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1981 Physics C Solutions

1981 Physics C Solutions - Solution Distribution of Points...

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Unformatted text preview: Solution Distribution of Points 1981 C: Mech l. a) 4points A/ F flN m 9 1 point for each of the four arrows. Directions must be correct. Any reasonable label is acceptable. 4 points b) 7 points 2? = m: or facsimile 1 point fk = p.N or facsimile lpoint From summing forces perpendicular to the plane N=mgcosB+Fsin0 I lpoint From using Newton’s second law in a direction parallel to the plane a,.=Ecosfl — gsin0 —Ll\l Luefl. or 1 pt. 1 pt. 1 pt. 3 points F—f‘ ” [J— a..=Ecos0 — gsin0 -(,rgcoso+flsino) m 111 Also, for correct algebra to get expression. 1 point Solution Distribution of Points c) 4 points v = const =) a.l = 0 1 point F = "18(H) ' [point F>0=>cos0>psina lpoint tan 6 < l 1 point It Total m 2. 15 points During downswing energy is conserved (2M)gh = '/2(2M)v‘ 2 points with h = L/2 1 point v = s/g_L for speed at bottom. 1 point During upswing energy is conserved Vsz,‘ = MgH 2 points with H = L (1 — fi/z) lpoint v, = x/g_L - (m ) for speed of swing after child jumps 1 point During jump momentum is conserved 3 points 2Mv = Mv, + Mvc 2 points vc=\/§T.-(2-‘/i-:72') Zgoints Total 15 points 3. a) 5 points From conservation of linear momentum m,v = m2 ( —_2—V) + M.v’ 4points v’ = 2flv l ' t 2 M. ' pm" b) 4 points From conservation of angular momentum L - L l . m,v 3 = m2 - 7" - 3 + EMU» 3 pomts 6 m v = __z_ . (0 ML 1 pomt c) 6 points AK = KEN - KB”, 2 points 1 = yum": + lAla.»z + l/2MI (v’)2 - V2m,v’ 3points 3 21 m 1 = 3;sz _ gfivz lpoint Total 15 points 1981 C: E&M l. a) 6points For statement of Gauss’s law —E, There must b integrand. Sketch of selleric parallel to ds. cw = a... + cm... = §C. 2. a) 2 points _. E to the right if axial Solution e a clear indication that a scalar product of vectors is involve a1 Gaussian surface or sketch of radial E field or statement that? is Distribution of Points 3 points 1 point 1 point 1 point 2 points 1 point 1 point 1 point I point 2 points 1 point _____._. Total 15 points 1 point 1 point b) 5 points (15:le 41m, 3.1 + [)1 dE‘= l £1030 41reo a‘ + b1 _l_flL. - 416:, a1 + b2 E = ”E JUJ&_ 415° (212 + b1)” c) 2 points [email protected] 21r d) 2 points Etc the left Eaxial e) 4 points ( b ¢—TTE? From Biot—Savart law, Nucladfi dB = ”(a1 + b’) a de=dB'——— W B E0 coal 9 =nw+wn 3. a) 3 points Solution arrow or statement for counterclockwise b) 5 points The induced emf 8 by Faraday’s law is _:£91 8— dt where (b. = B-A andA = s(s - x) 50,8 = B-s'v Distribution of Points 1 point I point I point 1 point I point 2 points 1 point 1 point 2 points I point 1 point Total _ 15 points 3 points 2 points 1 point 1 point 1 point c) 3 points From definition of R, HE R 1 d)4points P = 8i = 3% =i1-R = B’s‘v’ P R Alternate method: P = F-v where F = Bis B’s’v‘ .P= 50 R 1982 B l. a) 4 points For the first 2 seconds, while acceleration is constant, 5 = 'Aat‘ Solution Substituting the given values 5 :- 10 meters. t = 2 seconds gives 25 _ 2-10 a=——— =5ms" t' 2’ b) 4 points The velocity after accelerating from rest for 2 seconds is given by V = at sov = 5-2 =10ms" Alternate method: v: = 285 sov = x/Zas = Jz-s-lo =' lOms" c) 4 points The displacement, time, and constant velocity for the last 90 meters are related by S=Vt. Distribution of Points 2 points I point 3 points 1 point A lternate Points (1 point) (2 points) (I point) Total 15 points 2 points 2 points 2 points 2 points Alternate Points (2 points) (2 points) 2 points ...
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