Unformatted text preview: AP® Physics B
2005 Scoring Guidelines The College Board: Connecting Students to College Success
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2005 SCORING GUIDELINES
General Notes About 2005 AP Physics Scoring Guidelines
1. The solutions contain the most common method(s) of solving the freeresponse questions and the
allocation of points for these solutions. Other methods of solution also receive appropriate credit for
correct work. 2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is
correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.
One exception to this may be cases when the numerical answer to a later part should be easily
recognized as wrong, e.g., a speed faster than the speed of light in vacuum.
3. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a
particular concept is worth one point, and a student’s solution contains the application of that equation to
the problem but the student does not write the basic equation, the point is still awarded.
4. The scoring guidelines typically show numerical results using the value g = 9.8 m s 2 , but use of 10 m s 2 is of course also acceptable.
5. Numerical answers that differ from the published answer due to differences in rounding throughout the
question typically receive full credit. The exception is usually when rounding makes a difference in
obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that
should have five significant figures and that differ starting with the fourth digit (e.g., 20.295 and
20.278). Rounding to three digits will lose the accuracy required to determine the difference in the
numbers, and some credit may be lost. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 1
10 points total
(a) Distribution
of points 4 points The velocities can be found from the slope of the position graph.
For showing a positive velocity of magnitude 1.5 m s (i.e., 12 m 8 s) between
0 s and 8 s inclusive
For showing zero velocity between 10 s and 18 s inclusive
For showing a negative velocity of magnitude 2.4 m s (i.e., 12 m 5 s) between
20 s and 25 s inclusive
For showing two nonvertical transition regions; between t = 8 s and 10 s and between
t = 18 s and 20 s 1 point
1 point
1 point
1 point Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points
(b)
(i) 3 points For a definition or equation for average acceleration
aavg = Du Dt OR u = u0 + at 1 point For the correct substitution from part (a)
aavg = (0  1.5 m s) 2 s 1 point For the correct answer including units and sign
aavg = 0.75 m s2 1 point (ii) 1 point For a correctly drawn vector, with or without a label (c) 1 point 2 points
The acceleration is zero, so the normal force (apparent weight) is equal to the
gravitational force.
For a correct relationship leading to a calculation of apparent weight
N = W = mg OR N  W = ma ( Wapp = (70 kg ) 9.8 m s2 1 point ) For the correct answer with units 1 point Wapp = 686 N (or 700 N using g = 10 m s )
2 Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 2
10 points total
(a) Distribution
of points 2 points For each correctly drawn and labeled tension, with arrowhead in right direction
One point was deducted for each of the following until score reached zero:
No force of gravity
Each extraneous force
Any missing labels
Drawing all forces along correct lines with labels but no arrowheads received only
one point.
Components of the tension in the pendulum string could be included in addition to or instead
of the net tension, as long as they were clearly labeled as such. (b) 1 point each 4 points
For any indication that the net force is zero
For an attempt to determine the components of the tension in the pendulum string
For correctly determining these components
Th = T p sin 30∞ 1 point
1 point
1 point mg = T p cos 30∞ Th
sin 30∞
=
= tan 30∞
cos 30∞
mg
Th = mg tan 30∞ ( ) Th = (1.8 kg ) 9.8 m s2 tan 30∞
For the correct answer with units
Th = 10 N 1 point Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 2 (continued)
Distribution
of points
(b) (continued)
Alternate solution
Alternate points
For indicating that the addition of the three force vectors gives a 306090 right triangle
2 points
For a correct trigonometric relationship between forces
1 point
Th
= tan 30∞
mg
For the correct answer with units
1 point
Th = 10 N
