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Unformatted text preview: SOLUTIONS 1989 Physics C Distribution
of points
Mech l.
(a) 3 points
Fbr any statement of conservation of energy 1 point Fbr an equation containing three correct terms: mgh  mghc + %mvcz 1
W (4 mgs)2
hhc+EO.5m+ 2(9.8m/s)
h  1.3 m 1 (b) 3 points Fbr one correct force 2
IV Fbr second correct force 1
""3
1 point was deducted if any extraneous forces were shown.
(c) 3 points
i 2
For centripetal force equation: Fc = E%— 1
Fbr recognition that ”B = UC = 4 m/s 1
Fc  (0.1 kg)(4 m/s)3/(O.5 m) = 3.2 N 1
Only 2 points were awarded for a correct calculation of
the net force, instead of the force exerted by the track.
(d) 3 points
Fbr any statement of applicable equation: Ufz — viz = 2a Ay 1
Fbr recognition that vi = ”C sin 30° = 2 m/s 1
v12 U12
AV ‘ ‘ 2—; 7g
Ay  (2 m/s)Z/2(9.8 m/sz) = 0.2 m o_r y = 0.7 m 1
Alternate Solution I (Alternate
vf  ”i — gt = 0 (1
”i a ”C sin 30° = 2 m/s (1
Vi 2 s
t = g = 575157;? = 0.2 s
1 2 Y=Yi+Vit—§gt  0.5 m + (2 m/s)(0.2 s) — %(9.8 m/sz)(0.2 s)2 (1 y  0.7 m ‘93 Ay  0.2 m point point points
point point point point point point point
points)
point) point) point) 1989 Physics C Distribution of points
Mech 1. (continued)
Alternate Solution II
For conserving energy: %m(viy2 + ”ixz) + mghc  mgy + %muixz (1 point)
vi?  ”C sin 30°  2 m/s (1 point)
%(2 m/s)2 + (9.8 m/sz)(0.5 m)  (9.8 m/sz)y
y  0_7 m (1 point)
If part (a) was solved correctly, conservation of energy
is used to solve (d), and the answer obtained is y  1.3 m,
only 1 point was awarded for (d). If (a) was incorrect, this
same solution to (d) was awarded 2 points.
(e) 3 points
Fbr any indication that the work equals some change
in energy 1 point
W  EC  EA = mghc + %mvcz  mghA 1 point
W = (0.1 kg)[(9.8 m/sz)(0.5 m) + %(4 m/s)2 — (9.8 m/sz)(2 m)]
 ~ 0.7 J 1 point
Alternate Solution (Alternate points)
W = AE (1 point)
= mghpart(a) — mghpart(e) (1 point)
w — (0.1 kg)(9.8 m/s2)[1.3 m  2 m]   0.7 J (1 point)
Mech 2.
(a) 3 points
For Newton's 2nd Law: 2F = ma 1 point
Fbr applying Newton’s 2nd Law to block A:
2F = 2Mg — Tu 1 point
2Mg — Tu  2Ma
Tv = 2M(g — a) or equivalent 1 point
(b) 5 points
For relating torque to rotational motion: Zr = la 1 point
For relating a to a: a = aR 1 point
For relating torque to tension: 1  TR 1 point
For calculating the net torque: Zr = TvR — ThR 1 point
TUR — ThR = I; = 3MRa
Th = %(TUR — 3MRa) = 2M(g — a) — 3Ma
Th  2Mg — SMa 1 point 1989 Physics C Distribution of points
Mech. 2. (continued)
(c) 4 points
f  pN 1 point
For use of correct normal force: N  4Mg 1 point
For correctly applying 2nd Law to block B:
Th — f  3Ma 1 point
2Mg  5M3  Ang  3Ma
Ang  2Mg  8Ma 1 point
I‘  (2g ‘ 8a)/4g
 [2(9.8 m/sz)  8(2 m/sz)]/4(9.8 m/sz)
p  0.1
(d) 3 points
For including only the frictional force in 2nd Law: f = mac 1 point
For using m  4M 1 point
4Mpg  4Mac
ac  #g‘  (o.1><9.8 III/52>
ac  1 e'm/s.2 1 point
Mech. 3
(a) 3 points
mgh  %mv2
For either or both equations 1 point
v  2gb
For correct substitution: v  JZ(9.8 m/sz)(0.45 m) 1 point
v  3.0 m/s 1 point
Alternate Solution (Alternate points)
Ay  égtz
u  gt:
v2  2g Ay (1 point)
For correct substitution (1 point)
For correct answer (1 point)
(b) 3 points
2n m .
