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Unformatted text preview: 1987 Physics C Distribution
of Pbints
Mech. l.
(a) and (b) 7 points
:5‘ 7. If vector diagram shown.
. 1 point was awarded for 3 points
I each vector correctly
FL drawn and labeled.
W
Summation of forces equal to zero. Taking components:
(1) in ydirection: T cos 0 — W  0 For correctly
(2) in xdirection: T sin 0 — PH  0 obtaining any 2 points
(3) in direction of T: two of these
T  W cos 9 — F” sin 9  0 three equations
(If vector diagram not shewn, 5 points awarded for two
of the 3 equaxions)
For correct simultaneous solution for T and PH of 2 points
any two of the equations (1), (2), or (3)
acmle from (1) r W
a p ' cos 0
Substituting into (2) cos a sin 0  PH  0
FHWmnﬂ
(1 point for using wrong functions, 1 point for
answers not in terms of V and 0)
(c) 3 points
1—
Fbr vector diagram 1 point
I
l
/
h!
Summation of forces in direction of T equals zero 1 point
T  w cos 0 1 point (Full 3point credit given for simply stating the
correct answer) 1987 Physics c Mach. 1 (continued) (d) 5 points ac  E— 1 point
I  W  mac
U2
(1) T  W  mi: 1 point
From conservation of energy,
1 z
(2) mgL(1  cos 0)  Emu 2 points
1 point 1 point
letting mg  W, and solving (1) and (2) £5: T, gives
T  W(3  2 cos 0) 1 point
Mach. 2.
(a) 3 points
Pbints of equilibrium are:
x5  2 m 1 point
x5  S m 1 point
No extraneous points indiented 1 point
(b) 3 points
Etot  K + U  a J 1 point
At x  2.0 m: K  h  U(2) i A — 1  3 J 1 point
At x  5.0 m: K  4  U(A)  a  3 ’1 J 1 point
(c) 2 points
No, the particle cannot reach x  0.5 n 1 point The total energy is insufficient. 1 point 1987 Physics c
Mech. 2 (continued)
(d) 2 points
Yes, the particle can reach x  5.0 m. The total energy is sufficient (e) 5 points :
2
fix) 5 (newtons)  3 
27 40 x(memn) For rough slopes, or a shape indicating change
between positive and negative values at 2_places For having'signs correct (i.e., positive values of F,
then negative values, then positive values again) For correct placement of zeros at x  2 m and x  5 n Fbr constant values.(zero slopes) at appropriate places For correct numerical values of F for regions of
zero slope (If no graph, +1 point was awarded for F  dU/dx.
or +2 points were awarded for F  dU/dx)
Hech. 3. (a) 4 points
K.E. of rod at bottom of swing  P.E. of rod at top %qu  mgh 1 11112 Z i
5(le  “[2] z _ 25 _(3)(10) _
“’ 2 1.2 25 5 radians
” ' sec (1 point for numerical answer. 1 point for
Lnits of radians/second or seconds“) 1 point 1 point 1 point 1 point 1 point
1 point 1 point 1 point 1 point 2 points 1987 Physics c Mech. 3 (continued) (b) a points K.E. before collision  K.E. after collision %‘”0"02 ' %m0”2 + 3 points 1 2
51w
1 pt. 1 pt. 1 pt. 2
§<n<10>2  ﬁmuz + (5)2 SO%uz+l8
11264 v  8 nVs 1 point (c) 4 points
L  mvr sin 0 25 L  mo: 25 f  F x F 2 points
L  (1)(10)(1.2)
L  12 kgmz/s 2 points (1 point for numerical answer. 1 point for correct units) (d) 3 points
Angular momentum is conserved during collision. so
L (before)  L (after) 1 point about  ub(u cos 9)! + Iw . 1 point 12 ' (1)(5)(1.2) cos 0 + (l.44)(5) 12  9.6 cos 0 + 7.2 cos 0  0.5 c  6°. 1 point Alternatelz, angular momentum of ball after collision
may be written maul sin ¢, where o is the angle between
u and 1. The solution yields o  30', but 0  90' — ¢  90' — 30'  60'. 1987 Physics C
E & M 1.