If no forces were drawn in part (a), one point could be earned for writing the component
equations T = mg sin q and T = mg cos q , and two points for T = mg tan q . (c) 4 points
For any indication of conservation of energy
For any indication of the need to use a change in height
12
1
mgh0 + mu0 = mgh f + mu 2
f
2
2
For setting u0 = 0 1 point
1 point 1 point 12
mu f = mg Dh
2
u f = 2 g Dh Dh = L  L cos 30∞
u f = 2 gL (1  cos 30∞) ( ) u f = 2 9.8 m s2 (2.3 m ) (1  cos 30∞)
For the correct answer, with units
u f = 2.5 m s 1 point 2
A solution that used the kinematic equation u 2 = u0 + 2as could only receive full credit if
f the student explained how the equation is equivalent to conservation of energy. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 3
15 points total
(a) Distribution
of points 4 points
For use of the correct formula for the electric field E
q
q
1
E=
E=k 2
OR
4p 0 r 2
r
For adding the two E vectors from the charges
qˆ
Ê 2q ˆ Ê
EO = Á + k 2 ˜ + Á  k 2 ˜
Ë a¯ Ë a¯
For the correct (positive) expression for the magnitude of the field
q
EO = k 2
a
For any indication that the field is in the +ydirection (b) 1 point 1 point 1 point 1 point 3 points
For use of the correct formula for the potential V
q
1q
V =k
V=
OR
r
4p 0 r
For adding the two potentials
2q
q
VO = k
+k
a
a
For the correct (positive) answer
3kq
VO =
a (c) (i) and (ii) 1 point 1 point 1 point 4 points For determining the distance r between charges in terms of x0 and a in either part (i) or (ii) 1 point 2
r = x0 + a 2
For use of Coulomb’s law in either part (i) or (ii) 1 point F=k q1 q2 r2
For both correct substitution of charges into Coulomb’s law and correct substitution of the
previously determined expression for r in either part (i) or (ii)
For correct (positive) answer for both part (i) and (ii)
For part (i): Fq = k
For part (ii): Fq = k ( q ) (q )
2
x0 + a 2 =k (  q ) (2 q )
2
x0 + a 2 1 point
1 point q2
2
x0 + a 2 =k 2q2
2
x0 + a 2 Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 3 (continued)
Distribution
of points
(d) 4 points For having all three vectors on the same side of the xaxis (i.e., recognizing that the
vertical components should all be in the same direction)
For a single resultant vector from A pointing down and to the right, toward a point
below the xaxis and above +2q
For a single resultant vector from C pointing down and to the left, toward a point
below the xaxis and above +2q
For a single resultant vector pointing straight down from B 1 point
1 point
1 point
1 point Magnitudes of the vectors and any components that may have been drawn were not
evaluated. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 4
15 points total
(a) Distribution
of points 2 points
For checking any lengthmeasuring device and not the stopwatch
Only one point was awarded for checking any lengthmeasuring device and the stopwatch.
No points were awarded if no lengthmeasuring device was checked or all the equipment
was checked. (b) 2 points 3 points
For a diagram including at least a laser, a slide, and a screen
For correctly labeling the equipment
For correctly identifying the two distances to be measured, with symbols 1 point
1 point
1 point Example diagram: Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points
(c) 3 points Points could only be earned in the following order, with each successive point dependent
on earning the previous point.
For a central maximum
For a continuous graph with approximately equally spaced maxima
For nothing else incorrect (e.g., no negative intensity) (d) 1 point
1 point
1 point 4 points
For appropriate use of equipment (including all items checked in (a))
For indicating a measurement of the distance from the slits to the screen
For indicating appropriate measurements of the maxima (e.g., the distance between
adjacent maxima or the distance of successive maxima from the center)
For a complete and clear description
For example:
Set up the laser to shine on the slide, and set the screen far away on the other side of the
slide.
Measure the distance L from the slide to the screen with the tape measure.
Use the ruler to measure the distance x between adjacent maxima. 1 point
1 point
1 point
1 point Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points
(e) 3 points
Points could only be earned in the following order, with each successive point dependent on
earning the previous point.
