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For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP PHYSICS B
2004 SCORING GUIDELINES
General Notes about 2004 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 © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1
15 points total
(a)
i. Distribution
of points 1 point For correctly locating point P at the low point between the hill and the loop
ii. 1 point 3 points
For any indication that energy is conserved
Setting the kinetic energy at P equal to the potential energy at A:
1
mu
= mgy A
2 max
umax = 2 gy A
For correct substitution ( ) 1 point 1 point umax = 2 9.8 m s 2 (90 m )
For the correct answer
umax = 42 m s 1 point
2 (or 42.4 m s using g = 10 m s ) A maximum of 2 points could be earned for applying the kinematics equation
2
umax = u 2 + 2a Dy without an explicit statement that since energy is conserved,
A
anything “falling” from A to P will achieve the same speed as something that is
dropped. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points (b) 3 points
For any indication that energy is conserved
Equating the total energy at point B to the potential energy at point A:
1
mu + mgy B = mgy A
2B
Solving for the speed at point B:
uB = 2 g ( y A  yB )
For correct substitution ( ) 1 point 1 point uB = 2 9.8 m s 2 (90 m  50 m )
For the correct answer
u B = 28 m s
(or 28.3 m s using g = 10 m s 2 ) 1 point As in (a) ii above, a maximum of 2 points could be earned for using the
kinematics equation.
(c)
i. 3 points For each correctly drawn and labeled vector
For no extraneous vectors, given that the two vectors above are drawn correctly
ii. 1 pt each
1 point 3 points
For determining the correct value of the weight
mg = (700 kg ) 9.8 m s 2 ( 1 point ) mg = 6860 N (or 7000 N using g = 10 m s 2 )
For indicating there is a centripetal force on the car, either separately or by setting
the sum of the forces equal to the centripetal force
2
muB
N + mg =
r
2
muB
(700 kg ) (28 m s )2
 mg =
 6860 N
20 m
r
For the correct answer
N = 20, 580 N (or 21,000 N using g = 10 m s 2 ) 1 point N= 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points (d) 2 points
For any clear description of both a modification that will lower the height of
point B and a correct justification
For example: Flatten out the track on either side of the loop so the bottom of the
loop is on the ground, and thus point B is lower. The work done by friction will
reduce the total mechanical energy available at point B, so if the kinetic energy at
point B is to remain the same, the potential energy at that point must be reduced. 2 points Only one point was awarded for an answer that showed some understanding but was
not totally clear or complete. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2
15 points total
(a) Distribution
of points 2 points
For work showing pabsolute  patm in an appropriate equation or calculation
pg = pabsolute  patm 1 point pg = 413 atm  1 atm
For the correct answer
pg = 412 atm 1 point Note: An answer with no work shown only received 1 point total
(b) 3 points
For showing p = r gh in any of the following equations or during calculation
p = p0 + r gh OR pg = Dp = r gh OR p = r gh 1 point For correct substitutions in any of these equations
(412 atm ) 1 ¥ 105 N m 2 i atm
pg
=
For example, D =
rg
1025 kg m3 9.8 m s 2 1 point ( ( )( ( ) )) For answer consistent with (a), with a reasonable number of significant figures (1 to 4)
D = 4100 m (or 4020 m using g = 10 m s 2 ). Any negative sign was ignored.
Note: A range of answers was possible depending on the value used for g
( 9.8 or 10 m s 2 ) and on the conversion factor used to convert atmospheres to 1 point N m 2 (the approximate value in the equation sheet or the more precise value
found in some calculators).
(c) 2 points
For correct substitution of numerical values into a correct relationship
F = pg A = ( 412 atm ) 1 ¥ 105 N m 2 i atm 0.0100 m 2 ( ( )) ( 1 point ) Note: Since “force due to the water” might have been interpreted as due to the total
pressure instead of the gauge pressure, 413 atm was also accepted for the pressure.
For the correct answer with units consistent with calculation using 412 atm, 413 atm,
or answer to (a)
F = 4.12 ¥ 105 N
Note: In the absence of explicit indication of numerical substitution, a correct answer
with a correct equation could earn 2 points.
Also accepted was F = rVg , where r = 1025 kg/m3 and V = 0.0100 m3 (answer to (b) ) ( 1 point ) Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2 (continued)
Distribution
of points (d) 2 points
For substitution in the correct equation OR for the correct numerical answer
Negative sign was ignored.
For a correct numerical answer with correct units
a = D u t = (10.0 m s  0 m/s ) (30.0 s ) 1 point
1 point a = 0.333 m s 2 (e) 2 points
For correct substitution into a correct equation
For correct answer (using acceleration from (d) in the first two of the following solutions)
u 2 = u0 2 + 2 a D x 1 point
1 point d = u f 2 2a = (10.0 m/s )2 2 (0.333 m/s ) = 150 m OR
12
at
2
1
d=
0.333 m s 2 (30.0 s )2 = 150 m
2
d= ( ) uavg = Dx t OR d = uavg t = (30 s ) (10.0 m/s + 0 m/s ) 2 = 150 m
Note: In the absence of explicit indication of numerical substitution, a correct answer
with a correct equation could earn 2 points. Negative sign was ignored.
