Download Student Class ______ Date ______ MULTIPLE

Student _________________
Class ___________
MULTIPLE-CHOICE QUESTIONS
1. Scalar is to vector as
(1) speed is to velocity
(2) displacement is to distance
(3) displacement is to velocity
(4) speed is to distance
2. If a car accelerates uniformly from rest to
15 meters per second over a distance of
100. meters, the magnitude of the car’s
acceleration is
(1) 0.15 m/s2
(2) 1.1 m/s2
(3) 2.3 m/s2
(4) 6.7 m/s2
3. An object accelerates uniformly from 3.0
meters per second east to 8.0 meters per
second east in 2.0 seconds. What is the
magnitude of the acceleration of the object?
Date _____________
EXPLAIN WHY
1. Speed is a scalar, while velocity is a
vector. Therefore the answer is (1). Also,
distance is a scalar, while displacement is a
vector.
2. Formula from the Reference Table is:
vf2= vi2 + 2ad
From rest it means vi=0 m/s; vf=15 m/s and
d=100. m, therefore:
(15m/s)2 = 2(a)(100.m)
or
a=1.125 m/s2 or a=1.1 m/s2
(2 sig figs)
Therefore, the answer is (2).
3. Formula from the Reference Table is:
a=(vf-vi)/t
a = (8.0 m/s – 3.0 m/s)/(2.0s) =2.5 m/s2
(1) 2.5 m/s2
(2) 5.0 m/s2
(3) 5.5 m/s2
(4) 11 m/s2
4. A rock is dropped from a bridge. What
happens to the magnitude of the acceleration
and the speed of the rock as it falls?
[Neglect friction.]
(1) Both acceleration and speed increase.
(2) Both acceleration and speed remain the
same.
(3) Acceleration increases and speed
decreases.
(4) Acceleration remains the same and
speed increases.
5. A soccer ball kicked on a level field has
an initial vertical velocity component of
15.0 meters per second. Assuming the ball
lands at the same height from which it was
kicked, what is the total time the ball is in
the air? [Neglect friction.]
(1) 0.654 s
(3) 3.06 s
(2) 1.53 s
(4) 6.12 s
Therefore, the answer is (1).
4. The acceleration due to gravity remains
constant, ag = -9.8 m/s2 (- because it points
downward). If a rock falls it speeds up,
which means that the speed increases.
Therefore, the answer is (4).
5. For the vertical component of the
velocity, when the ball reaches the top:
vfy=0 and the time is tup=t/2.
Formula from the Reference Table is:
vf = vi+at
which for this case is written as:
vfy = viy-9.81tup
so:
0 = 15.0 m/s-(9.81m/s2)(tup)
Time to go up to the top is:
tup = 15/9.81 s = 1.53 s
so total time is:
t=2tup = 3.06 s
Therefore, the answer is (3).
Student _________________
Class ___________
6. A student is standing in an elevator that is
accelerating downward. The force that the
student exerts on the floor of the elevator
must be
(1) less than the weight of the student
when at rest
(2) greater than the weight of the student
when at rest
(3) less than the force of the floor on the
student
(4) greater than the force of the floor on the
student
7. The magnitude of the centripetal force
acting on an object traveling in a horizontal,
circular path will decrease if the
Date _____________
6. Since the student is accelerating
downward that means the net force acting on
the student has to point downward as well.
Fnet=Fg-F
For Fnet to have the same direction as Fg,
that means F (the force that the students
exerts on the floor of the elevator) has to be
less than Fg. Therefore the answer is (1).
7. Formula for the centripetal force from the
Reference Table is:
Fc =m v2/r
(1) radius of the path is increased
(2) mass of the object is increased
(3) direction of motion of the object is
reversed
(4) speed of the object is increased
Fc and r are in inverse proportion, so if the
radius r is increased then the centripetal
force Fc is decreased. Therefore the answer
is (1).
