Exam1 Fall09 - PHY 317K September 24, 2009 Signature:

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Unformatted text preview: PHY 317K September 24, 2009 Signature: ....................................... .. Exam 1 — Unique number 59400 Instructions: 0 No notes, textbooks, or similar aids are permitted. 0 Use the scantron answer sheet to previde the answers. Follow exactly the directions how to mark it. Write your name, UT EID, course number, unique number. Sign and date it. Mark your scantron using #2 pencil. NO PARTIAL CREDIT will be given. 0 There are 16 problems on the exam. Every correctly answered problem is worth 6.25 points. An incorrectly answered problem is worth 0 points. 0 Assume g = 10m/s2 for all problems on this exam. 0 Any questions you may have about the test have to be directed to the instructor or TA. No conversations and/ or collaborative work are permitted. 0 You may use the blank sides of the handout for notes and calculations. 0 Some equations and other information that may be useful on this exam are on the last page of this exam. You may not ask questions about those sheets, however. 0 Sign the exam handouts and return them with the scantron form. You must return ALL pages of this handout with your scantron. 1. A 2.0 kg block attached to an ideal spring with a spring constant of 50 N/ In oscillates on a horizontal frictionless surface. The total mechanical energy is 50 J. The greatest extension of the spring from its equilibrium length is about: (a) 0.5m (Total mechanical energy [.5 Con SCH/(gt (b) 0.7m (5 (“face ['5 ftfction RI; and Spn’nj [j @1411; Zola/)- Kin: lit-C energy of Mass I3 0 (d) 2-0111 When Sfi'mj Maj maximum exfensu‘on. (e) 2.4 III géo tall : Kmajsj. USP”? ,__’., 5’0 :5 USPnh (When exfengon I-Sjrtd‘rfj Us’pn‘nJ gal—((22 #) 50 :lL—xa’mxl afar; I/jsl’éi ml 2. An Object is released from rest from a window which is 45.0 m above the ground. The time it takes the object to reach the ground is about (remember, assume 9 = 10 m/ 82): (3)3.53 hfZO/léza/a=-j (mugs hf=Ag+L¢f+3LaH n, 0:45-éc2f1fi455552 3.05 (a, wit 35! 3. The vector ii. is given by: if = Bi + 33' -— What is the magnitude of A“? "" 2. 2 l (a) 14 wajmm of A :/,4/=/,4;+,4Cy+/;Z)é (b) 4 2 2. .—— (c) m ~—> (HI: [2 +32%“) Jig: (dismal/2 : 1/14 (d) 3 none of these 4. The position of a particle is given by I(t) = A sin2(wt), where A and w are constants. What is the velocity of the particle at t = 1r/w ? (‘9 2”“ wan/95m 2/0173 2 “b 0 _ (c) 2A2w Velocity :Wfl: %Z(1L):_j% [/11 5m (000] (3:11:22 F) Wt): A X 2 SIMM‘) x (105(th X ngfiwé‘rmwzfflflwf Elm/mfg wt) af kg .' VLZIT) :Qflwsr‘nmjcqm) 2 shark—0w W175.) ; 0 Refer to the following situation in the next three problems. A cannon, which is sitting on the ground in a region where the ground is perfectly flat and horizontal, fires a. cannon ball into the air at an angle of 30 degrees with respect to the horizontal direction. Ignore air resistance. The initial speed of the cannon ball is 500 m/s. In your calculations remember to take 9 = 10 m/s2. 5. About how long will the cannon ball be in the air before striking the ground? (a) 5 sec "TA, / compo/9m; O Ve hardy. dale/mm” 4 b (b) 10 sec Verfi‘ca/ ma 5764. We have [V] :500 run/5 , 30 , r: (c) 25sec V 1”] Sim? :500 SMGO”) , . 5056': y / /'//,. z e sec 2 j (HOD (15¢ l’lf=/z£"rV£'f+:21-61f when: V :3Vl5in30" -— MBA—0 \/‘—V a=« : _, 1‘ r L- I o :0 + 2502i ’51le t2 P>jt(,z§0»5t) = 0 —-.