physics 111 spring 2019 final exam: may 8, 2019; 4:30pm...

PHYSICS111SPRING2019

FINALEXAM:May8,2019;4:30pm-6:30pmName(printed):______________________________________________RecitationInstructor:_________________________Section#_______INSTRUCTIONS:Thisexamcontains30multiple-choicequestion,eachworth3points,foranominalmaximumscoreof80pointsplus10extracreditpoints.Chooseoneansweronlyforeachquestion.Choosethebestanswertoeachquestion.Answerallquestions.Allowedmaterial:Beforeturningoverthispage,putawayallmaterialsexceptforpens,pencils,erasers,rulersandyourcalculator.Thereisaformulasheetattachedattheendoftheexam.Othercopiesoftheformulasheetarenotallowed.Calculator:Ingeneral,anycalculator,includingcalculatorsthatperformgraphing,ispermitted.Electronicdevicesthatcanstorelargeamountsoftext,dataorequations(likelaptops,e-bookreaders,smartphones)areNOTpermitted.CalculatorswithWiFitechnologyareNOTpermitted.Ifyouareunsurewhetherornotyourcalculatorisallowedfortheexam,askyourTA.Howtofillinthebubblesheet:

Useanumber2pencil.DoNOTuseink.Ifyoudidnotbringapencil,askforone.YouwillcontinuetousethesamebubblesheetyoualreadyusedforExam1-3.Bubbleanswers64-93onthebubblesheetforthisexam.Only,ifforsomereasonyouarestartinganewbubblesheet,writeandfillinthebubblescorrespondingto:- Yourlastname,middleinitial,andfirstname.- « « YourIDnumber(themiddle9digitsonyourISUcard)« « - SpecialcodesKtoLareyourrecitationsection.Alwaysusetwodigits(e.g.01,09,11,13).

Pleaseturnoveryourbubblesheetwhenyouarenotwritingonit.Ifyouneedtochangeanyentry,youmustcompletelyeraseyourpreviousentry.Also,circleyouranswersonthisexam.Beforehandinginyourexam,besurethatyouranswersonyourbubblesheetarewhatyouintendthemtobe.Youmayalsocopydownyouranswersonapieceofpapertotakewithyouandcomparewiththepostedanswers.Youmayusethetableattheendoftheexamforthis.Whenyouarefinishedwiththeexam,placeallexammaterials,includingthebubblesheet,andtheexamitself,inyourfolderandreturnthefoldertoyourrecitationinstructor.Nocellphonecallsallowed.Eitherturnoffyourcellphoneorleaveitathome.Anyoneansweringacellphonemusthandintheirwork;theirexamisover.

Bestofluck,Dr.SoerenPrell

64. Asshowninthefigure,abimetallicstrip,consistingofmetalGonthetopandmetalHonthebottom,isrigidlyattachedtoawallattheleft.ThecoefficientoflinearthermalexpansionformetalGisgreaterthanthatofmetalH.Ifthestripisuniformlyheated,itwill

A) Curvedownward.B) Remainhorizontal,butgetlonger.C) Neitherchangeshapenorlength.D) Remainhorizontal,butgetshorter.E) Curveupward.

SolutionBothmetalswillexpandwhenheated.MetalGwillexpandmorethanmetalHandthebimetallicstripwillcurvedownward.

65. Atemperaturechangeof20C°correspondstoaFahrenheittemperaturechangeof

A) 11FoB) 36FoC) 18FoD) 68FoE) 293Fo

Solution

TF =

95TC + 32 F!⇒ΔTF =TF −TF ,0 =

95TC + 32 F!

⎛

⎝⎜

⎞

⎠⎟−

95TC ,0 + 32 F!

⎛

⎝⎜

⎞

⎠⎟=

95ΔTC =

95

20 C!( ) = 36 C!

66. Aquantityofanidealgasiskeptinarigidcontainerofconstantvolume.Ifthegasisoriginallyatatemperatureof19°C,atwhattemperaturewillthepressureofthegasdoublefromitsoriginalvalue?

