2022年-南师江宁高一期末物理满分冲刺宝典【解析版】

Unit1ForceandMotion.........................................................................................................................2

考点1：movementandposition..................................................................................................2

考点2：forcesandshape.............................................................................................................5

考点3：forcesandmovement....................................................................................................9

考点4：momentum....................................................................................................................12

考点5：theturningeffectofforces..........................................................................................15

Unit2Electricity.....................................................................................................................................18

考点1：mainelectricity..............................................................................................................18

考点2：currentandvoltageincircuits......................................................................................21

考点3：electricalresistance.......................................................................................................24

考点4：electriccharge...............................................................................................................28

Unit3Waves...........................................................................................................................................30

考点1：propertiesofwaves.......................................................................................................30

考点2：theelectromagneticspectrum.....................................................................................32

考点3：lightwaves......................................................................................................................34

考点4：sound..............................................................................................................................38

Unit4：ENERGYRESOURCESANDENERGYTRANSFER.....................................................................41

Unit1ForceandMotion

考点1：movementandposition

202106IP

4Thediagramshowsavelocity-timegraphforacarfromthetimethedriverseesan

obstacleintheroaduntilthecarcomestorest.

Velocity

inm/s

(a)(i)Calculatetheaccelerationofthecarbetween1.8and8.0seconds.

(3)

acceleration=...........m/s2

(ii)Calculatethebrakingdistanceofthecar.

(3)

brakingdistance=

(iii)Explaintheeffect,ifany,ofincreaseddrivertirednessonthethinkingdistance

andonthebrakingdistanceofthecar.

(4)

thinkingdistance.................................................................................................................................................................

brakingdistance

(b)Whichoftheserepresentsthedistance-timegraphforthecar?

DistanceDistance

□A□B

DistanceDistance

□C□D

(TotalforQuestion4=11marks)

【答案】

4(a)(i)useofacceleration=changeinvelocity/time;seenanywherein3

working

allowclearindication

thataccelerationis

gradient

substitution;

evaluation;ignoreminussign

eg-

acceleration=changeinvelocity/time

acceleration=(-)30/6.2

(acceleration=)(-)4.8(m/s2)allow

(-)4.8to(-)5.0(m/s2)

(ii)clearindicationthatdistanceisareaunderline;3

understandingbrakingdistanceisareaoftriangle

sectiononly;

evaluation;

54(m)=1mark

147(m)=2marks

eg

distance=area

distance=0.5x30x6.2

(distance=)93(m)

acceptalternative

methodusingecf

answerfrom(a)(i)and

v2=u2+2asgiving

93.75(m)

(iii)thinkingdistance:4

increaseinthinkingdistance;

(dueto)increasedreactiontime;

brakingdistance:

noeffectonbrakingdistance;

(dueto)noeffectonbrakingtime/brakingforce;allowideathatbraking

distancedoesnot

dependonhuman

factors

(b)A；1

Bisincorrectbecauseitdoesnotshowdeceleration

Cisincorrectbecausethedistancecannotchange

abruptlyandthecarismovingthroughout

Disincorrectbecausethefirstportionshowsthat

thecarisnotmoving

考点2：forcesandshape

202111IP

10Astudentusesthisapparatustoinvestigatethestretchingofarubberband.

clampstandrubberband

weights

Thisisthestudent'smethod.

•attachthe12cmlongrubberbandtoaclampstand

•hanga1Nweightfromtheotherendoftherubberband

•determinetheextensionoftherubberband

Thestudentrepeatsthismethod,increasingtheweightby1Neachtimeuntilthe

weightis10N.

(a)Describehowthestudentcoulddeterminetheextensionoftherubberband.

⑶

(b)Thegraphshowsthestudentsresults.

Weight

inN

(i)ExplainhowthegraphshowsthattherubberbanddoesnotobeyHooke'sLaw.

(2)

(ii)Theareaunderthecurveonthegraphisequaltotheincreaseintherubber

band'selasticenergystore.

Estimatetheincreaseintherubberband'selasticenergystorewhenthe

rubberbandhasbeenextendedby20cm.

(4)

increase=.............................................

