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Unformatted text preview: Distribution
' of points
1 point
1 point
1 point
1 point ds to two or three decimal places T2 (52) For correctly computing the period of each pendulum using the time for 10 oscillations For correctly computing the square of each period
For expressing the periods and the squares of the perio For completing the data table 2000 Physics C Solutions
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1 1 1 point
1 point
I point a1 number of data points above and below the line not have to do so)
For drawing this line to approximate a best ﬁt with about an equ For correctly plotting the square of the period as a function of the pendulum length
For drawing a single straight line through the data points (could go through (0,0) but did 2000 Physics C Solutions Mech. 1 (continued) (c) 4 points
For computing the slope of the straight line
slope : 912—
M Credit was awarded for a calculated value that was consistent with the line drawn. A
typical correct value was 4.50 szlm. Most values were between 4 and 5 szlm.
For recognizing that the motion of each pendulum is simple harmonic or for using the equation for the period of a simple pendulum T=27rJZ
E For using this relationship and the slope from the graph to determine gap . 2 . . 2 47!2 471:2
For example, the relation between T and f 15 given by T = —Tg 2, so slope =
exp
w 47:2
gexp " Slope Credit was awarded for a calculated value that was consistent with the slope determined
above. For example, for a slope = 4.50 szlm, gm: 8.77 m/sz. I”: For using appropriate units for the computed acceleration (a) 2 points For correctly stating whether 9.8 ml 52 is within 14% of the experimental value gexp .
Credit was awarded if the answer was consistent with the value obtained for gap. For
example, for gexp == 8.77 m/sz, the experimental value is not in agreement with g. For justiﬁcation, either by displaying the range of acceptable values Within 14% of the
value of gm, (e.g., approximately 8.42 rn/s2 to 9.12 rn/s2 , forgexp= 8.77 this2 , Q}; by computing the percent difference between the gexp and 9.80 rn/s2 (e.g., 9.80  8.77 8 77 x 100 = 11.7 %, which is greater than 4%, for gap: 8.77 m/sz) (e) 2 points For correctly computing the acceleration of the elevator From Newton’s 2”“1 law for objects in the elevator: mg~ N z ma,whereN = mgcxp,soa = g— g:mp For correctly stating whether the acceleration of the elevator is upward or downward Credit for magnitude and direction of acceleration given for answers consistent with
gm. For example, for gm: 8.77 mist a = 9.80  8.77 = 1.03 m/sz, directed downward. Distribution
of points l point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 2000‘Physics C Solutions Distribution of points
Mech. 2 (15 points) "
(a) 3 points
7’
l1), 49v) Force,
L%)m3 swuﬁk’c) or
Force. a"? G mv‘gi/j
For a vector arrow pointing downward 1 point
For a vector arrow pointing upward 1 point
For correct force labels on both vectors 1 point
For any extra vectors drawn, deduct 1 point
Ch) 3 points
For indicating that the acceleration decreases 1 point
For a correct explanation that includes a correct mention of forces. 2 points
‘ c For example, as the ball approaches terminal speed, the velocity increases, so the drag
force increases and gets closer in magnitude to the gravitational force. The resultant
force, which is the difference between the gravitational and drag forces, gets smaller,
and since it is proportional to the acceleration, the acceleration decreases.
Partial credit of 1 point given for only a statement including a basic deﬁnition of
terminal velocity (e.g., at terminal velocity v = constant, so a must decrease from
98 We2 to zero) '
(c) 2 points
For an expression for the resultant force on the ball 1 point
F = mg  15v2
. .. .. .42. a). .. .. 2
SmceF~ ma ~— m dwthenm d!  mg b1)
For a correct differential equation 1 point
.42 .. _. .22. v2
dt 8 g m Students did not need to use the convention + and  for up and down, respectively, but
they did have to be consistent in their Sign notation for credit. The integral form of
the differential equation was also acceptable. 200013hysics C Solutions « Distribution of points
Mech. 2 (continued)
(d) 3 points
For recognition that acceleration is zero at terminal speed 1 point
For setting the drag force equal to the gravitational force 1 point
m = b 2
8' vi
For a correct solution for v, 1 point
m = la
Full credit also given for writing answer only with no other work shown
(e) 4 points
For a correct statement of workenergy, recognizing that the energy dissipated by the 1 point
drag force is equal to the initial energy minus the ﬁnal energy
For correct recognition of both initial potential energy mgh and ﬁnal kinetic energy 1 point
3. mu 2
‘\ e: 2 t
AE = mg]: w— 32va
For correct substitution of 1), from part (d) 1 point
AE z mgh — 3 EggJ
For correct answer . 1 point
— .. .1".