(f) 4 points
For computing the distance Dy that the ship falls at constant velocity using D from
part (b) and d from part (e)
Dy = D  d = 4100 m  150 m
Dy = 3950 m
For consistent substitution in a correct equation to find t2 , the time the ship falls at
constant velocity
Dy = u f t 2 1 point 1 point t2 = Dy u f = (3950 m ) (10 m s ) = 395 s
For finding the total time by adding t2 to the given time t1 to reach terminal velocity
ttot = t2 + t1 = 395 s + 30 s
For the correct total time
t = 425 s (or answer consistent with previous answers) 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 3
15 points total
(a) Distribution
of points 3 points
For using the correct expression for the magnetic flux
f = BA
For correct substitution 1 point f = (0.030 T ) (0.20 m )
For the correct answer
f = 1.2 ¥ 103 T i m 2 (b) 1 point 1 point 2 4 points
For using the correct expression for the magnitude of the emf
e = Df
Dt
For recognizing that one needs to calculate a change in the magnetic field or the flux
For a correct determination of the change in magnetic field or the flux
B
e = DDt A 1 point 1 point
1 point 2
030
e = (0.20 T  0.0.50 T ) (0.20 m )
s For the correct answer
e = 0.014 V (with or without a negative sign)
(c)
i. 1 point 2 points
For using a correct expression for Ohm’s law
I=eR
For correct substitution
I = (0.014 V ) (0.60 W )
I = 0.023 A 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 3 (continued)
Distribution
of points (c) (continued)
ii.
3 points
For correctly indicating that the current is counterclockwise
The remaining points were only awarded if the point above was earned
For indicating that Lenz’s law or a hand rule applies
For correctly explaining how Lenz’s law or a hand rule leads to the answer
For example: The magnetic field is increasing into the page. Current will be induced
to oppose that change. By the righthand rule, to create a field out of the page the
current must be counterclockwise.
(d) 1 point
1 point
1 point 2 points
For any description of a correct method to induce a current in a constant magnetic field
For example: Change the area of the loop
Pull the loop out of the field
Rotate the loop about an axis in the plane of the loop
No points were awarded if items such as batteries, capacitors, etc. were added to the loop Units:
1 point
For correct units on at least two of the three answers to parts (a), (b), and (c) 2 points 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 4
15 points total
(a) Distribution
of points 3 points
For a correct equation
u= fl
l=u f
For correct substitutions
l = (343 m s) (2500 Hz )
For the correct answer with units
l = 0.1372 m ª 0.14 m (b) 1 point 1 point
1 point 3 points
For demonstrating a correct approach to the problem using any of the following methods.
Method 1
1 lL
ml L
=2
=
xm ª
d
d
Y = xm = 0.457 m 1 2 1 point (0.1372 m ) (5.0 m )
0.75 m Method 2
d sin q = ml and Y = L tan q
Ê 1 (0.1372 m ) ˆ
ml
= sin 1 Á 2
= 5.25∞
q = sin 1
d
0.75 m ˜
Ë
¯ () Y = L tan q = (5.0 m ) tan (5.25∞) = 0.459 m
Method 3
Computing the actual path difference and setting it equal to l 2 : ( Y+ d
2 ) 2 + L2  ( Y d
2 ) 2 + L2 = l
2 (Y + 0.375 m )2 + (5 m )2  (Y  0.375 m )2 + (5 m )2 = 0.1372 m
2 By algebraic solution or by using a calculator
Y = 0.460 m
Assignment of points for each of the three methods:
For using m = 1 2 or for using path difference = l 2
For substitution of l from (a), d = 0.75 m, and L = 5.0 m 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points (c) 3 points
For correct identification of another minimum
Examples:
Y = 0.46 m along the line PQ but on the opposite side of P than Q
OR
At 3, 5, 7, … times the answer from (b) along the line PQ. Could be on either side
of P, but it was not necessary to mention which side.
For a complete explanation of the placement of the minimum cited
This explanation may include a clear statement of the symmetry of the problem, a
description of the path differences for sound from the speakers, or a mathematical
derivation of the new distance from P to the new minimum. One point only was
awarded for justifications that were not complete or less clear. (d)
i. 1 point 2 points 3 points
For an indication that Y increases
For a clear correct justification
Examples:
A statement that the distance d between the speakers decreases and Y is inversely
proportional to the separation of the speakers
OR
A mathematical calculation of the value of Y for a value of d < 0.75 m
Note: The justification points were awarded based on the quality of the explanation.
One point only was awarded for justifications that were not complete or less clear. ii. 1 point
2 points 3 points
For an indication that Y decreases
For a clear correct justification
Examples:
A statement that includes the inverse proportion between wavelength and frequency
and the direct proportion between Y and wavelength
OR
A mathematical calculation that demonstrates these relationships starting with
f > 2500 Hz and ending with Y < 0.46 m .
