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Unformatted text preview: AP® Physics B
2003 Scoring Guidelines
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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 (Form B)
Question 1
15 points total
(a) Distribution
of points 3 points One point for each correctly drawn and appropriately labeled force
For no incorrect forces
(b) 2 points
1 point 5 points
For use of the correct equation relating acceleration and velocity
a = ∆v t
For correctly calculating the magnitude of the acceleration
65 m s − 0 13
a=
=
m s 2 (or 2.17 m s 2 )
30 s
6
For using an equation relating distance and acceleration
1
2
d = d0 + K0t + at 2 OR K 2 = K0 + 2a ,d
2
For substituting the calculated acceleration
1 æ 13
2
2
ö
æ 13
ö
d= ç
m s 2 ÷ ( 30 s )
OR d = ( 65 m s ) 2 ç
m s2 ÷
2è 6
è6
ø
ø
For the correct answer
d = 975 m
Alternate solution
For use of the correct equation for average speed
Kavg = K f + K0 2 ( 1 point
1 point 1 point 1 point 1 point
Alternate points
1 point ) For correctly calculating the average speed
Kavg = ( 65 m s + 0) 2 = 32.5 m s 1 point For using the appropriate equation relating distance to speed
d = Kavg t 1 point For substituting Kavg into the equation 1 point d = ( 32.5 m s )( 30 s ) For the correct answer
d = 975 m 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 2 AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 1 (continued)
Distribution
of points (c) 5 points
For using the correct x and y components of the tension
Tx = T sin G and Ty = T cos G 1 point For the correct equation relating the forces along the xaxis
T sin G = ma
For the correct equation relating the forces along the yaxis
T cos G = mg
For combining these two equations to eliminate the tension
tan G = a g 1 point For the correct answer
G = 12.5° (or 12.2° using g = 10 m s2 ) 1 point 1 point 1 point tan G = ( 2.17 m s 2 ) (9.8 m s 2 ) (d) 2 points
For indicating that one would need to know the mass of the airplane, with some attempt to
give an explanation that relates to the mass.
For a correct explanation
Example: The kinetic energy is the only form of mechanical energy in this case. The
velocity is known, but you need the mass to calculate the kinetic energy. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 3 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 2
15 points total
(a) Distribution
of points 3 points
For correct equation for power
P = IV
For the correct answer
P = 3 mW (or 0.003 W)
One point was subtracted for incorrect or missing unit (b) 2 points 3 points
For the correct equation for work or energy
W (or energy) = Pt
For correct substitution of power from part (a)
W (or energy) = (0.003 W)(60 s)
For the correct answer consistent with substitution of power from part (a), with correct units
W (or energy) = 0.180 J (c) 1 point 1 point
1 point
1 point 5 points
For the correct efficiency equation
W
efficiency = o
Wi
For correct substitution of Wi (work done in 60 s) from part (b)
For indicating that the work output Wo equals the change in gravitational potential energy
Wo = mg ∆h
For correct calculation of work output in 60 s
Wo = ( 0.012 kg ) ( 9.8 m/s 2 ) (1 m ) = 0.12 J (or same answer using g = 10 m/s 2 ) For correct calculation of efficiency consistent with calculation made in part (b).
0.12 J
= 66.7% (or 65.3% using g = 9.8 m/s2 and unrounded value for Wo )
efficiency =
0.18 J
Alternately, full credit could also be obtained by calculating efficiency using the ratio of
power output to power input, in which case
2
Po ( 0.012 kg ) 9.8 m/s (1 m ) 60 s
= 65.3% (or 66.7% using g = 10 m/s 2 )
eff = =
Pi
0.003 W
Similar point allocations were assigned using this method. ( ) Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 4 1 point 1 point
1 point
1 point 1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 2 (continued)
Distribution
of points (d) 4 points For a calculation or notation that 6 V was the voltage drop across the resistor needed in order
to reduce the voltage across the motor from 9 V to 3 V.
6.0 V
= 6000 Ω
Series resistance needed to produce this voltage drop =
1.0 mA
For the selection of a 1000 Ω and a 5000 Ω resistor
For the placement of the 1000 Ω and 5000 Ω resistor in series
For an appropriate sketch of the system with appropriate symbols and labels for resistors Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 5 1 point 1 point
1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 3
15 points total
(a) Distribution
of points 4 points One point for each correctly drawn ray passing through the lens (maximum of 2 points)
For the intersection of the rays occurring at 30 cm ± 5 cm
For correctly drawing the image
(b) 2 points
1 point
1 point 3 points
For using the lens equation
111
+=
so si f
For correct substitutions
1
1
1
+=
15 cm si 10 cm
For answer consistent with substitutions
si = 30 cm 1 point 1 point 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 6 AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 3 (continued)
Distribution
of points (c) 3 points
For correct equation relating ratio of image to object heights to ratio of image to object
distances
h
s
hi
s
= − i OR i = i
ho
so
ho
so
For consistent substitution
hi
hi
30 cm
30 cm
=
OR
=−
5 cm
15 cm
5 cm 15 cm
For answer consistent with substitutions
hi = −10 cm OR hi = 10 cm
Note: Since the minus sign in the first answer just indicates the image is inverted, it was not
necessary for full credit, since the question could be interpreted as asking only for the
actual size of the image.
