Unformatted text preview: AP® Physics B
2004 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
2004 SCORING GUIDELINES (Form B)
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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2 AP PHYSICS B
2004 SCORING GUIDELINES (Form B)
Question 1
15 points total
(a) Distribution
of points 4 points
For any indication of conservation of energy
For a correct conservation equation for this situation
12
1
mu + mgh0 = mu 2 + mghA
A
20
2
2
u 2 = u0 + 2 g ( h0  hA )
A
For the correct substitutions u2
A ( 1 point
1 point 1 point ) = (1.5 m s ) + 2 9.8 m s (2.0 m  1.9 m )
2 2 2
u A = 4.2 ( m s )
(or 4.3 ( m s )2 using g = 10 m s 2 )
For the correct answer
u A = 2.0 m s
(or 2.1 m s if u 2 was not rounded or if g = 10 m s 2 was used)
A
2 (b) 2 points For each correctly drawn and labeled force
One point earned for correct forces was deducted for having any extraneous forces.
(c) 1 point 1 pt each 3 points
For indicating that the sum of the forces equals the centripetal force
mu 2
A
F=
Â
r
For having the correct relative signs in the application of the above equation
mu 2
A
mg  N =
r
For solving for the normal force, substituting, and solving correctly
mu 2
A
N = mg r ( N = (0.50 kg ) 9.8 m s N = 2.7 N 2 ) 1 point 1 point 1 point (0.50 kg ) (4.21(m s )2 )
0.95 m Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 1 (continued)
Distribution
of points (d) 3 points
For any indication that the work done must equal the kinetic energy at point A in the
absence of friction
1
Wfriction =  mu 2
2A
For correct substitution
1
Wfriction =  (0.50 kg ) 4.2 ( m s )2
2
For the correct answer, including the negative sign
Wfriction = 1.1 J ( (e) ) 1 point 1 point
1 point 3 points
Method 1:
For indicating that one should decrease the radius of the curve of the second hill
mu 2
A
£ 0 . Decreasing the radius of the curve
The cart will lose contact when mg r
will cause the second term to increase and thus meet the condition.
mu 2
A
For referring to the equation N = mg r
For indicating that the cart will loose contact when N £ 0 Method 2:
For indicating that one should make the initial hill higher or the second hill lower
mu 2
A
£ 0 . Increasing the difference in
The cart will loose contact when mg r
heights will cause the speed at A to increase, and thus the second term will
increase and the condition will be met.
mu 2
A
For referring to the equation N = mg and indicating that the cart will
r
loose contact when N £ 0
For indicating that the speed at A will increase 1 point 1 point
1 point
(1 point) (1 point)
(1 point) Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 2
15 points total
(a) Distribution
of points 4 points
For calculating the length of time t1 that the diving bell is accelerating (i.e. the time it
takes to reach the constant speed uc )
uc = at1 t1 = uc a ( 1 point ) t1 = (2.0 m s) 0.10 m s2 = 20 s
For calculating the distance the bell descends while accelerating
12
d1 = at1
2
1
2
d1 =
0.10 m s2 (20 s) = 20 m
2
The bell therefore descends a distance d 2 = 80 m  20 m = 60 m at the constant speed.
For calculating the time to descend this 60 m
d2 = uc t2 ( 1 point ) 1 point t2 = d 2 uc t2 = (60 m ) (2.0 m s) = 30 s
For calculating the total time
ttot = t1 + t2 = 20 s + 30 s
ttot = 50 s (b) 1 point 3 points
For a correct expression for the weight of the water above the crosssectional area A
w = rVg = r Ahg
For correct substitution
w = 1025 kg m3 9.0 m 2 (80 m ) 9.8 m s 2 1 point For the correct answer, including units
w = 7.2 ¥ 106 N (or 7.4 ¥ 106 N using g = 10 m s 2 ) 1 point ( )( ) ( ) 1 point Full credit could also be earned for determining the pressure due to the water alone,
and using F = PA to determine the weight
A maximum of 2 points could be earned for only determining the mass of the
water, or for calculating the absolute pressure and using F = PA . Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 2 (continued)
Distribution
of points (c) 3 points
For indicating that the pressure at a depth h has a term r gh , or using P = w A
and the value of the weight from part (b)
For indicating that the atmospheric pressure must be added (in correct SI units –
otherwise the r gh term must be converted to atmospheres for full credit)
P = ( w A) + Patm ( 1 point
1 point ) P = 7.2 ¥ 106 N 9.0 m 2 + 1.0 ¥ 105 N m 2
For the correct answer, including units
P = 9.0 ¥ 105 N m 2
(or 9.2 ¥ 105 N m 2 using g = 10 m s 2 )
(d) 1 point 3 points
For a correct expression for the force on an area Ahatch
F = PAhatch
For a correct expression for the area of the hatch
Ahatch = p r 2
Since there is one atmosphere pressure inside the diving bell, the net pressure
corresponding to the minimum applied force is that from the water only.
