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Exam 2 solutions (phy317k)

Exam 2 solutions (phy317k) - Name T 15HY317K Signature...

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Unformatted text preview: Name: T( 15HY317K October 16, 2008 Signature: 50LUTION$ Exam 2 Unique number 60735 Instructions: a NOTICE: On this exam you should always assume g = ID m/sz. 0 fig notes, textbooks, or similar aids are permitted. You may use a calculator. II Use the scantron answer sheet to provide the answers. Write and bubble your name and UT ETD carefully. Mark the answer sheet using #2 pencil. Mark only one answer per problem. If you need another scantron, ask the instructor. 0 There are l5 problems on the exam. A correct answer is worth 100/15 = 6.67 points. An incorrectly answered problem is worth 0 points. a You must complete this exam during the class period. 0 Any questions you may have about the test have to be directed to the instructor or a TA. No conversations and/or collaborative work are permitted. a You use the back sides of the exam pages as scratch paper for calculations as needed. I Some equations and other information that may be useful on this exam are provided on the last sheet of the handout. You may not ask questions about anything on this sheet. You may detach it from the exam, but will have to return it with the rest of the exam handout at the end of the exam. 0 You must turn in this complete exam handout (all pages. including the equation sheet) at the end of the exam. Put your printed name and signature in the indicated areas on the cover page. Page I 1. Starting from rest, a disk begins to rotate about its central axis It has constant angular acceleration 4.0 rad/s3 for 5.0 s, then immediately begins to slow its rate of rotation with a constant angular deceleration —4. 0 rad/s2 until it stops During the period when the disk ts rotating, what IS the full angular displacement? B 200 d . 1. c; 400 2d L; 1°”: — 1(4-Drd/9Xsm)‘ a 50 m; D) 5001111 M Jc= 5-05, w :1.) +a{.—.(4.=,,on1/,)(so.,) lord/5 E) 800 rad 1+ ”M has J-M’dur 5.01 a.» 11 +. 911:? #9; ' 6' J' “J [5"‘3'ai401‘5/s‘Xs OJ" " 50 “J +100p‘A—5‘o ,1; 5 1w (“0.3 2. An ice skater with rotational inertia Io is spinning with angular velocity :1) 0. She pulls 1 her arms in thereby decreasing her rotational inertia to —10. Her angular velocity' IS then: No no” +¢rquc is aveiic: 4'1: ”HM. ska-tar A 1331:1012 5" Vt" 011131401" MOMw‘i'UM is eons-iao-h (3)2011 _._. .. __ _ ,L o) @014 _> L“ I°w° ' I“ ’ (410)” "$134600 fiugl‘ldn 3 A block of mass 10 0 kg is lifted vertically from the floor to a position 1.5 m above the floor. The work doneb by gravity during this motion is about: -——-,~A) “1501 _ a. B) —751 3 W3 - F3- 3 C) 751 5 D) 1501 e=130° _ F5; “’56 E) none ofthese " “4346.056 = Mjé «25180‘ F3 5 -M31-_1 = F150 '3‘ 4. Assuming the Bath is a uniform sphere, its center of mass is: A) at the north pole B) atthe south pole The. uni-er; °F'M¢55 5F 0. 111.113ng C) on the equator 5 am is 01+ +1.13 D) in New York City F: ““41" 0F ‘Hm HE) at the center 5'? re. Page 2 60735 5. A particle of mass M = 2.0 kg moves to the right with velocity 12 mfs. as shown. Its path is along a line that is 0.50 m from point A. What is the magnitude of the angular momentum of the particle with respect to point A: M .. ....._‘.‘. Afluu.u.uu.uu ‘3 . V = 12 mis 0.50 m A A) zero L: rxP B) 5.0 kg - mils A z ' $6) ”kg-“12’s [Ll r? sine D) 24 kg - mzfs : my“) she 13) none of these '-’- (0.5“)(20 kj)(12.Ml5) : 1?. k3 M75 6. A block of mass 4.0 kg is attached to an ideal spring with spring constant k = 400 Mm. It slides back and forth on a horizontal frictionless surface. The total mechanical energy of the spring-mass system is 50 .I. As the block slides back and forth, the greatest extension of the spring from its equilibrium length is: A) 0.10m A+ HA; Semi-28+ eXJreneion as? dete spring, B) 0.25m -->C) 0.50m ‘Hne speeg a"; 'Hne. block is. ‘V=O and. on 1)) 0.75m Hue energy is poteniiol entry]: Jkaa E) [.0111 Since. +lne medaanical energy is mnSerUeA, Eweek=yé +KE : ékxl -'- '50 I o i73<= $3? 