Test 1 Spring 2009 - PHYS 2211 Test 1 February 4th 2009...

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Unformatted text preview: PHYS 2211, Test 1, February 4th, 2009 Name (print) _______________ "_<_ gay/fl___________________________r ______________________________________ Instruct ions 0 Read all problems carefully before attempting to solve them. 0 Your work must be legible, and the organization must be clear. 0 You must show all work, including correct vector notation. 0 Correct answers without adequate explanation will be counted wrong. 0 Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything you don’t want us to read! 0 Make explanations correct but brief. Don’t write a lot of prose. 0 Include diagrams! —3 6 0 Show what goes into a calculation, not just the final number, e.g.: “+3 = (3:3; £823 ; = 5 X 104 C 0 Give standard SI units with your results. Unless specifically asked to derive a result, you may start from the formulas given on the formula sheet, including equations corresponding to the fundamental concepts. If a formula you need is not given, you must derive it. If you cannot do some portionvof a problem, invent a symbol for the quantity you can’t calculate (explain that you are doing this), and use it to do the rest of the problem. Honor Pledge “In accordance with the Georgia Tech Honor Code, I have neither given nor received unauthorized aid on this test.” Sign your name on the line above PHYS 2211 Do not write on this page! Problem 1 (20 pts) Problem 2 22 pts ( ) Problem 3 (20 pts) Problem 4 (20 pts) Problem 5 (18 pts) Problem 1 (20 Points) In a recent lab, you wrote a program to predict the motion of a fancart. The initial lines of your code were similar to the following: from __future__ import division from visual import * track = box(pos=vector(0,—0.05,0‘)’, size=(2.0, 0.05, 0.10), color=color.white) cart = box(pos=vector(1.0,0,0), size=(0.1, 0.04, 0.06), color=color.blue) mcart = 0.80 pcart = mcart*vector(-O.5, 0, 0) deltat = 0.01 t=0 while t < 7.0: rate(100) t = t + deltat Ea}; :vggl'or( O-l,0,00 ‘Fl‘ Force ,‘5 ‘l’o ‘llne. m‘glnl' Pcari' = PCOH‘l‘ ”l“ E01”. *cfléllq'l' fl” fir : l: + EU tAt Peg +(chrlr/MCqF‘l51rJP/ll‘fil' # 7:5 : f3 + GAt L Car‘l’. P03 : C0,,» Jr, (a 16pts) Write (above) the remaining code required to predict the following motion: Starting at the right end of the track, the cart moves to the left while gradually slowing down, comes to a stop momentarily, and then moves to the right while speeding up. Take the magnitude of the fan force to be 0.1 N. (b 4pts) When the cart is at its leftmost position for the situation described in part (a), draw an arrow that represents the direction of A}? for the cart. (If you think the cart’s A}? is zero, say so.) Give a brief justification for your answer. Conglolfif min lAleK/ql glarl'l‘nj Wlnek fine cart is ql’ i‘eSl’. Al {PO , «15; I a kjms". SGMQ A'L ‘l’hL Cmf‘l' latter, is mouly to We m‘ghl so 713:0 O<<I,0,0§, TLeceéore A]: 22.7% If: : .y—————? Problem 2 (22 Points) Two thin hollow plastic spheres, about the size of a ping—pong ball with masses (m1 = m2 = 2 X 10'3 kg) have been rubbed with wool. Sphere 1 has a charge q1 = —3 X 10‘9 C and is at location < 40 x 10—2, —20 X 10‘2,0 > m. Sphere 2 has a charge (12 = ——6 X 10'9 C and is at location < —30 X 10—2, 50 X 10‘2,0 > m. (a 2pts) Draw a diagram of the situation showing the positions of the charges. 