hand side is 600 kpa on the ight hand side 00 kpa Theefoe the valve is closed An adiabatic compession happens until the pessue in the ight side eaches 600 kpa (κ = 5/3) 00 V = 600 hence the volume on the ight side (when the valve opens): V = 38 dm 3 Fom the ideal gas equation the tempeatue is on the ight side at this point pv T = = 55K nr Duing this phase the whole wok pefomed inceases the intenal enegy of the gas: W = (35 J/gK) ( g) (55 K 73 K) = 760 J Next the valve opens the piston is aested The tempeatue afte the mixing has been completed: T = = 33K 4 Duing this phase thee is no change in the enegy no wok done on the piston An adiabatic compession follows fom 38 = 50 dm 3 to dm 3 : hence = T 3 33 T 3 = 38 K The whole wok done inceases the enegy of the gas: The total wok done: W 3 = (35 J/gK) (4 g) (38 K 33 K) = 3000 J W total = W W 3 = 4760 J The wok done by the outside atmospheic pessue should be subtacted: W atm = 00 kpa dm 3 = 0 J The wok done on the piston by us: W = W total W atm = 3640 J 8

9 Poblem 3 Somewhee in a glass sphee thee is an ai bubble Descibe methods how to detemine the diamete of the bubble without damaging the sphee We can not ely on any value about the density of the glass It is quite uncetain The index of efaction can be detemined using a light beam which does not touch the bubble Anothe method consists of immesing the sphee into a liquid of same efaction index: its suface becomes invisible A geat numbe of methods can be found We can stat by detemining the axis the line which joins the centes of the sphee and the bubble The easiest way is to use the tumble-ove method If the sphee is placed on a hoizontal plane the axis takes up a vetical position The image of the bubble seen fom both diections along the axis is a cicle If the sphee is immesed in a liquid of same index of efaction the spheical bubble is pactically inside a paallel plate (Fig ) Its boundaies can be detemined eithe by a micomete o using paallel light beams Along the axis we have a lens system consisting of two thick negative lenses The diamete of the bubble can be detemined by seveal measuements and complicated calculations Figue If the index of efaction of the glass is known we can fit a plano-concave lens of same index of efaction to the sphee at the end of the axis (Fig ) As ABCD foms a paallel plate the diamete of the bubble can be measued using paallel light beams A B A R ϕ ψ B C D d Figue Figue3 Focusing a light beam on point A of the suface of the sphee (Fig 3) we get a diveging beam fom point A inside the sphee The ays stike the suface at the othe side and illuminate a cap Measuing the spheical cap we get angle ϕ Angle ψ can be obtained in a simila way at point B Fom we have sin ϕ = and R d sinψ sinϕ = R sinψ sinϕ sin ψ = R d sinψ sinϕ d = R sinψ sinϕ 9

10 The diamete of the bubble can be detemined also by the help of X-ays X-ays ae not efacted by glass They will cast shadows indicating the stuctue of the body in ou case the position and diamete of the bubble We can also detemine the moment of inetia with espect to the axis and thus the diamete of the bubble Expeimental poblem The whole text given to the students: At the wokplace thee ae beyond othe devices a test tube with V electical heating a liquid with known specific heat (c 0 = J/g C) and an X mateial with unknown themal popeties The X mateial is insoluble in the liquid Examine the themal popeties of the X cystal mateial between oom tempeatue and 70 C Detemine the themal data of the X mateial Tabulate and plot the measued data (You can use only the devices and mateials pepaed on the table The damaged devices and the used up mateials ae not eplaceable) Heating fist the liquid then the liquid and the cystalline substance togethe two timetempeatue gaphs can be plotted Fom the gaphs specific heat melting point and heat of fusion can be easily obtained Liteatue [] W Gozkowski: Poblems of the st Intenational Physics Olympiad Physics Competitions 5 no pp [] R Kunfalvi: Collection of Competition Tasks fom the Ist though XVth Intenational Physics Olympiads Roland Eötvös Physical Society in coopeation with UNESCO Budapest 985 [3] A Nemzetközi Fizikai Diákolimpiák feladatai I-XV Eötvös Loánd Fizikai Tásulat Középiskolai Matematikai Lapok 985 0

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