国际物理林匹克竞赛试题ipho_Problemsof2ndand9thIPhO

Problemsofthe2ndand9thInternationalPhysicsOlympiads(Budapest,Hungary,1968and1976)PéterVankóInstituteofPhysics,BudapestUniversityofTechnicalEngineering,Budapest,HungaryAbstractAfterashortintroductiontheproblemsofthe2ndandthe9thInternationalPhysicsOlympiad,organizedinBudapest,Hungary,1968and1976,andtheirsolutionsarepresented.IntroductionFollowingtheinitiativeofDr.WaldemarGorzkowski[1]Ipresenttheproblemsandsolutionsofthe2ndandthe9thInternationalPhysicsOlympiad,organizedbyHungary.IhaveusedProf.RezsőKunfalvi’sproblemcollection[2],itsHungarianversion[3]andinthecaseofthe9thOlympiadtheoriginalHungarianproblemsheetgiventothestudents(myowncopy).Besidesthedigitalizationofthetext,theequationsandthefiguresithasbeenmadeonlysmallcorrectionswhereitwasneeded(typemistakes,smallgrammaticalchanges).Iomittedoldunits,wherebotholdandSIunitsweregiven,andconvertedthemintoSIunits,whereitwasnecessary.IfwecomparetheproblemsheetsoftheearlyOlympiadswiththelastones,wecanrealizeatoncethedifferenceinlength.Itisnotsoeasytojudgethedifficultyoftheproblems,butthesolutionsaresurelymuchshorter.Theproblemsofthe2ndOlympiadfollowedthemorethanhundredyearstraditionofphysicscompetitionsinHungary.ThetasksofthemostimportantHungariantheoreticalphysicscompetition(EötvösCompetition),forexample,arealwaysveryshort.Sometimesthesolutionisonlyafewlines,too,buttofindtheideaforthissolutionisratherdifficult.Ofthe9thOlympiadIhavepersonalmemories;IwastheyoungestmemberoftheHungarianteam.TheproblemsofthisOlympiadwerecollectedandpartlyinventedbyMiklósVermes,alegendaryandfamousHungariansecondaryschoolphysicsteacher.Inthefirstproblemonlythedetailedinvestigationofthestabilitywasunusual,inthesecondproblemonecouldforgettosubtracttheworkoftheatmosphericpressure,butthefully“open”thirdproblemwasreallyunexpectedforus.Theexperimentalproblemwasdifficultinthesameway:incontrasttotheOlympiadsoftodaywegotnoinstructionshowtomeasure.(Inthelastyearstheonlysimilarlyopenexperimentalproblemwastheinvestigationof“Themagneticpuck”inLeicester,2000,areallyniceproblembyCyrilIsenberg.)Thechallengewasnottoperformmany-manymeasurementsinashorttime,buttofindoutwhattomeasureandhowtodoit.Ofcourse,theevaluatingofsuchopenproblemsisverydifficult,especiallyforseveralhundredstudents.Butinthe9thOlympiad,forexample,onlytencountriesparticipatedandthesamepersoncouldread,compare,gradeandmarkallofthesolutions.22ndIPhO(Budapest,1968)TheoreticalproblemsProblem1Onaninclinedplaneof30°ablock,massm2=4kg,isjoinedbyalightcordtoasolidcylinder,massm1=8kg,radiusr=5cm(Fig.1).Findtheaccelerationifthebodiesarereleased.Thecoefficientoffrictionbetweentheblockandtheinclinedplane=0.2.Frictionatthebearingandrollingfrictionarenegligible.SolutionIfthecordisstressedthecylinderandtheblockaremovingwiththesameaccelerationa.LetFbethetensioninthecord,Sthefrictionalforcebetweenthecylinderandtheinclinedplane(Fig.2).Theangularaccelerationofthecylinderisa/r.Thenetforcecausingtheaccelerationoftheblock:Fgmgmamcossin222,andthenetforcecausingtheaccelerationofthecylinder:FSgmamsin11.Theequationofmotionfortherotationofthecylinder:IrarS.(Iisthemomentofinertiaofthecylinder,Sristhetorqueofthefrictionalforce.)Solvingthesystemofequationsweget:221221cossinrImmmmmga,(1)2212212cossinrImmmmmgrIS,(2)m1m2Figure1m2gsinFigure2FFm2gcosSm1gsinr32212212sincosrImmrIrImgmF.(3)Themomentofinertiaofasolidcylinderis221rmI.Usingthegivennumericalvalues:2sm3.25gmmmmmga3317.05.1cossin21221,N13.012122115.1cossin2mmmmmgmS,N0.19221125.1sin5.0cos5.1mmmgmF.Discussion(SeeFig.3.)Theconditionforthesystemtostartmovingisa0.Insertinga=0into(1)weobtainthelimitforangle1:0667.03tan2121mmm,81.31.Forthecylinderseparately01,andfortheblockseparately31.11tan11.Ifthecordisnotstretchedthebodiesmoveseparately.WeobtainthelimitbyinsertingF=0into(3):6.031tan212Irm,96.302.TheconditionforthecylindertoslipisthatthevalueofS(calculatedfrom(2)takingthesamecoefficientoffriction)exceedsthevalueofcos1gm.Thisgivesthesamevaluefor3aswehadfor2.Theaccelerationofthecentersofthecylinderandtheblockisthesame:cossing,thefrictionalforceatthebottomofthecylinderiscos1gm,theperipheralaccelerationofthecylinderiscos21gIrm.Problem2Thereare300cm3tolueneofC0temperatureinaglassand110cm3tolueneofC100temperatureinanotherglass.(Thesumofthevolumesis410cm3.)Findthefinalvolumeafterthetwoliquidsaremixed.Thecoefficientofvolumeexpansionoftoluene1C001.0.Neglectthelossofheat.r,ag0306090F,S(N)12=31020FSraFigure34SolutionIfthevolumeattemperaturet1isV1,thenthevolumeattemperatureC0is11101tVV.Inthesamewayifthevolumeatt2temperatureisV2,atC0wehave22201tVV.FurthermoreifthedensityoftheliquidatC0isd,thenthemassesaredVm101anddVm202,respectively.Aftermixingtheliquidsthetemperatureis212211mmtmtmt.ThevolumesatthistemperaturearetV110andtV120.Thesumofthevolumesaftermixing:212201102202011010221120102122112120102010201020101111VVtVtVtVVtVVdtmdtmVVmmtmtmdmmVV