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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 & M1 (a) 3 points
fEdX=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 & M1 (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,EO 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 & M2 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
f5d§=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: M3 (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 CyFair College.
 Spring '09
 Park
 Physics

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