1979 Physics C Solutions

1979 Physics C Solutions - 1979 C: Mach. 1 (a) Solution 4...

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Unformatted text preview: 1979 C: Mach. 1 (a) Solution 4 points Conservation of energy provides the relation mgh = '/2mv’ which can be solved for v, the speed. v = x/Efi (This expression is the speed just before collision at P,. Since the collision is elastic, it is also the speed after the collision.) The direction of the ball's motion after the collision at PI is, “horizontal to the right." (b) & (c) 8 points (combined) (d) Horizontal and vertical motions are considered separately. During the flight from P1 to P2, the ball maintains a horizontal speed of VZgh and travels a horizontal distance L / thus d = vt yields L 75= v2ght During the same time t, the ball travels the same distance vertically, given by 7% = V23!Z Equating the two expressions for L / {2- gives 2gb t = ‘/2gt2 and cancelling times (which means recognizing they are equal), we obtain Jig—h = l/2gt leading to the answer to part (b) of t= 2 JET/g = m Substitution of this value back into either the horizontal or vertical expression for L / (/5 gives L/x/i = VZgh ngh/g O!‘ L=4Jih 3 points The speed just before striking P; may be found from conservation of energy mgh + mgL/fi - '/2mv§ Substituting L = 4x/ih gives mgh + 4mgh - l/zmv§ from which v2 = m Distribution of Points 1 point 1 point 2 points 2 points 2 points 1 point 1 point 1 point 1 point 1 point 1 point 1 point Total 15 points 1979 C: Mech. 2 (a) 5 points Linear momen turn is conserved during collision; therefore Mlvinitial = (Ml + M2)Vfinal Substituting appropriate values, one obtains M2 = 105 (b) 5 points x=fvdt x = f Ze‘“ dt = —‘/ze“‘ + C Att=0,x=0,thusC = '/2 Thus as t—+ as, xfina. -+ '/2 meter (or full credit for correct limits of integration) (c) 5 points The retarding force FR can be calculated directly from Newton’s second law and the expression for the velocity. Solution F ‘2 R—mdt d l -| FR ma(3+flt) 1 _2 FR—-mB(§+Bt) =—vaZ 1979 C: Mach. 3 (a) 5 points The answer is 0 force and centripetal acceleration. F = ma kx = mwzr k(92 - 91) - k2. Q2 (b) 5 points PE = '/zk)(2 —- mwtzi’zz =k—mw}, _ 1/1 ml _ 52,)2 KB a 1/2 mv2 - ‘/2 mom? ET = PE + KE - 1/; mm, (c) 3 points L = In) L ‘5 ngwo kill 2 )“° Lamp—7 k— mwo btained from Newton’s second law and knowledge of the spring Distribution of Points 3 points 2 points 1 point 2 points 1 point 1 point 1 point 3 points 1 mint Total 15 points 1 point 2 points 1 point 1 point 2 points 2 points 1 point 1 point 1 point 1 point Distribution Solution of Points (d) 2 points Time 2 points Time Total 15 points 1979 C: E & M-1 (a) 3 points fE-dX=Q/eo . Ipogm E (41rr2) - 0/60 1 point E = Q/41reorz or kQ/r2 l pomt (b) 3 points . r > c, E = Q/Zweor2 or ZkQ/r2 I point b<r<c,E=0 lpomt r<a,E=0 lpoint ' Distribution S olutzon of Points (d) 2 points Time 2 points Time Total 15 points 1979 C: E & M-1 (a) 3 points f E’ . dz? = Q/co _ 1 point E (47rr2) == Q/eo 1 point B = Q/4reor1 or kQ/r2 1 point (b) 3 points r > c, E = Q/21reor2 or 2kQ/r2 1 point b<r<c,E=O lpoint r<a,E-O lpoint Solution 3 points E=Ofor0<r<a andb<r<c Curves are decreasing and concave for a<r<bandc<r. E(c) lies above extrapolated value at point c of curve for a < r < b. Vsconstant for 0<r<aandb<r<c Curves are decreasing and concave for a<r<bandc<r. Continuous curve Alternate Solution V = q/47reor q = 2Q and V = Q/Zfl'toc or kq/r r=c 2kQ/c Distribution of Points l point 1 point 1 point 1 point 1 point 1 point I point l point I point 1 point 1 point 1 point Total 15 points 1979 C: E & M-2 Distribution Solution of Points (a) 2 points For direction 1 point For approximate 1 point magnitude (b) 5 points For any statement that there are no horizontal components I point For statement of Gauss‘s law f5-d§=q/eo lpoint Choose a “pill box" with a top and bottom each of area A. Let the top face pass through PI and the bottom through P2. By symmetry, E is perpendicular to both ends, is directed outward, and is of equal magnitude at both ends (and is parallel to the sides). Thus, by Gauss’s law ZEA = AZpa/eo 2 points E = pa/eo 1 point (c) 3 points 3 points (d) 5 points For any statement that there are no vertical components I point For statement of Ampere’s law 1 point f E’. (ii = poi Choose a rectangular path of length L and width 2a with the top and bottom passing through PI and P2. By symmetry, B is parallel to both top and bottom, is in the same direction as path increment, and is of equal magnitude at the top and bottom (and is normal to the ends). 2 BL = uoi 1 point = #023L} 1 point B = poaj l flint Total 15 points 1979 C: E 8: M-3 (8) (b) (0) Solution 6 points _> Experiment I demonstrates that B is in the plane of the paper because F is perpendicular to both—V's: B. Experiment ll demonstrates that? ma_lges an angle of —60° in the plane_9f the paper since it must be perpendicular to F2, which is in the direction of Vx B. 5 points The conditions for a circular orbit are that the B field is constant and perpendicu- lar to the velocity. These conditions are met in case 11 only. Thus, motion in case ll is a circle. F; = var r = vaF2 4 points _. Since the velocity and the B field_gtre not perpendicular in experiment I, the component of velocity parallel to B produces no force and hence no change in motion. The perpendicular component of velocity _produces circular motion, thus the resulting motion is a helix (spiral) about the B vector. Partial credit: helix (2 points), axis of helix along§(l point) Distribution of Points 3 points 3 points 2 points I point 1 point 1 point 4 points Total 15 points ...
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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.

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1979 Physics C Solutions - 1979 C: Mach. 1 (a) Solution 4...

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