AP® Physics 1 2022 Scoring Guidelines Practice Exam 2
Question 1: Short Answer
(a)
7 points
Horizontal component: For a non-zero horizontal line
1 point
Vertical component: For a non-vertical line with a negative slope that crosses the
horizontal axis near t f 2 and reaches t f
1 point
Scoring note: Any part of the sketched graph beyond t f is ignored.
Total for part (a)
(b)
(i) For a gravitational force with correct direction and label
2 points
1 point
1 point
For a tension force with correct direction and label
Scoring note: A maximum of one point may be earned if there are any extraneous forces.
(ii) For a vector pointing downward representing the gravitational force with no extraneous
forces
1 point
Scoring note: A label is not required, since the use of labels is assessed in part (b)(i).
(iii) For attempting to find the horizontal component of velocity
1 point
Scoring note: See examples below. This point is earned for the wrong use of trig
functions if it is clear it is being used to find the horizontal component, but not awarded if
it’s not clear which component the trig function is referring to, or for stating that the
vertical component is zero.
1 point
For a correct numerical answer for horizontal velocity with correct units
Example response:
0 points: v sin θ = 13 m s
(No clear indication of an attempt to find the horizontal
component)
1 point:
=
vx v=
sin θ 13 m s (An attempt to find the horizontal component is implied
by the vx term)
1 point: v cos θ
(An attempt to find the horizontal component is implied
by the cos θ term)
(Correct numerical answer with no units)
1 point: v cos θ = 7.5
1 point: 7.5 m s
(Correct answer with units, but no work shown)
2 points: 15cos 60 = 7.5 m s
(Correct answer with units)
2 points: 15cos 60 m s
(Correct answer with units)
Total for part (b)
5 points
Total for question 1
7 points
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AP® Physics 1 2022 Scoring Guidelines
Question 2: Experimental Design
(a)
12 points
For a valid statement relating the speed of cart and dissipation of mechanical energy
1 point
Examples of valid statements:
More damage when going faster
Slower collision means compression and decompression during collision happen
without permanent deformation
Examples of invalid statements:
Higher fraction of K is lost at higher speeds
Because more kinetic energy is lost
Total for part (a)
(b)
1 point
1 point
For measuring the speed of both carts before and after collision
Scoring note: It is not necessary to measure the initial speed of the cart that is at rest.
For measuring the mass of the carts (or calculating mass ratio using conservation of
momentum, which may be stated later)
1 point
For equipment and measurements consistent with procedure as described or drawn in
diagram (e.g., motion detectors in the right places to measure speed of both carts)
1 point
For varying the speed (making speed the independent variable) of the first cart in a
feasible experiment that could be done in a school lab.
1 point
Scoring note: “Varying the speed” means at least 2 trials (to test hypothesis), and
“feasible” means that equipment is used appropriately for each measurement.
For attempting to reduce uncertainty, e.g., multiple trials for a given initial speed of cart 1,
or trials involving at least three different initial speeds
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AP® Physics 1 2022 Scoring Guidelines
For attempting to reduce uncertainty, e.g., multiple trials for a given initial speed of cart 1,
or trials involving at least three different initial speeds
1 point
Example response:
Quantity to be Measured
Initial speed of heavy cart (H)
Final speed of heavy cart
Mass of heavy cart
Final speed of light cart (L)
Mass of light cart
MS1
H
Symbol
viH
vf H
mH
vf L
mL
Equipment for Measurement
Motion sensor (MS1)
Motion sensor (MS1)
Triple beam balance
Motion sensor (MS2)
Triple beam balance
L
MS2
Measure the mass of each cart with the triple beam balance. Set the carts and motion
sensors on the track. With the motion sensors recording, push the heavy cart toward the
light cart so that they collide. Record the motion sensors readings for the speeds of the
heavy cart immediately before the collision, and the speeds of both carts immediately
after the collision. Repeat 8 times, varying the force with which the heavy cart is pushed
so that a wide range of initial velocities for the heavy cart are used.