Fmin = r ghAhatch
Substituting:
Fmin = 1025 kg m3 9.8 m s 2 (80 m ) p (0.25 m )2 1 point For the correct answer
Fmin = 1.6 ¥ 105 N 1 point ( )( ) ( 1 point ) Since the question could have been interpreted as asking for the minimum total
force required, full credit was awarded for including atmospheric pressure.
(e) 2 points
For a correct modification that would reduce the force necessary to open the hatch
For a correct justification
Examples:
Increase the pressure inside the diving bell. The internal pressure would then
compensate for atmospheric pressure plus some of the pressure of the water,
reducing the force needed.
Reduce the size of the hatch. Pressure times a smaller area yields a smaller force.
Use a tool that gives a mechanical advantage, such as a lever. Such a tool is
designed so that one exerts a smaller force over a larger distance than one would
without it. 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 3
15 points total
(a) Distribution
of points 4 points
There must be a node at the liquid surface and an antinode near the top of the tube,
so the length of consecutive resonances differ by a half wavelength.
For equating the difference in air column lengths to the difference in the number
of wavelengths
l
L2  L1 =
2
l = 2 ( L2  L1 )
For correct substitution and answer
l = 2 (0.80 m  0.25 m )
l = 1.1 m
A maximum of 3 points was awarded for using only one of the air column lengths to
calculate a wavelength
A maximum of 2 points was awarded for equating the difference in air column lengths
to one or onequarter wavelength (b) 1 point 2 points
For using the relationship between wavelength, speed, and frequency
f = uair l
For correct substitutions (using l from part (a)) and a corresponding correct answer
f = (343 m s) (1.1 m )
f = 312 Hz (c) 3 points 1 point
1 point 3 points
For indicating the need to use the values of the wave properties appropriate for water
lwater = uwater f water
For indicating that the frequency in water is the same as in air
For correct substitution and answer
(1490 m s )
lwater =
312 Hz
lwater = 4.8 m 1 point
1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 3 (continued)
Distribution
of points (d) 3 points 1
L3 is approximately 1 wavelengths, which is onehalf wavelength longer than L2
4
For any indication that L3 equals L2 + l 2 or L1 + l
1.1 m
L3 = 0.80 m +
2
For the correct answer (using whatever wavelength was determined previously)
L3 = 1.35 m 2 points 1 point A maximum of 2 points were awarded for using L3 = 5l 2 or L3 = L2 + l
One point was awarded for a good drawing of the standing wave in the air column with
no calculation.
(e) 3 points
For indicating that the calculation of frequency was too low
For a correct justification
For example: As temperature increases, the speed of sound in air increases, so the
speed used in part (b) was too low. Since f = uair l , that lower speed of sound
yielded a frequency that was too low. 2 points
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 4
15 points total
(a) Distribution
of points 4 points
For a correct expression for the magnitude of the emf per turn in the coil
e = D f OR e = B u
Dt
D
e = ADtB = w ( B t 0) OR e = BDtw
D
For correct substitutions in either equation (both yield the same value)
e = (0.25 m) (0.15 m) (0.20 T )
0.50 s
For correct calculation
e = 0.015 V per turn
For recognizing that with 20 turns the total emf in the coil is 20 times the emf per turn
etot = 20 (0.015 V per turn ) = 0.30 V (b) 1 point 1 point
1 point 2 points
For correct expression for Ohm’s law, substitution from part (a), and magnitude
V = IR
0.30 V
V
=
= 0.06 A
I=
5.0 W
R
For correct direction, i.e., counterclockwise (c) 1 point 1 point 1 point 3 points
For a correct expression for the power dissipated in the coil
P = IV OR P = I 2 R OR P = V 2 R
For correct substitution of answers from (a) and/or (b) into one of these expressions
(Must recognize that V = etot ) For example, P = (0.06 A ) (0.30 V )
For correct calculation including correct units
P = 0.018 W 1 point
1 point 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 4 (continued)
Distribution
of points (d) 3 points
The force on the right side is zero, and forces on the top and bottom sides cancel,
so the net force is that on the left side.
