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Unformatted text preview: AP® Physics B
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For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP PHYSICS B
2003 SCORING GUIDELINES
Question 1
15 points total
(a) Distribution
of points 3 points For a free body diagram that includes any one of the three forces on Student A (weight,
normal, or tension)
For correctly including the other two forces on Student A
For a correct freebody diagram for Student B (including both weight and tension)
One point was deducted for each extraneous vector, up to a maximum of the number of point
already earned (b) 1 point
1 point
1 point 3 points
For equating the tension in the rope to the weight of Student B
T = mB g
For a correct expression for the sum of the forces on Student A
ΣFA = T + N − mA g = 0
Eliminating T and solving for N
N = mA g − mB g 1 point For the correct answer
N = 98 N (or 100 N using g = 10 m s 2 ) 1 point 1 point N = ( 70 kg − 60 kg ) ( 9.8 m s2 ) = 686 N − 588 N Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 2 AP PHYSICS B
2003 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points (c) 3 points
For applying Newton’s 2nd law to Student B
ΣFB = mB a
For a correct expression for the sum of the forces on Student B
ΣFB = T − mB g
Solving for T and substituting:
T = mB g + mB a = 588 N + ( 60 kg ) ( 0.25 m s 2 )
For the correct answer
T = 603 N (or 615 N using g = 10 m s 2 ) (d) 1 point 1 point 2 points
For a correct response of “No”
For a reasonable explanation
Example: To lift student A off the floor, the tension must be greater than the students’ weight
of (70 kg)g
An answer of “Yes” was acceptable IF the answer to (c) was greater than the weight of
Student A AND the justification was consistent. (e) 1 point 1 point
1 point 3 points
For applying Newton’s 2nd Law to Student B
ΣFB = mB a
For a correct expression for the sum of the forces on Student B
ΣFB = T − mB g
Solving for a
T
a=
−g
mb
The minimum tension required to lift Student A is the student’s weight
Substituting:
686 N
a=
− 9.8 m s 2
60 kg
For the correct answer
a = 1.63 m s2 (or 1.67 m/s2 using g = 10 m s 2 ) For indicating correct units in parts (b), (c) and (e) 1 point
1 point 1 point 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 3 AP PHYSICS B
2003 SCORING GUIDELINES
Question 2
15 points total
(a) Distribution
of points 3 points
For the correct formula for total capacitance in series
CC
1
1
1
OR CT = 1 2
=
+
CT C1 C2
C1 + C2
For correct substitution
(12 µF)( 6 µF)
1
1
1
OR CT =
=
+
CT 12µF 6µF
12 µF + 6 µF
For the correct numerical answer
CT = 4 µF (b) 1 point 1 point 3 points
The capacitors are fully charged so current flows through the resistors but not
the capacitors.
For calculating the total resistance in series
RT = R1 + R2 = 10 Ω + 20 Ω = 30 Ω
For use of correct form of Ohm's law
V
6V
I= =
R 30 Ω
For the correct answer
I = 0.2 A (c) 1 point 1 point
1 point 1 point 3 points
Potential difference between A and B is the voltage across the 20 Ω resistor
For a correct form of Ohm's law
V = IR
For correct substitutions for I from part (b) and for R
V = ( 0.2 A )( 20 Ω )
For the correct answer
V=4V
Note: 1 point was subtracted for indicating a wrong unit.
Alternately, full credit could be obtained for finding the voltage across the 10 Ω resistor and
subtracting it from 6 V or for recognizing that the voltages across the 20 Ω and 10 Ω
resistors would add to 6 V and be in a 2/1 ratio, and using this to obtain the correct answer. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 4 1 point
1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 2 (continued)
Distribution
of points (d) 4 points
For using the correct formula to determine the charge
Q = CV
For correct substitution of value of C from part (a)
For correct substitution of value of V from part (c)
Q = ( 4 × 10−6 F ) ( 4 V ) For the correct answer
Q = 16 × 10−6 C = 16 µC 1 point
1 point
1 point 1 point Alternately, full credit could be obtained by first determining the voltage across the 6 µF
capacitor (which can be done one way by recognizing that the voltages across the two
capacitors are in a 2/1 ratio and sum to 4 V) and substituting this value for V and the value
for C of 6 µF into the equation Q = CV . (e) 2 points
For checking the box “remains the same”
For a reasonable justification
Example: No current is flowing from A to P to B. Therefore breaking the circuit at point P
does not affect the current in the outer loop, and therefore will not affect the potential
difference between A and B (or across the 20 Ω resistor). Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 5 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 3
15 points total
(a) Distribution
of points 3 points
For using the correct kinematic equation
1
x − x0 = K0 t + at 2
2
For using Newton’s second law to determine the acceleration
F = ma , so a = F m
Both x0 and K0 equal zero.
