2012 exam solutions - bhs physics - homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf ·...

        The following pages offer suggested solutions to the  2012 SACE Stage 2 Physics final  examination. These solutions are not the official set  of solutions used by the examiners of the SACE Board.  

Tension

F mv2

r 0.035x2.42

0.32 0.63 N

vH vsin 25sin 40 16.1 ms1

v2 vo2 2as v 0 max height

0 16.12 2x9.8s

s 16.12

2x9.813.2 m

Total height above the ground = 13.2 + 1.5 = 14.7 m

COMMONERRORThereleaseheightmustbeaddedasthequestionasksforthemaximumheightabovetheground.

Increasingthelaunchheightincreasestherange.Thejavelinspendsmoretimeintheairasitfallstheextraheight.Sincetherangeistheproductofthehorizontalcomponentoftheinitialvelocity(vH)andthetimeofflight.Thehorizontalcomponentoftheinitialvelocityisconstantsoanincreaseinthetimeofflightincreasesintherange.

FN VerticalcomponentFV

HorizontalcomponentFH

Theverticalcomponentofthenormalisequalinmagnitudebutoppositeindirectiontotheweight,whereweightW=mg.

ThehorizontalcomponentofthenormalforceFH,providesallthecentripetalaccelerationforuniformcircularmotion.

tan FH

FV

mv2

rmg

v2

rg

Tan v2

rg

v tanxrg tan 42x26x9.8 15 ms1

ForceontheMoonduetotheEarth

FME GmM mE

rmE2

6.67x1011 x7.35x1022 x5.97x1024

(3.85x108 )2

ForceontheMoonduetotheSun

FMS GmM mS

rmS2

6.67x1011 x7.35x1022 x1.99x1030

(1.5x1011)2

Force on the Moon due to the Earth

Force on the Moon due to the Sun

6.67x1011 x7.35x1022 x5.97x1024

(3.85x108)2

6.67x1011 x7.35x1022 x1.99x1030

(1.5x1011)2

5.97x1024

(3.85x108 )2x

(1.5x1011)2

1.99x1030

0.455

Sincespeedofasatellitevisgivenby v GM

Ro

whereMisthemassofEarthand

GistheGravitationalconstantthen v 1

rwhereristheradiusoforbitofthe

satellite.ThisisbecausebothMandGareconstant.TheQuickBirdsatellitehasalargerradiusoforbitandwillthereforehaveaslowerspeedthantheinternationalspacestation.

Asexplainedabove,themassintheequationforthespeedofasatelliteisthemassoftheEarthnotthemasofthesatellite.Thedifferentmasseshavenoeffectontheirspeeds.

Imageswithhigherresolutionareproduced.

MEANINGAlow‐altitudeorbitmeansthatthesatelliteisclosertotheground.Theimagesproducedshowmoredetail.Werefertothisasresolution.

ThemomentumvectorofBafterthecollision

ThetotalinitialmomentumvectorisjustthatofAasofBwasinitiallystationarybeforethecollision

Usingthelawofconservationofmomentum,thetotalinitialandfinalmomentumisthesame(bothmagnitudeanddirection).Sincepi=pfthenpAi=pAf+pBfThefinalmomentumvectorisshownonthediagramabove.

Thechangeinmomentumpisgivenbythefinalmomentumofthephotonsubtracttheinitialmomentum.Thefinalmomentumofaphotonthatisabsorbediszerobutthefinalmomentumofaphotonthatisreflectedisthesameastheinitialmomentumbutintheoppositedirection.Thechangeinmomentumofaphotonthatisabsorbedisgivenbyp pf p

i 0 p p whichishalfthatofaphotonthatisreflected

p pf pi p p 2 p .

Usingthelawofconservationofmomentum,thechangeinmomentumexperiencedbythesolarsailisequalbutoppositeindirectiontothechangeinmomentumexperiencedbythephoton.

SinceFsail psail

t,thenitfollowsthatsolarsailsthatreflectphotonswill

experienceagreaterforce(ifthecollisionoccursoverthesameamountoftime)

andhencegreateraccelerationsince asail F

msail

p 2 p

p p

Photonisabsorbed

p

p

p

HINTTheelectricfieldisuniform.Theelectricfieldlinesneedtobeevenlyspaced.Theelectricfieldlinespointawayfromtheconductorastheyindicatethedirectionoftheforceonapositivetestcharge.

