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Unformatted text preview: 1999 Physics (3 Solutions Distribution of points
Mesh. l (15 points)
(a) 5 points
For cemervatien of momentum 1 point
my0 r: (m + [W0 )1)
v m me)O
m +MO
For conservation of mechanical energy l point
K +1] 2 + U” or é—MWM m Mmm gh
Fer calculating I: in terms of E I point
11 m [(1  COS 69)
For substituting for Mmml and v in the energy equafjon 1 point:
2
5(1‘7’1 + A40 3 (m + .MO W COS For the correct: answer 1 point;
m +M
yo 7: 0 .lZngwCOSG)
m
(b) 4 points
For Newton’s seeend law (not awarded if net force was set equal to zero) 1 point;
If“: = ma
For any equatien that indicated that the tension minus the weight is £191; zero l point
Y "” Nltotalg : Muﬁala
For an expression fer the centripetal foree '1 point
a l. 2: or 313.
w r 3?
U2
T ”"‘ A4ng 3 Mtoml T "‘
For curreetly substituting for Mom; and 1) Ll. point w 2
1
("2+MO)?(W> =T—(m+M0)g Solving for T
T : (i?1+M0)g(3m2COSI9) 1999 Physics C Solutions
Mech. 1 (continued)
(c) 4 points For all of the following:
‘1) A practical procedure that uses some or all of the apparatus
listed and would work
2) Recognition of any assumptions that must be made
3) Indication of the proper mathematical computation using the
variables measured Two points were awarded if the description of the procedure was not complete
but it would work, or if the mathematical work did not clearly specify
the variables used, or any combination of the above. No points were awarded if the procedure would not be feasible in a laboratory
situation with the apparatus listed, or if the procedure was merely a repeat
of that outlined in part (a) (d) 2 points Fm mma mwbo Ii ., . . a , t do
For expressing the acceleration as the time derivative of the speed, a = —« t) l’ 212
no 0
'2) hr
in —» a: « 
1)“ m
U m Uggwbt/m or a general expression for the length of the dart in the block as a function
of time or for the expression for the total distance L
M mun wet/m
i? —. Mb (I u— e ) “.2332”
L“ 32 Distribution
of points 4 points 1 point 1 point 1999 Physics C Solutions Distribution
0f points Mach” 1 (continued)
((11) (continual)
Alremate Solution 1 Alternate paints Fur indicating that work equals the: change in kinetic energy I point
If? dx 2 1' m2); 2
m £339., n) 4.4 average L A
b) q
Iii dxzimvg
2 2 0 Zn) 1 »
wiLx—mﬁ 2 2
For the correct expression f‘nr the: total distance L I paint mp0 3:: L: Alternate SG/z'ution 2 Alternate poinm Fm At a Ap
For the, above: expmssien on f ~51) d: = ~va
0 I paint d ~ 1 . .
Using 1) :3— and the fact that 3 goes from zero to L as nme goes dt
from zero to inﬁnity \ L
j ~22 (IIS :2 mm
0 «('71) = «mu0
Fm the: conect expression for the total distance L I paint
ml)0 b Lm 1999 Physics C Solutions Distribution of points Mean. 2 (15 points)
(a) 3 points For indicating that the equation for gravitational force is applicable 11 point F x m Gmélm2 r
For using the proper expression for the mass of the planet enclosed by the radius 1 point
Gm pim3
17‘ m .— 3
P2
For proper cancellation of terms to ShOW the ﬁnal result 1 point
4 v F x  ~3~ mirsz r (b) 4 points
F
S For drawing a straight line from the origin to a distance R, and not going past R 1 point For having the maximum magnitude occur 21th 1 point For having the curve from R to SR decreasing in magnitude with proper curvature and appearing to reach an asymptote 1 point For recognizing that the force is always negative, Le. the graph is always below
the xaxis 1 point 1999 Physics C Solutions Distribution of points
Mech. 2 (continued)
to) 3 points
In this and all subsequent parts, either C or i;— nGpm could be used.