8. The centripetal force acting on the space
shuttle as it orbits Earth is equal to the
shuttle’s
(1) inertia
(3) velocity
(2) momentum
(4) weight
8. The Earth acts on the space shuttle with
the force of gravity. This force of gravity
(weight) serves as the centripetal force.
Therefore, the answer is (4).
9. As a box is pushed 30. meters across a
horizontal floor by a constant horizontal
force of 25 newtons, the kinetic energy of
the box increases by 300. joules. How much
total internal energy is produced during this
process?
(1) 150 J
(3) 450 J
(2) 250 J
(4) 750 J
9. The work done to push the box is:
W=Fd = (30. m)(25N) = 750 J
If 300. J of this work goes for the increase
of kinetic energy, the rest: 750 J-300 J=450J
is produced as internal energy. Therefore,
the answer is (3).
10. What is the power output of an electric
motor that lifts a 2.0-kilogram block 15
meters vertically in 6.0 seconds?
(1) 5.0 J
(2) 5.0 W
(3) 49 J
(4) 49 W
10. Formula for the power from the
Reference Table is:
P=Fv
Since the motor is lifting the block up,
F=Fg=mg
P=mg(v)=2.0kg(9.81m/s2)(15m/6.0s)=49W
Therefore, the answer is (4).
Student _________________
Class ___________
11. Four identical projectiles are launched
with the same initial speed, v, but at various
angles above the level ground. Which
diagram represents the initial velocity of the
projectile that will have the largest total
horizontal displacement? [Neglect air
resistance.]
Date _____________
11. The formula for the horizontal
displacement of a projectile is: Dx=vixt.
Using this formula, some people think that
the largest vix means the largest Dx, therefore
making choice (1) the answer. The trick is
that total time t a projectile is in the air
depends on the initial angle as well. As a
result, this problem is a bit more
complicated. However, to make the long
story short, it is a known fact that 45o gives
you the largest horizontal displacement,
always that is. Simple as that! Therefore the
answer is (2).
12. Two forces act concurrently on an object
on a horizontal, frictionless surface, as
shown in the diagram below.
What additional force, when applied to the
object, will establish equilibrium?
(1) 16 N toward the right
(2) 16 N toward the left
(3) 4 N toward the right
(4) 4 N toward the left
13. As shown in the diagram below, an open
box and its contents have a combined mass
of 5.0 kilograms. A horizontal force of 15
newtons is required to push the box at a
constant speed of 1.5 meters per second
across a level surface.
12. The resultant of the two given forces is:
10N-6N=4N to the right. To establish
equilibrium (Fnet=0) we need 4N to the left,
therefore the answer is (4).
Student _________________
Class ___________
The inertia of the box and its contents
increases if there is an increase in the
(1) speed of the box
(2) mass of the contents of the box
(3) magnitude of the horizontal force applied
(4) coefficient of kinetic friction between the
box and the level surface
14. Which statement describes the kinetic
energy and total mechanical energy of a
block as it is pulled at constant speed up an
incline?
(1) Kinetic energy decreases and total
mechanical energy increases.
(2) Kinetic energy decreases and total
mechanical energy remains the same.
(3) Kinetic energy remains the same and
total mechanical energy increases.
(4) Kinetic energy remains the same and
total mechanical energy remains the same.
15. Which diagram represents the electric
field lines between two small electrically
charged spheres?
Date _____________
13. This is a typical Regents question. You
are given lots of information that is simply
trying to distract you. Inertia is the ability of
an object to resist the motion. More mass
means more inertia, therefore the answer is
(2).
14. This is a tricky question. Remember that
the block is being pulled up the incline
plane, not moving by itself. This means that
we cannot say that the mechanical energy of
the system is conserved. (If you like to use a
formula, you could say that: W=ET, which
means that work that you do, equals the
increase in the total mechanical energy of
the block). If the speed is constant that
means the kinetic energy is constant. If
height is increasing that means the potential
energy is increasing as well. Therefore, the
kinetic energy remains the same, while the
total mechanical energy of the system
increases (don’t forget you are doing work
to maintain the constant kinetic energy),
therefore the answer is (3).