AE  mg(h 2b)
Alternate partial solution (for maximum credit of 2 points) (Alternate points)
For a correct integral for work (I paint)
szpd: 9;; W=dex
For correct substitutions for P or F (I point) Wzlbv3drgngjbvzdxggw=jku2dx 2000 Physics C Solutions Distribution
of points
Mech. 3 (15 points)
(a) 2 points T 1’
z“ ' "I
a i
 I: 13mg
For a correct freebody diagram QB recognition that a = 0 _Q_R__ correct use of Newton’s 1 point
second law
21F = ma
T  2mg = O
For the correct answer 1 point
T = 2mg
Full credit was also awarded for writing the correct answer with no other work shown.
(12),,
r 2 points ‘1;
r ' '°
V ‘
x r
\ ‘ g
1 ‘. 105 .3
x ;
I t
1, .. J
i 3%
BF == ma .
For correct substitutions into Newton‘s second law 1 pomt
g
3mg  23 r: 3m(—3—)
For a correct solution for T3
T3 = 27722: 1 point 2000 Physics C Solutions Distribution of points
Mech. 3 (continued)
(b) (continued)
ii. 2 oints
P T‘
‘5
0.: "‘
T 3
lmg
2F = ma
For correct substitutions into Newton’s second law 1 point
Y1  me =3 For a correct solution for T1 1 point
T = i mg
1 3
iii. a 4 points
For 1 = [let 1 point
_ i _ __g__ 1 point
For a ~ R1  3R1
For 1 = (T3 — 7})R1 1 point
For correct substitutions into 1 = 110 and solution for 11 1 point
4 _ .5.
(2mg * 3mij1 m 11(3R1)
I} '53 2mR12
Alternate Solution (alternate points)
Use conservation of energy, AE 2 AK + AU = O
For AK z ~AU (I point)
___1_ 2 l 2 i 2 ___1_)__ (Ipoint)
ForAK «— va + 2(37:01) + 2 11a) ,wherea) —— R1
2 2 I ‘ t
For AU = mgh  3mgh = ~2mgh , where h = g; = Egg—— ( pom)
For correct substitutions and solution for I; (I point) 11 z: 2mm,2 2000 Physics C Solutions Evieehw 3 (continued) For recognition that the speed of the cord or the tangential speed of the pulleys is the (c)
i. Zpoints
same for both pulleys
wai = wzkz “ 0329131)
For the correct answer
a, .. 2.1..
2 ‘ 2
ii. Lipoint For correct substitutions in L = [w and correct solution L; =(1611>[5’2—‘~)
L2 = 811591 ‘ iii. 2points For a correct expression for the kinetic energy as the sum of the kinetic energies of the two pulleys ".1.
K’z For correct substitutions and solution 3.
K“2
K=—5 2 116012+ %12022 2
1
110) 12+ E 110212 0’1 Distribution
of points 1 point 1 point 1 point 1 point 1 point 2000 Physics C Solutions
E&M. l (15 points) (a) 4 points Since brightness is proportional to the power dissipated by a bulb, the answer may be
found by solving the circuit to determine the power dissipated by each bulb. For example,
l__ l “L__ 3 . . . ,
RP  ~—————12 Q + 6 Q — W12 9 , where Rpis the resmtance of the parallel combination of
resistors
19:45)
_.___«‘:3___._i.._._i?~_Y_._._
IA “RAHep “109+4n'3A
B RB RE 129
I zflz JAR}; 33.4(3A)(4r2)=2A
C RC RC so PA = 1,312A = (3 A)2(10 r2) = 90W
«r ». P3 = IBZRB = (1A)2(12 o) = 12W PC : ICZRC == (2 A}2(6 o) = 24 w For correct ordering, i.e., bulb A is brighter than bulb C, which is brighter than bulb B (Partial credit of 1 point given for incorrect answer but with an indication that bulb A is
brightest or that bulb C is brighter than bulb B.) For a correct explanation, which can be by a quantitative solution for the currents and
powers as above, or by a qualitative approach that notes that all the current in the circuit flows through bulb A, then branches in such a way that bulb C receives more
current than bulb B. (b)
i. 3 points Immediately after the switch is closed there is no current in the inductor so the circuit
consists of resistors A and B in series with the source of emf. For IC = For recognition that I A = 13 and they are nonzero For correct numerical answers for I A and 13, i.e., I A = [B a m z 1.91 A Distribution
of points 3 points 1 point 1 point
1 point
1 point 2000 Physics C Solutions E&M. 1 (continued) (b) (continued) 3 points A long time after the switch is closed, the potential difference across the inductor is
zero, so the circuit is essentially the same as in part (a)
For recognizing that VL : 0
For correct currents, the same as in part (a), i.e., [A = 3 A, I 3 z 1 A, I = 2 A
(if currents not computed in part (a), they could be computed here.) Unit point: For expressing all currents in Cb) in correct units of amperes (c) 2 points Attributes of correct curve:
1. Starts at a nonzero but ﬁnite point on the vertical axis
2. Smooth
3. Concave upward
4. Has asymptote equal to zero
