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1990 Physics C Solutions

# 1990 Physics C Solutions - SOLUTIONS 1990 Physics C...

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Unformatted text preview: SOLUTIONS 1990 Physics C Distribution of points Mech l. (a) 4 points For an expression of Newton's Law, Fnet = ma 1 point Fnet = *kv 1 point For calculating the correct acceleration corresponding to the initial velocity: mag = —kv0 kvb . a0 = —--;;- 1 pelnt For the correct direction, indicated by a negative sign or the words "to the left" 1 point (b) 6 points For the recognition that a = 3% 1 point QB — E3 1 oint dt m p 93 =—5 dt: v m Upon integration, £n v = — Et + 2n C m u = Ce'kt/m v = v0 at t = 0 gives 0 = no u = voe'kt/m 1 point L! ”o 0 z —l)0 For showing v = v0 at t = O 1 point For a curve reasonably representative of exponential decrease 1 point For the curve asymptotically approaching zero as t approaches infinity 1 point 1990 Physics C Distribution of points Mech 2. (a) 4 points For use of conservation of energy 1 point 1 mgH = -2-mu02 For correct expression for potential energy 1 point For correct expression for kinetic energy 1 point ‘hz . H = —EE 1 paint Alternate solution (Alternate points) a = —g sin 9 (1 point) v2 — uoz = 2ad (or equivalent relevant kinematics) (1 point) (d is the distance traveled up the incline) d = H/sin 6 (1 point) v2 — yo? = —2(g sin ”(H/sin a) = -2gH v = O at highest point Ibz . H = '2? (l p01nt) (b) 5 points For relating kinetic energy K, potential energy U, and work done by frictional force Wf: K = Wf + U Wf = Ffd Ff = p mg cos 6 d = h/sin 0 Therefore, Wf = (pmg cos €)(h/sin 0) %mv02 = (pmg cos 9)(h/sin 0) + mgh — mgh(p cot 0 + 1) U02 H h = 2g<u cot 0 + l) = p cot 0 + 1 F‘ P‘ P‘ P‘ point point point point point 1990 Physics C Distribution of points (c) 4 points u = \$5 = voe'kt/m 1 point Upon integration, x = v0[— %]e'kt/m + C muo x = 0 at t = 0 gives C = _2— nm _E9 [l — e'kt/"J 1 point (These points were awarded if a correct integration was performed using an incorrect expression for u) X For a reasonable concave—down curve 1 point For X = 0 at t = O and approaching an asymptote as t approaches infinity 1 point (d) 1 point For t —~ m, e'kC/m —* 0 mo X = T 1 point 1990 Physics C (C) Distribution of points Alternate solution (Alternate points) F = mg sin 9 + pmg cos 0 (one point for each term) (2 points) a - —g(sin 9 + p cos 6) (1 point) I)2 — v02 = Zad d = h/sin 9 (1 point) —v02 = —2g(sin 0 + y cos 0) h/sin 9 1’02 sin a H sin 0 H . h _ 2g(sin H + p cos 9) = sin 0 + p cos 9 ~ 1 + p cot 6 (l p01nt) 4 points For including both translational and rotational kinetic energy in an equation for conservation of energy: Ktrans + Krot = méh' 1 point 1 2 . Krot 51w 1 p01nt I = 121R2 1 point w = 3 R l 2 l 2 El 2 . 2mv0 + 2(mR )[ R = mgh %mv02 + %mv02 = mgh' 1 point 2 h' = 1—)”— = 2H g Alternate solution (Alternate points) For an expression relating torque to angular acceleration: ET = Ia (1 point) Taking the torque about the point of contact with the incline: (*mg)(R)(sin 9) - 1'0: 1’ = I + me? = 2an2 (1 point) a = E (1 point) (This point was awarded only if some indication was present that the point of contact was used as the reference point.) 1990 Physics C Distribution of points . 28 —ng Sin 6 = ZmR E a = ~§ :ln 9 v2 — v02 = Zad . h' I —v02 = —g Sln 6 sin 0 = —gh 2 h' = 39— = 2H (1 point) g (d) 2 points Rotational kinetic energy does not change. Therefore, 1 n . Emvoz = mgh l paint ll U0 2 ' h — 2g ~ H l pelnt Full credit was awarded for merely saying that the answer is the same as in part (a). Mech 3. (a) 2 points F = kAX 1 point = ﬁg = (8 kg)(9.8 m/sz) = . Ax R (1,000 N/m) 0.078 m g: 0.08 m l p01nt (b) 3 points For any one of the following equations: 1 k k m . f = E; E or w — m or T — 2w]; 1 p01nt For correct substitution in any of the above: _ 1_ 1,000 N/m _ 1,000 N/m _ 8 kg . f ‘ 2n 8 kg 95 w ‘ . 