7 (b 14pts) What is the relative position vectoi’pointing from ql to (12? .3 r: rz-r. : <'3c>xto'2,50xzo‘f 0% ~ «muff doxldz o> m T" = «0.2, 07, o> .44 What is the distance between q1 and q2? "A -’ Z 2 ‘2 7. i z 1r\—lr; +q+r£l :iCO?) +(O?l+<o) Mr- What is the unit vector vector 7‘"? Ye: F/r : <~0«?.0.?.0W ~ {—0307 0.707 o§ ”I 002% l ’ ' What is the magnitude of the gravitational force exerted on q2 by ql? .J ii: \2 6mm : (6.67xlo"'t3ts-'sv2)(zxio'it\(zxw'gt) m LO~°I‘1)1 [2 #2x10 /\J What is the gravitational force (vector) exerted on qz by ql? F5: 461:: = —(2.92xzo"’°M\403010307.05 t3 : < M2 “0"", - 1.612xlo'“ o> M I iii -; “LE Qfiz : (ixlolwwl/C‘W (-ero‘QCV—éx'o’qci cm ° (W W ”Eel : l'éS XIDJM What is the electric force (vector) exerted on q2 by ql? E, = Mil ? = (LASxm’lm <«o>r0?,0.707,o> (c 3pts) What is the ratio of the magnitude of the electric force to the magnitude of the gravitational force? lEl 1.691051: 0 '5‘“: " _, = )é.07xlo° f \jl 2-?ZXIO b/O (d 3pts) If the two masses were 5 times farther away (that is, if the distance between the masses were 5|FI, what would be the ratio of the magnitude of the electric force to the magnitude of the gravitational force now? 1+ would 5?— ‘er Sou/me ‘° —'_ 91,641 I“? 2 (17160 TF3? _ (1! a7. iEfii m -' qneo Gym‘mz ES a'flOiePQAdMi~ o; \f‘ IFV Problem 3 (20 Points) A spring has a relaxed length of 0.19 m, and its stiffness is 13 N/m. The spring sits vertically on a table. You place a block of mass 0.05 kg on the spring and push down on the block until the spring is only 0.13 m long. You hold the block motionless on the compressed spring, then remove your hand. (a 2pts) Choose the block as the system, and use the usual axis system, with the +y axis running vertically up. Just after you release the block, which external objects are interacting with the chosen system? Ewing TM EMH‘ 4140i i’lne SP""‘j are [nicraCi'n'nj Wl+L\ TM. UGOQ- Fern, (b 12pts) You remove your hand from the block without giving the block a shove, so that it is initially at rest even after you remove your hand. Therefore, what is the initial momentum of the block? J as) At the instant you release the block, what is the force exerted by the spring on the block? (Consider the 1n1t1a1 stretch of the spring.) Sei' 444,; o “‘3'," 0, ]~ ‘i’ha. here. 0 (- M {Fr '73 Erlgz_k(:-E°> : _ (IgN/w§ ((0,0.u3,o§m - mama») : - (l5 N/w) <0, ”0.06,0>W\ E”: <0, oaaw) M At the same Ins ant, w a 1s the orce exerted by the Earth on the block? Now let’s make the approximation that air resistance is negligible. What is the net force on the block at this instant? Q (—5 ___‘ Flue—i: E + l: PH?) tnri’k <0, mach/u + (o, ,0.qu0> A} H At a time 0.030 seconds after releasing the block, approximately what is the momentum of the block? (Assume the netpforce on the block is approximately constant during this interval). ”H z i). + Fwd At 2 (atolov kjl’HS—i + (0,0.2‘1“,0> (00%0‘3B Now that you know the final momentum, what is the final velocity? I A P3, k P‘ w :1 «mm/0,9 9”“ <o.om.0>M"' 0x05 1:5 (c 2pts) Using the final velocity, calculate how far the block moves in the +y direction during the time interval of 0.030 seconds after releasing the block. A? : \Znt = <o.om o» w (0.0305) 1 = (o, 5.2xzo’3l o» m 2? AV : 5.2x10'3w‘ (d 2pts) Answer the following questions about how you would do the calculations for a second time step. The gravitational force exerted by the Earth is the same as before, but because the position has changed, the force exerted by the spring has changed. Calculate the new force exerted by the spring. .J F : ~1<(EZ~Z03 SKI“) I 1 : ~ l< (EF‘AQC/ Ate) ~ (WU/w) ((0.0.I3,o>uq + (Mumfor- <o.o. "MONA : l<0,0.7l,o> All M (e 2pts) You would use the net force (spring force plus gravitational force) to update the momentum (and then use the final velocity to update the position). What would you use for the initial momentum in this second time step? it 3 Pm = l<o.8.? mafwkj ms“ ‘ ‘ /1 lmi'l‘fil Memento», T 9? Se cowl pfep \Clml WoMM‘i’V/m "90““ Previous siep Problem 4 (20 Points) A ping—pong ball whose mass is 0.0027 kg is acted upon by the Earth and the air (there is air resistance, and there is also a strong wind). This table gives the position of the ball at several times: Time 12.35 s < 3.17,2.54, ~9.38 > m 12.40 s < 325,250, —9.40 > m 14.35 s < 8 13 0 14 —10.20 > m 14.40 s < 8.30 0 06 —10.22 > m (a 6pts) What was the average momentum of the ball