1985 Physics C Solutions

# 1985 Physics C Solutions - 1985 Physics 0 Solutions...

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Unformatted text preview: 1985 Physics 0 Solutions Mechanics Distribution of points Mech. I. (a) 5 points "or = (50 m/s) 5i" 37. = (50 "1/5) (-60) = 30 m/s 1 point y =yo + toy! + %at2 3' point 0 = (35 m) + (30 m/s)! — g (10 m/sz);2 2 points i 1 point for y = O. I point for correct substitutions on right—hand side) For corretion solution. I = 7 s 1 point Full credit was given for alternate equivalent solutions. (b) 3 points R = 11),: I point R = (50 ms) (cos 37‘) (7 5) (Le, for correct substitution) 1 point For correct solution. R = 280 m I point (c) 5 points 1:4 = PM = 40 ms 1 point 1'5 = l'o = 50 m S 1130111! For [(2 For any mention of energy I point For conserving energy I point I ~. I s ; mrc' = ; mro' + mgh — 2 PC _- V, r0 + :gh = V (50 ms)‘ + 2H0 m s-)(35 m) = v/3200 m‘,’sz rc = 56.6 m 's 1 point Full credit was given for alternate equivalent solutions (e.g.. if time of travel from point B to point C is known from part (a). kinematic equations can be used) (d) 2 points For any use of momentum or center of mass 1 point .HIO kg) = (30 m) (6 kg) x = 18 m I point Mech. 2. (a) 5 points f1 = “r‘v l POint f, = mg sin 0 or w sin 0 I point .‘V = mg cos 0 or w cos 0 I point _ L — m2 - - u, — N - mg cos 9 - tan 9 2 pomts Alternate solution using F = ma: /.4/Iernale points) ' .V - mg cos 0 = 0 (1 point) If, — mg sin 0 = ma (I point) f, = u,.V (I point) u}mgcosB-mgsin9=ma =mgsinlﬂwnaz a a 11, mg cos 6 tan 0 + g cos 0 (l pomt) u = 0. therefore it, = tan 0 (1 point) 57 1985_Physics C Solutions Distribution of points (b) 5 points AE = AK + AU The initial and ﬁnal points are at the highest point on the incline and the point of full spring compression. respectively. AK = 0 I point AL' = til/gm"), + AUSW,g Ab’gmuy = mgh = mg(d 1- x) sin 9 2 points Amp“g = - £16K: 2 points AE = mg(d + x) sin 6 — %k.\:2 I point was deducted il'the student did not write down the ﬁnal expression for AE. 2 points were awarded if the student tried to solve this part using frictional force and showed AE =j;(d + .v) or AE = WNW + x) or AE = pkmg cos 6(d + x) [a (c) 5 points ' Wm = A5 =ft-(d + x) I point fk = ilk-V I point tit-AV“! + x ) = mg sin 0 (d + x) — ékﬁ 2 points 1 point was deducted if the factor of(d + x) was missing from left hand side. .V = mg cos 0 mg sin 0 (d + x) - ékx2 “k = mgcosBldﬁ-x) __ , _ kx2 . pt - tan 8 2—~——-————(d + x)mg c059 . or I pomt _ kx: “k _ “i _ 1(d + .r )mg cos 9 Since M < u.. the minus sign in the above expression is important. If the signs were interchanged in this expression. I point was deducted. 1985 Physics C Solutions Mach. 3. (a) 4 points T I T2 2m mg 2mg For indicating tension vectors For indicating weight vectors For indicating different tensions on each block For indicating different weights for each block (The last 2 points were also awarded if such indications were made in the solution of subsequent parts.) (b) 5 points i. For applying F = ma T. - mg = ma 2mg - T2 = 2ma The sign of the acceleration is arbitrary in these two equations. ll. [1 = (T: "' Ti), I point for torque = (at I point for torque = (T; - T.)r (c) 3 points a 1 = " , For using the system of three