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Unformatted text preview: AP® Physics B
2007 Scoring Guidelines The College Board: Connecting Students to College Success
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General Notes About 2007 AP Physics Scoring Guidelines
1. The solutions contain the most common method of solving the freeresponse questions and the
allocation of points for this solution. Some also contain a common alternate solution. 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 1 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. However, when
students are asked to derive an expression, it is normally expected that they will begin by writing one or
more fundamental equations, such as those given on the AP Physics exam equation sheet. See pages
21–22 of the AP Physics Course Description for a description of the use of such terms as “derive” and
“calculate” on the exams, and what is expected for each.
4. The scoring guidelines typically show numerical results using the value g = 9.8 m s2 , but use of 10 m s 2 is of course also acceptable. Solutions usually show numerical answers using both values
when they are significantly different.
5. Strict rules regarding significant digits are usually not applied to numerical answers. However, in some
cases answers containing too many digits may be penalized. In general, two to four significant digits are
acceptable. Numerical answers that differ from the published answer due to differences in rounding
throughout the question typically receive full credit. Exceptions to these guidelines usually occur 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. © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 1
15 points total (a) Distribution
of points 2 points
For using a correct equation relating distance, speed, and time
x = u Dt
x
Dt =
u
21 m
Dt =
2.4 m s
For the correct answer
Dt = 8.75 s
Note: Only 1 point was awarded for the correct answer with no supporting work. (b) 1 point 3 points For each correct force that is correctly labeled, is attached to the dot, and has an arrowhead
pointing in the correct direction, 1 point was awarded.
For each incorrect vector, a point was deducted, with the minimum possible score being 0. (c) 1 point 3 points 3 points
For recognizing that the sum of the forces upon the sled is zero
ÂF = 0
Writing the equation for the forces acting along the slope,
Â F = mg sin15∞  f = 0 , where f represents the force of friction
For equating the force of kinetic friction with the component of weight that acts down
the slope
f = mg sin15∞ ( 1 point 1 point ) f = ( 25 kg) 9.8 m s2 sin15∞
For the correct answer 1 point f = 63.4 N (64.7 N if g = 10 m s is used)
Note: Only 1 point was awarded for the correct answer with no supporting work.
2 © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points
(d) 3 points
For equating the force of kinetic friction to the product of the coefficient of friction and
the normal force
f = mN
Â F = mg cos15∞  N = 0
For equating the normal force to the component of the sled’s weight that is normal to
the slope
N = mg cos15∞ f
mg sin15∞
=
= tan15∞
N
mg cos15∞
For the correct answer or for an answer consistent with the friction force obtained in (c)
m = 0.27
Note: Only 1 point was awarded for the correct answer with no supporting work. 1 point 1 point m= 1 point (e)
(i) 2 points
For implicitly or explicitly stating that the velocity of the sled decreases
For explicitly stating that the acceleration of the sled is constant (ii) 1 point
1 point 2 points For sketching a horizontal nonzero line
For sketching a line of constant negative slope that begins at the righthand end of the
previously drawn horizontal line and also indicating the time t on the graph
Note: The second point was awarded only if the first point was awarded. 1 point
1 point © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 2
10 points total
(a) Distribution
of points 1 point
For correctly indicating that the direction of the field is INTO the page or in
the –z direction. (b) 2 points
For stating that the (magnetic) force is perpendicular to the velocity, or the force is
centripetal, or an equivalent concept.
Note: The use of the phrase “centripetal force,” without stating its source, earned no credit.
For stating that the (magnetic) force or field changes the direction of the velocity but not
the speed, or an equivalent concept. (c) 1 point 1 point 1 point 4 points
For a correct expression indicating that the magnetic force provides the centripetal force 1 point 2 mu
= qu B
R
For a correct calculation of the radius of the trajectory
x 1.75 m
R= =
= 0.875 m
2
2
For correct substitutions into a correct expression
2 1.60 ¥ 10 19 C ( 0.090 T )( 0.875 m )
qBR
u=
=
m
1.45 ¥ 1025 kg
For a correct answer with units ( ) 1 point 1 point 1 point u = 1.74 ¥ 10 m s
5 (d) 3 points
For a correct expression indicating an equivalence of work and energy or electric potential
energy and kinetic energy
1
qe = mu 2
2
For correct substitutions into a correct expression
25
2
1m 2
1 1.45 ¥ 10 kg
1.74 ¥ 10 5 m s
e= 2 qu = 2
19
2 1.60 ¥ 10
C 1 point For a consistent answer with units
e = 6860 V 1 point ( ( )
)( 1 point ) © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 3
15 points total Distribution
of points (a)
(i) 2 points
For ranking I A as the greatest 1 point For ranking IC as the second greatest and I B as the third greatest 1 point (ii) 2 points For a correct statement justifying that I A is greatest. (For example: The total current 1 point flows through R A and gets divided between the other two resistors.)
