C2000 - Distribution ' of points 1 point 1 point 1 point 1...

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Unformatted text preview: Distribution ' of points 1 point 1 point 1 point 1 point ds to two or three decimal places T2 (52) For correctly computing the period of each pendulum using the time for 10 oscillations For correctly computing the square of each period For expressing the periods and the squares of the perio For completing the data table 2000 Physics C Solutions Mech. l {15 points) (a) 4 points (15) ”- 3 points 1x1: xxwuaxinsuxnniuxtnxthfltaxtul . . n . . . . . . a . _ T x 3t?lLlITILIIfIL§t?ILlIT!LIITt . w w . . . . . _ . . _ nu T. , -4-.d..-I-+..++++ a a; 3 . . . . . . . _ a a . . _ 114x tthatawxauxvlazxwxatlwxaxsru , _ . _ . . . _ . . . . . raul :ur:erract:rlcttrxus'rxcxxr: . . a . . . . a _ . . m . .xau xx.x.zu.r.xt.xaxx.t_ll.:.luwl .1 4 q. 4 WJ .1 4 14 fl 4 w. a a . _ _ . . . . _ . _ . Txxx rial;TuattrxaztrtaiaTxarlTs . . . . _ a . _ . . . . . a . rtultrsux rnaxzrluuartuxlrtutlrtuttru fl“ _ . . . . . . . _ . . a . . . . . . . . _ _ . . . . . _ filaxtfitaxn tantfixulxnzaszfilanufisqranx . _ _ _ . _ . . . _ . . . . Txanrxntx LtIT:Lsnrxaxlrlalrrstxxrx . . . _ . _ . . . . . _ . . FluII—‘ILII llrlkxxrlLIXFxLlthLll—l. . _ . . _ . . _ . . . . . . . . a . . _ . . . a _ fixaxlwlusxwéa wrutzfiluliflxalxwtatxnt . . _ . . . . . . . . . _ . . ‘ TiensrzLaxraL r L Irvuxtrxstlrrsnxrx MW . . . . . n . . . . . . . . I:T++J a +,++,++-++ . . . _ . . . . . a . . _ . Txaxr1iatlwrax aviaatTuaxnrtaxxwl . . _ . _ . . . . . . _ . rletrrthrzut i rILxIrxLxxrxLxsrr . . . . . _ . . . . . . . _ _ .laxx.x.ll.x. :. . .n.11.1.11.|.2x.x .1 J .x A 1 1.1 fils. .1 A .1 «— .1 A q. . _ _ . . . _ . . . . _ _ . 5 rlaxthastrxxtirxa laztrlaxxraaixr: 1. . _ . _ _ . . . . . . _ . . _ rlulnrxutlruulrrxua r LtlrIleryLsrrz _ . . . . . . . . . _ . . _ . _ _ . . . . . . . . . . . . a 114x1fitJttfllJllfilJllfi l5fixdllfll41x1: _ . . . . . _ . . . . . . _ _ Ttaulrxaxrrletrlaxlrrn Latrltrtrs :Criccfafp . _ . . . + f . . . a . + . . . _ . _ . . _ . . . . . . . O neuaX1341zwsalxwzaliwxa Jux1zaxt1s 1; _ . . _ _ . . _ . . _ ,. _ _ . rig:xrxttxrxLIrrxalsrxgxirlaxzrnnerr: _ _ . . . n _ . . . _ . . . . T a + +4. ,7 T+ T; + a +4. + . . . . . . . . . . . . . . _ 1:4;17xausrsasxwxalthaxxwxaxxwxa1‘1; , . . . . . _ _ a . _ _ . n . rzcxsrxuzxrtuazrxuxxrszsr»Lx\rsLxlrt :IL.:::: . a,{dzlfligllwlullfltqztfiluts 14X! 37.11... 5 . . . a . _ . . . _ a _ . Tsatvrlatxv:a:31ialarxats tax: lasttt . . , . . _ a . _ . . . rautxrauxxrauttruuztrtuit set: snr: . . _ a . . . _ _ a . _ _ . a _ . . _ . a . . . nruzxnnuaznraxa :Jxlfirc 4:) 1: a . _ . . w _ . . . . Tlaxurxaxxraast (axxrua an: r: _ . . _ . . . . _ . . 5 0. 1 1 1 point 1 point I point a1 number of data points above and below the line not have to do so) For drawing this line to approximate a best fit with about an equ For correctly plotting the square of the period as a function of the pendulum length For drawing a single straight line through the data points (could go through (0,0) but did 2000 Physics C Solutions Mech. 1 (continued) (c) 4 points For computing the slope of the straight line slope : 91-2— M Credit was awarded for a calculated value that was consistent with the line drawn. A typical correct value was 4.50 szlm. Most values were between 4 and 5 szlm. For recognizing that the motion of each pendulum is simple harmonic or for using the equation for the period of a simple pendulum T=27rJZ E For using this relationship and the slope from the graph to determine gap . 2 . . 