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Unformatted text preview: 1991 Physics C Solutions Distribution of points
Mech l.
(a) 3 points
For some statement of conservation of momentum 1 point
For a correct equation:
mun  3m: 1 point
u  110/3 1 point
(b) 5 points
For some statement of conservation of energy 1 point
For correct equation containing potential and kinetic energy:
Ui + If " Uf + Kf 1 point
For correct substitution of all terms:
1 Va 2
0 + 3(3117)  3mg: + Kf 2 points
(If only some terms correct, one point was deducted for
each incorrect term, up to two points)
2 V
Kf  Eg— — 3mgr or equivalent expression 1 point
(Full credit was awarded if student solved for the final
velocity instead of the kinetic energy)
(c) 7 points
Ui + K,  Uf + Kf
' u  2
For correct Ki: X;  %(3m)[ man] 1 point
For correct Uf: Uf  3mg(2r) 1 point
1 U ‘ 2 l 2
o + E(3m)[ “’3“]  3mg(2r) + E(3m)vt°p
Force equation must be used to solve for yup:
Fnormal + Foravity ' Fcentripetal
, For recognition that in the limiting case Fnomal — 0 1 point
112
3mg  3&3 1 point
viop  rg or vtap  Jrg 1 point
v  2
%(3m)[ my] — 3mg(2:) + %(3m)(rg) 1 point
m2~ 5
mm  6111" + 21:: r  l—zzgr
6 ‘3‘ 2 g 2 “min "' 3453: 1 point
EH 1991 Physics C Solutions Mach 2.
(a) 3 points
For recognition that Ztorque  O or equivalent
"1231’: ' a1151'1 (01r mzrz ‘ “71751) _ mm _ (20 kg)(0.5 112)
m2 r2 (1.5 m) mz'jkg (b) and (c) 8 points For a correct dynamical e
r  Io quation for the torque:
For a correct application for the two cylinders:
Tri  (45 kgm2)o of the torque equation For Newton's second law:
F  am For a correct applicatio n of Newton's law for mass m1:
(20 kg)g — T  (20 kg)a For recognizing that
attempting to solve s T  (20 kg)(g  a) aar T  (20 kg)(g  on) parts (b) and (c) are coupled (i.e.,
imultaneous equations) Substituting into torque equation: (20 kg)(g — a:,)  (as kg~m2)a
(20 kg)gr1 — (20 kg)ar,2 — (as kgm2)a
am kgmz + (20 kg)r1Z]  (20 kg)gr1 :2 — (20 kg)(9.8 m/sa)(0.5 m) / [45 kgmz + (20 kg)(0.5 m)2] a e 2.0 rad/52 :r  (20 kg)[9.8 m/s2 — [2.0 L:§](o.5'm)] T  180 N Distribution ~“‘\ of points 1 point 1 point 1 point 1 point 1 point [‘4 noint 1 point 1 point 1 point a; 73 1991 Physics C Solutions Mech 2. (continued) (d) 3 points
For applicable kinematic equation(s):
uz  2as 0R 5  %at2 OR
and
u  at
a  a: t  u/a v  J2ar1s
For correct substitution: v  u  1.4 m/s (Alternate solution) For at least one answer with correct units Using conservation of energy: lziz
mgs zmu + zIw w  u/r l2 1‘12
mgs  zmv + 2E2:  §u2[m + 5?] 2(2.0 rad/s )(O.5 m)(l m) Distribution of points
wz  2&5
and 1 point
v  ur
v  :1225
95
r
2a:
v _ ___
1 point
1 point (Alternate Points) (1 point) (1 point) 2 _ _2 _ 2(20k )(9.8msz)(l m)
U mugs/(m + U‘ ) [20 kg + (as kgmé)/(O.5 1:02)} u  l.h m/s , and no (I point) 1 point incorrect units 1991 Physics c Solutions Distribution
of Points “RV
Mech 3.