For a correct statement justifying that IC is second and I B is third. 1 point (For example: RB and RC share the current in the parallel segment, and that current
divides between RB and RC so that the smaller resistor RC carries the most current.)
(b)
(i) 1 point
For the correct ranking VA , VB , VC = 1, 2, 2 (ii) 1 point 2 points For a correct justification that VA is the greatest. (For example, because no resistor is 1 point greater than R A , and R A has the full current through it. Or, because R A is greater
than the equivalent parallel resistance of RB and RC .)
For a correct justification that VB = VC . (For example, since the voltage across the 1 point parallel resistors RB and RC is the same.)
(c) 3 points
Let RBC be the resistance of the parallel combination of RB and RC .
1 point For one correct form for determining RBC 1
1
1
1
1
1
1
3
=
+
=
+
=
+
=
RBC
RB RC
2R R
400 W 200 W
400 W
2R
400 W
=
= 133 W
3
3
For one correct form for determining the total resistance
2R
Rtot = RA + RBC = 2 R +
= 400 W + 133 W
3
For the correct numerical value
Rtot = 533 W
RBC = 1 point 1 point © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 3 (continued)
Distribution
of points
(d) 3 points
For one correct form for the current I A , using Rtot from part (c) e IA = Rtot = 1 point 12 V
= 0.0225 A
533 W For the correct value of VC
VC = e  VA = e  I A RA = 12 V  ( 0.0225 A ) ( 400 W) = 3.0 V 1 point VC
3V
=
RC
200 W IC = 1 point For the correct numerical value of IC
IC = 0.015 A Alternate solution
Using RB = 2 RC and VB = VC so that I B RB = IC RC Alternate points 2 RC
RB
I=
I = 2I B
RC B
RC B IC = 1 point For the correct numerical value for I tot
I tot = e Rtot = 0.0225 A For correctly relating I tot to IC 1 point IC
3I
+ IC = C
2
2
For the correct numerical value of IC 1 point I tot = I B + IC = 2 Itot
2
= ( 0.0225 A ) = 0.015 A
3
3 IC = (e) 2 points
In the new circuit, I B = 0 at equilibrium, so the total current goes through each of
the two resistors
e = e = e = 12 V = 0.02 A
I tot =
2R + R
3R
600 W
RA + RC
For the correct value of the voltage across the capacitor
VC = I tot RC = ( 0.02 A ) ( 200 W ) = 4.0 V
Q = CVC ( 1 point ) Q = 2.0 ¥ 10 6 F ( 4.0 V )
For the correct numerical value of Q
Q = 8.0 ¥ 10 6 C 1 point © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 4
10 points total
(a) Distribution
of points 2 points
For any indication that the volume rate of flow is defined as volume/time
Define the symbol V for the volume flow rate V = 7.2 ¥ 10 4 m3 [( 2.0 min ) (60 s min )]
For the correct answer, including units
V = 6.0 ¥ 10 6 m3 s (b) 1 point 2 points
For a correct relationship between volume flow rate, speed, and area V = uA
u =V A ( u = 6.0 ¥ 10 6 m3 s 1 point ) (2.5 ¥ 106 m2 ) For the correct answer, including units
u = 2.4 m s
Note: An attempt to use kinematics with the distance x and height d could earn a maximum
of 1 point.
(c) 1 point 1 point 3 points
For applying Bernoulli’s Equation, either to points at the top of the liquid and the hole or
to points just inside and outside of the hole, and recognizing the specific conditions
for one of the three variables (pressure, speed, or height)
For recognizing the conditions for the remaining two variables
Top and hole
Inside and outside
12
12
12
12
Pin + rgyin + ruin = Pout + rgyout + ruout
Pt + rgyt + rut = Ph + rgyh + ruh
2
2
2
2
Pt = Ph = Patm
Pin = Patm + rgh , Pout = Patm
ut = 0 1 point u in = 0 yt = h , yh = 0 1 point yin = yout = 0 Both cases simplify to the same equation, where ue is the exit speed
2
rgh = rue 2 2
h = ue 2 g = ( 2.4 m s) 2 ( 2 9.8 m s2 ) For the correct answer, including units
h = 0.29 m 1 point Notes: Solutions that begin with the equation rgh = ru 2 2 could earn 2 of the 3 points.
