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Multiple Choice Final – AP Physics B
1. If a is acceleration in m/s2 and v is velocity in m/s, and x is position in m, which of
the following equations DOES end up with the same units on each side of the
equation?
A. at = 2x
B. v/t = x
C. xa = v2
D. vt = a
C
2. Which of the following expressions gives units of Joule? M is measured in kg, x is
measured in m, v is measured in m/s and t is measured in seconds.
A. Force multiplied by time
B. Force multiplied by distance
C. ½ Mv2
D. Both A and C
E. Both B and C
E
3. The best way to describe acceleration is which of the following:
A. how much ground is being covered per unit of time
B. as a quantity that gets greater with greater speed
C. the rate at which velocity changes
D. the direction of an object’s motion
C
4. Which of the following correctly relates acceleration to force?
A. It’s guaranteed to be in the direction of net force
B. It’s only in the direction of net force when something is slowing down.
C. It’s only in the direction of net force when something is speeding up.
D. None of the above because acceleration is not directional.
A
5. When velocity is negative and acceleration is negative, which of the following is
true?
A. Kinetic energy is decreasing.
B. Object is speeding up.
C. Net force is in a direction opposite of the acceleration.
D. The acceleration cannot be constant.
B
6. A baseball catcher throws a ball vertically upward and instead of catching it, lets
it fall to the ground. At what point in the ball’s vertical motion does it experience
zero displacement and NONZERO velocity at that very same time?
A. The instant it returns to its launch level.
B. Midway on the way up.
C. At the top of its trajectory.
D. When it hits the ground at the catcher’s feet.
A
Figure A: You see an image of a ball that is allowed to rise to a maximum height and
then fall to points lower than where it started. The upward launch is at t = 0. The time
listings to the left of the ball refer to instants while on the way up. The time listings
on the right refer to instants during downward motion. For the three of the positions,
the ball is at that position twice, once while on the way up and again while on the way
down. One of these three positions is launch level; the ball is there twice: once
initially and again at 1.2 s while on the way down. At no time in the diagram is there a
surface that the ball hits. After 1.4 s, it keeps moving downward,. The time at which it
reaches apex is 0.6 s. The ball is in a state of free-fall in a vacuum for all times shown.
t = 0.6 s
t = 0.4 s
t = 0.8 s
t = 0.2 s
t = 1.0 s
t=0s
t = 1.2 s
launch level
t = 1.4 s
7. During which time intervals is acceleration in the same direction as velocity?
A. From 0 to 0.6 seconds
B. From 0.6 seconds to 1.4 seconds
C. From 0 to 1.4 seconds
D. Acceleration cannot have a direction.
B
8. At what instant or intervals is the acceleration equal to zero?
A. From 0 to 0.6 seconds
B. From 0.6 seconds to 1.4 seconds
C. From 0 to 1.4 seconds
D. At the 0.6 second instant.
E. At no time shown.
E
9. Find the launch speed
A. 0 m/s
B. 6 m/s
C. 10 m/s
D. 14 m/s
10. Find the height of apex above the launch level.
A. 6 m
B. 10 m
C. 3.6 m
D. 1.8 m
E. 16 m/s
D
B
Figure B: Same time instants as before. Only now the launch is not vertical. The instant
defined by 0.6 s is still the apex. It’s the same planet as figure A. g = 10 m/s/s.
t = 0.6 s
t = 0.4 s
t = 0.2 s
launch level
t=0s
t = 1.4 s
11. If the arrow shown on the last image in the new diagram is the vertical component
of the ball’s velocity at that instant, how does the arrow’s magnitude compare to
the speed at 1.4 s in Figure A?
A. Greater than #14
B. Less than #14
C. Equal to #14
D. More information would be needed to answer this.
C
12. Consider the altitudes of the ball in Figure A compared to Figure B. Can you say
whether the altitudes shown in B are the same as A, greater or less?
A. They are clearly the same, because the time to reach apex is the same.
B. B’s are lower, because the launch angle is less than 90 degrees.
C. B’s are higher, because the launch angle is less than 90 degrees.
D. Without being given launch speeds, it is impossible to tell.
A
13. How does the launch speed of figure B compare to that of A?
A. B’s launch has greater speed.
B. B’s launch has lesser speed.
C. They are equal.
D. It depends on knowing the launch angle.
A
Figure B covers through Question 18.
The range is defined as the horizontal distance traveled upon return to launch level. The
range of the one in Figure B is 1.5 m. Determine the following:
14. The horizontal component of launch velocity.
1.25 m/s
15. The launch speed.
6.13 m/s
16. The launch angle.
78˚
17. The speed at apex.
1.25 m/s
18. The horizontal distance covered from launch to the 1.4 s instant.
1.75 m
19. If identical twins, each with mass 50 kg run up a staircase, starting at the same
place and ending at the same place, but twin A with constant speed of 4 m/s, and
twin B at constant speed 8 m/s, what is true of the work they do?
A. They do the same work, and it’s not zero.
B. Twin B does double the work of Twin A.
C. Twin A does double the work of Twin B.
D. Since speeds are constant, they both do zero work.
A
20. What is true of the power they apply?
A. They apply the same power, and it’s not zero.
B. Twin B applies double the power of Twin A.
C. Twin A applies double the power of Twin B.
D. Since speeds are constant, no power is applied.
B
21. Which of the following is true of potential energy for Twin B?
A. The difference in potential energy between top and bottom will be a
known value no matter where one puts the reference level.
B. If the bottom of the staircase is the reference level, the top of the staircase
is a place with PE equal to 3,000 J.
