1987 Physics C Solutions

1987 Physics C Solutions - 1987 Physics C Distribution of...

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Unformatted text preview: 1987 Physics C Distribution of Pbints Mech. l. (a) and (b) 7 points :-5‘ 7. If vector diagram shown. . 1 point was awarded for 3 points I each vector correctly FL drawn and labeled. W Summation of forces equal to zero. Taking components: (1) in y-direction: T cos 0 — W - 0 For correctly (2) in x-direction: T sin 0 — PH - 0 obtaining any 2 points (3) in direction of T: two of these T - W cos 9 — F” sin 9 - 0 three equations (If vector diagram not shewn, 5 points awarded for two of the 3 equaxions) For correct simultaneous solution for T and PH of 2 points any two of the equations (1), (2), or (3) acmle- from (1) r- W a p ' cos 0 Substituting into (2) cos a sin 0 - PH - 0 FH-Wmnfl (-1 point for using wrong functions, -1 point for answers not in terms of V and 0) (c) 3 points 1— Fbr vector diagram 1 point I l / h! Summation of forces in direction of T equals zero 1 point T - w cos 0 1 point (Full 3-point credit given for simply stating the correct answer) 1987 Physics c Mach. 1 (continued) (d) 5 points ac - E— 1 point I - W - mac U2 (1) T - W - mi: 1 point From conservation of energy, 1 z (2) mgL(1 - cos 0) - Emu 2 points 1 point 1 point letting mg - W, and solving (1) and (2) £5: T, gives T - W(3 - 2 cos 0) 1 point Mach. 2. (a) 3 points Pbints of equilibrium are: x5 - 2 m 1 point x5 - S m 1 point No extraneous points indiented 1 point (b) 3 points Etot - K + U - a J 1 point At x - 2.0 m: K - h - U(2) i A — 1 - 3 J 1 point At x - 5.0 m: K - 4 - U(A) - a - 3 -’1 J 1 point (c) 2 points No, the particle cannot reach x - 0.5 n 1 point The total energy is insufficient. 1 point 1987 Physics c Mech. 2 (continued) (d) 2 points Yes, the particle can reach x - 5.0 m. The total energy is sufficient (e) 5 points : 2 fix) 5 (newtons) - 3 ------- 27 -40 x(memn) For rough slopes, or a shape indicating change between positive and negative values at 2_places For having'signs correct (i.e., positive values of F, then negative values, then positive values again) For correct placement of zeros at x - 2 m and x - 5 n Fbr constant values.(zero slopes) at appropriate places For correct numerical values of F for regions of zero slope (If no graph, +1 point was awarded for F - dU/dx. or +2 points were awarded for F - -dU/dx) Hech. 3. (a) 4 points K.E. of rod at bottom of swing - P.E. of rod at top %qu - mgh 1 11112 Z i 5(le - “[2] z _ 25 _(3)(10) _ “’ 2 1.2 25 5 radians ” ' sec (1 point for numerical answer. 1 point for Lnits of radians/second or seconds“) 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 2 points 1987 Physics c Mech. 3 (continued) (b) a points K.E. before collision - K.E. after collision %‘”0"02 ' %m0”2 + 3 points 1 2 51w 1 pt. 1 pt. 1 pt. 2 §<n<10>2 - fimuz + (5)2 SO--%uz+l8 112-64 v - 8 nVs 1 point (c) 4 points L - mvr sin 0 25 L - mo: 25 f - F x F 2 points L - (1)(10)(1.2) L - 12 kg-mz/s 2 points (1 point for numerical answer. 1 point for correct units) (d) 3 points Angular momentum is conserved during collision. so L (before) - L (after) 1 point about - ub(u cos 9)! + Iw . 1 point 12 ' (1)(5)(1.2) cos 0 + (l.44)(5) 12 - 9.6 cos 0 + 7.2 cos 0 - 0.5 c - 6°. 