(d) 1 point 1 point 1 point 3 points
Method 1: Ray diagram One point for each correctly drawn ray passing through both lenses (maximum of 2 points)
For the image location at 16.6 cm ± 3 cm Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 7 2 points
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 3 (continued)
Distribution
of points Method 2: Mathematical approach using the lens equation
The image produced by the first lens becomes the virtual object for the second lens.
For the correct object distance to substitute into the lens equation for the second lens
so′ = − ( 30 cm − 10 cm ) = −20 cm (The minus indicates that the object is virtual.)
1
1
+=
so′ si′
1
+
−20 cm 1 point 1
f
1
1
=
si′ 10 cm For the final image location with respect to the second lens
si′ = 6.7 cm 1 point For the final image location on the scale shown
xi = si′ + 10 cm = 16.7 cm 1 point (e) 2 points
For checking spaces consistent with answer to part (d) and a correct explanation
Explanation had to either refer to ray diagram, or, if a mathematical approach was used in
part (d), had to be consistent with answer to part (d).
No points were awarded for students who checked spaces without any explanations. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 8 2 points AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 4
15 points total
(a) Distribution
of points 4 points
i. (1 point) For a correctly drawn and labeled vector for E, directed vertically upward as
shown in the diagram above 1 point ii. (1point) For a vector directed vertically downward representing the force on an electron
in the field, as shown in the diagram above 1 point iii. (2 points) For a curved path from A to B
For the path being symmetrical about the midpoint of d Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 9 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 4 (continued)
Distribution
of points
(b) 2 points
For using the equation that relates the force on a charged particle to the electric field
F = qE
Equating this force to the net force in Newton's second law, Fnet = ma
qE = ma
For the correct answer
qE
a=
m (c) 1 point 1 point 4 points
In the electric field the electron is accelerated vertically downward.
K y (t ) = K y − at , where K y is the vertical component of the electron's velocity at point A
The time t1 to reach maximum vertical displacement occurs when K y ( t1 ) = 0 . Solving for t1 :
t1 = Ky a
For using the correct component for K y 1 point K y = K sin G 1 point For indicating that the total time is twice t1
2K sin G
ttot = 2t1 =
a
For correct substitution of a from part (b)
2K sin G
ttot =
qE
m
For the correct answer
2mK sin G
ttot =
qE 1 point 1 point 12
at
2
= −K sin G , and y ( ttot ) = y0 . Alternately, the kinematic equations K y (t ) = K sin G − at OR y ( t ) = y0 + ( K sin G ) t −
could be used. When the electron reaches B at time ttot , K y ( ttot ) Substituting these values and the expression for a from part (b) into the respective equations
2mK sin G
. This approach
above and solving either equation for ttot gives the answer, ttot =
qE
also received full credit with awarding of points similar to those above. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 10 AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 4 (continued)
Distribution
of points (d) 3 points
While the electron is in the field the horizontal component of its velocity is constant.
