PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm...

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PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm - 6:30pm Name (printed): ______________________________________________ Recitation Instructor: _________________________ Section #_______ INSTRUCTIONS: This exam contains 30 multiple-choice question, each worth 3 points, for a nominal maximum score of 80 points plus 10 extra credit points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, e-book readers, smart phones) are NOT permitted. Calculators with WiFi technology are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet: Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. You will continue to use the same bubble sheet you already used for Exam 1-3. Bubble answers 64-93 on the bubble sheet for this exam. Only, if for some reason you are starting a new bubble sheet, write and fill in the bubbles corresponding to: - Your last name, middle initial, and first name. - «« Your ID number (the middle 9 digits on your ISU card) «« - Special codes K to L are your recitation section. Always use two digits (e.g. 01, 09, 11, 13). Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over. Best of luck, Dr. Soeren Prell

Transcript of PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm...

Page 1: PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm ...courses.physics.iastate.edu/phys111/exams/Exam... · 64. As shown in the figure, a bimetallic strip, consisting of metal

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

Page 2: PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm ...courses.physics.iastate.edu/phys111/exams/Exam... · 64. As shown in the figure, a bimetallic strip, consisting of metal

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

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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

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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

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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.

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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

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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

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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.

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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

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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

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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

Page 12: PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm ...courses.physics.iastate.edu/phys111/exams/Exam... · 64. As shown in the figure, a bimetallic strip, consisting of metal

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.

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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.

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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= × =

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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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