15. For a system of two unlike charges the
electric field lines always start from the
positive charge and point toward the
negative charge; in other words these
electric fields lines seem to “like” each
other, therefore they attract. For a system of
two like charges (two positives or two
negatives) the electric field lines seem to
“dislike” each other, therefore they repel.
For a single positive charge the electric field
line go out of the positive charge (think of a
fountain). For a single negative charge the
electric field lines go toward the negative
charge (think of a sink). Therefore, the
answer is (2).
Student _________________
Class ___________
16. Two metal spheres, A and B, possess
charges of 1.0 microcoulomb and 2.0
microcoulombs, respectively. In the diagram
below, arrow F represents the electrostatic
force exerted on sphere B by sphere A.
Date _____________
16. Both spheres appear to have positive
charges. Since the like charges repel, sphere
B will exert the same force F (action equals
reaction in magnitude but in opposite
direction) that points towards the left.
Therefore the answer is (1).
Which arrow represents the magnitude and
direction of the electrostatic force exerted on
sphere A by sphere B?
17. The diagram below represents a
positively charged particle about to enter the
electric field between two oppositely
charged parallel plates.
The electric field will deflect the particle
(1) into the page
(2) out of the page
(3) toward the top of the page
(4) toward the bottom of the page
18. What is the total amount of work
required to move a proton through a
potential difference of 100. volts?
(1) 1.60 × 10–21 J
(2) 1.60 × 10–17 J
(3) 1.00 × 102 J
(4) 6.25 × 1020 J
17. The electric field of a parallel-plates
capacitor starts from the positive charge and
goes toward the negative charge, so for the
given capacitor the electric field is
downward. A positive charge will move
along the electric field, therefore the positive
charge will be deflected toward the bottom
of the page. Some people think of it this
way: Positive will attract with the negative,
therefore the answer is (4).
18. From the Reference Table formula for
work is: W=qV. We substitute: for a proton
q=e=1.6x10-19 C and V=100. V, and we get:
W=(1.6x10-19 C)(100.V)= 1.6x10-17 J
therefore the answer is (2).
Student _________________
Class ___________
19. A baseball player runs 27.4 meters from
the batter’s box to first base, overruns first
base by 3.0 meters, and then returns to first
base. Compared to the total distance traveled
by the player, the magnitude of the player’s
total displacement from the batter’s box is
(1) 3.0 m shorter
(2) 6.0 m shorter
(3) 3.0 m longer
(4) 6.0 m longer
20. A motorboat, which has a speed of 5.0
meters per second in still water, is headed
east as it crosses a river flowing south at 3.3
meters per second. What is the magnitude of
the boat’s resultant velocity with respect to
the starting point?
(1) 3.3 m/s
(3) 6.0 m/s
(2) 5.0 m/s
(4) 8.3 m/s
21. A car traveling on a straight road at 15.0
meters per second accelerates uniformly to a
speed of 21.0 meters per second in 12.0
seconds. The total distance traveled by the
car in this 12.0-second time interval is
(1) 36.0 m
(2) 180. m
(3) 216 m
(4) 252 m
22. A 0.149-kilogram baseball, initially
moving at 15 meters per second, is brought
to rest in 0.040 second by a baseball glove
on a catcher’s hand. The magnitude of the
average force exerted on the ball by the
glove is
(1) 2.2 N
(3) 17 N
(2) 2.9 N
(4) 56 N
23. Which body is in equilibrium?
(1) a satellite moving around Earth in a
circular orbit
(2) a cart rolling down a frictionless incline
(3) an apple falling freely toward the surface
of Earth
(4) a block sliding at constant velocity
across a tabletop
Date _____________
19. If the player overruns first base by 3.0
meters and then returns to first base, that
means the displacement from the first base
is zero (player is on first base). However,
distance the player traveled (from and to
first base) is 3.0m+3.0m=6.0 m. This means
the magnitude of displacement is 27.4 m,
while the distance is 27.4m +6.0m.