2 47!2 471:2 For example, the relation between T and f 15 given by T = —Tg-- 2, so slope = exp w 47:2 gexp " Slope Credit was awarded for a calculated value that was consistent with the slope determined above. For example, for a slope = 4.50 szlm, gm: 8.77 m/sz. I”: For using appropriate units for the computed acceleration (a) 2 points For correctly stating whether 9.8 ml 52 is within 14% of the experimental value gexp . Credit was awarded if the answer was consistent with the value obtained for gap. For example, for gexp == 8.77 m/sz, the experimental value is not in agreement with g. For justification, either by displaying the range of acceptable values Within 14% of the value of gm, (e.g., approximately 8.42 rn/s2 to 9.12 rn/s2 , forgexp= 8.77 this2 , Q}; by computing the percent difference between the gexp and 9.80 rn/s2 (e.g., 9.80 - 8.77 8 77 x 100 = 11.7 %, which is greater than 4%, for gap: 8.77 m/sz) (e) 2 points For correctly computing the acceleration of the elevator From Newton’s 2”“1 law for objects in the elevator: mg~ N z ma,whereN = mgcxp,soa = g— g:mp For correctly stating whether the acceleration of the elevator is upward or downward Credit for magnitude and direction of acceleration given for answers consistent with gm. For example, for gm: 8.77 mist a = 9.80 - 8.77 = 1.03 m/sz, directed downward. Distribution of points l point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 2000‘Physics C Solutions Distribution of points Mech. 2 (15 points) " (a) 3 points 7’ l1), 49v) Force, L%)m3 swufik’c) or Force. a"? G mv‘gi/j For a vector arrow pointing downward 1 point For a vector arrow pointing upward 1 point For correct force labels on both vectors 1 point For any extra vectors drawn, deduct 1 point Ch) 3 points For indicating that the acceleration decreases 1 point For a correct explanation that includes a correct mention of forces. 2 points ‘ c For example, as the ball approaches terminal speed, the velocity increases, so the drag force increases and gets closer in magnitude to the gravitational force. The resultant force, which is the difference between the gravitational and drag forces, gets smaller, and since it is proportional to the acceleration, the acceleration decreases. Partial credit of 1 point given for only a statement including a basic definition of terminal velocity (e.g., at terminal velocity v = constant, so a must decrease from 98 We2 to zero) ' (c) 2 points For an expression for the resultant force on the ball 1 point F = mg - 15v2 . .. .. .42. a). .. .. 2 SmceF~ ma ~— m dwthenm d! - mg b1) For a correct differential equation 1 point .42 .. _. .22. v2 dt 8 g m Students did not need to use the convention + and - for up and down, respectively, but they did have to be consistent in their Sign notation for credit. The integral form of the differential equation was also acceptable. 200013hysics C Solutions « Distribution of points Mech. 2 (continued) (d) 3 points For recognition that acceleration is zero at terminal speed 1 point For setting the drag force equal to the gravitational force 1 point m = b 2 8' vi For a correct solution for v, 1 point m = la Full credit also given for writing answer only with no other work shown (e) 4 points For a correct statement of work-energy, recognizing that the energy dissipated by the 1 point drag force is equal to the initial energy minus the final energy For correct recognition of both initial potential energy mgh and final kinetic energy 1 point 3. mu 2 ‘\ e: 2 t AE = mg]: w— 32-va For correct substitution of 1), from part (d) 1 point AE z mgh -— 3- Egg-J For correct answer . 1 point — .. .1". AE - mg(h 2b) Alternate partial solution (for maximum credit of 2 points) (Alternate points) For a correct integral for work (I paint) szpd: 9;; W=dex For correct substitutions for P or F (I point) Wzlbv3drgngjbvzdxggw=jku2dx 2000 Physics C Solutions Distribution of points Mech. 3 (15 points) (a) 2 points T 1’ z“ ' "I a i - I: 13mg For a correct free-body diagram QB recognition that a = 0 _Q_R__ correct use of Newton’s 1 point second law 21F = ma T - 2mg = O For the correct answer 1 point T = 2mg Full credit was also awarded for writing the correct answer with no other work shown. (12),, r 2 points ‘1; r ' '-°| V ‘ x r \ ‘ g 1 ‘. 