> \I‘ 6 (W: (9120‘ #1: 6:0 so/utfon) Mach/[cos 3o 6. About how far from the cannon will the cannon ball strike the ground? (a) 6'800‘” The displacsz [h x 0/1): Jim “5"”; V1 . Mrffa/ P035719“ , 7’9 1 ’5 @22,000m 0 fi 2 in, v.1: lV/Cosé) = 500 Cosmo )= Sow—:2— ‘f33 5 ((938,000111 ofifigfl/D‘Ft/lfi; NZQOOOM (‘5 defermi'nrcf fez/(eh {Ll 0 . 7. If the r-axis and y-axis are defined so +2: measures horizontal distance from the can- non and +1; measures vertical distaHCe above the ground, approximately what are the components (111,113,) of the cannon ball’s velocity when it is at its highest point of its trajectory? “‘7‘ _ ‘ - V (/ rushes 66 are (a) (+430m/s,+250m/s) — Hf h’j/Wf Pamf j a f (b) (—430m/s,0m/s) Chanjmj 5/7 14/ muffle Va; 0. (c) (+250m/s, —250 m/s) A LL ® (+430m/s,0m/s) F [/x ha; jam mg mfg-Edi QZL h/J/JCS (e) (+250m/s,0m/s) PM f m (-5. A“ 0f #11 Sadat is (ensign; é’hlflrzOI’e Vx: ' 3 8. A block moves with constant velocity on a flat horizontal surface in the positive :1:— direction. Two horizontal external forces acting along the m-axis are being applied to it: a 100 N force in the negative x—direction and a 130 N force in the positive :r-direction. There is also a horizontal frictional force exerted by the surface on the block. The frictional force is: Black {-5 mam-n {Afl'fh (Onsfanf - \d' (a) 0 UNI/“at?! -——‘) a :0 . l (b) 30 N, in the positive x—djrection Newfghq [flu/45‘] .' flier» 3 ma 5 0 X . h . _. . @30 N,1nt enegativezdirect10n_) [BOEIOOA'LF ,0 ‘30” (d) not enough information given I Friction—— b, (e) none of the above ’7‘} écn-Cfi-On ;- _ / (LI-t. [n fl)! 17¢?ch {Tl/t X aficch'an) 9. A car of mass 1, 500 kg rounds a 200 m radius curve at a constant speed of 20 m/s. The magnitude of the centripetal force acting on the car is: (3)180N The Ear expenémes Ccnfn‘pr/a 2 BOON acgflr 2L— / acCe/E/af/bfl .' 10. A block of Weight 100 N sits motionless on an inclined plane. The inclined plane has inclination of 45° with respect to horizontal. What is the magnitude of the frictional force on the block that is parallel to the inclined plane? 71N flop/é, Newton’s gal/aw [0 force Components of ((1); :23: [fl {VCdTC’D’l (Para/5d ['0 (ha/[m] .' (d)35N f :_m SingJ’f/x’ I :ma (8) 40‘5N / Nlt *7 fanOfl 9L mgsina FI-(Iclflon ax :0 6006:: Mare/3 F10 MD [70” ; 5m : (0097145 0 fiction ———.) ’ _..— FNflf 5 O 9 6" : [00 X g, ’9 Emcth : 4 11. Two blocks are connected by a string passing over a pulley. The mass of block A is 3.0 kg, and the mass of block B is 6.0 kg. Assuming that the string and pulley are massless, and ignoring any friction, the magnitude of the acceleration of each block is about: (Remember, use 9 = 10 m/s2.) (a) 0.3 m/s2 (b) 0.7 m/s2 (c) 1.5 m/s2 (d) 2.3 m/s2 'd . w Newton’s Zn (owl to? block H =mfiCk (l) -Neotom 2nd W {or blbclLG: ol T-MBEJ = —m8(l (21 Subtract (.1) {mm (0: 12. Referring to the situation in previous problem, what is the magnitude of the tension in thestring? ‘ I ‘ (a) 21N U56 9:7. (U iFrom P/ei/IOMS Iowa/em and pm? Ln ax 4ON A M J” (c)10N Timfljzmflafl‘f: Mfl(a+3)__mfl J—fi +1 fl (d)35N 2mflm8 02,60 60 may” 8 30+“) (—40A/ 13. The position 2: of an object has the dimensions of [L] (i.e., length). Its velocity v has the dimensions of [L] (i.e., length/time) and its acceleration a has the dimensions of [L]/ [T]? Which of the following quantities has the same dimensions as v? (a) a12 (/16: dim to (kid/rare "drmfl 5/021 of ” .' (bl “2/35 (302/1; w’mmmumi aim/a) {AMT} , f“: dim (/27) :[agm(14)] {(2 =Zélr'mOLJOU/MMU /2 I = (féleLJU’J—l} ’V2;([Ll2(TJ‘2)l/2 JLJETJ‘ : oil/M (V) 14. A horizontal force of 100 N is used to drag a crate of mass 200 kg at a constant velocity across a fist horizontal surface. The coeflicient of kinetic friction is about: (a) 0.25 Corigf‘ant Vt/Ofllfy "'5 50 antt’iafl loo N (b) 0.50 , nd ' evi‘fli: (c) 0.10 Newton 5 '2 “WI 77 '/"/ (d) 0-75 Gut : ma :0 ’7) {0 0' an'clbfon 9‘0 —) Gin—d?