F

B 1

40

QqB

r2 9x109 x8x1019 x2.56x1018

0.082 2.9x1024 N (south)

ForceduetoqA 2.9x10‐24Neast ForceduetoqB 2.9x10‐24Nsouth

F FA

FB

F (2.9x1024 )2 (2.9x1024 )2 4.1x1024 N

tan1(2.9x1024 )

2.9x1024 450

F 4.1x1024 N SE

2.9x10‐24 N

2.9x10‐24 N

F

HINTThetriangleisrightangled–Pythagorasapplies.

ThemagneticfieldduetoI1actsintothepageinthepositionwhereconductor2lies.Usingtherighthandruletheforceonconductor2duetoconductor1isuptheplaneofthepage.ThemagneticfieldduetoI2actsoutofthepageinthepositionwhereconductor1lies.Usingtherighthandruletheforceonconductor1duetoconductor2isdowntheplaneofthepage.Theconductorsattracteachother.

HINTUsetherighthandruleforthemagneticfieldaroundastraightconductor.

HINTAgainusetherighthandruleforthemagneticfieldaroundastraightconductor.

F Bqvsin 2.5x102 x1.6x1019 x1.45x106 xsin 90 5.8x1015 N

E V

d 6x104

0.451.3x105 Vm1

F Eq 1.3x105 x(1.6x1019 ) 2.1x1014 N

vv sv

t t sH

vH

0.8

2 0.40 s

Maximumdeflectioninthedirectionoftheelectricfieldis 0.45

2 0.225 m

Thiscanbeusedtofindthemaximumacceleration.

s vot 1

2at 2 vo 0

a 2s

t2 2x0.225

0.42 2.81 ms2

a Eq

m m Eq

a 1.3x105 x1.6x1019

2.81 7.4x1015 kg

REASONINGThiscomponentofvelocityisperpendiculartotheuniformelectricfieldbetweentheplatesandisconstant.Theequationforconstantspeedapplies.

Vertical

v f v

f 3x108

64.25x106 4.67 m

COMMONERRORSIunitsarerequired.MHztoHzx106

REASONINGTheplaneofpolarisationisdefinedastheplaneoftheelectricfield.Theplaneofthemagneticfieldisperpendiculartotheplaneoftheelectricfield.

Anincandescentglobeemitswhitelight.Whitelightconsistsofallcolours/wavelengthsrangingfromredthroughtoviolet(ROYGBIV).Monochromaticlightconsistsofonecolour/wavelengthonly.Thelightfromanincandescentglobeisnotmonochromatic.

kmax hf W 6.63x1034 x2x1015 (3.66x1.6x1019 ) 7.4x1019 J

Kmax 1

2mv2 v 2K

m 2x7.4x1019

9.11x10311.3x106 ms1

HINT/COMMONERROReVneedstobeconvertedtoJ(x1.6x10‐19)

SincethedistancebetweenthedoubleslitsandthescreenissomuchlargerthatthedistancebetweenthedoubleslitsthentheangleS1XS2isapproximately90o.UsingthetriangleS1XS2then

sin path difference

d

Foramaxima,thepathdifferenceism

Itfollowsthat

sin path difference

d m

d

ie d sin m

Pathdifference

X

d sin m

sin1(md

) sin1(3x4.7x107

1.8x104)

0.450

Thelaserlightiscoherentwhereasthebluelightsourcewasn’t.Thismeansthatthesingleslitisn’tneeded.

Sincethefringeseparationy L

d,thenthefringeseparation(distance

betweenadjacentmaxima)isproportionaltothewavelength.Changingthewavelengthfrombluetoredmeansthatthewavelengthisgreater.Agreaterwavelengthmeansagreaterfringeseparation/distancebetweenadjacentmaxima.Thecolourofthebrightfringeswillalsochangefrombluetored.

REASONINGThepurposeofthesingleslitistoproducetwocoherentlightsourcesatthedoubleslits.