Therefore, the answer is (2).
20. This question in about addition of two
perpendicular velocities. Here is the picture
for this:
Vmotorboat = 5.o m/s east
Vriver=3.3 m/s south
Vresultant
Using Pythagorean theorem we get:
vresultant 2 =(5.0m/s)2+(3.3m/s)2 =36 m2/s2
So, vresultant =6.0 m/s
Therefore, the answer is (3).
21. Formula from the Reference Table is:
d  v  t , where v 
vi  v f
is the average
2
velocity in the accelerated motion (this
formula is not on the Reference Table). The
average velocity is:
15.0m / s  21.0m / s
v
 18.0m / s
2
So, the total distance is:
d  18.0m / s 12.0s  216m
Therefore the answer is (3).
22. Formula from the Reference Table is:
Ft=p. In the meantime: p=mv=m(vfvi)= 0.149kg(0m/s-15 m/s)=-2.235 kgm/s.
Solving for F we get: F=p/t = (-2.235
kgm/s)/( 0.040s) = -56N. The negative sign
means this force stops the motion. Question
is asking for the magnitude, therefore
answer is (4).
23. Constant velocity means acceleration
a=0, so Fnet=0 (equilibrium). Answer is (4).
Student _________________
Class ___________
24. As shown in the diagram below, a
student standing on the roof of a 50.0-meterhigh building kicks a stone at a horizontal
speed of 4.00 meters per second.
Date _____________
24. This is a question about the horizontal
projectile. In the vertical direction (y-axis is
positive going up) the initial vertical
velocity of the stone is viy=0. The distance
formula from the Reference Table is:
1
d  vi t  at 2
2
For the vertical component of the motion
this formula can be written as:
1
ayt 2
2
We substitute: dy = – 50.0 m (– sign
because of the downward displacement),
viy=0 (initial vertical velocity is zero) and
ay=– 9.81 m/s2 (acceleration due to gravity
is pointing downward, hence the – sign).
d y  viy t 
 50.0m 
How much time is required for the stone to
reach the level ground below? [Neglect
friction.]
(1) 3.19 s
(2) 5.10 s
(3) 10.2 s
(4) 12.5 s
25. On the surface of Earth, a spacecraft has
a mass of 2.00 × 104 kilograms. What is the
mass of the spacecraft at a distance of one
Earth radius above Earth’s surface?
(1) 5.00 × 103 kg
(2) 2.00 × 104 kg
(3) 4.90 × 104 kg
(4) 1.96 × 105 kg
26. A student pulls a 60.-newton sled with a
force having a magnitude of 20. newtons.
What is the magnitude of the force that the
sled exerts on the student?
(1) 20. N
(2) 40. N
(3) 60. N
(4) 80. N
1
(9.81m / s 2 )t 2
2
Solving for t we get: t=3.19s, therefore the
answer is (1).
25. Mass is an intrinsic quality of matter that
never changes, regardless of the position of
the object (Earth, moon, other planets etc.).
People get confused because they think of
the weight (Fg=mg), which depends on mass
m but also depends on gravity. Weight does
change if you go somewhere else. Therefore
the answer is (2).
26. Based on Newton’s 3rd law, every action
has an equal but opposite direction. If the
student pulls the sled with 20. Newtons, the
sled exerts exactly the same magnitude of
force the other way. Therefore the answer is
(1).
Student _________________
Class ___________
27. The data table below lists the mass and
speed of four different objects.
Date _____________
27. Inertia is the ability of an object to resist
the motion. More mass means more inertia,
therefore the answer is (4). Some people get
confused about the velocity. A physical
quantity that includes both mass and
velocity is momentum: p=mv. If the
question was asking about the object with
the greatest momentum (momentum is the
amount of motion in an object), then the
answer would be (2).
Which object has the greatest inertia?