For the correct equation relating distance to time
d = K xttot
For using the correct component for Kx
K x = K cos G
For correct substitutions for Kx and total time
d = ( K cos G ) æ
è
d= (e) 1 point
1 point 1 point 2mK sin G ö
qE ø 2mK 2 sin G cos G
qE 2 points
For an indication that the distance d would be less
For any reasonable explanation
Example: The additional gravitational force downward would increase the downward
acceleration thus decreasing the total time the electron would be in the field. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 11 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 5
10 points total
(a) Distribution
of points 2 points
For a statement of the ideal gas law
pV = nRT 1 point ( 200 N/m ) ( 20 m ) = (1 mol) (8.32 J ( mol ⋅ K) ) T
2 3 For the correct answer
T = 481 K
(b) 2 points
For indicating that the work W done on the gas is equal to the area enclosed by the
cycle or for W = − p ∆V
1
1
W = area of triangle enclosed by cycle = bh = ( 60 m3 − 20 m3 )( 400 N/m 2 − 200 N/m 2 )
2
2
For the correct answer
W = 4000 J (c) 1 point 1 point 1 point 2 points
i. (1 point)
For indicating that heat is removed from the gas during one complete cycle
ii. (1 point)
Using the first law of thermodynamics
∆U = Q + W
Recognizing that ∆U = 0 for a closed cycle
Q = −W
For the correct answer consistent with part (b)
Q = −4000 J
Note: Since the question could be interpreted as asking for the magnitude of the heat added to
or remove from, the minus sign was not necessary for full credit. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 12 1 point 1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 5 (continued)
Distribution
of points
(d) 2 points
For indicating that the internal energy of the gas after one cycle is the same as before
For a reasonable justification
Example: The internal energy of the gas is a function of the temperature and the temperature
is the same at the beginning and end of each cycle. 1 point
1 point (e) 2 points
For indicating that the entropy of the gas after one cycle is the same as before
For a reasonable justification
Example: The entropy is a function of the state of the gas, and after one complete
cycle the gas has returned to its original state. Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 13 1 point
1 point AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 6
10 points total Distribution
of points (a) 6 points
i. (3 points)
For correct use of the equation relating work to the distance raised
W = mgh
For correct use of the equation relating mass to density and volume
m = ρV
Combining the two relationships
W = ρVgh = 1000 kg/m 3 0.35 m3 9.8 m / s2 ( 50 m + 35 m ) ( )( )( ) For the correct answer
W = 290,000 J (or 300,000 J using g = 10 m/s 2 )
ii. (2 points)
For correct use of the equation relating power to work and time
W
P=
∆t
290,000 W
P=
( 2 hr )( 60 min/hr )( 60 s/min )
For the correct answer
P = 40 W (or 41 W using g = 10 m/s 2 ) 1 point
1 point 1 point 1 point 1 point (b) 4 points
i. (3 points)
For correct use of equation of continuity
υ1 A1 = υ 2 A2
For using the radius of each pipe as half the diameter
Substituting the given values:
2 0.03 m ö
æ 0.0125 m ö
÷ = υ 2π ç
÷
2ø
2
è
è
ø
For the correct answer
υ 2 = 2.88 m/s ( 0.50 m/s ) π æ
ç 1 point
1 point 2 1 point ii. (2 points)
For indicating the need to use Bernoulli's equation
For an explanation of how to use Bernoulli's equation
1
1
Example: p1 + ρ gh1 + ρυ12 = p2 + ρ gh2 + ρυ 2 2
2
2
If the subscript 1 represents quantities at the pump and subscript 2 represent quantities at
the house, then all the quantities are known except the pressure at the house, so the
equation can be solved for this pressure.
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 (Form B)
Question 7
10 points total
(a) Distribution
of points 7 points
There were two possible “other” or intermediate energy levels asked for in part iii., so there were two
possible correct energy level diagrams depending on which “other” energy level was used.
i. (1 point)
For drawing a horizontal line on the
diagram at −5 eV 1 point ii. (3 points)
For correct equation(s) to calculate the
energy of a 400 nm photon
E = hf and f = c/λ, so
E = hc λ ( E = 1.24 × 103 eV ⋅ nm ) ( 400 nm) For the correct answer
E = 3.1 eV
For drawing a horizontal line on the
diagram at −1.9 eV to represent the
first excited state, which is 3.1 eV
above the ground state OR 1 point 1 point
1 point iii. (3 points)
Using the same equations to calculate the
energy of a 600 nm photon:
hc 1.24 × 103 eV ⋅ nm
E=
=
600 nm
λ
For the correct answer
E = 2.1 eV
This photon could represent a transition
from the first excited state to an
intermediate state, in which case the
intermediate state is at E = −4.0 eV.
Or the photon could represent a
transition from an intermediate state to
the ground state, in which case the
intermediate state is at −2.9 eV.
For drawing a horizontal line on the
diagram at either −4.0 eV or −2.9 eV 1 point
1 point 1 point Note: Although the value of hc is given in the table of information, many students substituted h and c separately.
Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 15 AP PHYSICS B
2003 SCORING GUIDELINES (Form B)
Question 7 (continued)
Distribution
of points
(b) 3 points
If the intermediate state is at −4.0 eV, then the transition from the intermediate state to
the ground level at −5.0 eV is 1.0 eV. If the intermediate state is at −2.9 eV, then the
transition from the first excited state at −1.9 eV to the intermediate state is 1.0 eV.
For calculation of 1.0 eV by subtraction of energy levels
hc
Since E =
λ
hc 1.24 × 103 eV ⋅ nm
λ= =
1.0 eV
E
For calculation of wavelength
λ = 1240 nm
For statement that this wavelength is not seen because it is in the infrared or outside the
range of wavelengths for “white light” given in the problem Copyright 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 16 1 point 1 point
1 point ...
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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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