Increasingthefilamentcurrentwillreducetheexposuretime.ThisisbecausethefilamentcurrentreleaseselectronsthatcollidewithatargetmetaltoproduceX‐rays.Ifmoreelectronsarereleased,thenmorecollisionsoccurandmoreX‐raysareproducedreducingtheoverallexposuretime.

Exposuretoionisingradiationcanbereducedbywearingappropriateshieldingsuchasaleadlinedapron.

h

p h

mv 6.63x1034

9.11x1031 x4.36x1061.67x1010 m

Lowenergyelectronswerefiredtowardsacrystallatticeandtheywerediffractedatpreferredanglesratherthanbeingscatteredrandomly.Diffractionisawavephenomena.Inadditiontheequation d sin mwasusedtocalculatethewavelengthoftheelectronsandmatchedthatfoundusingthede

Broglierelationship h

p.

Thepositionoflinesinthespectraofhydrogenandlithiummatchthelinesinthespectrumofthemixture.Iehydrogenandlithiummustbepresent.

Thereisnoenergygapbetweenthegroundstateandthehigherenergylevelsthatmatches12.50eV.Thephotonwillnotbeabsorbed.

1.89eV

10.2eV

12.09eV

12.75eV

(b)(i)

(a)

REASONINGThesmallestjumpfromE3toE2willemitthesmallest‐energyphoton.

E3 E2 1.513.4 1.89eV

Boththenuclei(UandTh)havediscretenuclearenergylevels.WhenthenucleusdecaystoTh,itmaydecaytothegroundstateoroneofthehigherexcitedstates.Itfollowsthatasthenucleusdecaystheemittedalphaparticleswillhavearangeofdiscreteenergiessothattheoverall(discrete)energyofthisreactionisconserved.

After8alpha(8 2

4 )decaystheatomicnumberofproductnucleusbecomes92‐16=76andthemassnumberbecomes238–32=206.Lead(Pb)hasatomicnumber82.Usingthelawofconservationofcharge,itfollowsthat6betaminusdecaysmustoccur.

92238U 82

206 X 6 10e82

4 6 00

92238U 90

234Th 24

Thereexistsastrongnuclearforcebetweenthenucleons(protonsandneutrons)inthenucleusthatisattractiveandstrongerthantherepulsiveelectrostaticforcesbetweenthe82protons.

Thisforceactsovershortdistancesandquicklybecomesnegligibleatseparationsofmorethanafewnucleondiameters.Thisishowthenucleons(whichareveryclose)canbeheldtogetherinastablenucleus.

‐

neutrino

Theemittedpositroncancollidewithanearbyelectronandannihilate.Themassisconvertedintoenergyintheformoftwoidenticalphotonsthattravelinoppositedirectionssothatmomentumisconserved.

HINTThehalflifeis2days.Thisisconstant.

REASONINGThephotonstravelinoppositedirections.

Uranium‐238doesnotreadilyundergoinducedfissionwhereasuranium‐235does.Enrichingthesourcemeansthatalargerpercentageofuranium‐235isaddedtothefuel.Thisensuresthatenoughofthenucleiundergoinducedfissionandachainreactioncanoccur.

E hf hc

6.63x1034 x3x108

502x109 3.96x1019 J

COMMONERRORWavelengthneedstobeinSIunitsnmtommultiplyby10‐9

Massofproducts mcMg+mn= 3.8172x10-26 + 1.6749x10-27 = 3.98469x10-26 kgMassofreactant 2mC= 2x1.9921x10-26 = 3.9842x10-26 kg Themassoftheproductsisgreaterthanthemassofthereactants–energyisabsorbed.

m mproducts mreactants 3.98469x1026 3.9842x1026 4.9x1030 kg

E mc2 4.9x1030 x(3x108)2 4.41x1013 J

Lengthof

pendulum/stringL(cm)

PeriodT

(s)T2(s2)

20 0.94 0.8830 1.08 1.1740 1.36 1.8550 1.48 2.1960 1.54 2.37

Thelengthofthependulum/string.

REASONING:Theindependentvariableisthevariableintentionallychangedbytheexperimenter.