T  w  2njg 1 pelnt
. . , _ 2 kg = 2n .
For correct substitution. T Zn 200 N/m or T 10 Hz 1 p01nt T  0.63 s 1 point 1989 Physics C Mech. 3. (continued) (c) 3 points Fbr any one of the following: DEFo 2) Correct force diagram 3) a  0 when u is maximum 4) ky  mg ‘ 2
For correct substitution into 4) above: y  $2—%§%%2§%E?£§—1
y  0.098 m or 0.1 m (d) 3 points
For use of conservation of energy
For use of an appropriate reference level
for potential energy Fbr example: lkyz  ”130’ + 0.45 m) or may + lmuz 2 2 For use of the quadratic formula to solve above equation y  0.41 m or 0.42 m (e) 2 points
Fbr ony one of the following: 1) An indication of subtracting the answer to (c)
from the answer to (d) 2) Calculating both roots of the quadratic, and
taking half their difference 3) Indication that in the quadratic solution a i ﬂ, a
is the equilibrium point and ﬂ is the amplitude For correct answer: A = 0.31 m or 0.32 m (This answer must be the correct value  part (e) is where
credit is earned for correct math in solving quadratic
equation of part (d)) Additional 1 point awarded for correct use of units
and no incorrect units Full credit for part (c) can be obtained by: Solving (e) through second or third method Indicating that maximum speed occurs at equilibrium Finding the equilibrium point, using answer to (d) as necessary Distribution
of points 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point ____________________—___________.___————————— E & M l.
(a) 2 points E  0 because the net charge is zero (b) 2 points V = 0 because the net charge is zero 2 points 2 points 1989 Physics C Distribution of points
E & M 1. (continued)
(c) 5 points
For some statement of Gauss' Law: f Edz a q/co 1 point
For any attempt to calculate the portion of the negative
charge within the radius r 1 point
. . (4 3)1rr3 —Qr3
Portion of negative charge —Q (4/3)nR —§3
Tbtal charge inside r: Q — ggi 1 point
A  hnrz 1 point
1 Qr3
E(4nr2)  23 [Q — ﬁz’]
___9_1 _r  .
E hneo :7 E3 or equivalent 1 paint
(d) 6 points
V  ~ ff E.d; » 1 point
(D
R r
— Eng—I ad;
0 R
r> ’
 0 — Edr
IR
Fbr using correct limits, r and R, for non—zero
contribution to V 1 point
For correct substitution of E from part (c):
 _ Q r 1 _ r‘r .
V [met] fRFdr IRFdr 1 paint
Fbr correct integrations:
1 2 r
_ _ _Q__ _ _ _ r .
V (urea r 212 1 Punt
R
Fbr correct substitution of limits:
_ o 1; r2 132 
VzmoL KW?! m 1PM“
Q l r2 _ §_ , .
V 4neo r + §§3 2R 1 p01nt If both (a) and (b) were wrong, and (c) and (d) were not attempted,
points for Gauss"Law, correct area A etc. were awarded for
(c) and (d) if they were written in (a) and (b). 1989 Physics 6 Distribution
of points E & M 1. (continued) If no points were awarded for any of the above, the
equations E  kq/r2 and V  kq/r were awarded 2 points each when present.
_______________________________.______._..__—————— E & M 2.
(a)
i) 4 points
d<I> .