For using appropriate equations
For explicitly identifying, in some other part of the question, the symbols used here
For correct correlation of symbols in equations to symbols in diagram in part (b)
x
ml L
OR
d sin = ml and tan q = m (since sin q ª tan q for small angles),
xm ª
L
d
where m is an integer, xm is the distance of a maximum of order m from the central
maximum, l is the wavelength, L is the distance between slits and the screen, and d is
the slit separation
Solving for d
lL
dª
using distances x between adjacent maxima and Dm = 1
x
ml L
OR d ª
using distances from center of pattern
xm 1 point
1 point
1 point Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 5
10 points total
(a) Distribution
of points 3 points
For an equation that uses the ratio of densities rr rw to find the fraction of the total
volume (or height) submerged 1 point The weight of the raft equals the weight of the displaced water.
Wr = Ww mr g = mw g
rrVr g = rwVw g
Solving for the volume of displaced water, which equals the submerged volume of the raft
r
Vw = r Vr
rw
For recognizing that the submerged volume (or height) must be subtracted from
the total volume (or height)
Vsubmerged = Ah = Vr  Vw Ah = Vr h= Vr
A 1 point rr
rˆ
Ê
Vr = Vr Á 1  r ˜
Ë
rw
rw ¯ rr ˆ
Ê
Á1  r ˜
Ë
w¯ 650 kg m3 ˆ
1.8 m 3 Ê
1Ë
¯
8.2 m 2 Á
1000 kg m3 ˜
For the correct answer
h= 1 point h = 0 .07 7 m Some students misinterpreted the statement about the volume of the raft, taking it to mean
the volume of the part above the water instead of the total volume. If the solution to
part (b) showed work that demonstrated understanding of the concepts needed for part
(a), appropriate credit for this part was awarded. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 5 (continued)
Distribution
of points
(b) 4 points
For indicating that the buoyant force is equal to the weight of the raft
Fbuoy = Wr
For correct substitutions for calculating the buoyant force, either by directly calculating
the raft’s weight or calculating the weight of the displaced water ( ) (1.80 m ) (9.8 m s )
OR Fbuoy = rwVw g = (1000 kg m 3 ) (1.17 m 3 ) (9.8 m s2 ) Fbuoy = rrVr g = 650 kg m 3 3 1 point 1 point 2 For the correct answer with units 1 point Fbuoy = 1.15 ¥ 10 N (or 1.17 ¥ 10 N using g = 10 m s )
4 2 4 For indicating that the direction of the buoyant force is up (c) 1 point 3 points
The additional weight that can be carried is equal to the weight of water displaced by
the part of the raft now above water.
For indicating a correct equation for the net force
Waddl = Wextra water OR Waddl = FbuoyNEW  Wraft 1 point The first equation above yields Waddl = rwVtop g = rw Ahg
Substituting the algebraic expression for h from part (a) and simplifying yields
Waddl = rwVr g  rrVr g = Vr g ( rw  rr ) , which is equivalent to substitution
into the second equation above.
For correct numerical substitutions to get the weight (or mass) of the top of the raft ( Waddl = 1.80 m 3 ) (9.8 m s ) (1000 kg
2 m  650 kg m
3 3 ) Waddl = 6200 N (variation due to rounding earlier on was accepted)
For dividing the total weight (or mass) by the weight (or mass) of a person and
indicating the correct number of people that the raft can carry. (The final answer
must indicate a whole number of people.)
W
6200 N
n = addl =
= 8.4
mp g
(75 kg ) 9.8 m s2 ( 1 point 1 point ) A maximum of 8 people can be on the raft. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 6
10 points total
(a) Distribution
of points 2 points
For writing the ideal gas equation of state
PV = nRT
For an indication that V = AH
PAH = nRT
nR
H=
T
PA
Note: Simply writing PAH = nRT or the equivalent earned both points. If the student tried
to rearrange the equation, algebra mistakes were not penalized in this part. (b) 1 point
1 point 4 points
For labeling both axes with linear numerical scales
For having neither axis labeled with its scale starting at zero (no penalty for showing
zero at the end of an axis and a “break” in the axis)
For accurately plotting five data points that closely fit a straight line with a positive slope
One point was lost if some points were inaccurately plotted.
