1983 Physics C Solutions

1983 Physics C Solutions - 1983 Physics B Solullons (c) 5...

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Unformatted text preview: 1983 Physics B Solullons (c) 5 points The frequency of the light is _c_ 3x IO8 __ f—I—SX l0_7—0.6>< lO'SHz The. energy of a photon of thi hf= (4 x 10‘”) (0.6 x10”) the work function, so no phot ; ~ or If one sketched the graph in part (b), one can observe that the frequency is less than the threshold frequency of 0.75 x l0” Hz, where the graph intersects the x-axis. ns will be emitted. 1983 Physics C Solutions Mechanics Mech. l. (a) 9 points i. The velocity vector is tangent to the path. Since v, = constant, a,r = 0, and the acceleration vector has only a positive y-component as shown below. v “’1 0 x ii. The y-component of velocity is v). = g; - 42 = 4X 3 By the cham rule, d! dx dt Sincey =lx2, %=x dx and v}, =x7! = Cx iii. The acceleration is given by ay = so a), = C% = C 2 Alternate solution for ii and iii: ii. Sincey =%x2 and x = CI, y =%C2t2 42- 2 Thusvy dt Ct Since Ct=x,vy=Cx duv’ d iii. ay = —~ = ——(C2t) = C2 Distribution of points 2 points 3 points 1 point 2 points 1 point 1 point I point 1 point 1 point I point (Alternate points) (I point) . (2 points) (1 point) (2 points) 1983 Physics C Solutions Distribution of points (b) 6 points i. The speed is given by v = . /v,‘2 + vyz' 1 point . d d dx . By the chain rule, 0, = a); = i a? = xvx - l pomt I 2 Sov=‘/vx2(l+x2 = C (1+x2)=C l + x2 I point ii. Again the velocity vector is tangent to the path. I point Since the speed is constant, there is no component of5’along the path, so E’is centripetal, perpendicular to 3215 shown. 2 points y 5’ v S 0 X Mech. 2. (a) 4 points Support Force T T "'1" mlg On the block, the forces are weight = ng I point and tension T I point 0n the disk, there are tension T 1 point' and the pair: mg and supporting force 1 point (b) i. and ii. 8 points "On the disk, torque = F = Ia: = %m1R2az 1 point with F=T~Rand lpoint acceleration of the block = a = Ra, so a: = a/R 1 point For the block, Em = mza 1 point with Fnet = mg - T 1 point Solving simultaneously gives 1 point _ Zng d 1 ‘ a — —-—2m2 + m! an pomt T = mm”; 1 point 2m2+ml 1983 Physics C Solutions (b) iii. 3 points By definition. angular momentum L 4— [w Angular velocity u; = 1! Since I= émle and z=%. - l 2 _ 2 , _ l L—ZmlR R t—zmlaRt Substituting the expression for a from part (b) gives L __ mlngt — 2m2 + ml Mech. 3. (a) ll points ii. iii. Energy is conserved as the particle slides down the sphere. so K + U = constant. where gravitational potential energy U = mgh. If h is measured from the center of this sphere. 0+ng=K+ngcosO so that K = ng (1 -cos 0). The centripetal acceleration is v- (1‘. = K From part i: émr2 = ng(l - cos 6) .2 . so at. =5- =2g(1—c050) '\ R The only force with a ___‘ tangential component ""4 \\ mg .m u is the gravitational I / \\ force mg. As the diagram I / shows. its tangential I/W’ component is mg sin 0 me By Newton's second law, may = mg sin 9. so a1 = g sin 0 (b) 4 points The particle leaves the sphere when the normal force N has decreased to zero. . At that point the radial component of the weight provides the centripetal acceleration, so mg cos 6 = mac Therefore g cos 0 = ac = 2g(l - cos 0) and cos 0=§ _ 2 or 9— arc cos 3 Distribution afpoims l point 1 point 1 point 2 points I point I point 1 point 1 point 1 point 1 point 1 point 2 points 1 point I point 1983 Physics C Solutions Electricity and Magnetism E&M l. (a) (b) E&M 2. (a) 6 points Gauss‘s Law states that (D = 5E . d3 = qgeniiosedl 0 Apply _it to a sphere of radius r (a < r < b). Since I; is uniform and directed outward, §E - dA = E - (area of sphere) = E - 4an The enclosed charge is Q. Therefore E = —Q— 4n£0r3 5 points The potential V0 equals the potential difference between the spheres, which may be expressed as fed: bdr Therefore V0 = Tge—J F5 0 (1 Evaluating the integral gives _ Q _l 1’ Vo-4neol: r] 4 points .Substitution of the expression for V0 from part (b) gives C=_Q_=fl“_ _Q_ b-a b-a o 47:60 ab 4 points Initially there is no potential drop across the capacitor, so 8 = [QR and 8 = 10(10’6)(2 x 106) = 20 volts Distribution of points 2 points 2 points 1 point 1 point 1 point 1 point 2 points I point 2 points 2 points 2 points 2 points 1983 Physics C Solutions Distribution of points (b) 5 points Current 1' and charge Q are related by . dQ . . l: —d— or Q = 51d: lpomt t (1) so that Q = fioe 6 dt 1 point Evaluation of the antiderivative gives (1) Q = —- 6i0e " + C 2 points Since Q = 0 when t = O, the constant of integration C must equal fig, and " I Q = 6io<l —- 6(7)) 1 point (c) 3 points Q Qmax I 0 Q=0att=0 [point As t—+oo, Q —->Q,m 1 point Exponential shape 1 pomt (d) 3 points For a charging or discharging capacitor, the time ' constant is RC. . I paint Here, the numerical value of the time constant is 6 seconds, so 2 points 1983 Physics c Solutions E&M 3. (a) 6 points The field at M is down. ’-‘ / ’ \ . . / \ The path of integration / \ is a Circle around I. _ \ Agipere’s Law states that I \ §B . ai = no] l 18) M Applied to the circular \ |<—R path, it gives \ / B - 2m = no! \ / _ H01 \\ // 50 that B - m \.___ z . if (b) 9 pomts i. 3 points At point 0, the B field is zero because the field caused by the right current cancels the field caused by the left current. ii. 6 points Bresultant I®<._a—>.<——Ol a——>®I Each current contributes a field vector of magnitude 3=__i101_ Zak/aZ-Ir-yz The vector sum of the two is directed to the right. Bresultant = 2 ' Bx =2-B-c059 = 2 - B - <——y—), giving the final result ‘ /a2 + y2 poly , directed to the right. Bresultant = W Distribution of points I point 1 point 2 points l point 1 point 2 points 1 point 1 point 1 point 1 point 1 point I point , I point ...
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1983 Physics C Solutions - 1983 Physics B Solullons (c) 5...

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