(a) a points Gauss's law for r > R I 3555  Q/qJ 1 point E fdA  Q/eo 1 point sum?)  Q/eo 1 point
1 . E  Z;:; 32 I paint (1 point for answer in terms of R, 3 points
for answer only without showing details) (b) 5 points
Show or mention Gaussian surface 1 point
as inside sphere with radius r. Charge enclosed by Gaussian surface 3
Q  [ZQ—] [gxr3]  9E3 2 points
3
31R
\.~w~./ ‘—~r—’
1 pt. 1 pt.
From Gauss's law, :3
2  .9.  9.
E(Axr ) ‘0 (053 1 point
_ _Q_ r
E “wen E3 1 point
Alternate solution in terms of density p (Alternate Points)
Show or mention Gaussian surface as above (1 point)
,  .J—a (1 point)
~«R3
3
lb 3 "ﬁr
q  pV  p g" (1 point)
EUHrzZ)  geo— .. %[§ir3] (1 point)
E _ E. (1 point) (1 point for using p without defining it, 3 points
for answer only without showing details) 1987 Physics c E & M 1 (continued) (c) 2 points Q
V  ARCOR 2 points
(1 point for answer in terms of r)
(d) 4 points 0
Vcenter "' Vsurfece + I E'd—E (1 P01“: for sum) 2 points
w
1 pt.
0
 A. .9...’ l’_2
hueok + AneoR 2 R 1 p01“:
 _Q_ _9_ _ .29_
. hneoR + SneoR SteoR 1 p01“:
E & M 2.
(a) 4 points
¢  BA (or o  f3.d3 or o  f~cd3) 1 point
o  (2e"t)(.09) 2 points
‘—~r~—/ L‘f"
1 pt. 1 pt.
4»  0.13Mc , 1 point
.
(However, no points were awarded for going from Igdﬁ
to a time integral such as fe“c dt)
(b) 1 point
For statement or arrows indicating counterclockwise 1 point
direction
(c) a points
do
5  " d—c 1 point
 0.729“c 1 point
i % 1 point 1  0.12.2“ 1 paint 1987 Physics 0 E & H 2 (continued) (d) 6 points
PiZR  (ggPsZ/RggPis) 1!) W  f izRﬂc (+1 point for indication of integral. If
0 integral is shown, then +1 point for correct limits) 1”
w  f 0.864 6‘5ch
0  0.108 e‘8t
0 W  0.108 J l E & H 3. (a) 3 points he switch is closed, there is no Immediately after t
its impedance is infinite. current in the inductor; Rm  10 +9o1ooo E 20
i  RtOt  100  0.2 A the potential difference across the 90ohn resistor,
V90. is given by v."J  m  (0.2)(90)  16 v 11. _£__1_3.
‘dc‘ L 0.5 365/5 (Students who used the emf of the battery, 20 V.
and obtained dl/dt  40 A/s received 2 points) 1 point 2 points 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1987 Physics C E & M 3 (continued)
(c) 2 points After a long time, the inductor acts as a short circuit Rt"  10 O 1 point
. s _ 29 _
l  Rt“ 10 2 A 1 point (:1) 2 points Ehergy  %12L 1 point
 %(4)(o.5)  1.0 J 1 point (e) 2 points
Immediately after the switch is opened, the current in the inductor is the same as in part (c). All the current
nust go through the 90ohm resistor. V90 ' Vinductor  LR " (2)(9°) 1 point
' 180 V 1 point (f) 2 points
Potential Difference [90 V
0 Tune:
lo ‘
For curve decreasing in value and concave upward 1 point
For 180 V starting point g any carefully derived 1 point answer to part (e)
Extra 1 point For obtaining three out of five correct
units for answers to parts (a) and (e) 1 point ...
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 Spring '09
 Park
 Physics

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