(1) A
(2) B
(3) C
(4) D
28. The diagram below shows a horizontal
12-newton force being applied to two
blocks, A and B, initially at rest on a
horizontal, frictionless surface. Block A has
a mass of 1.0 kilogram and block B has a
mass of 2.0 kilograms.
28. The mass of the system A+B is:
m=1.0kg + 2.0 kg = 3.0kg
The net force acting on the system A+B is
F=12N. Using the Newton’s 2nd law we get:
a=F/m = 12N/3.0kg = 4.0 m/s2
This is the acceleration for the whole system
A+B, so for block B as well. Therefore the
answer is (4).
29. Formula from the Reference Table is:
The magnitude of the acceleration of block
B is
(1) 6.0 m/s2
(2) 2.0 m/s2
(3) 3.0 m/s2
(4) 4.0 m/s2
29. A ball is thrown vertically upward with
an initial velocity of 29.4 meters per second.
What is the maximum height reached by the
ball? [Neglect friction.]
(1) 14.7 m
(2) 29.4 m
(3) 44.1 m
(4) 88.1 m
vf2= vi2 + 2ad
This formula written for the vertical
direction (y-axis is positive in the upward
direction) is:
vfy2= viy2 + 2aydy
Initial velocity viy=29.4 m/s; final velocity
at the top vf=0m/s (when the ball reaches the
top the height becomes maximum as well)
and ay=-9.81 m/s2, so we get:
(0m/s)2= (29.4 m/s) 2 + 2(-9.81 m/s2)dy
Solving for dy we get: dy=44.1 m, therefore
the answer is (3).
Student _________________
Class ___________
30. The diagram below represents a mass, m,
being swung clockwise at constant speed in
a horizontal circle.
Date _____________
30. Centripetal force Fc always is directed
toward the center of the circle (acceleration
in the circular motion ac also is always
directed toward the center of the circle).
Therefore the answer is (3). If the question
was asking about the direction of velocity at
the given point, the answer would be (D).
That is because the velocity in the circular
motion is always directed toward the tangent
of the circle at the given point.
At the instant shown, the centripetal force
acting on mass m is directed toward point
(1) A
(2) B
(3) C
(4) D
31. A 3.1-kilogram gun initially at rest is
free to move. When a 0.015-kilogram bullet
leaves the gun with a speed of 500. meters
per second, what is the speed of the gun?
(1) 0.0 m/s
(2) 2.4 m/s
(3) 7.5 m/s
(4) 500. m/s
32. A wound spring provides the energy to
propel a toy car across a level floor. At time
ti ,the car is moving at speed vi across the
floor and the spring is unwinding, as shown
below. At time tf, the spring has fully
unwound and the car has coasted to a stop.
31. This question is about the law of
conservation of momentum. Formula from
the Reference Table is:
pbefore = pafter
Before is before the shot (gun at rest, bullet
at rest) so pbefore =0.
After is after the shot (bullet is moving let’s
say to the right, while is gun is moving to
the left, opposite of the bullet).
0=mbvb+mgvg
0=(0.015kg)(500. m/s) +(3.1kg)(vg)
Solving for vg we get: vg= - 2.4 m/s. The
question is asking for speed, therefore the
answer is (2). The – sign in velocity means
that the gun moves in the direction opposite
of the bullet.
Student _________________
Class ___________
Date _____________
32. At ti the car has both kinetic energy (the
car is moving with speed vi) and elastic
potential energy of the spring (the spring is
unwinding). At time tf the car has stopped,
which suggests that the two form of energy
the system had at ti (kinetic energy of the car
and the elastic potential energy of the
spring) have converted to heat (which is a
form of internal energy). Therefore the
answer is (3).
Which statement best describes the
transformation of energy that occurs
between times ti and tf?
(1) Gravitational potential energy at ti is
converted to internal energy at tf.
(2) Elastic potential energy at ti is converted
to kinetic energy at tf.
(3) Both elastic potential energy and
kinetic energy at ti are converted to
internal energy at tf.