Solutions that begin with the equation mgh = mu 2 2 could earn 1 of the 3 points. © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 4 (continued)
Distribution
of points
(c) (continued)
Alternate solution 1
For explicitly stating by name that Torricelli’s theorem applies
For writing the correct expression for the theorem
u = 2 gh h= (2.4 m s)
u2
=
2g
2 9.8 m s2 Alternate points
1 point
1 point 2 ( ) For the correct answer, including units
h = 0.29 m 1 point Alternate solution 2
For relating the pressure difference across the hole to the acceleration of the liquid
through the hole
F = ma = DP A
DP = rgh
a = rghA m = ghA V , where V is the volume of the hole
For applying an appropriate kinematics equation and substituting the expression for
acceleration
2
2
u = u0 + 2 a , where u0 = 0 and is the thickness of the container wall Alternate points
1 point 1 point u 2 = 2 ( ghA V ) = 2 gh
h= (2.4 m s)
u2
=
2g
2 9.8 m s2
2 ( ) For the correct answer, including units
h = 0.29 m
(d) 1 point 3 points
For correctly indicating that the liquid will hit to the left of the beaker
For an explanation that relates the decrease in water height to a decrease in the pressure
at the hole and a decrease in velocity exiting the hole
Other explanations, such as relating force and acceleration at the hole, describing
changes in potential and kinetic energy, or using a relationship from part (c), could earn
full credit.
Note: In the exam booklets, the container was erroneously referred to as the beaker in this
part. Answers indicating that the liquid would hit to the right of the beaker received full
credit if there was an explanation indicating that the student was now using the
container as the reference object. 1 point
2 points © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 5
10 points total (a) Distribution
of points 2 points
Using the relationship between pressure and force
P=F A
For correctly determining the area of the piston A = p R = p ( 0.20 m 2 )
2 F = Pabs A = Pabs p R 2
For correct substitution of values for pressure and area (or for correct answer in the
absence of explicitly showing the substitution) ( ) 1 point 2 F = 4.0 ¥ 10 5 Pa p ( 0.20 m 2 ) 1 point 2 F = 1.3 ¥ 10 4 N (b) 2 points
Using the ideal gas law
PV = nRT
nRT
V=
P
For correct substitution of at least three numerical values
(2.0 mol ) (8.31 J mol i K ) (300 K )
V=
4.0 ¥ 10 5 Pa
For the correct answer
V = 1.2 ¥ 10 2 m 3 (c) 1 point 1 point 2 points
Using the expression for the work done on the gas
Won =  P DV
The work done by the gas has the opposite sign
Wby = Pabs DV
For a correct expression for the change in volume
DV = Ax = p R 2 x 1 point Wby = Pabs p R 2 x
For substituting the correct pressure and a change in volume ( ) 1 point Wby = 4.0 ¥ 10 Pa p ( 0.20 m 2 ) ( 0.15 m )
5 2 Wby = 1.9 ¥ 103 J © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 5 (continued)
Distribution
of points
(c) (continued)
Alternate solution
W = Fx
For substituting the value of force from part (a)
For substituting the correct value for the distance ( Alternate points ) 1 point
1 point W = 1.3 ¥ 10 4 N ( 0.15 m )
W = 1.9 ¥ 10 3 J
Note: One point was deducted for any indication that the final value of the work done by the
gas is negative. (d) 3 points
For correctly indicating that heat is transferred to the gas.
For any indication that because the expansion occurs under constant pressure, the
temperature or internal energy of the gas increases.
For correctly applying the first law of thermodynamics to explain why heat is transferred
to the gas. For example: Since the internal energy goes up while the gas loses energy by
doing work, heat must be added. Units point
For including correct units in at least two of the answers in (a) through (c) 1 point
1 point
1 point 1 point Note: If a substitution was made using a value with units other than those given in the problem or in
the Table of Information, the units used had to be explicitly stated. An exception was part (b), where
the commonly used values of pressure in atmospheres and R in (liter ) (atmospheres ) ( moles ) ( K )
were acceptable. © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 6
10 points total (a) Distribution
of points 2 points
For a clear and complete method of estimating the focal length by focusing the image
of the tree on a screen and stating that the distance between the image and the lens
is the focal length
Partial credit:
1 point only was awarded if the method above was incomplete.