A) 273°CB) 38°CC) 122°CD) 91°CE) 311°C

Solution

T0 =19 !C = 19+273( )K = 292K; TP=T0

P0

⇒ T =T0PP0

= 2T0 = 2×292K = 584K = 584 −273( )!C = 311!C

67. Thefigureshowsagraphofthetemperatureofa

puresubstanceasafunctionoftimeasheatisaddedtoitataconstantrateinaclosedcontainer.IfLFisthelatentheatoffusionofthissubstanceandLVisitslatentheatofvaporization,whatisthevalueoftheratioLV/LF?

A) 7.2B) 3.5C) 4.5D) 5.0E) 1.5

SolutionThetimeittakestomeltthesubstanceis2unitsoftime(1sthorizontalsegment)inthegraphwhereasthetimeittakestoevaporateitisabout7units(2ndhorizontalsegment).Sinceheatisaddedataconstantratethetimesareproportionaltotherespectiveamountsofheatadded.SinceQ=mLandthemassmofthesubstanceisconstantLV/LF=3.5

68. Acamperisabouttodrinkhismorningcoffee.Hepours400gramsofcoffee,initiallyat75°Cintoa250-galuminumcup,initiallyat16°C.Whatistheequilibriumtemperatureofthecoffee-cupsystem,assumingnoheatislosttothesurroundings?Thespecificheatofaluminumis900J/kg•K,andthespecificheatofcoffeeisessentiallythesameasthatofwater,whichis4186J/kg•K.

A) 45oCB) 68oCC) 65oCD) 71oCE) 62oC

Solution

Qcoffee +Qcup = 0⇒ mcoffeeccoffee Tcoffee −T( ) + mcupccup Tcup −T( ) = 0

T =mcoffeeccoffeeTcoffee + mcupccupTcup

mcoffeeccoffee + mcupccup

= 68o C

69. Thethermodynamicprocessforanidealgasshownon

thepVdiagraminthefigureis

A) adiabaticB) isobaricC) isothermalD) isochoricE) notpossible

SolutionThere’snochangeinvolume.Thisprocesscouldbetheheatingupofafixedamountofgasinsideafixed-volumecontainer.Thepressureofthegasincreasesproportionaltotemperature.Constant-volumeprocessesarecalledisochoric.

70. Amonatomicidealgasundergoesanisothermalexpansionat300K,asthevolumeincreasesfrom0.010m3to0.040m3Thefinalpressureis130kPa.Whatisthechangeintheinternal(thermal)energyofthegasduringthisprocess?

A) 0.0kJB) –7.2kJC) +7.2kJD) +3.6kJE) –3.6kJ

SolutionTheinternalenergyofamonatomicidealgasisU=3/2nRT.Duringanisothermalprocessthetemperatureandtheinternalenergydonotchange.

71. ThefigureshowsapVdiagramforanidealgasthatiscarriedaroundacyclicprocess.Howmuchworkisdoneinonecycleifp0=4.00atmandV0=3.00liters?

A) 4850JB) 2420JC) 485JD) 2280JE) 1210J

Solution

W = (area enclosed by the path) = ΔpΔV = p0V0 = 4×1.01×105 Pa( ) 3.00×10−3 m3( ) = 1210 J

72. Anarchitectisinterestedinestimatingthe

rateofheatloss,ΔQ/Δt,throughasheetofinsulatingmaterialasafunctionofthethicknessofthesheet.Assumingfixedtemperaturesonthetwofacesofthesheetandsteadystateheatflow,whichoneofthegraphsshowninthefigurebestrepresentstherateofheattransferasafunctionofthethicknessoftheinsulatingsheet?

A)AB)BC)CD)DE)ESolution

Qt= kA

lΔT ∝ 1

l

73. Theprocessinwhichheatflowsbythemassmovementofmoleculesfromoneplacetoanotherisknownas

A)conductionB)radiationC)convectionD)sublimationE)dissipation

SolutionThethreeformsofheattransferareconduction,radiationandconvection.Nomatterisbeingtransportedinthefirsttwo.Onlyconvectioninvolvesthemovementofmolecules.

