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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(4Drd/9Xsm)‘ a 50 m; D) 5001111 M Jc= 505, w :1.) +a{.—.(4.=,,on1/,)(so.,) lord/5
E) 800 rad 1+ ”M has JM’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. skatar A
1331:1012 5" Vt" 011131401" MOMw‘i'UM is eonsiaoh (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. unier; °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 springmass 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; Semi28+ 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(VF __ V?) ———>E) v ZBI'O 2; 33511 V; : L‘3MI1)1+(4‘“{53L ’— ismt/S'”
D) 2501 V." = 25 “41/5“ E) none of these ﬁwncl ”"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 Earthball 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 reststenet "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¢ citrateHon, —+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 Page1
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 'lLe. Maximum 'l‘orque, ”mu“
311:2 hI‘M"1 Fané. {3 are at Maximum amci tel/tan
3 __ B
D) F4 610. Since 'Hae Forces are. all oi ~lLtg same,
E) 125 MAJnEiuéa, 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“. Tanyaas
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 Prismital emu37' 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 conserValion. =>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‘}
'' ((00 k1) v
Pageﬁ : (50 B)(SOMIS)
2 25 hBM[S ...
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