1991 Physics C Solutions

1991 Physics C Solutions - 1991 Physics C Solutions...

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Unformatted text preview: 1991 Physics C Solutions Distribution of points Mech l. (a) 3 points For some statement of conservation of momentum 1 point For a correct equation: mun - 3m: 1 point u - 110/3 1 point (b) 5 points For some statement of conservation of energy 1 point For correct equation containing potential and kinetic energy: Ui + If " Uf + Kf 1 point For correct substitution of all terms: 1 Va 2 0 + 3(3117) - 3mg: + Kf 2 points (If only some terms correct, one point was deducted for each incorrect term, up to two points) 2 V Kf - Eg— — 3mgr or equivalent expression 1 point (Full credit was awarded if student solved for the final velocity instead of the kinetic energy) (c) 7 points Ui + K,- - Uf + Kf ' u - 2 For correct Ki: X; - %(3m)[ man] 1 point For correct Uf: Uf - 3mg(2r) 1 point 1 U ‘ 2 l 2 o + E(3m)[ “’3“] - 3mg(2r) + E(3m)vt°p Force equation must be used to solve for yup: Fnormal + Foravity ' Fcentripetal , For recognition that in the limiting case Fnomal — 0 1 point 112 3mg - 3&3 1 point viop - rg or vtap - Jrg 1 point v - 2 %(3m)[ my] — 3mg(2:) + %(3m)(rg) 1 point m2~ 5 mm - 6111" + 21:: r - l—zzgr 6 ‘3‘ 2 g 2 “min "' 3453: 1 point EH 1991 Physics C Solutions Mach 2. (a) 3 points For recognition that Ztorque - O or equivalent "1231’: ' a1151'1 (01r mzrz ‘ “71751) _ mm _ (20 kg)(0.5 112) m2 r2 (1.5 m) mz'jkg (b) and (c) 8 points For a correct dynamical e r - Io quation for the torque: For a correct application for the two cylinders: Tri - (45 kg-m2)o of the torque equation For Newton's second law: F - am For a correct applicatio n of Newton's law for mass m1: (20 kg)g — T - (20 kg)a For recognizing that attempting to solve s T - (20 kg)(g - a) a-ar T - (20 kg)(g - on) parts (b) and (c) are coupled (i.e., imultaneous equations) Substituting into torque equation: (20 kg)(g — a:,) - (as kg~m2)a (20 kg)gr1 — (20 kg)ar,2 — (as kg-m2)a am kg-mz + (20 kg)r1Z] - (20 kg)gr1 :2 — (20 kg)(9.8 m/sa)(0.5 m) / [45 kg-mz + (20 kg)(0.5 m)2] a e 2.0 rad/52 :r - (20 kg)[9.8 m/s2 — [2.0 L:§](o.5'm)] T - 180 N Distribution ~“‘\ of points 1 point 1 point 1 point 1 point 1 point [‘4 noint 1 point 1 point 1 point a; 73 1991 Physics C Solutions Mech 2. (continued) (d) 3 points For applicable kinematic equation(s): uz - 2as 0R 5 - %at2 OR and u - at a - a: t - u/a v - J2ar1s For correct substitution: v - u - 1.4 m/s (Alternate solution) For at least one answer with correct units Using conservation of energy: lziz mgs zmu + zIw w - u/r l2 1‘12 mgs - zmv + 2E2: - §u2[m + 5?] 2(2.0 rad/s )(O.5 m)(l m) Distribution of points wz - 2&5 and 1 point v - ur v - :1225 9-5 r 2a: v _ ___ 1 point 1 point (Alternate Points) (1 point) (1 point) 2 _ _2 _ 2(20k )(9.8msz)(l m) U mugs/(m + U‘ ) [20 kg + (as kg-mé)/(O.5 1:02)} u - l.h m/s , and no (I point) 1 point incorrect units 1991 Physics c Solutions Distribution of Points “RV Mech 3. (a) 3 points For a correct expression for the spring force: 1'"s f- ch (either + or —) 1 point For a correct expression for the gravitational force: Fi - mg 1 point kD - mg [C - mg/D 1 point If a student attempted to solve the problem using conservation of energy, which is not correct, one point was awarded for Many“). - .ng and one for either W - AK or _AU5ravity _ AZIspring ' AK- " (b) 1 point For any reasonable explanation I point Some examples: When Dunn-ye - O, the separation is neither increasing or decreasnxg. When Urdu”: - 0, Ifrm is a minimum. Therefore UN!