1 point only was awarded if the distances so (tree to lens) and si (lens to image) were
measured or estimated, and then the numbers were used in the thin lens equation to
calculate the focal length. 2 points (b) and (c) 3 points For showing a functional setup involving an object, the lens, and a screen
For having the lens and the distances so and si labeled correctly on the diagram
For labeling on the diagram all the equipment checked in (b) (must have a functional setup
to earn this point) 1 point
1 point
1 point © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 6 (continued)
Distribution
of points
(d) 2 points For all data points from the table plotted correctly
For a bestfit straight line (i.e., two points above and two points below the line and/or
intercepts on axes close to 3.35 m 1 ) 1 point
1 point © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 6 (continued)
Distribution
of points
(e) 3 points
For a clear and accurate solution with the correct answer in the range 0.28 m to 0.32 m
with units (The two most common approaches to doing this are illustrated below.) 3 points Approach 1:
Pick a point on the bestfit line (not a data point)
For example, for the graph shown in part (d), one point on the line is (1.0, 2.3).
1
1
= 1.0 m 1 ,
= 2.3 m 1
so
si 1
1
1
=
+
= 1.0 m 1 + 2.3 m 1 = 3.3 m 1
f
si so
1
3.3 m 1
f = 0.30 m
f= Approach 2:
1
1
1
+
=
si so
f 1
1
1
=
si
f
so
So 1
1
1
is the y intercept of a graph of versus
si
so
f For example, for the graph shown in part (d) the y intercept is 3.3 m 1
1
So
= 3.3 m 1
f
1
f=
3.3 m 1
f = 0.30 m
Partial credit:
2 points only were awarded for a mostly complete solution where either units were missing
or it was unclear that the bestfit line was used (e.g., not using numbers from the bestfit
line if Approach 1 was used, or not showing the idea of using the intercept if Approach
2 was used).
1 point only was awarded either for a correct answer with units where it was not clear how
the answer was obtained, or for using the lens equation with data from the line where
there was no final answer. © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 7
Distribution
of points
(a) 3 points
Using the relationship between mass and energy
E = me c 2
For correct substitutions of the positron mass and the speed of light
For the correct value of energy in joules ( )( E = 9.11 ¥ 10 31 kg 3.00 ¥ 108 m s ) 2 = 8.20 ¥ 10 14 J For using the correct factor to convert from joules to eVs E = 8.20 ¥ 10 14 J 1.60 ¥ 10 19 1 point
1 point 1 point J eV E = 5.12 ¥ 10 eV
5 (b) 1 point
Since the electron and positron have the same mass, the energy before annihilation is
twice the value found in part (a). That energy goes into creating the two photons
of equal energy, so each photon has the energy equivalent of one of the particles.
For any indication that this is the same numerical answer as in part (a)
Eg = 5.12 ¥ 10 5 eV 1 point Note: Full credit was earned if the student’s answer to part (a) was zero and the correct
calculation shown for part (a) was done here.
(c) 3 points
For a correct equation relating energy and wavelength
hc
Eg = hf =
l
For substituting the value of energy from part (b), either in eV or joules
For substituting a value of h or hc in units that are consistent with those of the energy used ( l = hc Eg = 1.24 ¥ 103 eV i nm ) (5.12 ¥ 105 eV) 1 point 1 point
1 point l = 2.42 ¥ 10 3 nm = 2.42 ¥ 10 12 m Alternate solution
Alternate points
Using the relationship between wavelength and momentum
l=h p
1 point
For substituting values for h and p (but not earned if p = mu was used)
For using the value of p from part (d)
1 point
For substituting a value of h in units that are consistent with those of the momentum used
1 point ( l = 6.63 ¥ 10 34 J i s ) (2.73 ¥ 1022 kg i m s ) l = 2.42 ¥ 10 3 nm = 2.42 ¥ 10 12 m © 2007 The College Board. All rights reserved.
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2007 SCORING GUIDELINES
Question 7 (continued)
Distribution
of points
(d) 2 points l=h p
For correct substitutions, using the value of λ determined in (c)
6.63 ¥ 10 34 J i s
4.14 ¥ 10 5 eVis
p=hl=
or
2.42 ¥ 10 12 m
2.42 ¥ 10 12 m
For correct units in the final answer p = 2.74 ¥ 10 22 kg i m s ( or Nis or Jis m ) or p = 2.73 ¥ 10 (e) ) 22 ( ) kg i m s ( or Nis or Jis m ) or 1 point 0.0017 eVis m Alternate solution
E = pc
For substituting the value of energy from part (b)
8.20 ¥ 10 14 J
5.12 ¥ 10 5 eV
E
OR
p=
=
c
3.00 ¥ 108 m s
3.00 ¥ 108 m s
For correct units in the final answer ( 1 point Alternate points
1 point 1 point 0.0017 eVis m 1 point
For indicating that the total momentum is zero 1 point © 2007 The College 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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