1986 Physics C Solutions

1986 Physics C Solutions - 1986 Physics C SOLUTION 2 F = 0...

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Unformatted text preview: 1986 Physics C SOLUTION 2 F = 0 2T = (80 kg + 20 kg)(10 m/sz) = 1000 N T é 500 N (b) 4 points Again considering the forces on the person and platform, 2 F = II 2T - 1000 N = (100kg)(2 m/sz) 1 pt. 1 pt. 1 pt. T = 500 N (c) 5 points Considering the forces on the person only, 2 F = no Normal + 600 N - (80 kg)(10 m/szl = (80 kg)(2 m/sl) v W W \W‘/ 1 pt. 1 pt. 1 pt. 1 pt. Normal = 360 N (d) 2 points work ugh P‘s -ETEE- = t = (I9)V P = (1000 N)(O.4 m/s) = 400 N Distribution of points 1 point l-J points 1 point (4 points 1 point points 1 point point 1 point Physics C Mach 2. (a) 6 points P.E. = H.E. 1 point th = %Hv= + %Iu= 7 points Ly. W 1 pt. 1 pt. w = v/R 1 point th = =Hu2 + %[§HR=][§] . 7 . 2. any. «0- é-I'ILI. = Wye _ 10 v — ~7gh ‘ r = luv= = §M h 1 oint 1- ‘trans g 7 9 P .. , _ 1 . _ 1 2 _ 2 . 11. 'ert — Elw- - 5H1! — 7149’? 1 pmnt (b) b polnts i. v3 = 235 1 point _ g: = (10/7) gh = g A . a ' 2: 2(h/sin e) 7 9 5‘" ° 1 9°‘nt Alternate Solutions Method 1: (Alternate Points) Summing the forces parallel to the plane, 2 F = ma (1 point) Hg sin 9 - f = Ha - - £1 2 Mg sin 6 - gfia - Ha sin 9 (1 point) iwj Physics C mean 2, (b) i. (alternate solutions continued) Method 2: Summing the torques about the point of contact, 3 P = In (1 point) _ (Hg sin elk _ g sin 9 a — — -' 7 MR3 + EMR2 ;R o .1 a = GR .3 = “7’9 sin e (1 point) ii. P = In 1 point p = fR 1 point but a = % ’ 1 point 2 2 . so f - §HR% = §H[gg sin 9) f = §Hg sin 6 1 point Alternate Solution (Alternate Points) 2 F = ea (1 point) Hg sin 6 - f - Ha (1 point) F=Hgsin9-H[.$79sin e] (2 points) f = §Hg sin 9 (c) 1 point K“H - agh 1 point (d) 2 points The rotational kinetic energy 0‘ the hollow sphere i! 1 point greater than for the solid sphere. because the moment of inertia is greater. 1 point Mech 3. (a) 4 points r = - %% g; u = - I F(x)dx 1 point A u = ~1' (-kx3) dx 1 point 0 _ k.“ A . U — -3— 10 1 point - I’ U = f: 1 point (p) 4 points From conservation of energy, maximum kinetic energy equals potential energy at point of release % Hyman: = % kA‘ 2 points Vt "max: = 1 point ,, k . vnaI f= A- ~ 2, 1 paint (c) 4 points From conservation of energy, E‘o‘ = K + U when R = U , E‘o‘ = U + U = 2U _ l U ' Egtot ..b . t k: = % [kg 1 3 points A . x = ¢‘/‘ 1 point .(11) 3 points The period of oscillation decreases 1 point For a spring that obeys Hooke's law, F c x, and the period is independent of amplitude. For the springsin this problem, F c x', so the force and hence acceleration increase at a greater rate with displacement than for the Hooke's law spring. 2L 2 points For a Hooke's law spring v”M x A, and T a A I constant. Vac: For the spring in the problem, from part (b), v.0' I A3, and T c s c —%— Physics C 534” 1 . (a) (b) (c) (d) (e) (f) 7 a points All vectors drawn with correct sense (higher to lower potential) All vectors drawn perpendicular to equipotential lines For showing vectors at all three points (-1 point for missing vectors) 2 points Magnitude of electric field is greatest at point T, because equipotential lines are closest together near T. (Note: Some students misinterpreted the question to mean at which of the points L, N, and V, from part (a), is the field greatest. For these students, an answer of point U received credit.)- 4 points ='AZ E Ax For correct substitutions AV = 10 V .02 m 500 V/m Ax= E: 2 points V — V = 40 V - 5 V = II 3 35 V (Note: deviation of $2 in the answer was acceptable) 3 points H = «W H = (S x 10": C) (40 V - 30 V) H = S x 10‘** J 1 point No, 9; Same, 9; same as (a), 9: Work does not depend on path point point point point point point point point point points point points point Physics C EtM 2. (a) 4 points‘ V '= IR 1 point For correctly obtaining 20 n as the resistance of the 1 point combination of 10 fl and 30 n resistors RT = 5 9 +—20 fl = 25 Q 1 point - §§_V a IT ' 25 a 1 A IF = ézr = % A 1 point (D) 2 points 1R = % A’ 9: same as in part (a) 2 points (c) 4 points a = CAV 1 point AV = 10V 2 points 0 = 100 uC 1 point (d) 4 points f In the static situation, there is no potential difference across the inductor; the situation is the same as if the inductor were a resistanceless wire. .1. = 1 + 1 R,, 10 Q 30 fl 1 point RI‘ = 7.5 9 RT. = s + 217.5) a 20 n 1 point - '5‘ v 5 . IT - 33 a = z A I paint 1 = 31 = l§ A = o 9375 A 1 oint R 4 T 16 I ' p (e) 1 point The current through each 30 R resistor is glr. Since 2 I I 0 at point A, 3 1 1 5 1L = 1,, - IR - 317 - 317 - 5[; A] 1L 2 g A - 0.624 A 1 point Physics C Ean 3. (a) 3 points f B 0 d2 = uol, where I = :7 EA]. .. . B = 2', - points = unfit _ . B zfir 1 point (b) 1 point Induced current in loops is counterclockwise 1 point (c) 7 points {3 = I B ' d5 2 points b+a = u; . {B I 2', bdr 1 point a Ib a+b . 13 = 23;— 1n [ a ] , I = at 1 point 8 = -N 9% (Note: sign ignored in grading. Magni- 1 point tude of 8 was all that was required.) e = (1) -§ [ leQi—E ln [ azb ] ] 8 8 E3%2 ln [ ég- ] 1 point . E cb a+ . 1 8 E 3‘ E§;§ ln [ a J I paint (d) 1 point Force on the loop is away from the wire. However, the 1 point point was awarded only if (d) was consistent with (b). If (h) was incorrect (i.e., clockwise); but (d) was consistent (*orce toward wire), credit was given. (e) 3 points F I 12x3 ' 1 point For recognition that F,., I .F. - F... 1 point rn,‘ - 1:8, - 1t8,+. - 12(3, - a...) For correct substitutions for: I, §ron part (c) B, from part (a) for r - a and r - b + a 1 point z. using t I length of side of loop I b rm -1"[‘-?]][%-:1—b] ...
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This note was uploaded on 01/22/2012 for the course PHYS 2425 taught by Professor Park during the Spring '09 term at Cy-Fair College.

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1986 Physics C Solutions - 1986 Physics C SOLUTION 2 F = 0...

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