MS P2

Sec4PurePhysicsPrelimMarkingScheme(Paper2)SectionAQn Answer Marks Remarks1(a) AccordingtoNewton’sFirstLaw,iftheforcesactingontheballarebalanced,theballwouldeitherbestationaryormovewithconstantvelocity,withoutanyacceleration./AccordingtoNewton’sSecondLaw,theaccelerationoftheballistheresultofaresultantforceactingontheball.Hence,theforcesactingontheballareunbalanced. 1 Acceptexplanationbasedoneither1 st or2 nd Law. Nomarkawardedforjuststating“unbalanced” 1(b) a=(v–u)/t2.0=v/2.0v=4.0m/s 11(c) s=(1/2)x2.0x4.0=4.0m 1 1 Ecfmarkscanbeawarded 2(a) ForceexertedbyManB=100x10–600=400N 12(b) LetthedistancebetweentheCGoftherodandmanAbed.TakingmomentaboutmanA,1000xd=400x6.0d=2.4m or TakingmomentabouttheCGoftherod,600xd=400x(6.0–d)d=2.4m 11 11 Ecfmarkscanbeawarded

3(a) PressureofgasB=120000–13600x10x0.08=109000Pa(3s.f.) 11 1markcanbeawardedforcalculatingthepressuredifference3(b) H1 woulddecreasewhileH2 wouldincrease.Blackisabetterabsorberofinfraredradiationthanwhite,hencethetemperatureofgasAwouldincreasefasterthanthatofgasB,resultinginanevengreaterpressuredifferencebetweengasAandgasB, 1 11 4(a) InitialGPE=finalKEmgh=(1/2)mv 2 300x10x8=(1/2)x300xv 2 v 2 =160v=12.6m/s 1 14(b) averageretardingforcex0.5=300x10x8averageretardingforce=48000N 115(a) Theairmoleculesinthecylinderareincontinuousandrandommotion,collidingwiththewallsofthecylinder.Theaverageforceexertedbytheairmoleculesonaunitareaofthecylindergivesrisetoapressureinthecylinder. 1 1 5(b) Thevolumeofairinthecylinderdecreases,causinganincreaseinthenumberofairmoleculesperunitvolume.Theairmoleculescollidewiththewallsofthecylindermorefrequently,causingtheforceexertedperunitareatoincrease,andhenceincreasingthepressureinthecylinder. 1 1 6(a) sin27°/sinr=1.50,r=17.6°angleofincidenceatCD=angleofreflection=angleofincidencebackatAB=18°angleofrefractionatAB=27° 1 1 FortheangleofincidenceatAB,accept26°to28°

6(b) sinc=1/1.50,c=41.8°TheangleofincidencewhenthelightrayleavesABis,whichissmallerthanthecriticalangle.Hence,totalinternalreflectiondoesnothappen. 11 7(a) Whentheobjectdistanceistwicethefocallengthofthelens,theimagedistanceisalsotwicethefocallengthofthelens.Fromthegraph,theobjectdistanceandimagedistanceareequalat3.0cm.Hence,thefocallengthofthelengthis3.0/2=1.5cm. 1 1 Nomarkawardedifnoinformationisquotedfromthegraph7(b) Whenobjectdistance=2.0cm,imagedistance=4.2cm.Magnification=4.2/2.0=2.1 11 Acceptraydiagrammethod 8(a) Accordingtothequestion,thecorksmoveupanddownasthewavepasses.Thisshowsthatthewaterparticlesmoveinadirectionperpendiculartothedirectionofthewaterwave,andthisisacharacteristicoftransversewave. 1 Nomarkawardedifnoinformationisquotedfromthequestion 8(b)(i)Speed=wavelength/period=8.0/0.50=16cm/sor0.16m/s 18(b)(ii) Sameamplitudeandperiod.WhenAisatthecrest,Bisatthetrough. 1 9(a) 1.Ultrasoundwavesarelongitudinalwaves,whilemicrowavesaretransversewaves.2.Ultrasoundwavescannottravelinvacuum,whilemicrowavescantravelinvacuum.3.Ultrasoundwavesrequireamediumtotravel,whilemicrowavesdonotrequireanymediumtotravel.4.Ultrasoundwavestravelatabout330m/sinair,whilemicrowavestravelat3x10 8 m/sinair. 1 1 Donotaccept2and3astwodifferences.Acceptanyothervaliddifferences 9(b) Totaltime=6000/1500+(36000000+40000000)/3x10 8 =4.25s(3s.f.) 1 1

