PHYSICS 111 SPRING 2017

EXAM 2: March 7, 2017; 8:15 - 9:45 pm Name (printed): ______________________________________________ Recitation Instructor: _________________________ Section #_______ INSTRUCTIONS: This exam contains 20 multiple-choice questions plus 1 extra credit question, each worth 3 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. Devices (including calculators and smart watches) with WiFi technology are NOT permitted. If you are unsure if 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. Bubble answers 22-42 on the bubble sheet for this exam. If you did not take the first exam, 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

22. A spring scale is tied to two ropes that run

over pulleys as shown in the figure. The pulleys are ideal (massless and frictionless) and the masses of the strings and the scale are negligible. What force does the spring scale read?

A) 2.0 N B) 0.5 N C) 0.0 N D) 1.0 N E) More than 2.0 N

Solution The scale reads the tension in the ropes. It is 1.0 N. This would be the same, if one end of the scale would be tied to a wall.

23. A 10-kg block is held at rest on a frictionless horizontal table and is connected by a horizontal string to a 63-kg block that is hanging over the edge of the table. What is the magnitude of the acceleration of the 10-kg block when it is released? A) 8.1 m/s2 B) 9.0 m/s2 C) 8.5 m/s2 D) 7.5 m/s2 E) 6.6 m/s2

Solution 63 kg( )g −T = 63 kg( )a; T = 10 kg( )a

→ 63 kg( )g − 10 kg( )a = 63 kg( )a→ a = 63 kg( )73 kg( ) g = 8.5 m/s2

24. A 6.00-kg box is held at rest by two light wires that form 30° angles with the vertical. An external force of magnitude F acts vertically downward on the box. The tension exerted by each of the two wires is denoted by T. A free-body diagram, showing the four forces that act on the box, is shown in the figure. If the magnitude of force F is 410 N, what is the magnitude of the tension T? A) 376 N B) 271 N C) 470 N D) 235 N E) 188 N Solution

2T cos 30o( )−w − F = 0→ T = F +w2cos 30o( ) =

410 N( ) + (6 × 9.8 N)2cos 30o( ) = 271 N

25. Twofriskyottersslidedownfrictionlesshillsidesofthesameheightbut

differentslopes.Theslopeofthehillofotter1is30°(withthehorizontal),whiletheslopeofthehillofotter2is60°.Ifbothstartfromrest,whichotterismovingfasterwhenshereachesthebottomofherhill?

A) Otter1ismovingfaster.B) Otter2ismovingfaster.C) Bothottershavethesamespeedatthebottom.D) Theotterthattooktheshortertimeismovingfaster.E) Theheavierotterismovingfaster,nomatterwhichhillsheused.

Solution

PEf + KEf = PE0 + KE0 → mgyf +12mvf

2 = mgy0 +12mv0

2;

12mvf

2 = mg(y0 − yf )→ vf = 2g(−Δy) Iftheinitialheightisthesamethefinalspeedwillbethesameforthetwootters.

26. Anidealspringwithaforceconstantof15N/misinitiallycompressedby3.0cmfromitsuncompressedposition.Byhowmuchisthepotentialenergystoredinthespringchangedwhenthespringiscompressedbyanadditional4.0cm?

A) 0.012JB) 0.0068JC) 0.030JD) 0.024JE) 0.0034J

Solution

PEf − PE0 =12kx f

2 − 12kx0

2 = 12k x f

2 − x02( ) = 1

215 N/m( ) 0.07 m( )2 − 0.03 m( )2( ) = 0.030 J

27. Thehorizontalforcethatananimalexerts

onalargefruitithasfoundisobservedandshowninthegraphinthefigure.Ifthefruitwasinitiallyslidingonthefrictionlessgroundat5.5cm/swhentheanimalfirsttouchedit,byhowmuchdidtheanimalchangeitskineticenergyduringthisencounter?A) 50JB) 0JC) 25JD) 12.5JE) 22J

Solution Thechangeinkineticenergyofthefruitisequaltotheworkdoneonthefruit.Theworkdoneonthefruitistheareaundertheforce(F)versusdistance(x)curve.