Note: The justification points were awarded based on the quality of the explanation.
One point only was awarded for justifications that were not complete or less clear. 1 point
2 points Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 5
10 points total Distribution
of points (a)
i. 2 points
For a correct calculation of the work done on the gas
Won =  P DV ( )( Won =  600 N m 2 9.0 m3  3.0 m3 1 point ) Won = 3600 J
For recognition that the work done by the gas is the negative of the work done on the gas
Wby = 3600 J
ii. 1 point 3 points
For a correct expression or derivation of the expression for DU
3
DU = nR DT
2
For correct calculation of T’s or DT using the ideal gas law, PV = nRT
3
J
DU = ( 2 moles ) 8.31
(325 K  108 K )
2
mol K
3
3
OR since P DV = nR DT , DU = P DV = 600 N m 2 9 m3  3 m3
2
2
For the correct answer
DU = 5400 J ( ) ( )( 1 point 1 point )
1 point 3
nR DT can be derived from the expressions for
2
found in the equation sheet as follows: Note: The equation DU =
K avg and urms U = NK avg , where N = number of molecules in the gas = nN 0
3
kT
2B
k
3k BT
3RT
R
urms =
=
, so
=b
M
m
m
M
M
R = kB
= kB N0
m
3
U = n RT
2
3
DU = nR DT
2 U = nN 0 Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 5 (continued)
Distribution
of points (a) continued
iii. 1 point
For correct substitution of answers from parts i. and ii. into the first law
of thermodynamics
DU = Q + Won
Q = DU  Won
Q = 5400 J  ( 3600 J )
Q = 9000 J
Alternate Solutions for parts ii. and iii.
Solving part iii. first:
5
5
J
Q = ncP DT = n R DT = ( 2 moles ) 8.31
(325 K  108 K ) = 9000 J
2
2
mol K
For a correct equation
For correct calculation of T’s or DT
For the correct answer
Returning to solve part ii.:
DU = Q + Won = 9000 J + ( 3600 J ) = 5400 J
For correct substitutions into the first law of thermodynamics of answers
from parts i. and iii. ( (b) ) 1 point Alternate points 1 point
1 point
1 point 1 point 1 point For point C plotted and labeled correctly as above, and for a correct straight line from
point B to point C 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 5 (continued)
Distribution
of points (c)
i. 1 point For a correct curve from point C to point A. Curve must be concave upward.
ii. 1 point 2 points
For correctly indicating that heat is removed from the gas
For a correct justification such as explaining in words or symbols that the change in
internal energy is zero, so from first law of thermodynamics Q = W . Since the
work done on the gas is greater than zero, Q is negative. Therefore heat is removed
from the gas. 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 6
10 points total
(a) Distribution
of points 2 points
For indicating that the ammeter is M 2
For indicating that the voltmeter is M 1 (b) 1 point
1 point 2 points For correctly plotting all 4 points
For a correct straight line based on the points that were plotted
(c) 1 point
1 point 2 points K max = hf  f , so h equals the slope of the line
Taking two points from the graph, for example
( 6 ¥ 1014 Hz , 0 eV) and ( 7.5 ¥ 1014 Hz , 0.65 eV):
DK max
Dy
0.65 eV  0 eV
h=
=
=
Dx
Df
7.5 ¥ 1014 Hz  6.0 ¥ 1014 Hz
h = 4.3 ¥ 10 15 eVi s
DK max
K
Dy
y
For any indication of
or
or
or max
Dx
Df
x
f
For a value of h consistent with the plotted data.
Note: For correctly plotted points, a range of values from 3.73 ¥ 1015 eVi s to 1 point
1 point 4.55 ¥ 10 15 eVi s was accepted. This range is ± 10% of the actual value of h. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 6 (continued)
Distribution
of points (c) (continued)
Alternate Solutions Alternate points Method 1 By simultaneous equations
Set up two equations with two unknowns using K max = hf  f and two points on
the graph. Solve the equations for h.
For indicating K max = hf  f
For the correct value of h from the two equations 1 point
1 point Method 2 By calculator program
Enter the data into a calculator and run a program to determine the bestfit line
for K max as a function of f. Then recognize that h is the coefficient of f in
the equation.
For the correct equation y ª 4.2 ¥ 10 15 x  2.51 or K max ª 4.2 ¥ 1015 f  2.51 For the statement that h = 4.2 ¥ 1015 eVi s from the equation
(d) (1 point)
(1 point) 4 points
For a statement that the graph moves to the right or down, OR for a sketch of
a second parallel line to the right of the first graph and labeled as the
second graph.
Note: 1 point was deducted for an indication that the slope of the graph changes.
For a correct explanation that relates to a graph or to the physical situation, such as
one of the following:
• A larger work function means a larger yintercept (no penalty for not including
a minus sign before the yintercept)
• A larger work function means a larger xintercept or threshold frequency
• A larger work function means that greater energy is needed in order for an
electron to escape from the surface 2 points 2 points Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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 Physics, Energy, AP, College Entrance Examination Board, college entrance examination, Entrance Examination Board

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