For the correct expression for the force on a straight wire (force per turn in this
situation) in a magnetic field
F = BI sin q OR F = BI (recognizing that the wire is perpendicular to the field)
For correct substitutions
F = (0.20 T ) (0.60 A ) ( 0.15 m )
F = 0.0018 N
For recognizing that with 20 turns the total force on the left side of the coil is
20 times the force per turn
Ftot = 0.036 N
Alternate Solution For using one of the equations P = F u sin q W
Fd
OR P =
=
t
t P
Pt
OR F =
d
u sin q
For correct substitutions (both approaches yield same value)
(0.018 W ) (0.50 s)
Ftot =
0.25 m
For correct answer
Ftot = 0.036 N 1 point
1 point 1 point Alternate points
1 point F= (e) 1 point 1 point 3 points
For recognition that doubling the number of turns doubles the total emf, which
would tend to increase the current
For recognition that doubling the number of turns doubles the total resistance,
which would tend to decrease the current
For putting these together to show that the current is unchanged
e f 2ei ei
If =
=
=
= Ii
2 Ri Ri
Rf 1 point
1 point
1 point Note: The third point was awarded for an indication that the current is unchanged,
even if the justification was wrong. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 5
10 points total
(a) Distribution
of points 2 points
For recognizing that PA = PB
At point A: PV1 = nRT1
1
V
At point B: P 1 = nRT2
1
2
Dividing these equations and solving for T2 :
PV1
T
1
=1
V1
T2
P
1
2 1 point T1
T2
For the correct answer
T2 = T1 2
2= (b) 1 point 2 points
Answer can be obtained by either comparing points B & C or points A & C using
the ideal gas law at each point
For recognizing that TC = 2TB OR TC = TA 1point Comparing points B & C
V
T
At point B: P 1 = nR 1
1
2
2
V1
At point C: P2
= nRT1
2
Dividing the equations and solving for P2 :
V
T
P1
nR 1
1
2=
2
V1
nRT1
P2
2 Comparing points C & A P
1
1
=
2
P2
For the correct answer
P2 = 2 P
1 2P
1
=1
P2
For the correct answer
P2 = 2 P
1 At point A: PV1 = nRT1
1 V1
= nRT1
2
Dividing the equations and solving for P2 :
At point C: P2 PV1
1
=1
V
P2 1
2 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 5 (continued)
Distribution
of points (c) 2 points
For correct equation for work done on the gas (regardless of sign) OR for recognition
that work is the area under the line AB on the graph. (No work is done from B to C
because the change in volume is zero.)
W =  P DV
ÊV
ˆ
W =  P Á 1  V1 ˜
1
Ë2
¯
For the correct answer including correct sign
PV
W= 11
2 (d) 1 point 1 point 4 points
Heat was added to the gas in processes BC and CA, but not in AB.
For two or three correct indications of whether or not heat was added to the gas in
each process (The absence of a check mark was taken as in indication that heat
was not added.) 1 point For a reasonable justification for process AB. This point was awarded only if this
process was indicated correctly (i.e., not checked)
Example: The volume decreases so the work done on the gas is positive. The
temperature decreases so the change in internal energy is negative. Therefore
Q = DU  W is negative. Heat is expelled from the gas. (Note: Answer must
mention work for credit.) 1 point For a reasonable justification for process BC. This point was awarded only if this
process was indicated correctly (i.e., checked)
Example: There is no change in volume so no work is done. The temperature increases
so the internal energy increases. Therefore Q = DU  W is positive. (This point also
awarded for only referring to increasing temperature or for referring to increasing
speed of the molecules.) 1 point For a reasonable justification for process CA. This point was awarded only if this
process was indicated correctly (i.e., checked).
Example: There is no change in temperature or internal energy. The volume increases
so work is done by the gas. Heat needs to be added to the gas to do this work. 1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 6
10 points total
(a) Distribution
of points 3 points For a smooth curve passing through or very close to all plotted points
Partial credit points could be awarded for curves of lesser quality.
For example:
Showing all the data points connected by straightline segments earned 2 points.
Curves that generally fit the data points at small angles, but are concave upward for
the whole range earned 1 or 2 points depending on how badly the curve misses
the points at higher angles.
A single straight “best fit” line or really bad curve earned 1 point.
A curve that was monotonically decreasing earned no points. 3 points Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES (Form B)
Question 6 (continued)
Distribution
of points
(b) 3 points
For reading the wavelength shift from the graph
From the graph shown, the wavelength shift at 120∞ is 8.000 ¥ 10 17 m .
The photon loses energy, which means the wavelength increases.
For adding (not subtracting) the wavelength shift to the original wavelength
ls = 1.400 ¥ 10 14 m + 8.000 ¥ 10 17 m
For correct answer ls = 140.8 ¥ 10
(c) 16 m = 1.41 ¥ 10 14 1 point
1 point m 2 points
For using the de Broglie equation to calculate momentum (even if a value from (b)
that was actually Dl was substituted)
ls = h p s OR p = h l ps = h ls ( ps = 6.63 ¥ 1034 J is 1 point ) (1.408 ¥ 1014 m) For consistent answer including correct units
ps = 4.71 ¥ 1020 kg i m s
Note: The answer point could also be awarded if the value of ls = li  Dl was used
instead of the correct relationship.
(d) 1 point 1 point 2 points
For use of conservation of energy
DEnuc =  DE photon
DE photon = D ( hf ) = D ( hc ) OR DE
l 1 point photon = D ( pc ) = D ( hc )
l 1ˆ
Ê hc hc ˆ
Ê1
DEnuc =  Á
 ˜ = hc Á  ˜
Ë ls
Ë li
li ¯
ls ¯ ( ) 1
1
ˆ
DEnuc = 1.99 ¥ 10 25 J ◊ m Ê
Á
˜
14
Ë 1.400 ¥ 1014 m
1.408 ¥ 10
m¯
For the correct answer
DEnuc = 8.08 ¥ 10 14 J (or 510 keV) 1 point Note: The answer point was awarded for correct use of whatever ls was obtained from
part (b), but not for using the value of Dl . Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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