equations obtained in part (b) Tl = m(g + a) T;=2m(g-a) T;-T,=mg-3ma 2% = (mg - 3ma)r [tr-1 =gmr a=g ‘OIN (d) l point Ti =m<s+a>=mtg +§gi = 34mg Full credit was awarded for answer that was consistent with previous work. Distribution of points I point I point I point 1 point I point I point I point 2 points I point I point I point I point 1985 Physics C Solutions Distribution of points (e) 2 points N = 7mg + T1 + T; or a clear indication that the normal force is different from the static weight 1 point r = 2m — 3 — - m 2 g 93 9 g N=7mg+%mg+195mg=%§mg lpoint Electricity and Magnetism E&M l. (a) 4 going IE - dA = q to or 4nkq 2 points E(2n:rL) = gig, or 4nkq I point = _3__ 2_k9 - E lncorL or rL ! pomt (b) 5 points .b—e _. 1/ = 'i E - dE (V = [Edi was acceptable) 2 points b q dr . = - -- l t J; 22:501. r P°'" _ q a . V— —- 2n£0L [In rlb lpomt = __q_ ‘ ’2 1181 '2 - V MOL 1%) or L I a I pomt No deduction was taken for reversal of limits of integration. (c) 3 points _ 1 , Co — V 2 pomts _ ZmOL L . C0 ‘ ln (b a) 0' 2!: Who) ‘ P°"“ I point was deducted if the value of C 0 was negative. as would be the case if the expression contained In (a ,b). (d) 3 points C = C1 + C; or the capacitors are parallel I point C|=k%Co=§Co lpoint C2 = gco I point c = g Co Full credit was awarded for correct ﬁnal answer. regardless of the method used by the student. 1985 Physics c Solutlone E&M 2. (a) 3 points V = R or equivalent (may appear here or in part (c)) For correct substitutions. V = 3000 V and R = 5 x 105 9 For correct answer, i, = .006 A (b) 3 points V7 0 For starting curve at the origin For showing V2 asymptotic to a positive ﬁnal value For showing V2 monotonically increasing and exponential (c) 2 points i, = i; or % = 153—202Tv5 or a statement indicating the series nature of the circuit in this situation For correct answer. 1'; - .002 A (d) 2 points Q =CVand/or V-ZOOOV For correct answer. Q - .0l C (e) 2 points U - % QV or equivalent For correct answer. U - [0 1 Distribution of points 1 point I point I point I point I point I point I point I point I point I point I point I point 1985 Physics 0 Solutions 3 oints - . . . (n P ,2 Distribution of points For non-zero initial value 1 point lpoint l ' For showing 1'; asymptotic to zero hing the ﬁnal value For showing 1': monotonically approac and exponential. 1 point l point was deducted if parts (a) and (c) were both correct but no work was shown. ' l point was awarded if at least one graph showed an exponential form and no other points were available. E&M 3. (a) Spoints C = — ‘18? (negative sign not required) 2 points (I) = BA 1 point «19 , 42 4 d1 d1 ‘ A = m3 I point 8=‘f—g 1tr2=(60'l"s)n(.5m)z = 1511 v=47v lpoint (b) 4 points _. For showing E directed to the left at point P on diagram 1 point 8 = Ed l point = .nr l point __ § _ _§_ ‘_ 47 v __ . E—d_2nr_2n(.5m)—‘5v’m lpomt (c) 3points F = '1}: I point I“ = my ' I point Thus ’1"; = qt-B - 19 — i l.6 x l0 C .5 m l0 T = 8‘8106mgs ‘ point FEE: m 9.11 x 10‘" kg 1985 Physics C Solutlons (d) 3 points F = ma F = (15 Thus mg = (15 -19 , a=LE=1L6xlo CﬁlSVm2=2.6xlolzm/sl m 9.11x10'3’kg 1 point was awarded for merely stating that a = dv d: Distribution ofpoints '1 point 1 point I point ...
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