28. Apersoncarriesa2.00-Npebblethrough

thepathshowninthefigure,startingatpointAandendingatpointB.ThetotaltimefromAtoBis6.75min.HowmuchworkdidgravitydoontherockbetweenAandB?A) −30.0JB) 56.0JC) 30.0JD) −56.0JE) −36.0J

Solution Wg = −PE = PE0 − PEf = mg y0 − yf( ) = 2.0 N( ) 1.5 m −16.5 m( ) = −30 J

29. PersonXpushestwiceashardagainstastationarybrickwallaspersonY.Whichoneofthefollowingstatementsiscorrect?

A) BothdopositiveworkandpersonXdoesfourtimestheworkofpersonY.B) BothdopositiveworkandpersonXdoestwicetheworkofpersonY.C) Bothdothesameamountofpositivework.D) Bothdozerowork.E) Bothdonegativework.

Solution Withoutdisplacementnoworkisdone.

30. Arailroadcarcollideswithandstickstoanidenticalrailroadcarthatisinitially

atrest.Afterthecollision,thekineticenergyofthesystem

A) isonefifthasmuchasbefore.B) isonequarterasmuchasbefore.C) isonethirdasmuchasbefore.D) ishalfasmuchasbefore.E) isthesameasbefore.

Solution

mv0 = (2m)vf → vf =v0

2;

KEf

KE0

=

12

2m( )vf2

12mv0

2=

2vf2

v02 = 2v0

2 / 4v0

2 = 12

31. Givethexcoordinateofthecenterofmassfor

thesystemoffourpointmasseslocatedatthepositionsshowninthefigure.

A) 2.0mB) 2.3mC) 2.5mD) 2.7mE) 2.9m

Solution

xCM =(8 kg) 0 m( ) + (4 kg) 3 m( ) + (6 kg) 5 m( ) + (2 kg) 2 m( )

(8 + 4 + 6 + 2) kg= 46 kg m

20 kg= 2.3 m

32. Acargoesaroundacircularcurveonahorizontalroadatconstantspeed.Whatisthedirectionofthefrictionforceonthecarduetotheroad?

A) perpendiculartothecurveinwardB) tangenttothecurveoppositetothedirectionofthecar'smotionC) tangenttothecurveintheforwarddirectionD) perpendiculartothecurveoutwardE) Thereisnofrictiononthecarbecauseitsspeedisconstant.

Solution Friction provides the centripetal or radial force to keep the car on the curve. It points inward to the center of the circular curve.

33. AplanethasfourtimestheEarth'smassandtwiceitsradius.Ifajarofpeanutbutterweighs12NonthesurfaceoftheEarth,howmuchwoulditweighonthesurfaceofthisplanet?

A) 6.0NB) 36NC) 24ND) 3.0NE) 12N

Solution

Wp

WE

=GmMp

Rp2

G mME

RE2

=

Gm 4ME( )2RE( )2

G mME

RE2

= 1→Wp =WE

34. Whenafanisturnedoff,itsangularspeeddecreasesfrom13.7rad/sto6.3

rad/sin10.0s.Whatisthemagnitudeoftheaverageangularaccelerationofthefan?

A) 1.2rad/s2B) 0.86rad/s2C) 0.74rad/s2D) 11rad/s2E) 0.37rad/s2

Solution

ω =ω 0 +αΔt→ α = ω −ω 0

Δt=

6.3 rad/s( )− 13.7 rad/s( )10.0 s

= −0.74 m/s2 = 0.74 m/s2

35. AFerriswheelhasaperiodof120seconds,aradiusof

15meters,amassof12,000kgandrotatescounter-clockwise.Whatisthecentripetal(radial)forceonapersonwithmassof80kginaGondolaathalfthemaximumheight(seefigure)?

A) 784NB) 0NC) 3.3ND) 117x103NE) 42N

Solution

FC = mRω 2 = mR 2πT

⎛⎝⎜

⎞⎠⎟

2

= 80 kg( ) 15 m( ) 2π120 s

⎛⎝⎜

⎞⎠⎟

2

= 3.3 N

36. Auniformrodis2.0mlong.Itishingedtoa

wallatitsleftend,andheldinahorizontalpositionatitsrightendbyaverticalverylightstring,asshowninthefigure.Whatistheangularaccelerationoftherodatthemomentafterthestringisreleasedifthereisnofrictioninthehinge?(ThemomentofinertiaofarodwithmassMandlengthLrotatingaroundoneendis1/3ML2.)A) 3.3rad/s2B) 15rad/s2C) 11rad/s2D) 7.4rad/s2E) Itcannotbecalculatedwithoutknowingthemassoftherod.Solution