C. If the reference level is at a location halfway between top and bottom, the
top of the stair case is a place with 3,000 J more potential energy than the
bottom of the staircase.
D. Only Choices A and B are true
E. Choices A, B, and C are true.
Only A, unless h is 6 m, then E
22. In the diagram at the right, what is the magnitude of
the force F necessary for the acceleration of the block
F
to be 2 m/s2 to the left? The mass is 20 kg and each leftward force is 50 N.
A. 80 N
B. 120 N
C. 140 N
D. 60 N
E. none of these
D
The beam below is not to move sideways, up, or down and is not to rotate,. The length
from A to C is 8 m. There is a stand supporting things at point A and it provides a vertical
support force. The vertical line at B is a wire that goes up to a ceiling and that also
supports the beam through tension. The length from A to B is 6 m. If either of the two
supports at A or B were gone, this would rotate.
A
B
C
23. If the beam’s weight is 300 N, how strong must the support force at point A be?
A. 50 N
B. 100 N
C. 150 N
D. 200 N
E. 300 N
B
24. With the origin placed at point A, which of the following is true?
A. There are 2 torques and 2 forces.
B. There are 3 torques and 2 forces.
C. There are 3 torques and 4 forces.
D. There are 2 torques and 3 forces.
D
The system below is released so that nature can take its course. Note that there will be a
friction force of 20 N from the tabletop acting on the 5 kg
block.
5 kg
3 kg
25. What is the acceleration?
A. 0
B. 10 m/s2
C. 1.25 m/s2
26. What is the tension?
A. 20 N
B. 26 N
C. 30 N
D. 3.75 m/s2
D. 33 N
C
E. 35 N
B
1. (2) At the surface of planet Vulcan, where Spock is from, g was equal to 8 m/s2. What
was g at the following locations:
A. At a point in space above Vulcan where the distance to the center of the planet
was 40 times the planet’s radius.
(8 m/s2)/1600
B. On the surface of a planet whose radius is equal to Vulcan’s, but whose mass is
triple that of Vulcan.
24 m/s2
True/False: Half a point each
2. F: An orbiting moon in circular orbit has an acceleration vector that points tangent to
the path of the orbit.
3. F: The ancient Greeks and Roman astronomers were wrong when they said that we
see Mars move backwards in its orbit around Earth from time-to-time.
4. F: Johannes Kepler was wrong when he said that Mars and all the other planets move
in a single direction on elliptical paths around the Sun.
5. T: It is possible to use Newton’s equations to derive Kepler’s relationship that T2 is
proportional to R3 for any orbit around a common body, where T is the period of orbit
and R is the average radius.
6. T: At any instant, an orbiting moon has a velocity that is not in the direction of its
acceleration.
7. F: Two satellites orbit a common planet with the same average orbital radius. The
more massive satellite will have a greater speed than the less massive one.
8. T: The force of gravity exerted on an object by a planet gets stronger as the object
travels closer to the planet.
9. F: g is a number which is universally constant.
10. T: G is a number which is universally constant.
11. T: Newton succeeded in showing mathematically how far the moon falls (i.e. deviates
from a straight line path) in one second due to the force of gravity.
12. F: The acceleration of the moon toward earth is g which is 9.8 m/s2.
13. F: The force the earth exerts on the moon is Mg which is 9.8 m/s2 divided by
approximately 3600, and M is the mass of earth.
Problems 27 through 28:
Shown below is a forklift arm, and it is pivoted at point O. There is a hydraulic piston
shown that exerts an upward force P to lift the arm. The far right of the arm is where the
load of mass M sits. Suppose the arm is going to just barely lift the mass M, which is
2,000 kg (approximately 2 tons.) The arm itself is quite beefy and has a mass of 500 kg.
Distances from point O to various points are given and are in the diagram:
 Distance from O to the piston = 1.5 meters
 Distance from O to the center of the arm = 3.0 meters
 Distance from O to the center of the load = 6.3 meters
 Length of the arm (not counting the L bracket) = 6.0 meters
P
ARM
M
O
PISTON
(NOT the pivot point)
F1
6.3 m
1.5 m
3.0 m
6.0 m
Note: Please ignore the mass of the L bracket. Just think of it as part of the 2,000 kg
mass M.
42. If there is a missing force, where is it located?
A. 1.5 m from O
B. 3.0 m from O
C. 6.0 m from O
D. 6.3 m from O
E. There are no missing forces.
B
43. What is the value of the force P?
A. 94,000 N
B. 0
C. 141,000 N
A
(14) A 2 kg marble is moving horizontally at point B with a speed of 5 m/s. It rolls uphill,
perfectly through the circular-arc track to point A. This process causes no friction.
For the entire situation, assume that all forces are conservative. It means that no
mechanical energy is ever lost.
Point C
Point A
5 m/s
Point B
Ground
A. (4) Point B is 5 meters above the ground below. Point A is 1 meter above point B.
Using the ground as the reference level, calculate the total mechanical energy at
point B. This has multiple steps to break down and you must organize it clearly.
125 J
B. (2) Using the same reference level, what is the potential energy of the marble at
Point A?
120 J
C. What is the mechanical energy of the marble at Point A?
125 J
D. What is the kinetic energy of the marble at Point A?
5J
E. What is the speed of the marble at Point A?
2.24 m/s
F. Point C is the marble’s highest point in space during pure vertical motion from A
to C. Calculate C’s height above A.
0.25 m
G. After reaching apex at C and then falling through A and B again, the ball lands on
the ground at a point called D. Find the horizontal distance between B and D.
5m