(a) 3 points
For a correct expression for the spring force:
1'"s f ch (either + or —) 1 point
For a correct expression for the gravitational force:
Fi  mg 1 point
kD  mg
[C  mg/D 1 point
If a student attempted to solve the problem using conservation
of energy, which is not correct, one point was awarded for
Many“).  .ng and one for either W  AK or
_AU5ravity _ AZIspring ' AK "
(b) 1 point
For any reasonable explanation I point Some examples: When Dunnye  O, the separation is neither increasing
or decreasnxg. When Urdu”:  0, Ifrm is a minimum. Therefore UN!“ is
a maximum, which occurs at maximum compre551on. (c) 7 points i
For some statement of conservation of momentum ' 1 point
For correctly applying conservation of momentum between the 1
initial state and the state of maximum compression with
common speed V:
mun  312V 1 point
For some statement of conservation of energy 1 point
. . . 1 2 .
For the correct expreSSion for spring potential energy, 3210: 1 point
For correctly applying conservation of energy between the
initial state and the state of maximum compression x:
1 .
Emvoz  %k::3 + %(3m)'.72 1 paint 1991 Physics C Solutions Distribution
of points
Mach 3. (continued)
For attemptlng to solve simultaneous equations 1 point
_ 2n
V 3
1 u 2  kxz + an [Ea]
.3; 2 2 £1.12
‘ 2‘“ + 2m 3
l 2
kxz  mu02[1 — 3]  Emvoz
 u 2.!!! _ 22 _D
x 0 3k “0 mg
 v g2 ‘ l l ‘
x a 3g pOLnt
(d) 4 points.
For correctly applying momentum conservation:
mvo  muI + 2muII 1 point
For correctly applying conservation of energy:
Zimuoz  51211112 + ' %(2m)uII2 1 point
(This point also awarded for any other equation
applicable to elastic collisions)
For attempting to solve simultaneous equations 1 point
v1  vo  ZUII
v02  (U0 — ZUII)2 + ZUIIZ
 1102  4v°uII + (tulle + ZUIIZ
51.2112  4UO‘UII
UII  0 or 6UII  41:0
vII  2/3 no 1 point
E & M 1.
(a) 3 points
For a correct expression for the electric field magnitude
for a point charge:
k0 kQ .
5 E? or g? 1 pelnt
For recognition that the electric field is a vector 1 point
E  O 1 point as 744 1991 Physics C Solutions E & M 1. (continued) (5) (c) (d) (e) 3 points
For a correct expression for the electric potential for
a point charge: V  59 or 5g
r a For recognition that the potential is a scalar “31:9 a 3 points (a. 0) From figure r  a + b For the magnitude of E due to one of the charges: E  z£2—z
0:2 +b For recognizing the need to find the components of E
The xcomponents cancel, so only the ycomponents need
be calculated E  —259——; sin a  ;59——g ———3L————  —2—59§z—37T
Y a + b a + b J;2_1‘37 (a + b ) The yconponents add: 2ka
E? ' (a2 + b2)37z 1 point
For indicating that the particle will move toward the origin
or that it will oscillate about the origin 4 points
For indicating that the particle will move away from the origin For some statement of conservation of energy Distribution
of points ” \ 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1991 Physics C Solutions Distribution
of points
//‘\ E & M 1. (continued)
For an expression for the initial potential energy
of the particle: U  2k: °r qV 1 point 1 ' 2k imi ' Tog k .
vm  2 ~52 1 p01nt (Alternate solution) (Alternate Points)
For indicating that the particle ‘ will move away
from the origin (1 point)
émvmz  Wnet or fF dy (1 point)
For correctly setting up the integral
 °" vd .
I” d? ‘ Zqug I?;?f§¥;z§37? (I paint)
u  a2 + y2 > du  2y dy
/ a ‘
du 1 2kQ
J‘F dy  qu f 37:  mom—17”  43
a2 a2
l 2 42m
, 2amcu ‘ a
_ £93 .
