#### You've reached the end of your free preview.

Want to read all 10 pages?

**Unformatted text preview: **Physics 2210 NAME: Sam» /
Spring Semester 2016 UID NUMBER:
FINAL EXAM, Version A TA NAME: , Wednesday, May 4, 2016
Enter this number as “Special Code” on the Scantronsheet: 01
Academic Integrity: Giving assistance to or receiving assistance from another student or using unauthorized materials during a University Examination can be grounds for disciplinary action, up to and including dismissal from the
University. Honor pledge: On my honor as a University of Utah student, I have neither given nor received unauthorized assistance
on anypari of this exam. Name @lease prinu Signature Date Instructions 1) Use a pencil to ﬁll in the circles corresponding to your name and UID on the answer sheet.
2) In the space above, write your name, UID#, TA name,‘and circle your discussion days. 3) Sign and date the honor pledge above. 4) Use a pencil to ﬁll in the circles on the answer sheet that match your answer choices. 5) Each problem has only one correct answer, but you can select more than one answer for partial credit on questions that have 5
choices (MCS): see the grading scheme below. 6) Circle your answers in the exam booklet ALSO as a backup to the answer sheet.
7) Use the space provided in the exam booklet to work out your problems. If you need more space, we can provide scratch paper. 8} Bring your completed exam booklet and answer sheet to the front of the classroom along with your UID card. Your UID card
must match your face and the name on your exam to receive a grade. mm MCS: multiple—choicey‘ive-answer questions, each worth 8 points. Partial credit will be granted as follows: (a) If you mark only one answer and it is the correct answer, you earn 8 points. (b) If you mark two answers, one of which is the correct answer, you earn 4 points.
(o) If you mark no answers, or more than two, you earn 0 points. M03: multiple-ahaice-three—answer questions, each worth 6 points. No partial credit. Choose only one answer!
’1‘in true or false questions, each worth 4 points. No partial credit. Choose only one answer! Unless you are informed otherwise, assume that there is no air resistance and that g = +9.81 mlsz. Scenario 1: The next two questions pertain to the situation described below A skateboarder of mass m = 55 kg starts with some initial
speed v,- = 3.5 m/s and coasts down a hill that has vertical
height h 2 5.0 m. Friction and air resistance do a total of WNC = —447 J of work on the skateboarder during this
time. 12:29 vi m 6% 1. The net work Wm,t done on the skateboarder during this time is: [MC3—select only one answer] (6 points) A. 2.701s @ 2.251s C. 3.141s 2. The skateboarders ﬁnal speed vf at the bottom of the hill is: [MC3m—se1ect only one answer] (6 points) A. 11.251n/s B. 10.5ml's ©9.7mls l/l/ne-f : M/g “"Mc “/3 5 7%;4 WM: 3 ‘74717 - WM: (segg)(aﬁ%)(§.0m)H$/47’J : {62025737. AK 1: WM ==> ﬁmaﬁwvﬂ : rug-Lew”; vex/s: an t we
' ’ﬂk Scenario 2: The next four questions pertain to the situation described below ' A cannon of mass M = 650 kg (initially at rest on level ground) ﬁres a cannon ball of mass m = 15 kg
towards a cliff faCe that stands a distance of D = 1660 m away and has a height h = 280 m. Note: the
ﬁgure is not to scale! The initial speed of the cannon ball is V;- = 145 m/s and the ﬁring angle of
elevation is 6: 37° above horizontal. You may assume that the cannon ball is ﬁred from ground level
and ignore air resistance. 3. How long does it take for the cannon ball to reach the wali?
[MC3—select only one answer] (6 points) 14.3 seconds B. 19.0 seconds C. 8.90 seconds D _ lééom
D ' 160951929 é Aff— V; 60519 (HS‘Zvﬁoﬁ—F“ 4. What is the maximum height that the cannon bail reaches?
[MC3—select only one answer] (6 points) :l/ﬁi 3.: A. 454m B. 683m @388m
): [045% we] 2
%V; :e2gfmx gm [Va 5/le .09» _ 1’3”“
am W /%L)_ '
5. Does the cannon ball clear the cliff edge and make it to the top? ‘\
[MC3——select only one answer] (6 points) £1
A. Yes I 'No C. Not enough information to tell L§
vi Case
gm” " W 9”" _.___(’°)_1.%Q,( :D’ian 9 (a; (9%;1P
‘6‘ 60:9 am :(mmrwma?
6. The cannon is on wheels and rolls on the ground without slipping. What IS the re 1 alspeeclhﬁuilga
cannon after the cannon ball IS ﬁred? [MC3—select only one answer] ](6 points) 15‘}?! A. 3.35 m/s B. 2.01m @267 m/s
902‘ 5 P-F (Mid/@115) J9
—- ﬂy 60 ‘- §
MW +m acme =0 2? l/r: —--—-’--—--— 6%13)€$%)6€33?: [:wg Scenario 3: The next four questions pertain to the situation described below A Ferris wheel has a radius of 7 meters, and spins counterclockwise
with a constant angular velocity to = 0.75 radians/second.