74. Themotionofaparticleisdescribedinthevelocityvs.timegraphshowninthefigure.Overthenine-secondintervalshown,wecansaythatthespeedoftheparticle

A) RemainsconstantB) Decreasesandthenincreases.C) Onlydecreases.D) Onlyincreases.E) Increasesandthendecreases.SolutionInitiallythespeedishigh(about2.5m/s).Thenthespeedgoestozero,andthenincreasesagain(toabout4.0m/s).

75. Aballisthrownstraightup,reachesamaximumheight,thenfallstoitsinitialheight.Whichofthefollowingstatementsaboutthedirectionofthevelocityandaccelerationoftheballthehighestpointofitstrajectoryiscorrect?

A) Bothitsvelocityanditsaccelerationpointdownward.B) Bothitsvelocityanditsaccelerationpointupward.C) Itsvelocityiszeroanditsaccelerationpointsdownward.D) Itsvelocitypointsupwardanditsaccelerationiszero.E) Itsvelocityiszeroandanditsaccelerationiszero.SolutionAtthehighestpointthevelocityiszerosinceitchangesdirection.Theaccelerationatanytimeisg=9.8m/sanditpointsdownward(tothecenteroftheEarth).

76. Aboyjumpswithavelocityofmagnitude20.0m/satanangleof25.0°abovethe

horizontal.Whatisthehorizontalcomponentoftheboy'svelocity?

A)8.45m/sB)12.6m/sC)9.33m/sD)15.6m/sE)18.1m/sSolution

vx = vcosθ = 20.0 m/s( )cos 25.0!( ) = 18.1 m/s

77. Apersonwhonormallyweighs700Nisridinginaratherswiftelevatorthatismovingat

aconstantspeedof9.8m/s.Ifthispersonisstandingonabathroomscaleinsidetheelevator,whatwouldthescaleread?

A)morethan700NB)lessthan700N,butmorethanzeroC)0D)700NE)Itcouldbemoreorlessthan700N,dependingonwhethertheelevatorisgoingupordown.

SolutionTheelevatorismovingwithconstantspeed.Thus,thenormalforcebytheelevatoronthepersonequalstheirweightregardlessofthedirectionormagnitudeofthevelocity.

78. Inthefigure,whatdoesthespringscaleread?

Thepulleysareidealandthestringsandscalearealsomassless.

A)2.0NB)morethan19.6NC)0.0ND)1.0NE)0.5NSolutionThespringscalereadsthetensioninthestring,whichis1.0N.

79. A3.0-kgmassanda5.0-kgmasshangverticallyattheoppositeendsofaverylightrope

thatgoesoveranidealpulley(alsocalledAtwood’smachine).Ifthemassesaregentlyreleased,whatistheresultingaccelerationofthemasses?A)6.1m/s2B)3.7m/s2C)4.9m/s2D)2.5m/s2E)0.0m/s2

SolutionTheheaviermasswillacceleratedownwards.Choosethepositivedirectiontobeinthedirectionoftheacceleratio

(5.0 kg)g −T = (5.0 kg)a; T − (3.0 kg)g = (3.0 kg)a

⇒ (5.0 kg)g − (3.0 kg)g = (5.0 kg)a + (3.0 kg)a ⇒ a = 28

g = g4= 2.5 m/s

80. Aforceactsonanobject,causingittomoveparalleltotheforce.Thegraphinthefigure

showsthisforceasafunctionofthepositionoftheobject.Howmuchworkdoestheforcedoastheobjectmovesfrom6mto12m?