10(a) Whentheswitchisclosed,electronsfromLwillflowtotheearthasthepositiveterminalofthehighvoltagesupplyisconnectedtoit.Asaresult,Lbecomespositivelycharged. 1 110(b) Aftertheswitchisclosed,thepositivelychargedLwouldattracttheelectronsinS,causingthemtomovetoitsleftside.TheleftsideofSnowhasexcessivenegativecharges,andexcessivepositivechargesareonitsrightside.TheforceofattractionbetweenLandthenegativechargesonSisstrongerthantheforceofrepulsionbetweenLandthepositivechargesonS,therefore,SwouldmovetowardsL. 1 1 AcceptSwouldtouchLbyinduction 11(a) Method2: Method3: 1 11(b) Power=240 2 /40x2=2880W 1111(c) Theleastcostlymethodiswhenthetwocoilsareconnectedinseries.Power=240 2 /80=720W=0.72kWCost=0.72x1.75x20=25.2=25centsor$0.25 1 1 AcceptconversionfromJtokWh

12(a) B:northpole,C:southpole 112(b) Anticlockwise 1 Ecfmarkcanbeawarded12(c) UsingFleming’slefthandrule,theforefinger/indexfingerpointstotherightasthemagneticfieldisfromBtoC,thesecondfinger/middlefingerpointsintothepageascurrentflowsfromPtoQ,andthethumbpointsdownwards.TheforceactingonPQisdownwards,andtheforceactingonRSisupwardssincethecurrentisintheoppositedirectionasPQ,causingthecoiltoturnanticlockwise. 1 1 Ecfmarkcanbeawarded 12(d) Thesplit-ringcommutatorsreversethedirectionofcurrentinthecoilafteritturns180°.Asaresult,whenPQisrotatedtotheright,theforceactingonPQwouldbeupwards,andwhenRSisrotatedtotheleft,theforceactingonRSwouldbedownwards,andthecoilwouldcontinuetoturnintheanticlockwisedirection. 1 1 SectionBQn Answer Marks Remarks13(a) 2 1markforthepoint(12,0.25),1markforthecorrectshape 13(b) Thecurrent-voltagegraphforafixedresistorwouldbeastraightlinepassingthroughtheorigin.Thisisbecauseafixedresistorisanohmicconductorwithfixedresistance/obeysOhm’sLaw,thereforethegradientofthegraphisconstant.However,afilamentlampisanon-ohmicconductorwithvaryingresistance/doesnotobeyOhm’sLaw,thereforethegradientofthegraphisnotconstant. 1 1 Deduct1markifthegradientofthegraphisnotdescribed

13(c) Whenthebrightnessofthetorchlightincreases,theresistanceoftheLDRdecreases.Thiscausesthecurrentinthecircuittoincrease,hencethepotentialdifferenceacrossRincreases. 1 1 Acceptexplanationwithpotentialdividerformula,providedtheformulaandthetermsinitareclearlyspecified13(d)(i)Current=12/5=0.24A 1113(d)(ii)CurrentthroughR=0.24+0.25=0.49AResistance=(18–12)/0.49=12.2Ω(3s.f.) 11 Award1markifthep.d.acrossRiscorrectlycalculatedEcfmarkcanbeawarded14(a) 4 2markseach 1markfortheforces,1markforthelabels 14(b) Forcesinanaction-reactionpairmustactondifferentbodies. 1 14(c)(i)Acceleration=28/(3.0+4.0)=4.0m/s 2 1114(c)(ii)Force=3.0x4.0=12N 1114(d) Theblockswouldmovewithconstantvelocity. 1 15(a) Thealuminiumatomsneartheheatingelementgainthermalenergyandvibratemorevigorously.Theycollidewiththeirneighbouringparticles,transferringenergytothemandmakethemvibratemorevigorouslyaswell.Thefreeelectronsinaluminiumalsotransferthethermalenergytothecolderpartofthealuminiumrodthroughelectrondiffusion,causingthealuminiumrodtobeheatedquickly. 1 1 1

15(b) Power=230x9=2070W 11 15(c) Inonesecond,E=2070x1=2070JLetthemassofwaterbem.4200xmx70+2260000xm=2070m=8.10x10 -4 kg(3s.f.) 1 1115(d) Asthetemperatureofsteamisloweredtothecondensationpoint,thesteammoleculescontinuestoloseenergy,andgetmuchclosertoeachother.Thebondsbetweenthemoleculesarestrengthen,andthemoleculesareabletomovewithinthewaterbodyonly. 1 1