For the correct answer
x = Ft 2 2M
(or IDBt 2 2M using F = IDB ) (b) 1 point 1 point 1 point 3 points
For an appropriate kinematic or momentum equation
2
K 2 = K0 + 2a ( x − x0 ) (or other kinematic equations) OR M ∆K = Ft
If an equation containing t was used, the expression from part (a) can be used to find t
For indicating the correct distance
x − x0 = L
Substituting
K 2 = 0 + 2( F M ) L OR 1 point 1 point M K = F 2 ( x − x0 ) a = F 2 LM F For the correct answer
K = 2 FL M (or 2 IDBL M using F = IDB ) 1 point Alternately, full credit was also awarded for using conservation of energy and directly
1
relating the change in kinetic energy to the work done, i.e. mK 2 = FL .
2
(c) 3 points
For indicating that the energy supplied equals the final kinetic energy
1
E = mK 2
2
For substituting the value of υ determined in part (b)
2
1
E = m 2 FL M
2
For the correct answer
E = FL (or IDBL using F = IDB ) ( 1 point 1 point ) Full credit could also be earned for using the workenergy theorem and indicating that the
work equals FL
Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 6 1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 3 (continued)
Distribution
of points (d) 2 points
For indicating that the magnetic field points out of the plane of the page, or in
the +zdirection
For a correct explanation
For example, referencing the righthand rule or giving a verbal explanation of how it
applies to the situation (e) 1 point
1 point 4 points
For using the expression for υ determined in part (b)
K = 2 FL M
OR
2 IDBL M
For an indication that F = I lB
For an indication that l = D
For correctly substituting the values of D and L, and obtaining the correct
answer, with units
K = 2 ( 200 A ) ( 0.1 m) ( 5 T) (10 m) ( 0.5 kg ) = 63 m s
(or 60 m s with the correct number of significant figures) Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 7 1 point
1 point
1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 4
15 points total Distribution
of points Two possible situations fit the requirements: one forms a real image and the other a virtual image.
(a) 3 points
For placing the concave mirror on the optical bench
For a very good description of a procedure for locating the image
For example: “Place the mirror at one end of the bench and the candle more than 30 cm
from the mirror. Place the screen out beyond the candle and reposition it to get an image.
Measure the height of the image. Reposition candle and screen until image height is four
times object height.” 1 point
2 points No points were awarded if there was no reference to use of a mirror
Only one point could be earned if student used a convex mirror instead of a concave mirror
An imperfect or incomplete description of a useful procedure received 1 point
(b) 2 points
For checking a meter stick and/or ruler, plus everything else that they used in part (a)
For having no checks on items that they did not use in part (a) 1 point
1 point If part (a) is blank or earned no points, then part (b) earned no points.
If part (a) only earned 1 point, then part (b) could not earn more than 1 point.
If everything or nothing is checked, then part (b) earned no points.
(c) 6 points Note: Diagrams are drawn with curvature of mirror not to scale, in order to clearly illustrate the situation. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 8 AP PHYSICS B
2003 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points
Real Image (c) (continued) Virtual
Image For showing the concave mirror 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 9 AP PHYSICS B
2003 SCORING GUIDELINES
Question 4 (continued)
At least one ray from the candle is shown reflecting from the mirror
One point each for up to two correctly drawn principal rays
For showing the intersection of principal rays at a location such that the image distance
is larger than the object distance
(This point was only awarded if the correct mirror and rays were present.)
For a correctly oriented image drawn at the intersection of the principal rays
(This point was only awarded if the fifth point was earned.)
No points were awarded if there was no curved mirror used in the diagram. The curved
mirror could be indicated by drawing it curved, or labeling it, or showing rays
consistent with a curved mirror.
Only one point was awarded for using a convex mirror instead of a concave mirror. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 10 Distribution
of points
1 point
2 points
1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points
(d) 3 points
If the student drew at least one ray AND an image in part (c) that made it possible to
interpret the drawing, then part (d) was graded on the basis of a literal interpretation
of the drawing. All answers had to be consistent with the drawing.