A) 70JB) 0JC) 20JD) 30JE) 40J

Solution

W = 1

212 m − 6 m( ) 10 N( ) = 30 J

81. Agirlthrowsastonefromabridge.Considerthefollowingwaysshemightthrowthestone.Thespeedofthestoneasitleavesherhandisthesameineachcase.CaseA:Thrownstraightup.CaseB:Thrownstraightdown.CaseC:Thrownstraightouthorizontally.Inwhichcasewillthespeedofthestonebegreatestwhenithitsthewaterbelowifthereisnosignificantairresistance?

A)caseAB)caseBC)casesAandBD)caseCE)Thespeedwillbethesameinallcases.SolutionMechanicalenergyisconservedduringfreefallwithnoairresistance.Inallcasesaretheinitialkineticandpotentialenergiesthesame.Thefinalpotentialenergiesarealsothesame.Therefore,thefinalkineticenergiesandthusspeedshavetobethesameaswell.

82. Asshowninthefigure,agivenforceisappliedtoarodinseveraldifferentways.InwhichcaseisthetorqueaboutthepivotpointPduetothisforcethegreatest?

A)1B)2C)3D)4E)5SolutionForcase1,theforceisperpendiculartothelineconnectingthepointPandthepointwheretheforceapplies.Inaddition,thedistancebetweenPandthepointwheretheforceappliesisthelargest.Therefore,case1hasthelargestlevelarm.

83. Astoneinitiallymovingat8.0m/sonalevelsurfacecomestorestduetofrictionafterittravels11m.Whatisthecoefficientofkineticfrictionbetweenthestoneandthesurface?

A)0.13B)0.43C)0.80D)0.30E)0.25Solution

ΔKE =W = −Fs = − fks = −µmgs = −KE0 = − 1

2mv2 ⇒ µ = v2

2gs= 0.3

84. Ameterstickbalancesatthe50.0-cmmark.Ifamassof50.0gisplacedatthe90.0-cmmark,thestickbalancesatthe61.3-cmmark.Whatisthemassofthemeterstick?

A)89.7gB)178gC)32.6gD)73.4gE)127g

Solution

61.3 cm −50 cm( )m = 90 cm − 61.3 cm( ) 50 g( )⇒ m = 127g

85. Twoiceskaterssuddenlypushoffagainstoneanotherstartingfromastationaryposition.The45-kgskateracquiresaspeedof0.375m/srelativetotheice.Whatspeeddoesthe60-kgskateracquirerelativetotheice?A)0.38m/sB)0.50m/sC)0.28m/sD)0.75m/sE)0.00m/sSolution

ΔPtotal = 0⇒ m1v1 + m2v2 = 0⇒ v2 =

m1

m2

v1 = 0.28 m/s

86. Saltwaterhasgreaterdensitythanfreshwater.Ablockofwoodfloatsinbothfreshwaterandinsaltwaterandablockofironsinksinbothfreshwaterandinsaltwater.ThebuoyantforceonthewoodblockinsaltwateriscalledFW,SandinfreshwateritiscalledFW,F.ThebuoyantforceontheironblockinsaltwateriscalledFI,SandinfreshwateritiscalledFI,F.Whichoftherelationsbetweenthebuoyantforcesiscorrect?A)FW,S>FW,FandFI,S>FI,FB)FW,S=FW,FandFI,S>FI,FC)FW,S>FW,FandFI,S=FI,FD)FW,S=FW,FandFI,S=FI,FE)FW,S<FW,FandFI,S<FI,FSolutionThebuoyantforceonafloatingobjectisalwaysequaltoitsweight.Thebuoyantforceofanyfullyorpartiallysubmergedobjectisequaltotheweightofthedisplacedfluid.

87. TheintensityofthewavesfromapointsourceatadistancedfromthesourceisI.Whatis

theintensityatadistance2dfromthesource?

A)2IB)I/4C)I/√2D)I/2E)4I

Solution

I(d) = P4πd 2 ⇒ I(2d) = P

4π 2d( )2 = I(d)4

88. Anobjectofmassm=8.0kgisattachedtoanidealspringandallowedtohangintheearth'sgravitationalfield.Thespringstretches2.2cmbeforeitreachesitsequilibriumposition.Ifitwerenowallowedtooscillatebythisspring,whatwouldbeitsfrequency?