(Cart B is going up at the instant shown). 7. What ,is the speed of the cart A at the instant shown? +3?
[MOB—select only one answer] (6 points) \ _
. +x
C. 7.5 1111's , . ... _ _ A. o @25 m/s 8. What is the x—component of the acceleration of cart B at the instant shown
[MOS—select up to two answers] (8 points) A. -7.21n/32 acmi‘erwl—wyl ML ® {3 gal] m “WW K oLNtetuh
@‘3-9m’52 Unmet/d WW 0-9 Label) C.0 D. +3.9m/s2 - ac: ward T' LOFIK (219%)1(;2-M) 1: qui %
E. +7.21m’s2 Midway 9. What is the y—component of the acceleration of cart B at the instant shown
[MCS-mseleet up to twa answers] (8 points) A. ~72 m/sZ 3- 3.9111182 (lasted/Mr Moat-3 :5; wad-await Speed
@0 around Curd/e :1) a,” adder/“W WM (ﬁt/bits!" D. +3.9 III/s2 E. +7.2mf32 0% aha] 2% Do ment 0+ ® 10. What is the maximum force the Ferris wheel will exert on a 75 kg person as they go around the
ride? [MC3—se1ect only one answer] (6 points) A. 350 N B.681N Max.oc0ursai-©WWMWS+0
©1031N canCQQ airmail-«5 £2119] mum dearth/192W M) or, E9]: Mac: Media 3; U—mg :meK—g “tn/Mama’s) M : ngg)[a.s| Wise—F (Ms mama] T- ]IOSI I0 { Scenario 4: The next three questions pertain to the situation described below A man having mass m z 80 kg stands at the left end of a railroad car having mass M = 320 kg and a
length L = 12 m. The railroad car is at rest on a track but can roll without slipping in the usual way
(kinetic “rolling” friction may be neglected). Deﬁne a coordinate system that is attached to the ground
by taking the initial position of the man as x = 0 and the positive x direction to be to the right. 11. What is the (ac-coordinate) center of mass X GM of the system of man plus car?
[MC3—select only one answer] (6 points) A. XCM = 4.0 m B. XCM = 7.2 m ©CM = as m
" - mm +- MCL/z) :. @920 we My l’ﬁ 12. The man now starts walking on the car to the right with a co tant speed of 2.4 rule (with respect to _
the car). What is the man’s velocity with respect to the ground?
[MCS—seiect up to two answers} (8 points) A. "0.60m/s “0 Madame M2) T/ZM ”:0
:3 WLiji’MVcé:O 1:) ch’: D. +1.80m1's W0. ago We. Va; 2 Van + ij (a) (MUM, Q) t (2.):
ﬁ'ﬁvg : ch JVVMSL ¢~ch t Vang (1+ WM) ch:“vmc Va : Lt 1+ n/M =_ 51,. a) vm8 1%me Lil/((2.4%)
~ (”M/H) : 1.92. We.
13. The man continues walking all the way to the other end of the car and then jumps off of the . the moment just aﬁer he leaps, what is the center-of—mass velocity VCMof the man—car system?
[MC3—select only one answer] (6 points) €90 811! M m mama ace :9 “Vtm —-==o (still!) B. +1.9mis
C. -—l.9m/s_ ; Scenario 5: The next four questions pertain to the situation described below An elevator at a loading dock is used to lift heavy items from ground level to the level of the warehouse
ﬂoor. The mass of the elevator when empty is 11422.: 350 kg. It starts at rest from the ground at t“ — 0 and
accelerates uniformly upward with a= 0.15 111/32 for 2.0 s after which time the elevator continues to move upward at a constant speed for 10.0 s, after which it accelerates downward with —a— — —0. 15 m/s2
for 2. 0 s, which brings it to a stop at the level of the warehouse ﬂoor. A constant frictional force of 1500 N opposes the elevator’s motion. 14. How far above the ground is the warehouse ﬂoor?