105 .3 x ; I t 1-, .. J i 3% BF == ma . For correct substitutions into Newton‘s second law 1 pomt g 3mg - 23 r: 3m(—3—) For a correct solution for T3 T3 = 27722: 1 point 2000 Physics C Solutions Distribution of points Mech. 3 (continued) (b) (continued) ii. 2 oints P T‘ ‘5 0.: "‘ T 3 lmg 2F = ma For correct substitutions into Newton’s second law 1 point Y1 - me =3 For a correct solution for T1 1 point T = i mg 1 3 iii. a 4 points For 1 = [let 1 point _ i _ __g__ 1 point For a -~ R1 - 3R1 For 1 = (T3 — 7})R1 1 point For correct substitutions into 1 = 110 and solution for 11 1 point 4 _ .5. (2mg * 3mij1 m 11(3R1) I} '53 2mR12 Alternate Solution (alternate points) Use conservation of energy, AE 2 AK + AU = O For AK z ~AU (I point) ___1_ 2 l 2 i 2 ___1_)__ (Ipoint) ForAK «— va + 2(37:01) + 2 11a) ,wherea) —— R1 2 2 I ‘ t For AU = mgh - 3mgh = ~2mgh , where h = g; = Egg—— ( pom) For correct substitutions and solution for I; (I point) 11 z: 2mm,2 2000 Physics C Solutions Evieehw 3 (continued) For recognition that the speed of the cord or the tangential speed of the pulleys is the (c) i. Zpoints same for both pulleys wai = wzkz “ 0329131) For the correct answer a, .. 2.1.. 2 ‘ 2 ii. Lipoint For correct substitutions in L = [w and correct solution L; =(1611>[5-’2—‘~) L2 = 811591 ‘ iii. 2points For a correct expression for the kinetic energy as the sum of the kinetic energies of the two pulleys ".1. K’z For correct substitutions and solution -3. K“2 K=—5- 2 116012+ %12022 2 1 110) 12+ E- 110212 0’1 Distribution of points 1 point 1 point 1 point 1 point 1 point 2000 Physics C Solutions E&M. l (15 points) (a) 4 points Since brightness is proportional to the power dissipated by a bulb, the answer may be found by solving the circuit to determine the power dissipated by each bulb. For example, l__ l “L__ 3 . . . , RP - ~—-————12 Q + 6 Q -— W12 9 , where Rpis the resmtance of the parallel combination of resistors 19:45) _.___«‘:3___._i.._._i?~_Y_._._ IA “RA-Hep “109+4n'3A B RB RE 129 I zflz JAR}; 33.4(3A)(4r2)=2A C RC RC so PA = 1,312A = (3 A)2(10 r2) = 90W «r ». P3 = IBZRB = (1A)2(12 o) = 12W PC : ICZRC == (2 A}2(6 o) = 24 w For correct ordering, i.e., bulb A is brighter than bulb C, which is brighter than bulb B (Partial credit of 1 point given for incorrect answer but with an indication that bulb A is brightest or that bulb C is brighter than bulb B.) For a correct explanation, which can be by a quantitative solution for the currents and powers as above, or by a qualitative approach that notes that all the current in the circuit flows through bulb A, then branches in such a way that bulb C receives more current than bulb B. (b) i. 3 points Immediately after the switch is closed there is no current in the inductor so the circuit consists of resistors A and B in series with the source of emf. For IC = For recognition that I A = 13 and they are nonzero For correct numerical answers for I A and 13, i.e., I A = [B a m z 1.91 A Distribution of points 3 points 1 point 1 point 1 point 1 point 2000 Physics C Solutions E&M. 1 (continued) (b) (continued) 3 points A long time after the switch is closed, the potential difference across the inductor is zero, so the circuit is essentially the same as in part (a) For recognizing that VL : 0 For correct currents, the same as in part (a), i.e., [A = 3 A, I 3 z 1 A, I = 2 A (if currents not computed in part (a), they could be computed here.) Unit point: For expressing all currents in Cb) in correct units of amperes (c) 2 points Attributes of correct curve: 1. Starts at a nonzero but finite point on the vertical axis 2. Smooth 3. Concave upward 4. Has asymptote equal to zero For a correct curve with all four attributes Partial credit of 1 point for curve with flaws but at least two correct attributes Distribution of points 1 point 2 points 1 point 2 points 2000 Physics C Solutions Distribution ' of points E&M. 1 (continued) ((1) 2 points ‘33? O