For indicating the integral that needs to be calculated to determine the work 1 point
W 5 f1? air = j~Cr dr
For using the proper limits on the integral (R to zero, not 7') l point
0
W a j ~01» dr
R
2 O
W 3 ~61
2 R
For the correct answer ll point
C 2
W = R
2
Alternate Solution Alternate points
For recognition that the work is the area under the curve, which is triangular 1 point
For using the corrects limits (zero to R) 1 point
For the correct answer 1 point
2
W, :3 CR
2
(d) 2 points
For using conservation of energy or workcenergy relationship '1 point
AK 2 AU :2 W n
l 2 CR2
—mv =
2 2
For the correct answer 1 point
[CR2
1) z
m
An alternate solution indicating the potential energy as that of a harmonic
oscillator also received full credit:
(e) 2 points
For indicating that the ball will move from the center to the surface of the planet 1 point
For indicating that the ball will stop at the surface, return to the center,
and continue oscillating in this manner, with no damping i point Describing the motion as simple harmonic oscillation with no damping earned full, credit 1999 Physics C Soiutions Distribution of points
Mach. 2 (continued)
(1”) 1 point
For showing a propor application of N ewion’s first law 1 point,
F a ma
2,
Cr 3 mfg—E}:
dz Altornately, one could relate the time to the period of oscillation, T = 27: f 47;}? , Le. the time is one»fourth this period. The above equation was requirad;
a more general, form was not acceptable. 1999 Physics C Solutions
Mech, 3 (15 points)
(a) 5 points For indicating that the net torque is zero, or that the clockwise and
counterclockwise torques are equal
F or a correct expression for the torque exerted by the rod rm 2: ng sin 00 For a correct expression for the torque exerted by the block
7mm : 2mg (2]?)sin 00 = 4ng sin 00 For a correct expression for the torque exerted by the string
2" := TR string For adding the counterclockwise torques and setting the sum equal to the
clockwise torque (this point not awarded for just one torque)
TR := 4mg]? sin 60 + mg]? sin 60 T : 5mg sin (90 Only four points could be earned if the wrong trigonometric function was used. Only three points could be earned if no trigonometric function was used. 0))
i. 4 points For indicating that the rotational inertia is the sum of the inertias of the disk,
rod, and block For calculating the total rotational inertia I :Idisk lime "him 2 gm: + :j—mRQ + 2m(2R)3
3i i : 9—?— 2771?:
6
a we /I
For substituting the value of torque from part (a)
.3ng sin (90 01:}; ﬁg 65 v
m— ml?”
6
For an answer consistent with the values use for torque and rotational inertia
6 sin 6’
a 2 g o 13R Distribution
of points 1 point
1 point 1 point i point 1 point 1 point
1 point l point 1, point 1999 Physics C Solutions Distribution
of points
Mocha 3 (oontinuod) (b) (continued)
ii, 1 point
Expressing the linear acooloration in terms of the angular acceleration a :2 or
For“ substituting the value of 0: and the correct radius, 2R 1 point (a) :3 points For indicating that onorgy is conserved 1 point
For indicating that tho potential energy of two bodies (the rod and the block) ohangos l point
AU 2: iiighm Hughka
For the oorroct oxprossions for those two potential onorgios 1 point,
AU mg]? con 90 + 2/?‘2g(21€)cos 690
For indicating the correct kinetic energy when the rod is horizontal l point
i a
K x w In)”
'9 414 Equating the kinetic and potential onorgies, and solving for the angular spood m?3 )w2 :2 mg]? cos 00 + 4ng 003 00 A, 123(50th
(0m JW
13]? For using tho rolntionship botwoon linear and angular speed, and substituting
wand the contact radiusv 2R 1 point
2.2 x (or ‘ UT; l12gcon§o (213)24 lagi’toosﬁ3
13R 13 Alternate methodn of solution included use of the following proper integrations 25/3
(03 x 2 j a aft?
