PHYSICS 111 SPRING 2017

FINAL EXAM: May 2, 2017; 2:15pm - 4:15pm 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

64. Which two temperature changes are equivalent?

A) 1 C° = 1 K B) 1 F° = 1 C° C) 1 K = 1 F° D) None of the above E) All of the above

Solution The Celsius scale and the Kelvin scale are offset by 273.15 K, but a temperature difference of 1 C° is the same as a temperature difference of 1 K.

65. As shown in the figure, a bimetallic strip, consisting of metal G on the top and metal H on the bottom, is rigidly attached to a wall at the left. The coefficient of linear thermal expansion for metal G is greater than that of metal H. If the strip is uniformly heated, it will

A) Remain horizontal, but get longer. B) Remain horizontal, but get shorter. C) Curve out of the page. D) Curve upward. E) Curve downward.

Solution Both metals expand when heated. However, metal G expands more than metal H. Therefore the bi-metallic strip bends down.

66. Some properties of a certain glass are listed here: Density 2300 kg/m3 Specific heat capacity 840 J/(kg • °C) Coefficient of thermal expansion 8.5 × 10-6 (°C)-1 Thermal conductivity 0.80 W/(m • °C) A glass window pane is 2.7 m high, 2.4 m wide, and 9.0 mm thick. The temperature at the inner surface of the glass is 19 °C and at the outer surface 4 °C. How much heat is lost each hour through the window? A) 8.6 J B) 8.6 × 103 J C) 3.1 × 104 J D) 3.1 × 107 J E) 3.1 × 105 J

Solution

Q = kAΔTL

t =0.80 W/ m K( )( ) 2.7 × 2.4 m2( ) 15 K( )

9 ×10−3 m( ) 3600 s( ) = 3.1×10−7 J

67. Object 1 has three times the specific heat capacity and four times the mass of Object 2. The two objects are given the same amount of heat Q. If the temperature of Object 1 changes by an amount ΔT, the change in temperature of Object 2 will be

A) 3/4 ΔT B) 4/3 ΔT C) 6 ΔT D) 12 ΔT E) ΔT

Solution

Q = m2c2ΔT2 ⇒ΔT2 =Qm2c2

= m1c1m2c2

ΔT =4m2( ) 3c2( )m2c2

ΔT = 12ΔT

68. If 50 g of lead (of specific heat 0.11 kcal/(kg • C°) ) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/(kg • C°) ) at 0°C. What is the final temperature of the mixture?

A) 2.0°C B) 25°C C) 6.8°C D) 50°C E) −2.0°C Solution

T =mPbcPbTPb,0 +mwcwTw

mPbcPb +mwcw

=0.050 kg( ) 0.11 kcal/ kg K( )( ) 100 oC( ) + 0.075 kg( ) 1.0 kcal/ kg K( )( ) 0 oC( )

0.050 kg( ) 0.11 kcal/ kg K( )( ) + 0.075 kg( ) 1.0 kcal/ kg K( )( )= 6.8 oC

69. For water, the latent heat of fusion is 334 kJ/kg, the latent heat of evaporation is 2265 kJ/kg, and the specific heat capacity is 4180 J/(kg K). The specific heat capacity of ice is 2050 J/(kg K). You have 1.0 kg of water at 0 °C. After you remove 150 kJ from the water you have

A) 450 g of ice and 550 g of water at 0 °C B) 1,000 g of water at −36 °C C) 1,000 g of ice at −73 °C D) 67 g of ice and 933 g of water at 0 °C E) 550 g of ice and 450 g of water at 0 °C

SolutionThe heat removed from the water will freeze some of the water before the temperature drops below 0 °C. Only after all the water has frozen, will further heat removal result in decreasing the temperature. Removing 150 kJ freezes 0.45 kg of water to ice.

Q = mLfusion ⇒ m = QLfusion

=150 kJ( )

334 kJ/kg( ) = 0.45 kg

70. A sample of an ideal gas is originally at a temperature of 10 °C. What is the final temperature of the gas if both the pressure and volume are doubled?

A) 10 °C B) 40 °C C) 120 °C D) 620 °C E) 860 °C

Solution

PV = nRT ⇒ PVT

= constant

PFVFTF

= P0V0

T0

⇒TF =PFVFP0V0

T0 =2P0( ) 2V0( )P0V0

T0 = 4T0 = 4(283 K) = 1130 K = 860 oC

71. Dust particles in a grain elevator frequently have masses of the order of

1.0 × 10-9 kg. If, to a first approximation, we model the dust particles as an ideal gas, what would be the rms speed of such a particle in air at 300 K?