(TotalforQuestion10=9marks)

【答案】

Question

AnswerNotesMarks

number

10(a)determinestretched/originallength;setunstretchedlengthto3

zero(on)ruler

witharuler/tapemeasure;

extension=stretchedlength-originalreadingonrulerisextension

length/12(cm);

(b)⑴lineisnotstraight/lineiscurved;allowgradientisnotconstant2

(therefore)weight/forceandextensionarenot

proportionaltoeachother;

(ii)evidenceofusingareasonablemethodofe.g.countingsquares,4

estimatinganarea;splittingintorectangles,

approximatingtoatriangle

etc.

evaluationoftheareaofoneshape;mediumsquare=0.03125(J)

largesquare=0.125(J)

correctevaluationofareawithoutfactoringcmtoallow85-100

m;

correctfinalevaluation;allow0.85-1.00(J)

e.g.

countingsquaresapproachused"/1

V

75-

Wpht

inN

5.0

Z5-

/

/

ao-1

s101520

Extensionincm

1mediumsquare=(1.25x2.5=)3.125

totalarea=29squaresx3.125=91

increaseinenergy=(91/100=)0.91(J)

考点3：forcesandmovement

202106IP

9Thephotographshowsawhalejumpingoutofthesurfaceofthesea.

(Source:©AlexanderBaumann/Shutterstock)

(a)Atthetopofthejump,thewhale'svelocityis0m/s.

Thewhalefalls2.2mfromthetopofthejumptothesurfaceofthesea.

Calculatethevelocityofthewhalewhenithitsthesurfaceofthesea.

(4)

velocity=一.m/s

(b)Aresultantforcecausesthewhaletoslowdownwhenithitsthesurfaceofthesea.

(i)Drawanarrowtoshowthisresultantforce.

(1)

(Source:©EugeniaPetrovskaya/Shutterstock)

(ii)Theresultantforceactingonthewhaleis18000N.

Themassofthewhaleis4100kg.

Calculatetheaccelerationofthewhale.

(3)

acceleration=m/s2

(TotalforQuestion9=8marks)

【答案】

9(a)useofv2=u2+2as;seenanywherein4

working

substitution;allowuseofg=9.8,9.81

rearrangement;

evaluation;

allowalternative

methodusing

mgh=1/2mv2

egfinalanswerof44(m/s)

is2marksonly

v2=u2+2as

v2=(0)+2x10x2.2

v=/44

(v=)6.6(m/s)allow6.63...(m/s),

6.56...(m/s)

6.5scores3marksonly

(b)(i)verticalarrowdrawnupwards;ignorelabels1

rejectifmorethanone

arrowdrawnunless

resultantforceis

clearlylabelled

(ii)substitutionintoF=ma;3

rearrangement;

evaluation;-1forPOTerror

e.g.

18000=4100xa

a=18000/4100

(a=)4.4(m/s2)allow4.39...(m/s2)

考点4：momentum

3Thediagramshowsanairtrackthatcanbeusedtoinvestigatemotionwithoutfriction.

Aircomesoutthroughaseriesofsmallholesintheairtrack,whichliftsthegliders

slightlyabovethetrack.

Therearetwoglidersonthetrack.

Eachgliderhasamagnet.

0.045kgm/smagnetmagnet

4AB

airin

Thepolesofthemagnetsnearesteachotherarealike.

(a)ExplainthedirectionoftheforceactingonmagnetAfrommagnetB.

⑵

(b)Thegliderscollideandthemagnetscausethemtorebound.

Beforethecollision,themomentumofgliderAis0.045kgm/stotherightand

gliderBisatrest.

(i)StatethetotalmomentumofgliderAandgliderBafterthecollision.

(1)

totalmomentum=kgm/s

(ii)Afterthecollision,themomentumofgliderAis0.021kgm/stotheleft.

CalculatethemomentumofgliderBafterthecollision.

(2)

momentumofgliderB=kgm/s

(iii)ThetimetakenforgliderBtochangeitsmomentumis0.19seconds.

CalculatetheaverageforceongliderBthatcausesthischangeinmomentum.

(2)

averageforce=..........................N

(iv)GivethedirectionoftheforceongliderBfromgliderA.

(1)

(TotalforQuestion3=8marks)

【答案】：

Question

AnswerNotesMarks

number

3(a)(tothe)left;allowtowardsA,away2

fromB

ignorebackwards

(because)repulsion(duetolikemagneticpoles);rejectifmentionof

charge

(b)(i)0.045(kgm/s);allow'thesame,1

(ii)momentumofBaftercollision=0.045-(-0.021);2

evaluation;

allow0.024for1mark

eg

0.045--0.021allow0.045+0.021

(momentum=)0.066(kgm/s)

(iii)substitutionintoallowECFfrom(ii)2

F=changeinmomentum/timetaken;

evaluation;useof0.024from(ii)

gives0.126...(N)

e.g.

