1983 Physics C Solutions

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

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## 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.

### Page1 / 6

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online