For correct indication of the image as real or virtual
For correct indication of the image as upright or inverted (the orientation must be clearly
indicated by the image drawn)
For correct indication of the image as larger or smaller 1 point
1 point
1 point If part (c) was blank or didn’t have both an image and at least one ray, then part (d)
was graded of the basis of the two possible CORRECT answers. The first two points
above could be earned by indicating one of the correct pairs of answers: real & inverted
OR virtual & upright. The last point could be earned only by indicating that the image
is larger.
(e) 1 point
For an answer that said another type of image could be formed (e.g., virtual versus real
and/or upright versus inverted) or some other answer (e.g., adding a lens to the
concavemirror system) that went beyond simply restating the premise.
If student gave the correct explanation and one or more wrong explanations, zero points were
awarded. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 11 1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 5
10 points total
(a) Distribution
of points 1 point
For a correct calculation of the change in internal energy
U a − U c = ∆U c→ a = Qc → a + Wc→ a
∆U c→ a = 685 J − 120 J
∆U c→ a = 565 J (b) 3 points
i. (1 point)
For correct choice of heat removed from the gas 1 point ii. (2 points)
For recognition that the change in internal energy is opposite in sign from part (a) answer
∆U a →b →c = −∆U c→ a = −565 J
Calculating the heat:
Qc → d →a = ∆U c →d →a − Wc→ d → a
Qc →d →a = −565 J − 75 J
For the correct answer
Qc → d →a = −640 J
(c) 1 point 1 point 3 points
The total work done is the sum of the work for the two sections of the path
Wc→ d →a = Wc →d + Wd → a
For some indication that the work done along path c → d is zero
The work done along path d → a is the area under the curve
Wc→ d → a = 0 − P ∆V
For correct substitution
Wc→ d → a = − (6.0 × 105 Pa)(1.0 × 10−3 m3 − 0.75 × 10−3 m3 )
For the correct answer
Wc→ d → a = −150 J (d) 1 point 1 point 1 point
1 point 3 points
For correct choice of heat added to the gas
For a complete explanation that references the first law
Example: Since ∆U c→ d → a is positive ( i.e. 565 J) and work is done (i.e. 150 J), Q must
be positive.
An incomplete argument with correct relevant assertions and no mistakes earned 1 point.
An incomplete argument with irrelevant or incorrect assertions earned no points.
Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 12 1 point
2 points AP PHYSICS B
2003 SCORING GUIDELINES
Question 6
10 points total
(a) Distribution
of points 2 points
For any correct indication of the relation between pressure difference and depth in either part
(a) or part (b)
P − P0 = H gh
Gauge pressure is equal to the pressure difference from the surface to the given depth
kg öæ
mö
æ
Pgauge = ç1.025 × 103 3 ÷ç 9.8 2 ÷ ( 35 m )
m øè
sø
è
For the correct answer
Pgauge = 3.5 × 105 Pa. (b) 1 point 1 point
For any indication, in words or with a calculation, that absolute pressure is the gauge
pressure calculated above plus atmospheric pressure
Pabs = Pgauge + Patm = 3.5 × 105 Pa + 1.0 × 105 Pa = 4.5 × 105 Pa (c) 1 point 1 point 5 points
For any indication that the plate is in equilibrium
T + Fbuoy − mg = 0 (other statements such as a = 0 or Σ F = 0 also acceptable) 1 point For substituting correct values into the relations for the mass or weight of the plate
kg ö
æ
m = HV = ç 2.7 × 103 3 ÷ 1.0 × 2.0 × 0.03 m3 = 1.6 × 102 kg OR mg = HVg = 1.6 × 103 N
è
mø
For indicating the existence of an upward buoyant force, using a diagram, an arrow,
or an equation in which the buoyant force has the opposite sign of the plate’s weight
For substituting the density of the fluid into the appropriate relation for the mass of the
displaced fluid or the buoyant force
kg ö
mö
mö
æ
æ
æ
Fbouy = HfluidVg = ç1.025 × 103 3 ÷ 0.06 m3 ç 9.8 2 ÷ = ( 61.5 kg ) ç 9.8 2 ÷ = 6.0 × 102 N
è
è
è
mø
sø
sø
Solving for the tension and substituting:
T = mg − Fbouy = 1.6 × 103 N − 6.0 × 102 N 1 point ( ) ( 1 point
1 point ) For the correct answer
T = 1.0 × 103 N (or 9.9 × 102 N using g = 10 m s 2 )
Notes:
Full credit was awarded for a correct answer with relevant work that did not explicitly show
all of the steps above.