Therelationshipbetweenperiodsquared(T2)andlengthofthependulum.T2(s2)

0

0.5

1

1.5

2

2.5

3

0 10 20 30 40 50 60 70

LengthofpendulumL(cm)

line ofbestfit

rise

run

gradient rise

run 1.7 0.2

40 2 0.039 s2cm1

T 2 4 2 L

g

SincethestraightlinehasanequationT2=0.039L

Itfollowsthattheslope0.039=4 2

g

g 4 2

0.0391010 cms2 10.1ms2

Accuracyishowclosetheexperimentalvalueistotheactualvalueandcanbeimprovedbyreducingsystematicerrors.Anexampleofthiswouldbetocheckthatthelightgateiscorrectlycalibratedbyrepeatingtheexperimentwithadifferentlightgate.Precisionishowmuchscatterthereisinthedatacollected.Thiscanbereduced,byreducinganyrandomerrorsintheexperiment.Anexampleofthiswouldbetorepeatthemeasurementsofperiodforeachlengthatleastthreetimes.

Astheionsenterthedeestheydosoat90otoauniformmagneticfield.

Aconstantmagneticforcealwaysactsat90otothevelocityofthecharges.

Eventhoughthespeedremainsthesame,thedirectionofmotionchanges.

Bydefinitionthereisachangeinvelocityv

vf

v

iandhenceacceleration

givenbya

v

t.

Themagneticforcethereforeprovidesthecentripetalaccelerationforuniformcircularmotion.

X

X

X

X

X

X X

X

X

X

X

X

XX X

XX

X

X

X

X

XX

X

dees

Eachtimetheionscrossthegapbetweenthedees,theyareacceleratedbyauniformelectricfield.Everytimetheionscrosstheelectricfield,theworkdone(W)bytheelectricfieldisconvertedintokineticenergy.

Kineticenergyeachtimetheionscrossthegap=W=qV 1

2mv2 .Theionsdo

notgainenergyorspeedwhiletheyareinthedees.Itcanbeshownthatthefinalkineticenergyoftheionsastheyemergefromthe

cyclotronisgivenbyK q2B2r 2

2mwhereristheradiusofthecyclotron.

Itthereforefollowsthatthefinalkineticenergyisdirectlyproportionaltothesquareofthecyclotron’sradius.Thismeansthatiftheradiusofthecyclotronisdoubled,thekineticenergybecomesfourtimeslarger.Alternatively,iftheradiusis10timeslargerthekineticenergywillbe100timeslarger.Iftheradiusis3timessmaller,thekineticenergywillbe9timesetc

HINT22=4102=10032=9

Ifmonochromaticlightisincidentonametalsurfaceitcontainsasinglefrequency(f)only.Thelightconsistsofmanyphotons.Photonsarebundlesofdiscreteenergy.EachphotonhasadiscreteenergygivenbyE hf wherehisPlanck’sconstant(6.63x10‐34Js).Electronswithinthemetalareboundbydifferingamountsofenergy(dependingonhowdeepwithinthemetaltheyarepositioned).Usingthelawofconservationofenergy,theenergyofthephotonisusedtoreleaseandelectronandany‘leftover’energyisgivenupasthekineticenergyoftheemittedelectron.Thisproducesarangeofkineticenergiesuptomaximumgivenby:Kmax hf W whereWistheworkfunctionofthemetal.Theworkfunctionisdefinedastheenergyneededtoreleasetheleastboundorsurfaceelectrons.ElectronsinanX‐raytubearereleasedfromaheatedfilament.EachelectrongainsadiscreteenergygivenbyK qV astheyaccelerateacrossthepotentialdifferenceV .Whentheelectronscollidewiththetargetmetaltheydecelerateandlosekineticenergy.Thelawofconservationofenergyapplies.MostofthecollisionsproduceheatbutapproximatelyonepercentofthecollisionsproduceX‐rayphotonsofenergyequaltotheenergy‘lost’bytheelectronsduringthecollision.Theamountofkineticenergylostbythecollidingelectronsdependsonhowcloselytheycollidewiththenucleus.ThisproducesarangeofX‐rayphotonswitharangeofenergies.