[MC3—select only one answer] (6 points) 9:111 oi : Jia 36.1: 'LCOJ’SMZ: LW 0 3%
C. 4.2m 092/; \étﬁ. {CELL—I) {:1 —, C013 'W/S)Ci()$) -.. 3.0M Ois = v.13. - 2 lat; = (ceases) —- JiCOJSM/IZXZSf
lduf'dzﬂ'olg': 3,.gm 30.3%. 15. If the warehouse ﬂoor 15 a eig it above the ground, what is the net work done on a crate of mass m that is
moved from the ground to the warehouse ﬂoor with the elevator? [MC 3—seleet only one answer] (6 points) . net=m h - ___ _ ,
:wanet=(ﬂ:+m)ah MM :ZiK “0 (Kfﬂd‘ :0)
@W ‘ me gas at were .4an 0,747 we. 16. As the bottom of the elevator passes a height of 2.0 m above the ground, the elevator is moving upward with a
constant speed of 0.3 mls. How much work has been done by the elevator motor up to this point (on the
elevator and its cargo), if it is canying a ISO-kg crate? [MC3—se1ect only one answer] (6 points) 12.8 k] B. 9.8 kl C. 5.95 H Wﬂe+=4ik 5': Z,(mr/‘U(o,3l%)f Wne+ W3 +100;th
W5: -.-0om)5t M =- e/ga'opﬁz on) we}? 0 >0 _; we= A, [5359 ii) (a 2%) YﬂngZ¢affeJ/e.aM)vLK/§ZDU)(J whim 17. What' 13 the peak (maximum) power generated by the elevator motor in bringing a 150-kg crate up to the
Warehouse ﬂoor? [MUS—select up to two anSWers] (8 points) A. 6.41 kW .194 kW 0. 0.97m
1). 2.84 kW E. 1.49 kW ' 1:..- :1 E *(H+M)g..— Ec. arm-Mp =>(F.,)W:(H+m)g+a) w; Hildaw [mam “to o M. New ﬂow/s.
:iijiLiﬂ/Wi . Scenario 6: The next four questions pertain to the Situation described below A block of mass m = 2.5 kg is attached to a spring having spring
constant k = 15 N/m. The other end of the spring is attached to
mount that is ﬁxed to an incline (of angle 6 m 30°), as shown in
the diagram. The block has compressed the spring by an amount (1
= 1.5 m relative to its unstretched length and is being held at rest
in this position. Take the positive x-direction as upward along the
incline, as shown. 18. The block is released and starts moving upward along the
incline. Assuming the ramp to be frictionless, what is the
angular frequency of oscillation to of the mass—spring system?
[MC3—selec't only one answer] (6 points) ' k k'a . k
A.a)= , 13.0): m .m=\/; msmﬂ m WW7 If! Indtﬂwénf a? 62 Jens/arr/ 76:20 W ole/wit” d7! Maw"
.9239 :9 Mai 19. Still assuming the ramp i frictionless, how does the amphtudg A of the oscillation compare to the distance of?
(Both A and d are to be understood as positive non—zero values.) [MOB—select only one answer] (6 points) A. A >d 5/052 571mg 070W (kc! > angst/119)) So [7’ nos/r eras;
1" A = d W“:— b/cek's aﬂew/zéhém p0f97L ﬂit? 7’11 V03 ’P‘r
(9‘4“ Wag/y “MM/Chg g/rXCO age/$161 :k/Yo/ W’a/Jﬁkd 20. Still assuming the ramp is frictionless, which of the following expressions correctly describes the motion of :Q
the block, if' It is released at time t— — 0 ? [MCB—select only one answer] (6 points) A'Ql @350): Asin(cot-—- n/Z) 6’0 "—3 X” B. x(t)= A sin(wt + n/Z) C. x(t)mAcoswt jg”); (wﬁrfzj ﬁfL £20 1'; #5)” (_%J :_A 21. Now assume instead that when the mass is released, it does not move. What is the minimum value #5 for the
coefﬁcient of static friction between the mass and the incline that prevents the mass from moving? [MC S—select up to two answers] (8 points) A. Ju,=0.96 “ ' 2%: U : M36039
(;@:::.:: Raw €sz kol-mgsfnG ~15; :
:: :::::::: a"- a “it“ “‘3 (W ﬂ ”Mtg
V \ :9 I” @519: M“ 3%
mg n a far
é/J‘W '5 ————— —— 72M (9 _//$’” aM/ngg 7%30" #9,?in 0?. {k3 (fee/stigmﬂ Scenario 7: The next four questions pertain to the situation described below A satellite of mass m = 250 kg is in a geosynehronoas orbit, meaning that it is always located directly above the
same point on the Earth’s surface. For simplicity, let’s assume that the satellite moves in a circular orbit that is in
the plane of the Earth’s equator. It turns out that the radius R3 of this geosynchronous orbit is about 42,200 km.
The mass of Earth is ME = 5.97 x 1024 kg; the mean radius of the Earth is RE 2 6370 km. 22. True or False: The force exerted by the Earth on the satellite is equal and opposite to theforce exerted by the
satellite on the Earth. (4 points) @3 True I - ' t
B. False MW: .1341an MD” ./ 23. What is the period of this satellite’s orbit? [MC3—select only one answer] (6 points)
. A. 1440 seconds 13. 3600 seconds @ 86,400 seconds ﬂwﬂrﬂy/ P: gal/300:): lag/MI? 24. What IS the satellite 3 kinetic energy in such an orbit? [MC3—select only one answer ](6 points)
A. 4.72 x 1010 J B. 2.36 x 109 J _ @1. 18 x 109 J : 156293 WM 2W3- Yamwo‘mft )L/ 771x709? \. “WWW 25. How much work would have to be done on the satellite (say, by its thrusters) in order for it to leave
geosynchronous orbit and escape Earth’s gravitational inﬂuence completely?