A) 4.9 × 10-2 m/s B) 7.8 × 10-4 m/s C) 3.5 × 10-6 m/s D) 5.2 × 10-3 m/s E) 5.6 × 10-5 m/s

Solution

vrms =3kTm

=3 1.38 ×10−23 J/K( ) 300 K( )

10−9 kg( ) = 3.5 ×10−6 m/s

72. An ideal Carnot engine operating between a warm reservoir of unknown

temperature and a cold reservoir at 1.76 K has an efficiency of 40.0 %. What is the temperature of the warm reservoir?

A) 2.93 K B) 0.0400 K C) 0.0500 K D) 106 K E) 46.4 K

Solution

eideal = 1− TLTH

⇒TH = TL1− eideal

= 1.76 K1− 0.4

= 2.93 K

73. A car moves along the x-axis with zero acceleration. Which of the following

graphs of either position (x) or velocity (v) versus time (t) represent the motion of the car?

A) only graph a B) only graph b C) graphs a and b D) graphs c and d E) graphs b and c Solution (Instantaneous) acceleration is the slope of velocity with time curve. An object moving with zero acceleration is represented by a velocity versus time curve with slope zero. Zero acceleration is a special case of constant acceleration. For zero acceleration, the position versus time curve has constant slope i.e. it is a straight

line: x(t) = x0 + v0t +12at 2 = x0 + v0t

74. The eastward component of vector is equal to the westward component of

vector and their northward components are equal. Which one of the following statements must be correct for these two vectors?

A) The magnitude of vector must be equal to the magnitude of vector .

B) Vector is antiparallel (in the opposite direction) to vector .

C) The angle between vector and vector must be 90°.

D) Vector must be perpendicular to vector .

E) Vector is parallel to vector .

SolutionIfwedrawacoordinatesystemsuchthatthex-axispointsEastandthey-axispointsnorth,then

Ax = −Bx; Ay = By ⇒

!B = Bx

2 + By2 = −A( )x

2 + Ay2 = Ax

2 + Ay2 =!A

75. Awomanisstrainingtoliftalargecrate,butwithoutsuccessbecauseitistooheavy.Wedenotetheforcesonthecrateasfollows:Pisthemagnitudeoftheupwardforcebeingexertedonthecratebytheperson,Cisthemagnitudeoftheverticalcontactforceonthecratebythefloor,andWistheweightofthecrate.Howarethemagnitudesoftheseforcesrelatedwhilethepersonistryingunsuccessfullytoliftthecrate?

A) P=CB) P+C=W C) P+C>W D) P+C<W E) P–C=W

SolutionInthefree-bodydiagram,CandPpointupandWpointsdown.ThusfromNewton’s2ndlaw:C+P–W=ma=0andthereforeC+P=W

76. A50.0-kgcrateisbeingpulledalongahorizontalfrictionlesssurface.Thepullingforceis10.0Nandisdirected20.0°abovethehorizontal.Whatisthemagnitudeoftheaccelerationofthecrate?

A) 0.376m/s2 B) 0.0728m/s2 C) 0.188m/s2 D) 0.0684m/s2 E) 0.200m/s2 Solution

Fx = max ⇒ ax =Fxm

=10.0 N( )cos 20.0o( )

50.0 kg( ) = 0.188 m/s2

77. Anobjectattachedtoaspringispulledacrossahorizontalfrictionlesssurface.Iftheforceconstant(springconstant)ofthespringis45N/mandthespringisstretchedby0.88mwhentheobjectisacceleratingat1.4m/s2,whatisthemassoftheobject?

A) 36kgB) 24kg C) 31kg D) 28kg E) 44kg

Solution

F = ma = −kx ⇒ m = kxa

=45 N/m( ) 0.88 m( )

1.4 m/s2( ) = 28 kg

78. Howmuchkineticenergydoesa0.30-kgstonehaveifitisthrownat44m/s?

A) 580 J B) 290 J C) 510 J D) 440 J E) 370 J

Solution

KE = 12mv2 = 1

20.30 kg( ) 44 m/s( )2 = 290 J

79. Ablockslidesdownafrictionlessinclinedrampandexperiencesno

significantairresistance.Iftherampangleis17.0°abovethehorizontalandthelengthofthesurfaceoftherampis20.0m,findthespeedoftheblockasitreachesthebottomoftheramp,assumingitstartedslidingfromrestatthetop.