0.30M 7. If a wheel rotating counterclockwise at a constant rate completes 18 revolutions in 18 5, its angular velocity is: A) 1.0mm so 18 m; 21:; ms Bimadxs O=r= -*———‘=7:Itmg$ fl C) 2n rad/s ‘8 5 \ rt“ / D) 18rad/s E) 187: rad/s Page 3 60755 8. At time r= 0, a particle of mass 5.0 kg has velocity v]. =(3 FYI/S); — (4 misl/a. At r= 5.0 5. its velocity is vf = (—3 ms); + (4 III/51;. During this time interval the net work done . . _ .5. '2. Eithepartaclewas. wmd. : (E; "EX: Z—M(V-F __ V?) —-——>E) v ZBI'O 2; 33511 V; : L‘3MI1)1+(4‘“{53L ’— ismt/S'” D) 2501 V." = 2-5 “41/5“ E) none of these fiwnc-l- ”"0 . A ball is held at a height H above a floor. It is then released and falls to the floor. [fair resistance can be ignored, which of the five graphs below correctly gives the mechanical energy Ema: of the Earth-ball system as a function of the altitude y of the ball? I O H o v F .V i H II H m H 1' 3’ Iv H V H g; i] “0 Methanicai energy is \os—\' because‘ C) [-1-] 'i’idere. is no air rests-tenet "Ward-“om, D) w Ems. is constant, 50 'Hne JmPL‘ :5 ‘1 \norifiovfl'cxl lina. 10. A 2.0 N force is the only force acting on a 1.0 kg object that starts fi‘om rest. At the instant the object has moved 1.0 m fi'om its initial position. the rate at which the force is doingwork(i.e.,thepower)is: "'3 amé 3 ‘39-.me ‘n «H4?L 5¢M¢ citrate-Hon, —+A 4.0w 6 —- _. Bi sow 5° 63:0 ‘1‘“; P: F'” mum‘s ?=Fu c) 16W SW“ ”#0. ““14:- E)20kW t 1. ‘ 13 A=%*'§.a£ =34£1=IOM A Page-1 A g d?(?‘3)" F1) ; (2.0»)(2045) = 4.0 w ll. 12. 60755 A plank is attached at one end to a fixed hinge. The plank is free to rotate about the hinge. Five forces of equal magnitude are applied to the plank, at different locations and in different directions as shown in the figure. Which of these forces produces the greatest torque (i.e.. torque with the largest magnitude) about the hinge? A) F1 l’dfiir- F‘Fs‘me, so 'l-Le. Maximum 'l‘orque, ”mu“ 311:2 hI‘M-"1 Fané. {3 are at Maximum amci tel/tan 3 __ B D) F4 6-10. Since 'Hae Forces are. all oi ~l-Ltg same, E) 125 MAJnEi-uéa, one. need. only +9 Maximile. r a“; «sine. This occurs Ear Hag Rm: [-3. A block of mass M = 30 kg is attached to a cord that is wrapped around the rim of a wheel of radius R = 0i 0 m and hangs vertically. as shown. Assume the cord is massless. The rotational inertia of the wheel is 0.70 kg - m2. When the block is released and the cord unwinds, the magnitude ofthe downward acceleration of the block is: Neu‘i’ofis 542an L403 fi=mfi F: T’Mj '= "Mq :5th M3 -M.o. gum—tel Lam G“. Tanya-as A) 1.0 mfsz T g I: B) 2.0 m/s: KT=Io< - 359 33 13$}? 3’ T = ~14: E) 5.0m2 R Ms—Mq => (1 = MS a? Page 5 13. A ball of mass 2.0 kg is dropped from the window of a building. it strikes the sidewalk below at 30 mfs and rebounds up at 15 mz‘s. Assume the ball is in contact with the sidewalk for 0.10 s. The average force exerted by the sidewalk on the ball (while they ' tat ': 3,3130%)“ 5. .. 3: .. a—v. Merv.) B) 3mm 3 ‘5“: Art At C) 30N : 1.0 )[iSmls -('3°“Z‘-‘)] D) mm 040; E) 30“ : (Loxcitbisuh) DAD; = c[00M 14. A simple pendulum consists of a 2.5 kg mass attached to a string. It is released from rest at position X as shown. Its speed at the lowest point Y is: W A) 0.25 m/s . B) 1.5 mfg A+ paSi‘iIDIfl X, 1J=O emit a.“ '“ae. MQCLLanimi C) 2.5 m/s ' ~ . -—9D) 4.0 m/s energy, ‘5 Prism-ital emu-37' E) 8.0 mfs PE —— Mild = (ashlflfifimlsmflagow) : 20 I A+ Fosi‘fim Y! a“ “‘5 {V1373}! i5 kinc'i'ic. energy.- KE ‘= £1va ‘-= 7.0 :r by conserVa-l-ion. =>V=1/§l-.°;_3;3 = 4.0 MIS [5. Particle A has mass 5.0 kg and particle B has mass 1.0 kg. Initially. particle A is moving with speed 5.0 mfs, and particle B is stationary (i.e., not moving). Then particle A collides with particle B, and they stick together (i.e.. they collide completely inelastically). After the collision, what is the momentum of the combined (stuck- to ether oh'ect? _ A)g 5.0 kg this BEE)?! Coilblon: B) 101(ng Pew =MAVA + M3913 C) lSkgm/s :Mav since U = D) 20 kg m AH“ collision: A 3 0 '9 E) 25 kgm/s Momen‘i’uM unsettle; Phial = (ll/14+!” {0‘} '-' ((0-0 k1) v Pagefi : (5-0 B)(S-OMIS) 2 2-5 hBM-[S ...
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