A)0.52HzB)3.4HzC)0.20×10−3HzD)1.6HzE)21HzSolution

F = kΔx = mg; f = ω

2π= 1

2πkm

= 12π

gΔx

= 3.4 Hz

89. A0.588mlongstringwithamassof0.031kg,istightlyclampedatbothends.Iftheloweststandingwavefrequencyofthestringis326Hz,howfastdowavestravelonthisstring?

A)724m/sB)383m/sC)582m/sD)475m/sE)Impossibletotellwithoutknowingthetensioninthestring.

Solution

v = f λ = 326 Hz( ) 2× 0.588 m( ) = 383 m/s

Laboratoryfinalexamquestions

90. Inacertainexperiment,wemeasuretwoquantities:a=60unitsandb=30units,andweuse

thosevaluestoestimatetheratiobetweenthem:

2.0ab=

However,weknowthatourmeasurementscouldbeeachoffby3units.Whatisthentheassociatedmaximumerrorofourestimationofa/b?

A. Exactly0,becausetheerrorscancelout.B. 0.1C. 0.2D. 0.3E. 0.4

Theminimumandmaximumvaluesofthisratioare:

min

min max

max

max min

60 3 1.730 360 3 2.330 3

aab b

aab b

−= = =++= = =−

Therefore,thevaluesoftheratiocanbe0.3aboveorbelowthecentralvalueof2.0.

91. Theplotbelowshowsthepositionofacaralongastraightstreet.Asyoucansee,thedataisperfectforalinearfit.

Theslopeofthelinearfitrepresentsthe______________ofthecar,anditsintersectionwiththeverticalaxisrepresentsthe_________________ofthecar.A. acceleration;initialpositionB. acceleration;finalpositionC. velocity;initialpositionD. velocity;finalpositionE. velocity;initialtime

Theslopeofapositionversustimegraphisalwaysthevelocity.Theintersectionwiththevertical(position)axiscorrespondstothepositionvalueatt=0,sothisistheinitialpositionofthecar.

92. StudentstakingPHYS111onplanetXperformthefollowingexperimenttomeasuretheaccelerationofgravityonthesurfaceofthatplanet:theydropasmallrockfromdifferentheightshandmeasurethetimetherocktakestohittheground.Theythenplotthedatainthegraphsbelow.

Basedontheseresults,whatisthevalueofgonplanetX?A. 4.5m/s2B. 7.5m/s2C. 8.2m/s2D. 9.0m/s2E. 18m/s2

Forany1Dmotionwithconstantacceleration, 20 0

12y yy y v t a t= + + .Ifwetakey=0atgroundlevel

andpositivesup,thenwehavey0=h,y=0,v0y=0,anday=−g,whichyields:21

2h gt=

Indeed,thehvs.t2plotshowslinearbehaviorandintersectstheverticalaxisveryclosetotheorigin,soitlookslikeitfitsthisequation.Wecanthenidentifytheslopeofthelinearfitwithg/2.Thelinearfit(showninredonthegraph)containspoints(0,0)and(4s2,18m),sotheslopeis:

( )( )

22

18 0 m4.5m/s

4 0 sslope

−= =

−

Therefore, 22 9.0m/sg slope= × =

93. A0.50-kgpuckmovesatconstantspeedontheplaneofthepaperandleavesamarkonthesurfacewithafrequencyof60Hz.Eachofthesemarksisacircleinthefigurebelow.Thedistancebetweenthefirstandlastcircleis35cm.Findthelinearmomentumofthepuck.

A. 1.2kgm/sB. 1.3kgm/sC. 1.7kgm/sD. 2.2kgm/sE. 2.9kgm/s

( )( )

= = = =

= = =

35cmseparationbetweentwomarks 8 262.5cm/s 2.625m/s

1timebetweentwomarks s60

0.50kg 2.625m/s 1.3kgm/s

v

p mv

35cm

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