8 kg 95 T " 2“ 1,000 N/m 1 p01nt 1 1 .8 s’ or w = 11 s'1 1 point ll f (The third point was not awarded for T = 0.56 s) 1990 Physics C Distribution of points (c) 2 points If the spring pulls the 5—kg block downward such that a > g, the 5-kg block moves faster than the 3-kg block can fall, and they will lose contact. Therefore amax = g = 9.8 m/s2 2 points Mathematically: For 3-kg block: 3g ~ Fc - 3a, where Fc is the contact force Fc - 0 when the blocks lose contact, so amax " g (d) 3 points Maximum acceleration occurs at the extremes of motion, when x = A, the amplitude. amax — 5% or wZA 1 point For using amax obtained in part (c) 1 point A .. "Hmax a (8 kg)(9-.8 m/sz) or amax .3 9.8 rugs? k (1,000 N/m) —— wz (11.2 s' ) A = 0.078 m (or answer consistent with part (c)) 1 point (e) 4 points For use of conservation of energy, Kmax = Umax 1 point 1 2 . Kmax = Emu,“ax l paint 1 2 . Umax = EkA l p01nt kAZ 2 = __._ Umax m = ’(l,000 N/m)(0.078 m)2 ”max 8 kg umax - 0.87 m/s 1 point Alternate Solution (Alternate points) %mx . . . . vmax = wA or w , for Simple harmonic motion (2 p01nts) 2 um = (11.2 s")(0.078 m) 93 W (1 point) ”max = 0.87 m/s 2; 0.88 m/s (1 point) 1 point was awarded for correct units in at least three answers 1 point 1990 Physics C Distribution of points E & M 1. (a) 5 points For an expression of Gauss’s Law 1 point q fE‘dA = em: 6('1 4 3 , . Volume of a sphere = Ewr I paint Q3113 Qr3 <lent: = 4 = R3 1 point —7I'R3 3 (One of these last two points was awarded for an erroneous calculation which indicated a recognition that the enclosed change depends on r) Surface area of a sphere = Anrz 1 point 3 2 a, Qr E4nr 60R Qr ' . E W 1 p01nt (b) 2 points qenc = Q 1 point Ednrz = Q_ 60 E = _—Q——2 1 point Aﬁéor (Second point was not awarded if R was used instead of r) (c) 2 points 2R < r < SR is inside the conductor, therefore: 2 points E = 0 (Some students continued to treat this shaded area as an insulator. For an attempt to solve part (c) in this manner, evidenced by the use of r3 — R3, 1 point was awarded) 1990 Physics c Distribution of points (d) 4 points Since E a 0 for r inside the conductor, total charge enclosed by such a sphere must be zero. For an indication that the total charge on the inside surface, 41: has magnitude Q 1 point For indicating that this charge is negative 1 point QI . a = Area I paint For consistent use of formula for area 1 point a = ___7‘Q = __Z‘Q 4w(2R) l6nR (If student indicates that q = O, 1 point is awarded for indicating a = 0) (e) 2 points For r > 3K, E depends on total free charge enclosed, i.e., Q +3 q Since qenc = 0 inside the conductor, the net charge go on the outside surface is: go =-Q+ q lpoint For consistent use of formula for area 1 point _ Q + _ Q +, " 41mg)? ‘ 361ng E & M 2. (a) 1 point For a correct arrow (T) or a statement indicating the correct direction 1 point (b) 3 points Speed as particles enter region III equals speed at which they travel through region II in a straight line Felec - QE 1 point Fmag = QuB 1 point QvB - QE v = E/B 1 point 1990 Physics C (e) (d) (e) 3 points Region III is, in effect, a mass spectrometer, in which equating centripetal force to magnetic force allows the determination of mass mvz Fcen _ R 2 m 3 QBR U QBZR m E 4 points Distribution of points The accelerating potential brings the particle to the speed at which it moves through region II. Energy imparted by accelerating potential = QV -lz K — va _ l 2 QV 2mu V = mu2/2Q H r-—\ 0 m m N t_iE r~—\ WIN N N :0 RE V=2— 2 points U2 a acen = ﬁ- = l E 2 R B E2 ‘3:le Alternate solution = QBE m [5] ea E2 EEZ m H (Alternate (l (1 point point point point point point point point point Points) point) point) 1990 Physics C Distribution of points (f) 2 points t _ distance v = ZWR 2 = £5 1 point v v E - «R B «RB . t a _E_ 1 p01nt (For erroneous solutions, one point could be earned if n appeared in some sensible fashion that indicated an attempt to utilize the circular path) E & M 3. (a) 2 points For arrows or words indicating a clockwise direction 2 points (b) 3 points Fgravity ’3 Mg 1 point Fmag - I£B 1 point g as - I 2B 1 pelnt (c) 2 points 6 = IR 1 point = MgR . 5 28 l p01nt (d) 2 points find = d¢/dt where is the magnetic flux 1 point find a Bﬁv 1 point (e) 3 points I’ = Enet/R 1 point = (E — find)/R 1 point = ﬁg — 2&3 1 point Bi R (Two of the three points were awarded for calculating the induced current and not the net current) 1990 Physics C Distribution of points (f) 3 points Once the loop reachs terminal velocity, the net force on it is zero: (M — Am)g — 1’28 1 point ﬁg _ B£v [M —R ]M 2 2 a Mg - B 2 U 1 point R 2 2 Am = B ﬂ v 1 point ...
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