F=0.066/0.19

(F=)0.35(N)allow0.347...(N)

(iv)(tothe)right;allowtowardsB,away1

fromA

(TotalforQuestion3=8marks)

考点5：theturningeffectofforces

20210115

2Thediagramshowsametalblockontopofawoodenblock.

ThemetalblockisheldstationarybyforceF.

metal

block

4.3cm

wooden

block

forceF

(a)(i)TheweightofthemetalblockactsthroughpointG.

GivethenameofpointG.

⑴

(ii)Nameapieceofapparatusthatcouldbeusedtomeasuretheweightofthe

metalblock.

⑴

(b)(i)Statetheformulalinkingmoment,forceandperpendiculardistancefromthepivot.

(1)

(ii)Theweightofthemetalblockis0.68N.

ShowthatthemomentoftheweightofthemetalblockaboutpointPis

approximately2.9Ncm.

(1)

(iii)ForceFisappliedtothemetalblocktostopitfrommoving.

CalculatethemagnitudeofforceF.

forceFN

【答案】

Question

AnswerNotesMarks

number

2(a)⑴centreofgravity/centreofmass;allowcentreofweight1

(ii)balance/newtonmeter;allownewtonscales1

ignorescales

(b)(i)moment=forcex(perpendicular)distance;allowstandard1

symbolsand

rearrangementse.g.

moment=Fxd

ignoremformoment

(ii)substitution;1

eg

0.68x4.3(=2.924Ncm)

(iii)ideathatanti-clockwisemoment=clockwisestatedorimpliedby3

moment;working

rearrangement;

evaluation;allowECFfrom(ii)

eg

2.9=Fx11

F=2.9/11

(F=)0.26(N)allow0.2636...-

0.2658...

allow0.27

condone0.263

(TotalforQuestion2=6marks)

Unit2Electricity

考点1：mainelectricity

202101IP

4Thephotographshowsahairdryerpluggedintothemainssupply.

Thehairdryercontainsafuse.

©CristinaBernardo/Shutterstock

(a)Statewhichwireinthehairdryershouldbeinserieswiththefuse.

(1)

(b)Thefuseisanelectricalsafetyfeatureusedinmains-operateddomesticappliances.

Statetwootherelectricalsafetyfeaturesthatcanbeusedinmains-operated

domesticappliances.

(2)

2

(c)Thehairdryerhasacurrentof9.6A.

Themainssupplyvoltageis230V.

(i)Statetheformulalinkingpower,currentandvoltage.

(1)

(ii)Calculatethepowerofthehairdryer.

Givetheunit.

⑶

power=unit

(iii)Thehairdryercontainsacoilofwirewhichisusedtoheatairpassingthrough

thehairdryer.

Explainwhythecoilofwireheatsupwhenthereisacurrentinit.

(3)

(d)Explainhowafuseprotectsadomesticappliance.

(3)

(TotalforQuestion4=13marks)

【答案】

Question

AnswerNotesMarks

number

4(a)live/L;allowred/brownwire1

(b)anytwofrom:2

MP1.earthwire;

MP2.circuitbreaker;allowRCD,tripswitch,surge

protector

MP3,doubleinsulation;

MP4,insulatedcables;allowanymentionof

insulatedwires

(c)(i)power=currentxvoltage;allowstandardsymbols1

e.g.P=IxV

ignoreC,cforcurrent

(i>)substitution;3

evaluation;

unit;markindependently

2.2kW=fullmarks

e.g.

(P=)9.6x230

(P=)2200allow2208,2210

watts/WallowJ/s

(iii)allowwireforcoil3

throughout

coilhasresistance;

electronstransfer/loseenergy(astheyflow

throughcoil);

(dueto)electroncollisionswith(lattice)ionsintheallowatoms,particlesfor

coil;ions

(d)ideaofexcessivecurrent;e.g.”currentbecomestoo3

high"

meltsthefuse(wire);allowbreakingthefuse

condoneMblowsthefuse”

breakingthecircuit;allow"stopsthecurrent"/eq

TotalforQuestion4=13marks

考点2：currentandvoltageincircuits

202111IP

11Thediagramshowsacircuitthatincludesabattery,anammeter,avoltmeterand

threedifferentresistors.

(a)(i)Givethevoltmeterreading.

(1)

voltage=.V

(ii)Statetheformulalinkingvoltage,currentandresistance.

(1)

(iii)CalculatetheresistanceofresistorX.

⑶

resistance=

(b)(i)Givethereasonwhythereadingontheammeterwouldbe16mA.

⑴

(ii)CalculatetheresistanceofresistorY.