“ is a maximum, which occurs at maximum compre551on. (c) 7 points i For some statement of conservation of momentum ' 1 point For correctly applying conservation of momentum between the 1 initial state and the state of maximum compression with common speed V: mun - 312V 1 point For some statement of conservation of energy 1 point . . . 1 2 . For the correct expreSSion for spring potential energy, 32-10: 1 point For correctly applying conservation of energy between the initial state and the state of maximum compression x: 1 . Emvoz - %k::3 + %(3m)'.72 1 paint 1991 Physics C Solutions Distribution of points Mach 3. (continued) For attemptlng to solve simultaneous equations 1 point _ 2n V 3 1 u 2 - -kxz + an [Ea] .3; 2 2 £1.12 ‘ 2‘“ + 2m 3 l 2 kxz - mu02[1 — 3] - Emvoz - u 2.!!! _ 22 _D x 0 3k “0 mg - v g2 ‘ l l ‘ x a 3g pOLnt (d) 4 points. For correctly applying momentum conservation: mvo - muI + 2muII 1 point For correctly applying conservation of energy: Zimuoz - 51211112 + ' %(2m)uII2 1 point (This point also awarded for any other equation applicable to elastic collisions) For attempting to solve simultaneous equations 1 point v1 - vo - ZUII v02 - (U0 — ZUII)2 + ZUIIZ - 110-2 - 4v°uII + (tulle + ZUIIZ 51.2112 - 4UO‘UII UII - 0 or 6UII - 41:0 vII - 2/3 no 1 point E & M 1. (a) 3 points For a correct expression for the electric field magnitude for a point charge: k0 kQ . 5 E? or g? 1 pelnt For recognition that the electric field is a vector 1 point E - O 1 point as 744 1991 Physics C Solutions E & M 1. (continued) (5) (c) (d) (e) 3 points For a correct expression for the electric potential for a point charge: V - 59 or 5g r a For recognition that the potential is a scalar “31:9 a 3 points (-a. 0) From figure r - a + b For the magnitude of E due to one of the charges: E - -z£2-—z 0:2 +b For recognizing the need to find the components of E The x-components cancel, so only the y-components need be calculated E - —259——; sin a - -;59——g ———3L———— - -—2—59§z—37T Y a + b a + b J;2_1‘37 (a + b ) The y-conponents add: 2ka E? ' (a2 + b2)37z 1 point For indicating that the particle will move toward the origin or that it will oscillate about the origin 4 points For indicating that the particle will move away from the origin For some statement of conservation of energy Distribution of points ” \ 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1991 Physics C Solutions Distribution of points //‘\ E & M 1. (continued) For an expression for the initial potential energy of the particle: U - 2k: °r qV 1 point 1 ' 2k imi ' Tog k . vm - 2 ~52 1 p01nt (Alternate solution) (Alternate Points) For indicating that the particle ‘ will move away from the origin (1 point) émvmz - Wnet or fF dy (1 point) For correctly setting up the integral - °" vd . I” d? ‘ Zqug I?;?f§¥;z§37? (I paint) u - a2 + y2 --> du - 2y dy / a ‘ du 1 2kQ J‘F dy - qu f 37: - mom—17” - 43 a2 a2 l 2 42m , 2amcu ‘ a _ £93 . “Q 2 ma (1 pOLnt) (f) 1 point For indicating that the particle will move back toward the origin or that it will oscillate about the origin 1 point E & M 2. (a) 1 point Since the inductor prevents any sudden change in current: Ii - O 1 point (b) 3 points dI The voltage across the inductor is L E; , which is zero in the steady state condition. 