“Q 2 ma (1 pOLnt)
(f) 1 point
For indicating that the particle will move back toward the
origin or that it will oscillate about the origin 1 point
E & M 2.
(a) 1 point
Since the inductor prevents any sudden change in current:
Ii  O 1 point
(b) 3 points dI
The voltage across the inductor is L E; , which is zero in
the steady state condition.
5 ' 13:9: 2 points
” 50 v  1(100 n + 150 n)
I  0.2 A 1 point 45 75’ Distribution 1991 Ebysics C Solutions
of points ,,K§ E & M 2. (Alternate solution) (continued)
(Alternate Points) For an RL circuit with zero initial current:
I — g (1  e‘Rt/L) (2 points) As t —~ w, I  E/R I ‘ 250 a " 0'2 A (1 point)
(t) 3 points
Cuntnt
0 «‘\V
For beginning the graph at the origin 1 point
For a monotonically increasing curve 1 point
For an asymptotic aptroach to a constant value
' as t ——~ m 1 point
(d) 2 points
For an expression for the energy stored in an inductor:
U  éuz 1 point
1 2
U  5(1.0 H)(0.2 A)
U  0.02 J 1 point
(e) 2 points
Since the current in an inductor does not change abruptly, it
is equal to the steadystate current calculated in part (b)
I  0.2 A 2 points
(One point was awarded for either of these equations:
ail—E ~ IR  0, I  Io e‘RE/L) ‘\ 1991 E&M
(f) Physics C Solutions 2. (continued)
2 points Lg d:  IR VL Foz using correct value of R: R  lS VL  (0.2 A)(150 n) vLzov (g) 2 points 0 0 (One point was awarded for stating that the stored energy 'E & M 3. . "
(a) 3 points
92
6  d:
winCIA
dAV
E  —55E 6  —Buu (+ or — acceptable) (b). 5 points F  12 X B or F  IIB 112! For using 5 calculated in part (a) L;
2
F _ unzzi For indicating that the force is 6 Egg—ﬂags R (+ or — acceptable) ng.) opposite the direction of the velocity (including minus sign in above expression is sufficient) Distribution
of points 1 point 1 point 2 points _ is used to keep the current flowi 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point as W 1991 Physics C Solutions Dis tribution of points r
E. & H 3. (continued)
(c) 5 points
For an expression of Newton‘s second law:
F  ma 1 point
Using gxgression for F from part (b), with a generic velocity:
,uB 2 a  mg (+ or — acceptable) 1 point
For indicating that the acceleration is opposite the 1 point direction of the velocity (including minus sign above or in differential equation below is sufficient)
For a correct differential equation:
g _ _ 32,22 _
dt v ER 1 pelnt
dv _ __ 8222 d
v m}? u I 3232
In v v0  — mR c
ln 1"—   3212 t Do MR
2 2 A v  vae'B 1 t/mR 1 point (Alternate solution) For a statement of conservation of energy 3 (Alternate Points) (l'point) > For placing 12R term on the correct side of the energy equation (1 point) For using expression for E from part (a) with a BEU
R I  E/R  For a correct integral equation:
1 1 B 22112 among  5.72112 + I $— R d:
Try v  Ce‘Dc
2 2
%m,nz .. %mcz 2Dc + 2452 J's—2D: dz: 1 2 _ Emcee2D: _ E33: 02 [—1]e—ZDt + K0 5”” 2 R 2D Since this equation must hold for any time t:
.1. 2 lz 3212 2 1 K0 zmuo and 2mC R C 20 generic velocity: (1 point) (1 point) g5 lo/ 1991 Physics C Solutions E & M 3. (continued)
Therefore:
5212
0‘} v — no when t  0, so uo  Ce'a and C  on U _ HoeBzzzc/mk , he resistor will eventually
‘ ic energy of the rod (One point was aw arded for any other correct, relevant
statement reg arding energy or power) Distribution
of points (1 point) 2 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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