“ ; /OO/\/ but firm-on farmer's] War Mprmwf 0/ “Wm/KM and fllé -’ {iridium :mj [41K : MOUOKFK —) {00 :JOOO/‘K-m) PK:0‘05 l 15. A 10.0 kg block is lifted vertically 1.4 m. The work done by gravity during this motion isabout: 3 —-—:> ‘7) El 7 f s . _ F a (a) W: j Fjwolfl WWW C05 F v; if: it: i- we _140J all‘ép/aamehf 000‘ 1Com of 81/61le WU" OPPOSI'Q dlrccfi'onS—> 9:71 9;“ (e) none of these I - Li M v M W : ME) Cosflg Edy: 4003“? nit.) ‘—'—/\’ w: ‘lOXiDyl’éi ——)t-@ 16. On a frictionless horizontal surface a force of 40 N holds a mass of 1.0 kg on an ideal spring with a 400 N/rn spring constant in compression. When the force is released, the mass begins to oscillate. The speed of the mass when it is passing through the equilibrium point is: "‘3 (a) 0.5 rn/s ._,__ K A “X. (b) 1.0 m/s Ffiqu [($4wt’fifil (‘3 I'm/S 4” “I Elem/K ‘jgfom M” ‘ If“)! ./ '1 '/ ’// (e) 5.0 m/s Baffle 1%: force 15 rig/{0.94, all (mag/v [5. Stand in ‘ 2 the Spymj a1 pof'en fm/ mag/.- ;2L szzzLHOOWa/ng 2x7 Consfx'fafes 7% fofa/Wéhflm'a/ Mfr/~77 o/f/k mm; and Sony’j. 7L U89)??? : ’2 fiféfw-flérm‘“ ‘l Uspn'nj :0 ‘9/(Vnas; :ZJ j 6 but Km _’LMV'Z >v2=£5£ an - efm'l — 67M I ...--——) 3 - ‘40“! 2'0? {"703 Some Equations for Exam 1 sin0=cos%=sin7r=cos32—“=0; cosO=sin% = 1; coswr=sin§21=—1 sin 30° = % = 0.500; cos 30° = A? 0.866; tan 30° = 3333: = 715 = 0.577 2 sin45° = A? = 0.707; cos 45° = fl 0.707; tan45° = 331:: = 1 sin60° = A? = 0.866; c0360° = % = 0.500; tan60° = gagg: = J5 = 1.73 For vectors fl! = Ami + ij + Azli: and B = Bzi + By} + lei, |g|=A= /Ag+A§+Ag and §-§=ABcosfl=Asz+AyBy+Asz Iff(.7:) — amn' df — man—1 and [$2 f(:c)d.7: — a zen“ x2 _ ' dz: _ z, — n + 1 I, d d Also, a sin(a$) = acos(a.7:) and a; cos(a:c) = —a. sin(a:r:) d d (1 Chain Rule: If u is a function of 3: and :1: is a function of t, then d—T: = Here .1: and t can represent any variables (not just position and time). _ r d“ a d" at?" If a particle’s position is represented by F: 7'(t), v = E:- and a = (1—: = Therefore, 1130‘.) = fa3(t)dt + C and x(t) = fvz(t)dt + C, and similarly for y— and z—components. For a constant acceleration Ei, F(t) = F0 + fiat + 55112 and 17(t) = 170 + (it, where F0 and 170 are the position and velocity at time t = 0, respectively. . . . . . v2 , 2m: Umform Circular motion: centripetal acceleration a = — ; and period T = — r v , A (if ~ ~ Newton 5 Second Law: Fnet = ~d-tv; for in constant, FM; 2 ma. Static friction: f8 5 p.311; kinetic friction: f], = men (n is the magnitude of the normal force). _. d _. Work: W=/F-df'; power: P=d—P:/, 13:17-17 Kinetic energy: K = %mvz; work—energy theorem: Wnet = A K 3 Potential energy: AUAB = —WAB = —/ 1-3“ - (17"; for gravity AU = mgAy A Force due to a spring: F = —kx. Elastic potential energy AU = %k:c% —- %kx,-2 Conservation of Mechanical Energy: AK + AU = 0 (if all forces conservative). ...
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Exam1 Fall09 - PHY 317K September 24, 2009 Signature:

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