τ = Iα →α = τI=Mg L

2⎛⎝⎜

⎞⎠⎟

13ML2

= 3g2L

= 7.4 rad/s2

37. A15-kgchildissittingonaplaygroundteeter-totter,1.5mfromthepivot.Whatisthemagnitudeoftheminimumforce,applied0.30montheothersideofthepivot,thatisneededtomakethechildliftofftheground?

A) 66NB) 75NC) 44ND) 740NE) 23N

Solutionτ F ≥ τ child → F 0.3 m( ) ≥ 15 kg( ) 9.8 m/s2( ) 1.5 m( )→ F ≥ 740 N

38. Astore'ssignhasamassof20kgandis3.0m

long.Itisuniform,soitscenterofgravityisatthecenterofthesign.Itissupportedhorizontallybyasmalllooseboltattachedtothewallatoneendandbyawireattheotherend,asshowninthefigure.Whatisthetensioninthewire?

A) 460NB) 300NC) 230ND) 200NE) 120N

Solution

τ wire −τ weight = 0→ TL sin 25o( ) = mg L / 2( )→ T = mg2sin 25o( ) =

20 kg( ) 9.8 m/s2( )2sin 25o( ) = 230 N

39. Aballisthrownagainstawall;itbouncesoffandreturnswithspeedequaltoitsinitialspeedbeforestrikingthewall.Whichofthefollowingstatementsistruefrombeforetoafterthecollisionbetweentheballandthewall?

A) Thekineticenergyoftheballisthesame.B) Themomentumoftheballisthesame.C) Boththekineticenergyandthemomentumoftheballarethesame.D) Neitherthekineticenergynorthemomentumoftheballarethesame.E) Thecollisionisinelastic.

Solution Thekineticenergyoftheballisthesamebeforeandaftersincethespeedisthesame,andthusthecollisioniselastic.Themomentumisnotsincethedirectionisdifferent.

40. Whataverageforcewillstopahammerwithamomentumof48.0kgm/sinatimeintervalof0.030s?

A) 1600NB) 1440NC) 1.44ND) 2.88NE) Cannotbecalculatedwithoutknowingthemassofthehammer.

Solution

F = ΔpΔt

=48.0 kg m/s( )

0.030 s( ) = 1600 N

41. Consideraspringinitiallyhangingverticallyfromtheceilingwithnothingattachedtoit.Thespringisatrestinitsrelaxedstate.Thenamassiveobjectisattachedtothespringandgentlylowereduntilthespringholdstheobjectandthespringandtheobjectareatrest.Whichofthestatementsaboutthisprocessiscorrect?

A) Theelasticpotentialenergyofthespringincreasedandthegravitational

potentialenergyoftheobjectdecreased.B) Theelasticpotentialenergyofthespringdecreasedandthegravitational

potentialenergyoftheobjectincreased.C) Theelasticpotentialenergyofthespringandthegravitationalpotential

energyoftheobjectbothdecreased.D) Theelasticpotentialenergyofthespringandthegravitationalpotential

energyoftheobjectbothincreased.E) Neithertheelasticpotentialenergyofthespringnorthegravitational

potentialenergyoftheobjectchanged.

Solution The potential energyofaspringisthesmallestwhenthespringisintherelaxedstate.Gravitationalpotentialenergyincreaseswithheight.Thus,theelasticpotentialenergyofthespringincreasedandthegravitationalpotentialenergyoftheobjectdecreased.

42. TwoforcesF1andF2bothofmagnitude10Nareactingonabox(asshowninthefigure)andmovingit20meterstotheright.WhatisthenetworkdonebytheforcesF1andF2duringthisprocess?

A) 200JB) 0JC) 400JD) 350JE) 100J

Solution

W = F! s = 2cos 30o( ) 10 N( ) 20 m( ) = 350 N

Physics 111 Exam 2 - KEY

22D 32A 42D

23C 33E

24B 34C

25C 35C

26C 36D

27C 37D

28A 38C

29D 39A

30D 40A

31B 41A