A) 115m/sB) 7.57m/s C) 19.6m/s D) 10.7m/s E) 87.4m/s

SolutionKEf + PEf = KE0 + PE0 ⇒ KEf = PE0

12mv2 = mgh⇒ v = 2gh = 2 9.8 m/s2( ) 20.0 m( )sin 17o( ) = 10.7 m/s

80. Ablockofmassm=34kgandspeedVisbehindablockofmassM=81kgandspeedof0.50m/sasshowninthefigure.Thesurfaceisfrictionlessandtheblockscollideandcouple.Afterthecollision,theblockshaveacommonspeedof0.90m/s.Whatisthemagnitudeoftheimpulseonthe34-kgblockduetothecollision?

A) 32N•sB) 41N•s C) 73N•s D) 57N•s E) 14N•s SolutionMomentumisconservedinthecollision.Therefore,themagnitudeoftheimpulseonthelighterblockisthesameasthemagnitudeoftheimpulseontheheavierblock.J = Δp = m Δv( ) = 81 kg( ) 0.90 m/s − 0.50 m/s( ) = 32 kg m/s

81. AFerriswheelrotatingat20rad/sslowsdownwithaconstantangularaccelerationofmagnitude5.0rad/s2.Howmanyrevolutionsdoesitmakewhileslowingdownbeforecomingtorest?

A) 6.4 B) 20 C) 3.2 D) 40 E) 99

Solution

ω 2 −ω02 = 2αΔθ ⇒ Δθ =

ω 2 −ω02

2α=

20 rad/s( )2−0

2 5 rad/s2( )= 40 rad 1 rev

2π rad⎛

⎝⎜

⎞

⎠⎟= 6.4 rev

82. Let the orbital radius of a planet be R and let the orbital period of the planet be T.

What quantity is constant for all planets orbiting the sun, assuming circular orbits?

A) T/R2 B) T 2/R C) T/R D) T 2/R3 E) T 3/R2

Solution

Kepler’s 3rd law: T2

R3=4π 2

GM

83. A 0.25 kg harmonic oscillator has a total mechanical energy of 4.1 J. If the

oscillation amplitude is 20.0 cm, what is the oscillation frequency?

A) 1.4 Hz B) 2.3 Hz C) 4.6 Hz D) 3.2 Hz E) 0.96 Hz

Solution

KEmax =12mvmax

2 ⇒ vmax =2KEmax

m; vmax = Aω⇒ f = ω

2π=

2KEmax

m2πA

= 4.6 Hz

84. A1.1-kguniformbarofmetalis0.40mlongandhasadiameterof0.020m.

Whensomeonebangsoneendofthisbar,a1.5×106Hzshockwavetravelsalongthelengthofthebarandreachestheotherendin0.12×10−3s.Whatisthewavelengthoftheshockwaveinthemetal?

A) 3.8 mm B) 2.2 mm C) 3.0 mm D) 3.4 mm E) 2.6 mm

Solution

λ =vf=L / tf

=0.40 m( )

1.5×106 Hz( ) 0.12×10−3 s( )= 2.2×10−3 m

85. Ablockofwoodisfloatinginabathtub.Asecondblockofthesametypeof

woodsitsontopofthefirstblock,anddoesnottouchthewater.Ifthetopblockistakenoffandplacedinthewater,whathappenstothewaterlevelinthetub?

A) Itgoesup.B) Itgoesdown. C) Itdoesnotchange. D) Itcannotbedeterminedwithoutknowingthevolumesofthetwopieces

ofwood. E) Itcannotbedeterminedwithoutknowingthedensityofthewood.

SolutionThetwopiecesofwoodhavethesamedensity.Thebuoyantforceisequaltotheweightofthetwoblocksandalsoequaltotheweightofthedisplacedwater.Theweightoftheblocksdoesnotchangewhenthesecondblockisputinthewater.Neitherdoesthevolumeofthedisplacedwater.

86. Inahydraulicgaragelift,thesmallpistonhasaradiusof5.0cmandthelargepistonhasaradiusof15cm.Whatforcemustbeappliedonthesmallpistoninordertoliftacarweighing20,000Nonthelargepiston?Assumethepistonseachhavenegligibleweight.

A) 5.0×103NB) 2.9×103N C) 6.7×103N D) 2.2×103N E) 7.8×103N

SolutionFLAL

=FSAS

⇒ FS =ASAL

FL =πRS

2

πRL2 FL =

RSRL

⎛

⎝⎜

⎞

⎠⎟

2

FL =5.0 cm15 cm⎛

⎝⎜

⎞

⎠⎟

2

20 kN( ) = 2.2 kN

87. You are using a wrench to loosen a rusty nut. Rank the arrangements from most effective to least effective in loosening the nut?

A) 2 = 4, 1 = 3 B) 3 = 4, 1 = 2 C) 2, 1 = 3 = 4 D) 2, 1 = 4, 3 E) 2 = 4, 1, 3

SolutionBecause the forces are all the same, the only difference is the lever arm. The arrangement with the largest

( ) will provide the lever arm b largest . The lever arm, between line of torque

action and rotation axis, for (a) and (d) is the same and larger as for (c), since in (c) the force F is not perpendicular to the wrench.