Both points were lost if the data points were not visible, even if a line was drawn. 1 point
1 point
2 points Example answer shown below. The question did not ask for a bestfit line, and it was
not required for this part. However, a line is shown in the example, since it could be
used in the determination of n. Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 6 (continued)
Distribution
of points
(c) 4 points
nR
T
PA
For any clear indication that the student used more than one data point
For example: Setting up a slope calculation using subtraction, averaging calculated
values for n or the ratio H T , or using a linear regression on the calculator
For the correct slope of a bestfit line through the data points (this point not awarded
if slope method not used)
For correct substitutions into a correct expression containing n using consistent units
for the values of P, R, A, and H
Example using the line shown above, which happens to go through the first and last
data points
nR
Slope of line =
PA
(1.47  1.11) m 0.36 m
Slope =
=
(405  300 ) K 105 K
From part (a), H = n= PA
(slope )
R
1 ¥ 105 Pa 0.027 m 2 ( )( ) ( 1 point
1 point ) 0.36 m
105 K
8.31 J (moliK )
For a numerical answer that follows from substitutions into the correct expression above
n = 1.11 moles n= 1 point 1 point Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 7
10 points total
(a) Distribution
of points 4 points
For a correct calculation of the photon frequency
c
f=
l ( f = 3.00 ¥ 108 m s 1 point ) (1.219 ¥ 107 m) = 2.46 ¥ 1015 Hz For correct calculation of the photon energy in electronvolts or joules ( Eph = hf = 4.14 ¥ 10 15 )( eV is 2.46 ¥ 10 15 ) Hz = 10.2 eV OR 1 point 1.63 ¥ 10 18 J The two points above were also awarded for correctly using E = hc l or using
E = pc and the answer from part (b).
For indicating that the photon energy is the difference between the two energy levels
E4 = E2 + Eph 1 point E4 = 13.6 eV + 10.2 eV
For the correct numerical answer
E4 = 3.4 eV OR 5.44 ¥ 10 19 J 1 point Alternate solution
For use of energy levels and the Bohr model*
En = E1 n2
For identifying the ground state energy
E1 = 54.4 eV
For using the correct quantum number
n=4
For the correct answer
E4 = 3.4 eV Alternate points
1 point 1 point
1 point
1 point *Note: This equation is not on the equation sheet, nor is the Bohr model part of the Physics B
curriculum. However, students were given credit for this correct solution.
(b) 2 points p=hl
OR
p=Ec
For substitution of appropriate values into either of the above equations ( p = 6.63 ¥ 10 34 J is
) (121.9 ¥ 10 9 m ) OR ( p = 1.63 ¥ 10 For the correct answer with correct units p = 5.44 ¥ 10 27 kgim s (or 3.40 ¥ 10 18 J ) (3.00 ¥ 10 8 ms ) 1 point
1 point 8 eVis m ) Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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2005 SCORING GUIDELINES
Question 7 (continued)
Distribution
of points
(c) 2 points K max = Eph  f
For correct substitution of photon energy from part (a), or a calculation of it
K max = 10.2 eV  4.7 eV
For the correct answer in eV
K max = 5.5 eV (d) 1 point
1 point 2 points K max = W = qV
For using the definition of an eV as the work required to move a charge e through a
1volt potential difference
For the correct answer with units of volts
V = 5.5 V Alternate solution
For understanding the relationship between electrical potential and energy
V = K max q ( V = (5.5 eV ) 1.6 ¥ 10 19 J eV 1 point
1 point Alternate points
1 point ) (1.6 ¥ 1019 C) 1 point For the correct answer with units of volts
V = 5.5 V Copyright © 2005 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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