For a correct curve with all four attributes Partial credit of 1 point for curve with flaws but at least two correct attributes Distribution
of points 1 point
2 points 1 point 2 points 2000 Physics C Solutions Distribution
' of points E&M. 1 (continued) ((1) 2 points ‘33? O Attributes of correct curve:
1. Starts at zero
2. Smooth
3. Concave downward
4. Has ﬁnite but nonzero asymptote
‘ " For a correct curve with all four attributes 2 points Partial credit of 1 point for curve with ﬂaws but at least two correct attributes 2000 Physics C Solutions Distribution
of points
E&M. 2 (15 points) (a) i. 6 points B 0 C
§~—~ e ———————————oi
Figure 1 e a One point for each arrow drawn in the correct direction 3 points For not having all arrows approximately the same length, deduction of 1 point
For not having all arrows start at P1 , deduction of 1 point
For having one or more extra vectors, deduction of 1 point Figure 2 One point for each arrow drawn in the correct direction 3 points For not having lengths of arrows such that EC z E B > E A , deduction of 1 point For not having all arrows start at P2 , deduction of 1 point
For having one or more extra vectors, deduction of 1 point 2000 Physics C Solutions E&M. 2 (continued) (a) (continued) ii. 3 points I/ 5' One point for having check mark or other indicator in each correct box (b) 1 point For an indication that the xcomponents of the ﬁeld vectors due to particles C and B
cancel each other due to the symmetry created by having a vertex of the triangle on
the y~axis ‘I’: (c) 3 points For an indication that the potential is the sum of the potentials due to the individual
charges
V: 2.12;: (kegs...)
i ’1' 7A r B "c
For recognition that the terms due to the particles at B and C are equal
V : 15(2‘3. + YA 7‘ B
For correct substitutions for Q’s and r’s and correct answer 3 1 Q + 29
47550 «[375 __ 32 2
“'2 3’ ‘2’ + 3’ V , or equivalent Distribution
of points 3 points 1 point 1 point I point 1 point 2000 Physics C Solutions Distribution of points
E&M. 2 (continued)
((1) 2 points
Since E y r. ~— % V01) , to ﬁnd the y coordinates of the points on the y—axis at which
the electxic ﬁeld is zero, take the derivative of the expression in part (c) with
respect to )2, set the expression equal to zero and solve for y.
For recognition that E is a derivative of V ‘ 1 point
1 point . . dV __
For recognition that ‘29» — O 2000 Physics C Solutions E&M. 3 (15 points) (a) i. ii. (b) i. 3 points For a correct statement of Gauss’s law
§ E . 4A = .52. For expressing the permittivity of the oil in terms of the dielectric constant K' For a correct expression for the electriaﬁeld in the oil For a correct statement of Gauss’s law in the space outside the outer shell For stating that the electric ﬁeld is zero in this region 6 6 = K60
éL
E(27z‘rL)  K60
=_£L.
meorL
2 points
is  dA : 0
E = 0
3 points For an expression for the electric potential between the two shells
b
AV=Vb—Va ﬁnd;—
a
For substituting the expression for the electric ﬁeld between the shells AV: AV: Q 2713:6014 a I"
For a correct expression for the electric potential difference between the shells Q ZMEoL 523: b in(—
a J Distribution
of points 1 point 1 point I point 1 point 1 point 1 point 1 point 1 point 2000 Physics C Solutions Distribution
of points
E&M. 3 (continued) (b) (continued)
ii. 2points For an expression for the capacitance in terms of Q and AV 1 point 2.
C”AV Substituting the expression for AV from (b)i:
Q
C .— ” Q (‘11)
2M£0Lln a‘ For a correct expression for the capacitance 1 point
C __ 2717:60L we) (c)
i. 3 points For a correct statement of Ampere’s law 1 point §B a d6 == #01 For substituting the current through the inner shell 1 point
8 mm) = nob?) For a correct expression for the magnetic ﬁeld between the shells 1 point _ #08
B ' ZxrR 2 points For the correct substitution of the total current through both shells in to Ampere’s law 1 point B(2:rr) = no For a correct expression for the magnetic ﬁeld around the outer shell 1 point
_ 2M8
" IrrR ...
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This note was uploaded on 07/13/2009 for the course PHYS 1112 taught by Professor Leclair,a during the Summer '07 term at Cornell.
 Summer '07
 LECLAIR,A

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