One point was deducted for one incorrect numerical calculation with all other work correct.
Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 13 1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 6 (continued)
Distribution
of points (d) 2 points
For checking “increase”
For a valid explanation
Example: In (c) the net force is zero, but to accelerate there must be a nonzero net force.
Since the weight and the buoyant force are fixed, the tension must be greater to have a
nonzero net force upward.
Note: A valid explanation is not one that merely restates the conclusion that the tension
increases when the plate accelerates. It must go further by referring to Newton’s second
law and the fact that the net force must be upward when the plate accelerates upward.
Alternate interpretation
Some students clearly interpreted the question to be referring to a change in the tension over
time, rather than a change with respect to the situation in part (c). If the student checked
“remains the same” and gave an explanation stating that a constant acceleration implies a
constant force, they were awarded 2 points. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 14 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 7
10 points total
(a) Distribution
of points 2 points
For any indication of conservation of energy
1
1
1
mHeK0 2 = mHe K12 + mNe K2 2 + E2′ − E2 , where K0 is the minimum speed of the helium
2
2
2
atom before the collision, and K1 and K2 are the speeds of the helium and neon atoms
after the collision, respectively
The point is awarded for any of the following assumptions for K1 and K2 :
a. K1 = 0 and K2 = 0
b. K1 = 0 , K2 ≠ 0
c. K1 ≠ 0 , K2 ≠ 0 , and K1 ≠ K2
d. K1 = K2 ≠ 0
For any indication of conservation of momentum
mHeK0 = ± mHe K1 + mNeK2
The point is awarded for recognizing conservation of momentum, with any of the
following assumptions for K1 and K2 :
e. K1 = 0 , K2 ≠ 0
f. K1 ≠ 0 , K2 ≠ 0 , and K1 ≠ K2
g. K1 = K2 ≠ 0
To earn both points, assumptions for the two equations must be the same
The value of the final answer depends on the assumption used. ( (b) ) 1 point 1 point 2 points
Using the expression for the DeBroglie wavelength
h
h
l=
=
p
mu
For correctly substituting the answer obtained for K0 from part (a)
For correctly substituting mHe , including correct units
Students did not have to show conversions of units that would be needed to complete the
calculation correctly.
Both points were awarded if a value for momentum calculated in part (a) was substituted
directly. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 15 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES
Question 7 (continued)
Distribution
of points (c) 3 points
Using the equation relating energy and frequency for a photon:
f =E h
For substituting the correct difference in neon energy levels for the energy of the photon
20.66 eV − 18.70 eV
= 4.73 × 1014 Hz
f=
4.14 × 10−15 eV • s
Using the equation relating wavelength and frequency
l=c f
For substituting the calculated value of f and the correct value for the speed of light into the
above equation
l = ( 3 × 108 m s) 4.73 × 1014 Hz
For the correct answer, including units
l = 634 nm (or 633 nm if values are substituted directly into the equation l = hc E ,
using the value of hc in eV • nm from the Table of Information)
Alternate solution
For using the equation relating momentum and energy of a photon, including the correct
difference in neon energy levels
p = E c = E2′ − E1′ c ( 1 point 1 point
1 point Alternate points
1 point ) p = ( 20.66 eV − 18.70 eV ) 3 × 108 m s = 6.53 × 10−9 eV • s m For using the equation relating wavelength and momentum, and substituting the value of
momentum obtained above
l = h p = ( 4.14 × 10−15 eV • s) ( 6.53 × 10−9 eV • s m)
For the correct answer, including units
l = 634 nm (or 633 nm as explained above)
(d) 1 point 1 point 3 points
The total energy per pulse equals the power times the duration of the pulse
E = P ∆t
For substituting the values into the equation above
E = ( 0.50 W ) ( 20 × 10−3 s ) = 1× 10−2 J = 6.25 × 1016 eV 1 point The number of photons equals the total energy per pulse divided by the energy per photon
N = E E2′ − E1′ ( ) For correctly substituting the values of E1′ and E2′ into the equation above 1 point For the correct answer
N = 3.19 × 1016 1 point N = 6.25 × 1016 eV ( 20.66 eV − 18.70 eV ) Copyright 2003 by College Entrance Examination Board. All rights reserved.
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This note was uploaded on 02/25/2010 for the course PHY 15166 taught by Professor Mr.donald during the Spring '10 term at 4.1.
 Spring '10
 mr.donald
 Physics

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