2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf ·...
Transcript of 2012 EXAM SOLUTIONS - BHS Physics - Homephysicsbhs.weebly.com/.../4/9/5/...exam-solutions.pdf ·...
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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.
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Tension
F mv2
r 0.035x2.42
0.32 0.63 N
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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.
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Increasingthelaunchheightincreasestherange.Thejavelinspendsmoretimeintheairasitfallstheextraheight.Sincetherangeistheproductofthehorizontalcomponentoftheinitialvelocity(vH)andthetimeofflight.Thehorizontalcomponentoftheinitialvelocityisconstantsoanincreaseinthetimeofflightincreasesintherange.
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FN VerticalcomponentFV
HorizontalcomponentFH
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Theverticalcomponentofthenormalisequalinmagnitudebutoppositeindirectiontotheweight,whereweightW=mg.
ThehorizontalcomponentofthenormalforceFH,providesallthecentripetalaccelerationforuniformcircularmotion.
tan FH
FV
mv2
rmg
v2
rg
Tan v2
rg
v tanxrg tan 42x26x9.8 15 ms1
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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
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Sincespeedofasatellitevisgivenby v GM
Ro
whereMisthemassofEarthand
GistheGravitationalconstantthen v 1
rwhereristheradiusoforbitofthe
satellite.ThisisbecausebothMandGareconstant.TheQuickBirdsatellitehasalargerradiusoforbitandwillthereforehaveaslowerspeedthantheinternationalspacestation.
Asexplainedabove,themassintheequationforthespeedofasatelliteisthemassoftheEarthnotthemasofthesatellite.Thedifferentmasseshavenoeffectontheirspeeds.
Imageswithhigherresolutionareproduced.
MEANINGAlow‐altitudeorbitmeansthatthesatelliteisclosertotheground.Theimagesproducedshowmoredetail.Werefertothisasresolution.
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ThemomentumvectorofBafterthecollision
ThetotalinitialmomentumvectorisjustthatofAasofBwasinitiallystationarybeforethecollision
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Usingthelawofconservationofmomentum,thetotalinitialandfinalmomentumisthesame(bothmagnitudeanddirection).Sincepi=pfthenpAi=pAf+pBfThefinalmomentumvectorisshownonthediagramabove.
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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
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HINTTheelectricfieldisuniform.Theelectricfieldlinesneedtobeevenlyspaced.Theelectricfieldlinespointawayfromtheconductorastheyindicatethedirectionoftheforceonapositivetestcharge.
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F
B 1
40
QqB
r2 9x109 x8x1019 x2.56x1018
0.082 2.9x1024 N (south)
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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.
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ThemagneticfieldduetoI1actsintothepageinthepositionwhereconductor2lies.Usingtherighthandruletheforceonconductor2duetoconductor1isuptheplaneofthepage.ThemagneticfieldduetoI2actsoutofthepageinthepositionwhereconductor1lies.Usingtherighthandruletheforceonconductor1duetoconductor2isdowntheplaneofthepage.Theconductorsattracteachother.
HINTUsetherighthandruleforthemagneticfieldaroundastraightconductor.
HINTAgainusetherighthandruleforthemagneticfieldaroundastraightconductor.
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F Bqvsin 2.5x102 x1.6x1019 x1.45x106 xsin 90 5.8x1015 N
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E V
d 6x104
0.451.3x105 Vm1
F Eq 1.3x105 x(1.6x1019 ) 2.1x1014 N
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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.
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Vertical
v f v
f 3x108
64.25x106 4.67 m
COMMONERRORSIunitsarerequired.MHztoHzx106
REASONINGTheplaneofpolarisationisdefinedastheplaneoftheelectricfield.Theplaneofthemagneticfieldisperpendiculartotheplaneoftheelectricfield.
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Anincandescentglobeemitswhitelight.Whitelightconsistsofallcolours/wavelengthsrangingfromredthroughtoviolet(ROYGBIV).Monochromaticlightconsistsofonecolour/wavelengthonly.Thelightfromanincandescentglobeisnotmonochromatic.
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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)
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SincethedistancebetweenthedoubleslitsandthescreenissomuchlargerthatthedistancebetweenthedoubleslitsthentheangleS1XS2isapproximately90o.UsingthetriangleS1XS2then
sin path difference
d
Foramaxima,thepathdifferenceism
Itfollowsthat
sin path difference
d m
d
ie d sin m
Pathdifference
X
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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.
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Increasingthefilamentcurrentwillreducetheexposuretime.ThisisbecausethefilamentcurrentreleaseselectronsthatcollidewithatargetmetaltoproduceX‐rays.Ifmoreelectronsarereleased,thenmorecollisionsoccurandmoreX‐raysareproducedreducingtheoverallexposuretime.
Exposuretoionisingradiationcanbereducedbywearingappropriateshieldingsuchasaleadlinedapron.
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h
p h
mv 6.63x1034
9.11x1031 x4.36x1061.67x1010 m
Lowenergyelectronswerefiredtowardsacrystallatticeandtheywerediffractedatpreferredanglesratherthanbeingscatteredrandomly.Diffractionisawavephenomena.Inadditiontheequation d sin mwasusedtocalculatethewavelengthoftheelectronsandmatchedthatfoundusingthede
Broglierelationship h
p.
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Thepositionoflinesinthespectraofhydrogenandlithiummatchthelinesinthespectrumofthemixture.Iehydrogenandlithiummustbepresent.
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Thereisnoenergygapbetweenthegroundstateandthehigherenergylevelsthatmatches12.50eV.Thephotonwillnotbeabsorbed.
1.89eV
10.2eV
12.09eV
12.75eV
(b)(i)
(a)
REASONINGThesmallestjumpfromE3toE2willemitthesmallest‐energyphoton.
E3 E2 1.513.4 1.89eV
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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
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Thereexistsastrongnuclearforcebetweenthenucleons(protonsandneutrons)inthenucleusthatisattractiveandstrongerthantherepulsiveelectrostaticforcesbetweenthe82protons.
Thisforceactsovershortdistancesandquicklybecomesnegligibleatseparationsofmorethanafewnucleondiameters.Thisishowthenucleons(whichareveryclose)canbeheldtogetherinastablenucleus.
‐
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neutrino
Theemittedpositroncancollidewithanearbyelectronandannihilate.Themassisconvertedintoenergyintheformoftwoidenticalphotonsthattravelinoppositedirectionssothatmomentumisconserved.
HINTThehalflifeis2days.Thisisconstant.
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REASONINGThephotonstravelinoppositedirections.
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Uranium‐238doesnotreadilyundergoinducedfissionwhereasuranium‐235does.Enrichingthesourcemeansthatalargerpercentageofuranium‐235isaddedtothefuel.Thisensuresthatenoughofthenucleiundergoinducedfissionandachainreactioncanoccur.
E hf hc
6.63x1034 x3x108
502x109 3.96x1019 J
COMMONERRORWavelengthneedstobeinSIunitsnmtommultiplyby10‐9
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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
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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.
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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
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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.
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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
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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
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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.