E  HE  B21: 1 paint
For substitution of correct length: 8  Bhv 1 point
For a statement of Ohm’s Law: I  E/R 1 point
For correct answer: I = Bhu/R 1 point
ii) 3 points '
For FA  Ff'ield " I23 (or tu) 1 point
For correctly substituting I from i): FA  [%]33 1 point
For substitution of correct length: FA = [EEK] I13 1 point
Bzhzv \
FA ' R
(b)
i) 4 points
Counterclockwise
N
E
t:
:3
U
Clockwise
For correctly indicating ranges of x for which I  0 1 point
For constant values of I when w < x < 2w and SW < x < 4w 1 point
For positive I when w < x < 2W 1 point For negative I when 3w < x < 4w 1 point 1989 Physics C Distribution of points
E & M 2. (continued)
ii) 4 points
Ram
k
8 X
5 w 2w 3w 4w 5w
"
Lcﬂ
Fbr correctly indicating ranges of x for which F  O 1 point
Fbr constant value of F when w < x < 2w and 3w < X < 4w 1 point
Fbr positive F when w < x < 2w 1 point
Fbr positive F when 3w < x < 4w 1 point In both i) and ii), only 1 of the last 2 points was awarded
if the graph did not contain characteristic "breaks" at
multiples of w, e.g., CH” Also, a total of 3 points was awarded for graphs that were perfect
inversions of the correct graphs with respect to the xaxis. E & M 3. (a) 3 points
U  %CV2 1 point
For substitution: U  %(6 pF)(20 V)2 1 point
For correct answer: U = 1200 yJ 1 point The last point was not awarded for an answer of just "1200."
An indication of comprehension of the units "pF" was required
by the presence of "pJ" or a numerical answer indicating the
use of 6 x 10'6 F. 1989 Physics C E & M 3. (continued) 0)) 4 points Fbr realization that the charge is constant: Q  constant or CV  CIVI Fbr an indication that the work equals the change in stored energy: W  AU
Fbr correct new capacitance: C'  (1/4)C V’—£—Y4V C
1 W  %(c’v’2 — CV2  §<4cv2 — ch)  3[—cv2] 2
W  3600 pJ Alternate Solution I stFdxfggdx
For the factor of 1/2 . V 9
E  V x 15 constant since —   ———
0/ ° ’ x (,3: 60A 4X0
W % f (VD/xo)dx xo Fbr correct limits W = %3' (4X0 — X0) ' [gQVO] "‘ 3[%60V02] = 3(1200 uJ)  3600 ”J Alternate Solution II
wdefgdv For the factor of 1/2 Q Q
V=E ' (1173—6de 2 Co/4c1
W——Q—f 52
Go For correct limits
4
W _Q_2 _; co/
2 C ,. Q: [4 1] Lei
H = 3[lcov02] = 3(1200 [1.1) = 3600 pJ 2 6—0—07 3260 C0
2 Distribution
of points 1 point 1 point 1 point 1 point (Alternate points) (1 point)
(1 point) (1 point) (1 point) (1 point) (1 point) (1 point) (1 point) 1989 Physics C E & M 3. (continued) (0) (d) (e) 3 points
Fbr Ohm’s Law V  IR Vbltage across capacitor is 4V0  80 V I  V/R  (80 v — 20 V)/3oo,ooo n  2 x 104 A Alternate Solution Vb“,  —IR + (2/6” 112
CR R
I_£_V__Z
C'R R
20v '3oo,ooon (4 ' 1) I  2 x 10" A 3 points
Q  CV
Qi  (6 pF)(20 V)  120 pC Qf  [g pF](20 V)  3o pC AQlZOpC—30st90pC Alternate Solution
AQ  f I dt:
I  Io e't/RC a)
For correct integration: AQ  In I e't/RC dt = IORC
0 For either value AQ  (2 x 104' A)(300,000 mg [JF] = 90 [LC 2 points
AE ' AQ Vbatt AE  (9o pC)(20 V) = 1800 N Distribution
of points 1
l
1 (Alternate
(1 (1 (1 (1 (Alternate (1 (1 (1 point
point point
points)
point) point) point) point) point point point points) point) point) point) point point ...
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 Spring '09
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 Physics

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