Attributes of correct curve: 1. Starts at zero 2. Smooth 3. Concave downward 4. Has finite but nonzero asymptote ‘ " For a correct curve with all four attributes 2 points Partial credit of 1 point for curve with flaws but at least two correct attributes 2000 Physics C Solutions Distribution of points E&M. 2 (15 points) (a) i. 6 points B 0 C §~—---~ e -————-—-———-———oi Figure 1 e a One point for each arrow drawn in the correct direction 3 points For not having all arrows approximately the same length, deduction of 1 point For not having all arrows start at P1 , deduction of 1 point For having one or more extra vectors, deduction of 1 point Figure 2 One point for each arrow drawn in the correct direction 3 points For not having lengths of arrows such that EC z E B > E A , deduction of 1 point For not having all arrows start at P2 , deduction of 1 point For having one or more extra vectors, deduction of 1 point 2000 Physics C Solutions E&M. 2 (continued) (a) (continued) ii. 3 points I/ 5' One point for having check mark or other indicator in each correct box (b) 1 point For an indication that the x-components of the field vectors due to particles C and B cancel each other due to the symmetry created by having a vertex of the triangle on the y~axis ‘I’: (c) 3 points For an indication that the potential is the sum of the potentials due to the individual charges V: 2.12;: (kegs...) i ’1' 7A r B "c For recognition that the terms due to the particles at B and C are equal V : 15(2‘3. + YA 7‘ B For correct substitutions for Q’s and r’s and correct answer 3 1 Q + 29 47550 «[375 __ 32 2 “'2 3’ ‘2’ + 3’ V , or equivalent Distribution of points 3 points 1 point 1 point I point 1 point 2000 Physics C Solutions Distribution of points E&M. 2 (continued) ((1) 2 points Since E y r. ~— % V01) , to find the y coordinates of the points on the y—axis at which the electxic field is zero, take the derivative of the expression in part (c) with respect to )2, set the expression equal to zero and solve for y. For recognition that E is a derivative of V ‘ 1 point 1 point . . dV __ For recognition that ‘29-» — O 2000 Physics C Solutions E&M. 3 (15 points) (a) i. ii. (b) i. 3 points For a correct statement of Gauss’s law § E . 4A = .52. For expressing the permittivity of the oil in terms of the dielectric constant K' For a correct expression for the electriafield in the oil For a correct statement of Gauss’s law in the space outside the outer shell For stating that the electric field is zero in this region 6 6 = K60 -éL E(27z‘rL) - K60 =_£L. meorL 2 points is - dA : 0 E = 0 3 points For an expression for the electric potential between the two shells b AV=Vb—Va find;— a For substituting the expression for the electric field between the shells AV: AV: Q 2713:6014 a I" For a correct expression for the electric potential difference between the shells Q ZMEoL 523: b in(—- a J Distribution of points 1 point 1 point I point 1 point 1 point 1 point 1 point 1 point 2000 Physics C Solutions Distribution of points E&M. 3 (continued) (b) (continued) ii. 2points For an expression for the capacitance in terms of Q and AV 1 point -2. C”AV Substituting the expression for AV from (b)i: Q C .— ” Q (‘11) 2M£0Lln a‘ For a correct expression for the capacitance 1 point C __ 2717:60L we) (c) i. 3 points For a correct statement of Ampere’s law 1 point §B a d6 == #01 For substituting the current through the inner shell 1 point 8 mm) = nob?) For a correct expression for the magnetic field between the shells 1 point _ #08 B ' ZxrR 2 points For the correct substitution of the total current through both shells in to Ampere’s law 1 point B(2:rr) = no For a correct expression for the magnetic field around the outer shell 1 point _ 2M8 " IrrR ...
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C2000 - Distribution ' of points 1 point 1 point 1 point 1...

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