during the earlier time interval, from t = 12.35 s to 12.40 s? a .1 e’ , - *3 five": (NH/’1“ : W\ (_4:_E,> : PM ( Q ~ r¢> if ' ti. ‘ {0.0027123} (325,250.4-40» - <3.r+,2.sq,~7.ga7w IZ-‘iO s - [23$ 5 (b 6pts) What was the average momentum of the ball during the later time interval, from t = 14.35 s to 14.4? A r ’3 S PM: "A (”’r‘ a—& 3(0a002?i45\(<8*30)0‘0061'lo'22)m - <8-l3,0-W,’i-O.ZO‘)M J‘l-‘IDS - [H.555 (c 8pts) In the time interval fr0m__12.35_§ to 14.35 s, what was the average net force acting on the ball? fia‘AP FiAP‘ ’ Pf? t _ . t‘t‘ 'tz’flfi. T @miolpoihi ct fuck HM: p'n'i‘UVAl > , - - - -2 - - : (“LZXIO 3 ~?.tho 3 ~Hxlo 37%;" _, (Wylogl—zzuo ~1,(x103>1‘5m ‘ =5 Flue/r ' ’ ’ 14.3% s — 12.375: Problem 5 (18 Points) For each graph of van vs. t, choose the number (1—9) corresponding to the appropriate description of motion. Not all descriptions will be used. Assume the usual coordinate system (+x to the right, +y up, +z out of the page). E» W Answer i [ Answer Answer Answer (1) A cart moves to the left, gradually slowing down, stops, and moves to the right, speeding up. (2) A cart moves to the right at constant speed. (3) A cart moves to the left, gradually slowing down. (4) A cart moves to the left, gradually speeding up. (5) A cart moves to the right, gradually slowing down. (6) A cart moves to the right, gradually slowing down, stops, and moves to the left, speeding up. (7) A cart moves to the left at constant speed. (8) A cart remains stationary and does not move. (9) A cart moves to the right, gradually speeding up. This page is for extra work, if needed. Things you must know: Definition of average velocity The Position Update Formula Definition of momentum The Momentum Principle Definitions of particle energy, kinetic energy, and work The Energy Principle Other fundamental principles: AEA = Fnet,AAt —; Etrans,A = FA X I? (point particle) dLA _ c dt ' — Tnet,A TA=TAXF Multiparticle systems: a m1F1+m2F2+... Tam = m1 + m2 + . . . Eat % Mtitflcm (U << 0) Ktot = Ktmns ‘l‘ Krel Krel = Krot ‘l' Km’b Ktrans z EMtot'Ugm (U << C) I = mlriL + "71273; + ‘ ' ' L2 1 Krot = 27:? = EIWZ Ltrans,A = Tcm,A X Ptot LA = Ltrans,A ‘l‘ Lrot Other physical quantities: 1 2 '7 E _, 2 E2 _ (pc)2 = (mg) 1- (1') c _. mlmz A m1m2 Fgra/v = —G '7‘? 7‘ Ugrav = "G IF] 139m” 2 mg near Earth’s surface AUgmv z mgAy near Earth’s surface -' 1 (11(12 A 1 (1142 = __ U = __ else 47110 l7_‘12 7" elec 471'60 I711 _. 1 Fspring = less opposite to the stretch Usprmg = 5.79332 for ideal spring 1 U,- z 5193,32 — EM approx. interatomic pot. energy AEtheTmal = mCAT 13.6 V EN = — N: where N = 1, 2, 3. . . (Hydrogen atom energy levels) EN = N fiwo + E0 Where N = 0, 1, 2 . . . and wo = ”E93 (Quantized oscillator energy levels) ma d” 2 1 (a?) = 1% as :n—g— (if 1) << c) Where R = radius of kissing circle J. 2 k (4):?” at=Acoswt w: j A . k ' Y = £72,. (macro) Y = % (micro) speed of sound 1) = (1 721—3: f = (cos 91., cos 0y, cos 9;) unit vector from angles Moment of intertia for rotation about indicated axis (q + N — 1)! 1 BS (2 = —— E 1 —— E — ql (N — 1)! S k n” T a Q E AS = T (small Q) prob(E) cc 9 (E) 6 [CT Constant Symbol Approximate Value —.——__‘—‘-——'—'——g—_——-— Speed of light 0 3 x 10 m/s Gravitational constant G 6.7 x 10'11 N - m2/kg2 Approx. grav field near Earth’s surface 9 9.8 N/ kg Electron mass me 9 X 10‘31 kg Proton mass mp 1.7 x 10‘27 kg Neutron mass mm 1.7 x 10‘27 kg 1 Electric constant 4 9 x 109 N - m2/C2 7T60 Proton charge 6 1.6 x 10’19 C Electron volt 1 eV 1.6 X 10‘19 J Avogadro’s number N A 6.02 x 1023 atoms/mol Plank’s constant 6.6 x 10‘34 joule - second hbar = —h— 1.05 x 10‘34 joule . second 277 h 71 specific heat capacity of water C 4.2 J / kg Boltzmann constant k 1.38 X 10—23 J /K milli m 1 x 10-3 kilo K 1 x 103 micro p, 1 x 10‘6 mega M 1 X 106 nano n 1 x 10-9 giga G 1 x 109 pico p 1 x 10—12 tera T 1 x 1012 ...
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