Total for part (b)
(c)
5 points
For a correct indication of how initial and final total kinetic energy would be calculated
from the raw data
1 point
For using the difference between initial and final total kinetic energy to determine the
kinetic energy lost
1 point
For a description of how the calculated fractional energy losses would be used to confirm
or disconfirm the hypothesis
1 point
Example response:
For each trial, calculate the initial kinetic energy of the heavy cart using
Kinit = (1 2)mH viH 2 .
(1 2)mH vf H 2 + (1 2)mL vf L 2 .
Also calculate the final kinetic=
energy K f
K init
Make a graph of K lost =
( K init − K f )
K init as a function of viH .
If a positive trend is seen in the data, the hypothesis is confirmed.
Total for part (c)
3 points
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AP® Physics 1 2022 Scoring Guidelines
(d)
1 point
For using momentum conservation to find v f in terms of vi , or the ratio of postcollision to pre-collision speed
1 point
For a correct expression for final kinetic energy (or consistent with v f found from
conservation of momentum)
OR
a correct expression for the fractional energy loss in terms of m1 , m2 , the initial speed of
the heavy cart, and the final speed of the two-cart system
1 point
A correct expression for fraction of kinetic energy lost:
m2
m1
or 1 −
m1 + m2
m1 + m2
Scoring note: The above point is not earned if the answer includes any variables other
than m1 and m2
Total for part (d)
3 points
Total for question 2
12 points
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AP® Physics 1 2022 Scoring Guidelines
Question 3: Qualitative/Quantitative Translation
(a)
12 points
For an arrow directed upwards labeled Tension (or FT )
1 point
For an arrow directed downwards labeled Force of Gravity (or mg , or Fg ).
1 point
Scoring note: G or g are not acceptable labels for the force of gravity
1 point
For an arrow labeled tension that is larger than the arrow labeled Force of Gravity.
Scoring note: a maximum of two points can be earned if there are any extraneous
forces.
Example response for part (a)
Total for part (a)
(b)
3 points
For a sketch showing a non-linear path
1 point
For a sketch that is concave up
1 point
For a justification that addresses that the box is moving at a constant speed in the
horizontal direction and accelerating in the vertical direction which leads to a concave
up parabola.
1 point
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AP® Physics 1 2022 Scoring Guidelines
Example response for part (b)
Since the box is moving at a constant speed in the horizontal direction and accelerating
in the vertical direction, the path that the box takes should look like a concave up
parabola
Total for part (b)
(c)
1 point
For a kinematic equation correctly relating the horizontal position of the box to the
horizontal component of the speed of the box and the time
vx =
x
t
0.5 m/s =
t=
3 points
x
t
x
0.5 m/s
For a kinematic equation correctly relating the vertical position of the box as a function
of the vertical acceleration of the box and the time
1 point
1 2
ayt
2
1
y = (1.0 m/s 2 ) t 2
2
y=
1 point
For correctly combining the vertical and horizontal motions of the box into a single
equation.
1
 x 
y = (1.0 m/s 2 ) 

2
 0.5 m/s 
2
y = ( 2 m −1 ) x 2
Total for part (c)
3 points
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AP® Physics 1 2022 Scoring Guidelines
(d)
For an attempt to use functional dependence to describe how the equation and the sketch
in part B relate
1 point
For using functional dependence correctly to relate the equation derived in part (c)
consistently to the graph sketched in part (b).
1 point
Example response for part (d)
The equation I derived shows that the vertical position of the box is proportional to the
horizontal position squared and so the shape of the path traveled by the box should be a
parabola, as sketched in part (b).
Total for part (d)
(e)
2 points
1 point
One point for an arrow pointing upwards and to the right.
Example response for part (e)
Total for part (e)
1 point
Total for Question 3
12 Points
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AP® Physics 1 2022 Scoring Guidelines
Question 4: Paragraph-Length Response
(a)
(b)
7 points
For a labeled friction force vector in the correct direction in both diagrams
1 point
For labeled gravitational and normal forces in the correct directions in both diagrams,
with no extraneous forces
Total for part (a)
1 point
2 points
Correct Answer: tdown > tup
Scoring note: The response is graded even if an incorrect selection is made.