1 point Alternatelz, angular momentum of ball after collision may be written maul sin ¢, where o is the angle between u and 1. The solution yields o - 30', but 0 - 90' — ¢ - 90' — 30' - 60'. 1987 Physics C E & M 1. (a) a points Gauss's law for r > R I 35-55 - Q/qJ 1 point E fdA - Q/eo 1 point sum?) - Q/eo 1 point 1 . E - Z;:; 32 I paint (-1 point for answer in terms of R, -3 points for answer only without showing details) (b) 5 points Show or mention Gaussian surface 1 point as inside sphere with radius r. Charge enclosed by Gaussian surface 3 Q - [ZQ—-] [gxr3] - 9E3 2 points 3 31R \.~w~./ ‘—~r-—’ 1 pt. 1 pt. From Gauss's law, :3 2 - .9. - 9. E(Axr ) ‘0 (0-53 1 point _ _Q_ r E “wen E3 1 point Alternate solution in terms of density p (Alternate Points) Show or mention Gaussian surface as above (1 point) , - .J—a (1 point) ~«R3 3 lb 3 "fir q - pV - p g" (1 point) EUHrz-Z) - geo— .. %[§ir3] (1 point) E _ E. (1 point) (-1 point for using p without defining it, -3 points for answer only without showing details) 1987 Physics c E & M 1 (continued) (c) 2 points Q V - ARCOR 2 points (-1 point for answer in terms of r) (d) 4 points 0 Vcenter "' Vsurfece + I E'd—E (1 P01“: for sum) 2 points w 1 pt. 0 - A. .9...’ -l’_2 hueok + AneoR 2 R 1 p01“: - _Q_ _9_ _ .29_ . hneoR + SneoR SteoR 1 p01“: E & M 2. (a) 4 points ¢ - BA (or o - f3.d3 or o - f~cd3) 1 point o - (2e"t)(.09) 2 points ‘—~r~—/ L‘f" 1 pt. 1 pt. 4» - 0.13Mc , 1 point -. (However, no points were awarded for going from Ig-dfi to a time integral such as fe“c dt) (b) 1 point For statement or arrows indicating counterclockwise 1 point direction (c) a points do 5 - " d—c 1 point - 0.729“c 1 point i -% 1 point 1 - 0.12.2“ 1 paint 1987 Physics 0 E & H 2 (continued) (d) 6 points P-iZR - (ggP-sZ/RggP-is) 1!) W - f izRflc (+1 point for indication of integral. If 0 integral is shown, then +1 point for correct limits) 1” w - f 0.864 6‘5ch 0 - 0.108 e‘8t 0 W - 0.108 J l E & H 3. (a) 3 points he switch is closed, there is no Immediately after t its impedance is infinite. current in the inductor; Rm - 10 +9o-1ooo E 20 i - RtOt - 100 - 0.2 A the potential difference across the 90-ohn resistor, V90. is given by v."J - m - (0.2)(90) - 16 v 11. _£-__1_3.- ‘dc‘ L 0.5 365/5 (Students who used the emf of the battery, 20 V. and obtained dl/dt - 40 A/s received 2 points) 1 point 2 points 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1987 Physics C E & M 3 (continued) (c) 2 points After a long time, the inductor acts as a short circuit Rt" - 10 O 1 point . s _ 29 _ l - Rt“ 10 2 A 1 point (:1) 2 points Ehergy - %12L 1 point - %(4)(o.5) - 1.0 J 1 point (e) 2 points Immediately after the switch is opened, the current in the inductor is the same as in part (c). All the current nust go through the 90-ohm resistor. V90 ' Vinductor - LR " (2)(9°) 1 point ' 180 V 1 point (f) 2 points Potential Difference [90 V 0 Tune: lo ‘ For curve decreasing in value and concave upward 1 point For 180 V starting point g any carefully derived 1 point answer to part (e) Extra 1 point For obtaining three out of five correct units for answers to parts (a) and (e) 1 point ...
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1987 Physics C Solutions - 1987 Physics C Distribution of...

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