90
27/2
K 1: frth 90 1999 Physics C Solutions Distribution of points
E 84: M 21 {15 points)
(a) 4 points
For using the relationship between potential and charge I point
1 Q
V : ——
47mg r
Solving for Q:
Q = 47:60}?
For correct substitutions for the potential and radius .1 point
Q0 : Mean2000 'vxoczo m) or (wzooovxow m)/( 9 ><10'9 Null/c? )
Q0 : ~1600n'60C or — 4.4 ><10'8 C
For the correct magnitude of Q0 1 point:
For the negative sign 1 point
(13) 5 points
i, For indicating that the electric ﬁeld is zero '1 point
ii. The charge on the sphere can be treated as a point charge at its center
F a 1 a
M 471' £0 r3
Ms
E = ( 9 ><10g N'm‘Z/CZ )[w—W4'4 X120 C]
r
4 _ 396 E 400 E . . ~
13 — r2 C or r2 C Where ms 111 meters
For any of the above expressions for E 1 point
iii. For indicating that the electric ﬁeld is zero 1 point:
\
iv. For indicating that the electric ﬁeld is zero 1 point; For having all four answers correct OR for some mention of using the enclosed
charge OR for some mention of Gauss’ law 1 point 1999 Physics C Solutions Distribution of points
E & M 1 (continued)
(c) 3 points
ll
AV : Vb ~ Va m M} E Cir
F or recognition of the need to take the difference of the potentials at radii a and by
or for writing the deﬁnite integral (with limits) 1 point
:5
, Q0 Cir
mi "" 43mg i I?
a
 o ”‘
W Mateo r a
s _ Q0 ii 1
My} " 4mg (3 w For correct substitution of variables or numerical values for Q0 , a, and b l point
For the correct answer 1 point
50
lAVl : “‘0 or 1000 V
37(60
(Alternate solution) (Aflomafe points)
For recognition of" the need to take the difference of the potentials at radii a and b / pain!
AV 2 D; W Va
’ 47:60 rb 47(60 ra
For correct substitution of Q0 , a, and b I point
Q0 1 1"
AV ' 47r60 (it; m It“)
For the correct answer a 1 point
50 '
lAVl = “O or 1000 V
8x50
(Alternate solution) (Alternate points)
V 2 C
F or using the above relationship I point
For substituting Q from part (a) and C from part (d) alternate solution 1 point
For the correct answer I point
5
{AI/l : Q0 or 1000 V 87:60 1999 Physics C Solutions
E 6": M 1. (continued) ((5) 2 points Far using, thw above relationship
For gubstituting Q0 from part (a) and Amem part (c) (a m 4.4x}.0~8mg 1000 V
C a: 4.4x10”H F {Ahemme soluticm) For writing the aquation for the capacitance of the spherical capacitor
47:60:75 C m w
:5 w a
C, W m (0.02 m)(0.04 m)
' (9 x 10“) Nmz/c2 )(004 m  002 m)
For the correct answer
C m4u4x10‘“ F For correct units 011 two answers and no incorrect units Distribution
of points I paint
1 point (,4 [remove points)
/ point I paint I paint 1999 Physics C Solutions Distribution of Points
E 8a: M 2 (1:3 points)
(a) :3 points
For using Faraday‘s law for a loop l point
I
63‘ m w or 6‘ mm 5—3—93
dt AI
For relating magnetic flux to magnetic ﬁeld and area 1 point
05 ... 51%: m. at. _ A3;
a]: " dz A: A!