For stating that the magnitude of the net force on the block is greater when it is sliding
up the ramp than when it is sliding down the ramp because the direction of the frictional
force changes while the direction of the component of the gravitational force along the
ramp does not (this can be implied) OR a description of the net force consistent with the
free-body diagrams (FBDs) in part (a)
1 point
For stating that the magnitude of acceleration of the block while sliding up the ramp is
greater than that when sliding down, OR a description of acceleration consistent with the
FBDs in part (a)
1 point
For a justification that v f is less than v0 , OR average v up is greater than average v
1 point
down (e.g., speed changes more on way up than on way down because acceleration is
greater on the way up and the same distance covered and final/initial speed on way
up/down is zero), OR a description of final and initial speeds consistent with the FBDs
in part (a)
For a correct argument that, if v f is less than v0 or average speed up is greater than
1 point
average speed down, then tdown is greater than tup . (This argument could include a
kinematic equation.)
Scoring note: Student cannot earn this point even if justification is consistent with
incorrect part (a) FBD
For a logical, relevant, and internally consistent argument that addresses the required
argument or question asked, and follows the guidelines described in the published
requirements for the paragraph-length response
1 point
Scoring note: A maximum of 4 of 5 points (first 3 points plus 5th point) can be earned
if the FBD in part (a) is incorrect (e.g., friction force is in the same direction in both
FBDs)
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AP® Physics 1 2022 Scoring Guidelines
Alternate solution using work/energy reasoning for 1st two points of part (b)
For correctly applying work-energy to up/down ramp (or, indicating block-Earth system
has the same potential energy at the beginning and end)
1 point
For correct treatment of energy loss to friction
1 point
Total for part (b)
5 points
Total for Question 4
7 Points
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AP® Physics 1 2022 Scoring Guidelines
Question 5: Short Answer
(a)
7 points
1 point
For correctly relating the difference in maximum height of the ball to the kinetic energy
of the ball before and after the collision.
Example response for part (a)
If the collision between the ball and the floor were elastic, the speed just before the ball
hits would be equal to the speed of the ball just after it leaves the floor. Because of
conservation of energy, since air resistance is negligible, if the speed before and after the
collision were the same the ball should return to its starting height. Because it returns to
a lower height, we must conclude that there was energy lost in the collision, making the
collision inelastic.
Total for part (a)
(b)
1 point
For using conservation of energy or work energy principle to attempt to find the speed of
the ball just before the ball hits the force sensor.
1 point
For using conservation of energy or work energy principle to attempt to find the speed of
the ball just after the ball hits the force sensor.
1 point
For correctly addressing the vector nature of momentum (or velocity)
1 point
Scoring note: Responses do not have to correctly calculate the change in momentum. A
response must show that the momentum (or speeds) have different directions, and that the
difference in these two are the algebraic sum of the two magnitudes.
Example response for part (b)
∆=
p mv f − mvi
KE = U g
=
∆p ( 0.50 kg ) ( 4.7
1 mv 2 = mgh
5.1 Ns
∆p =
i
2
=
vi
2 gh
=
(
2 10
U g = KE
mgh =
=
vf
1
2
mv 2f
2 gh
=
(
2 10
m
s2
m)
) (1.5=
5.5
m)
) (1.1=
4.7
m
s2
m
s
) − ( 0.50 kg ) ( −5.5
m
s
m
s
m
s
Total for part (b)
3 points
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AP® Physics 1 2022 Scoring Guidelines
(c)
For attempting to calculate the area under the curve
1 point
For comparing the area under the curve to the value of the change in momentum for
part (b).
1 point
For a correct justification that the impulse shown from the force sensor is not large
enough to cause the ball to return to a height of 1.1 meters.
1 point
Example response for part (c)
The area under the force as a function of time graph is equal to the impulse or change in
momentum of the ball.
∆pobserved =
5.1 Ns
∆=
precorded
1 0.2
(
2
ms )( 4 kN ) + ( 4 kN )( 0.6 ms ) + 12 ( 0.2 ms )( 4 kN )
∆precorded =
3.2 Ns
The change in momentum recorded by the force sensor is significantly smaller than the
change in momentum determined from experimental data collected from the initial and
final heights of the ball. It is likely that the force sensor reading cannot be trusted.
Total for part (c)
Total for question 5
3 points
7 points
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