88. You stand a distance d away from a speaker and you hear a certain intensity I of sound. If you decrease your distance from the speaker to d/3, what is the sound intensity at your new position?

A) 3I B) 6I C) 9I D) 4I E) does not change at all Solution For a source of power P, the intensity is given by I = P/(4πd2). If the distance

, the intensity must its original value. decreases by a factor of 3 increase to 9 times

89. A mass attached to a vertical spring causes the spring to stretch and the mass to move downwards. What can you say about the spring’s potential energy (PES) and the gravitational potential energy (PEG) of the mass during that process?

A) both PES and PEG decrease B) PES increases and PEG decreases C) both PES and PEG increase D) PES decreases and PEG increases E) PES increases and PEG is constant

Solution The spring is stretched, so its elastic PE increases, because PES = ½ kx2. The mass moves down to a lower position, so its gravitational PE decreases, because PEG = mgh.

Laboratory final exam 90. Inacertainexperiment,weknowthatthemagnitudeofthevelocityofapuck

canbebetween40and50cm/s.Wealsoknowthattheanglebetweenthexaxisanddirectionofthisvelocitycanvarybetween30and40degrees.

Findthemaximumvalueofthexcomponentofthevelocityallowedbytheseranges.A. 31cm/sB. 32cm/sC. 38cm/sD. 43cm/sE. 45cm/s

Thexcomponentofthevelocityis θ= cosxv v .Therefore,themaximumvalueis:

( ) ( )θ= = ° =,max max mincos 50cm/s cos 30 43.3cm/sxv v 91. AccordingtoKepler’sthirdlaw,therelationbetweentheperiodTofasatellite

orbitingtheEarthonacirculartrajectoryandtheradiusroftheorbitis:

π=

32 4

Earth

rTGm

Ifwemeasuretheperiodandtheradiusforseveralsuchsatellites,whichofthefollowingplotsisexpectedtodisplayalineardependence?

32

2

3

A. versus

B. versus

C. versus

D. versus

E. versus

T r

T r

T r

r

r

T

T

Theequationabovecanberewrittenas π π= =

3324 4

Earth Earth

rT rGm Gm

,soTvsr3/2isa

straightlinethatgoesthroughtheoriginandwithslope π4

EarthGm

92. Acartmovesalongatrack.Itspositionismeasuredwithanultrasoundmotion

detectorclampedtooneendofthetrack.Thepositionandtimedataisfedontoacomputer,andnumericalderivativesareusedtoestimatetheinstantaneousvelocityofthecartasafunctionoftime.Theresultsareshowninthegraphbelow.

Whatistheaccelerationofthecartintheportionofthemotionatt=4.0s?

A. −1.0m/s2B. −1.5m/s2C. 1.0m/s2D. 1.5m/s2E. 4.0m/s2

Theaccelerationofthecartistheslopeofthisgraph.Intheintervalbetween2sand6s,theslope,andthustheacceleration,areconstant:

( )( )−Δ

= = = −Δ −

22 8 m/s1.5m/s

6 2 svat

93. Thefollowingapparatuswasusedinexperiment“SF–Staticfriction”.AblockofmassMrestsonahorizontalramp.Twolightstringsareattachedtotheblockasshown.Thestringsgothroughasystemofpulleysthathaveverylittlefriction,andweightsofmassesm1andm2hangattheotherendofthestrings.Whenthefollowingvaluesareused,theblockisatrest:

M=25gm1=4.5gm2=4.0g

Whatisthemagnitudeofthestaticfrictionbetweentheblockandtheramp?A. 0.012NB. 0.026NC. 0.039ND. 0.044NE. Thiscannotbeansweredwithoutthecoefficientofstaticfrictionbetween

theblockandtheramp.

Thefreebodydiagramoftheblockisshowntotheright.TheonlytwoforcesinthehorizontaldirectionareT1andthestaticfriction,fs.Sincetheblockisatrest,theymusthaveequalmagnitude.Ontheotherhand,T1mustbeequaltom1gbecausehanger1,withmassm1,isalsoatrest.Therefore,fs=m1g=0.044N.

Mg

N

Physics 111 Final Exam – KEY

64 A 74 A 84 B

65 E 75 B 85 C

66 D 76 C 86 D

67 D 77 D 87 D

68 C 78 B 88 C

69 A 79 D 89 B

70 E 80 A 90 D

71 C 81 A 91 E

72 A 82 D 92 B

73 C 83 C 93 D