For using the proper expression for the area of a loop 1 point
A z: m“:
8 mmﬁgﬁ or 8 m/rz‘géé
at A;
For using the correct radius re the radius of the ﬁeld 1 point
e from m)2 (0.40 T/s)
For commuting the correct answer 1 point
(‘3‘ r: O.45 V
(b) 3 points
For any statement of Ohm’s law I. point
V r: [R
Solving for the current:
1 2:: V/R :: 8/16
1‘ a (0.45 V)/(5.0 o)
For computing the correct answer l point
I r»: 0.090 A
For indicating 21 clockwise direction for the current 1 point
X
(c) 3 points
For relating the energy dissipated to the power in the resistor l point z: j P dz or [5 m 11);
For an expression for electric power I point
0 A V2
P mI‘It or w—« or [V
Example using P w 13R:
E m I 31%
1; = (0090 AV (5.0 rams s)
For computing the correct answer 1 point {M0013 1999 Physics C Solutions Distribution of Points
E 8:: M 2 (centinued)
(d) 3 points
Fer stating that the brightness of the bulb will be less 1 point
For indicating that the reduction in brightness is due to a decrease in
current or a decrease in the emf l point
For indicating that the decrease in current or emf, or the reduction in brightness,
is; due to a decrease in the area of the loop or a decrease in the changing flux 1 point For using correct unite with three numerical answers 1 point 1999 Physics C Solutions Distribution of points
E 62:: M 3 (133 points)
(a) 3 points
The charge on any section of the ring is equidistant from a point on the x~axisﬁ
so one can write an equation in terms ofthe single distance r
For a correct expression of the potential l point
1 dz; 1 WM ««««« i «we— or m
47560 r 47660 r
For a correct expression for the distance of the charge from location x l point
I” 2 \/X2 + 122
For the correct answer I point
m ‘1 Q
4x60 V262 "t 162
Alternate solution Alternate points
For correctly expressing the potential as an integral of the electric field 1 point
dV :2». m] dr
For a correct expreesion for the ﬁeld 1 point
Ox
M” :2 ml 1 » c/x
47% (x2 + 13*)“
F or correctly integrating to get the ﬁnal answer I point
m 1 Q
47% M + 122
(b)
i, 3 points
w dV
1: w M d!” ”
For using the above relationship 1 point
For taking the derivative with respect to x 1 point
For using the expression for V obtained in part (a) l point
F W w d [ 1 Q j
“x dx 47m0 .1353 + R2
1 0x V; a ~ 1999 Physics; C: Salutions Distribution of points
E & M 3 (continucd) (b) (cmtinued)
i. (ccntinucd) A llerrzate solumm
Calculating thc field by integraﬁon; E z jdEx = JdE c086? , where 9 is the angle between the x—axis and Alternate points the distance vector r
For using the hmrizcnta] component of the ﬁeld 1 point
For using a correct expression of Coulomb’s law 1 point 1 Q 2 Ea {'2
For indicating that, the intcgral is over the charge 1 point 4‘ W '1 dc]
Lx M I 47560 r2 E Substituting c036 : x/r and r 24x2 +2?2 1 0, Q “C x 4m:Q + R2)“
m 1 Qx x M 47r60 (x2 + REY/2 E E ii. 1 point Far any indication that the y and z~c0mponents are 2ch or cancel 1 point 1999 Physics C Solutions Distribution of points
E 82: M 3 (centimued) (c) i“ 1?: points
For taking the derivative of E with respect to x and setting 11: equal, to zero 1 point dEx Mei ‘1 W23...” __0
chat m dx 471's“3 (x2 +R2)3/2 “ Q 1 +6" 3‘) 2x2 "0
4m (x3 + R333” 2 (x2 + W” ” ‘ . 3x2
I a Z 2
x + R
3262 z x2 + R2 z: i and the maximum occurs at the positive value of x For the correct answer 1 point; gami— J?
ii, ‘1 point
For substituting the answer from part (0)1 into the given expression fer
the electric ﬁeld 1 peint ‘ g 1 M
xmax 47350 +R2)3/~ E a} Q
xmzm 1999 Physics C Solutions Distribution
of points E 84: M 3 (continued) (d) 3 points For a curve in the first quadrant displaying a single positive maximum. 1 point
For a curve passing through the origin, it point
For the negative reflection of the ﬁrst quadrant curve in the third quadrant 1 point
2 points For any statement that describes the subaequent motion as; oscillating, periodic: etc. 2 points; One point was awarded for a statement that only described the electron as moving toward
the ring or along the x—axis. ...
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 Summer '07
 LECLAIR,A

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