5 ' 13:9: 2 points ” 50 v - 1(100 n + 150 n) I - 0.2 A 1 point 45 75’ Distribution 1991 Ebysics C Solutions of points ,,K§ E & M 2. (Alternate solution) (continued) (Alternate Points) For an RL circuit with zero initial current: I — g (1 - e‘Rt/L) (2 points) As t -—~ w, I - E/R I ‘ 250 a " 0'2 A (1 point) (t) 3 points Cuntnt 0 «‘\V For beginning the graph at the origin 1 point For a monotonically increasing curve 1 point For an asymptotic aptroach to a constant value ' as t ——~ m 1 point (d) 2 points For an expression for the energy stored in an inductor: U - éuz 1 point 1 2 U - 5(1.0 H)(0.2 A) U - 0.02 J 1 point (e) 2 points Since the current in an inductor does not change abruptly, it is equal to the steady-state current calculated in part (b) I - 0.2 A 2 points (One point was awarded for either of these equations: ail—E ~ IR - 0, I - Io e‘RE/L) ‘\ 1991 E&M (f) Physics C Solutions 2. (continued) 2 points Lg d: - IR VL- Foz using correct value of R: R - lS VL - (0.2 A)(150 n) vL-zov (g) 2 points 0 0 (One point was awarded for stating that the stored energy 'E & M 3. . " (a) 3 points -92 6 - d: win-CIA dAV E - —55E 6 - —Buu (+ or — acceptable) (b). 5 points F - 12 X B or F - IIB 1-12! For using 5 calculated in part (a) L; 2 F _ unzzi For indicating that the force is 6 Egg—flags R (+ or — acceptable) ng.) opposite the direction of the velocity (including minus sign in above expression is sufficient) Distribution of points 1 point 1 point 2 points _ is used to keep the current flowi 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point as W 1991 Physics C Solutions Dis tribution of points r E. & H 3. (continued) (c) 5 points For an expression of Newton‘s second law: F - ma 1 point Using gxgression for F from part (b), with a generic velocity: ,uB 2 a - mg (+ or — acceptable) 1 point For indicating that the acceleration is opposite the 1 point direction of the velocity (including minus sign above or in differential equation below is sufficient) For a correct differential equation: g _ _ 32,22 _ dt v ER 1 pelnt dv _ __ 8222 d v m}? u I 3232 In v v0 - — mR c ln 1"— - - 3212 t Do MR 2 2 A v - vae'B 1 t/mR 1 point (Alternate solution) For a statement of conservation of energy 3 (Alternate Points) (l'point) > For placing 12R term on the correct side of the energy equation (1 point) For using expression for E from part (a) with a BEU R I - E/R - For a correct integral equation: 1 1 B 22112 among - 5.72112 + I $— R d: Try v - Ce‘Dc 2 2 %m,nz .. %mcz -2Dc + 2452 J's—2D: dz: 1 2 _ Emcee-2D: _ E33: 02 [—1]e—ZDt + K0 5”” 2 R 2D Since this equation must hold for any time t: .1. 2 l-z 3212 2 1 K0 zmuo and 2mC R C 20 generic velocity: (1 point) (1 point) g5 lo/ 1991 Physics C Solutions E & M 3. (continued) Therefore: 5212 0‘} v — no when t - 0, so uo - Ce'a and C - on U _ Hoe-Bzzzc/mk , he resistor will eventually ‘ ic energy of the rod (One point was aw arded for any other correct, relevant statement reg arding energy or power) Distribution of points (1 point) 2 points ...
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1991 Physics C Solutions - 1991 Physics C Solutions...

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