[MC3—select only one answer] (6 points) A. 4.72 x 1010 J _ B. 2.36 x 1091 ©1.18 x 109 j Z. .
GMEM am :9 J’MU— .quCM .1 K (Manse-KEVZMUL!) EEWZK'FUQ T: QHEM __ QUE—W) # #QHEW): :K1 :1: 3 .1. Fall. F“: 1&5, “l4: —|.I?K}OO‘T_ Se we Md aﬂmr—egcapel Scenario 8: The next three questions pertain to the situation described below ' A stick of length L and mass M which can rotate around a stationary
vertical rod can be used to measure the speed of a bullet. The stick is
mounted onto the rod by drilling a hole near the end of the stick of the
same diameter as the rod so that when the stick rotates, the two
surfaces slide against each other. If the coefﬁcient of kinetic friction
between these two surfaces is assumed to be constant, then there is a constant dissipative torque Tf acting on the stick while it rotates. The “if-1"“ ” f "
stick can rotate on the rod, but it cannot slide up or down the rod. The $3 . C]
bullet is assumed to travel at speed 17 horizontally and perpendicular ' to the rod, which starts at rest. During the collision, the bullet
embeds completely into the end of the stick. 26. If the collision time of the bullet is very short (compared, say, to the time it takes for the rod to rotate 1 radian
after the collision), for the system of stick+bullet, to a very good approximation during the collision:
[MC3——select only one answer] (6 points) ' angular momentum about the rod is conserved. B. linear momentum is conserved. NO / MUM 65f {mg/”MM gxgmlpd éa fibd/ C. both angular momentum about the rod and linear momentum are conserved. 27. Find an expression for the angular speed a),- of the stick+bullet systemlimmediately after the
collision. [MCSwselect up to two answers] (8 points) A; mi 2 [L]: a M = lLy-(Miigm)
3(M+ ) _ m B 1 (M+3nT).: E' 001 _ %(M+m) C wi—3( M ) I 7- yﬁML—Z T'MLZ.':€3—H+m)L-L
L M+3m . ‘ 28. Taking the moment of inertia of the Stick—bullet system to be I and its angular speed immediately
after the collision to be 00,-, ﬁnd an expression for the number of rotations N that the stick—bullet
system makes after the collision. [MC3—select only one answer] (6 points) A. N = 41r21wi2 O B N 2 27th»? @ N— _ loaf
“Cf Armrf Apczw a saw :29((27/’:J) 06: — TV: - z mwfgsw Scenario 9: The next three questions pertain to the situation described below A physical pendulIJm consists of a pair of identical thin uniform rods, each having
mass M = 1.20 kg and length L = 28.0 cm. The rods are welded together in a symmetric
T-shape, as shown, and hung upside down from a frictionless pivot. ‘ 29. What is the rotational inertia I of the pendulum about the pivot?
[MC5——select up to twa answers] (8 points) A. 0.047 kg-m2 a M 4m: placer +0 3“ 2r“? B. 0.225 kg-mz
C. 0.063 kg-m2 MM 1 3L M L} .01331qg-m2 _
E. 0.039kg-m2 WW '. 3'1;an + “MLZ WC; and ' 1% I
r, 2 _ ;_' pm we.” 'mm'
J, H JETHLZ’ 1" 1'21”” +W~ .5 CMﬂ’ ”at HQMLZ’ 7" CV2) ”(-2 ::(£?%z)0.aka)(o.aemf=10.133 “gt“? 30. What is the period P of this pendulum (for small angular displacements from equilibrium)?
[MCS—select up to two answers] (8 points) 6919:2142" B.P=211"—I- c.P=2nl—‘”— 3MgL MgL MEL
, 21 41 D. P—ZTL‘ M—gl: E.P—2T£ BMQL @Mrféa/éf (ﬁlial/(2), P: 2? j— : Mg/ 621‘
77?
9cm? Treat m piece? as m; @ I 3/1431"
all 53 We was: all-Emit @wwghte (907” %L
I _ Eon : 3
C;o.m— err" ’14,! tall: — /,_H_. m..- 31. If both of the rods in the T~shape are doubled in size to 2L (using the same material), the period of the
pendulum: [MC3—select only one answer] (6 points) (19 increases
B. decreases , ,
PA, I C. stays the same ' ...

View
Full Document

- Fall '09
- Physics