(4) Both kinetic energy and internal energy
at ti are converted to elastic potential energy
at tf.
33. What is the magnitude of the
electrostatic force between two electrons
separated by a distance of 1.00 × 10–8
meter?
(1) 2.56 × 10–22 N
(3) 2.30 × 10–12 N
(2) 2.30 × 10–20 N
(4) 1.44 × 10–1 N
33. The magnitude of the electrostatic force
between two charges q1 and q2 placed at a
distance r away from each other, is given by
Coulomb’s law, which can be found on the
Reference Table:
kq q
Fe  12 2
r
For two electrons: q1=q2=e=1.6x10-19 C
(constant on the Reference Table as well)
that are separated by a distance of
r=1.00x10-8 m we get:
Fe 
(8.99 x10 9 Nm 2 / C 2 )(1.6 x10 19 C )(1.6 x10 19 C )
(1.00 x10 8 m) 2
Fe=2.30x10-12 N
therefore the answer is (3).
Student _________________
Class ___________
34. The diagram below represents a 155newton box on a ramp. Applied force F
causes the box to slide from point A to point
B. What is the total amount of gravitational
potential energy gained by the box?
Date _____________
34. The total amount of gravitational energy
gained by the box is given by a formula in
the Reference Table:
PEg=mgh
PEg=(155N)(1.80m) = 279 J
Therefore the answer is (2).
Again, a question where extra information
is provided in order to throw you off.
(1) 28.4 J
(2) 279 J
(3) 868 J
(4) 2740 J
35. The diagram below represents the
electric field surrounding two charged
spheres, A and B.
What is the sign of the charge of each
sphere?
(1) Sphere A is positive and sphere B is
negative.
(2) Sphere A is negative and sphere B is
positive.
(3) Both spheres are positive.
(4) Both spheres are negative.
35. For a system of two unlike charges the
electric field lines always start from the
positive charge and point toward the
negative charge; in other words these
electric fields lines seem to “like” each
other, therefore they attract. Therefore the
answer is (2).
Student _________________
Class ___________
36. What is the approximate diameter of an
inflated basketball?
(1) 2 × 10–2 m
(3) 2 × 100 m
(2) 2 × 10–1 m
(4) 2 × 101 m
Date _____________
36. The approximate diameter of an inflated
basketball is 20 cm=0.2 m, therefore the
answer is (2). If you know the answer in
inches instead (approximately 8 inches),
multiply this by 2.5 to get the answer in cm
(1inch = 2.54 cm).
37. The graph below shows the relationship
between the speed and elapsed time for an
object falling freely from rest near the
surface of a planet.
37. In a graph of speed vs. time the distance
is given by the area under the graph. The
area under the graph for the first 3.0 seconds
is the area of the triangle:
1
1
A  bxh  (3.0s)(8.0m / s)  12m
2
2
Therefore the answer is (1).
What is the total distance the object falls
during the first 3.0 seconds?
(1) 12 m
(2) 24 m
(3) 44 m
(4) 72 m
38. A 75-kilogram hockey player is skating
across the ice at a speed of 6.0 meters per
second. What is the magnitude of the
average force required to stop the player in
0.65 second?
(1) 120 N
(2) 290 N
(3) 690 N
(4) 920 N
Even if you are not sure about the area under
the graph being the distance, try this:
Area is always: length  width, so:
(m/s)(s)=m, hence area is distance.
Using the same logic the slope of speed vs.
time would be: slope = (change of
y)/(change of x), so: (m/s)/s=m/s2 , hence
the slope is the acceleration.
38.
Formula from the Reference Table is: Ft=p.
In the meantime: p=mv=m(vf-vi)=
75kg(0m/s-6.0 m/s)= - 450kgm/s.
Solving for F we get:
F=p/t = (- 450kgm/s)/( 0.65s) = -692N
The negative sign means this force stops the
motion. Question is asking for the magnitude,
therefore answer is (3).
Student _________________
Class ___________
39. A child pulls a wagon at a constant
velocity along a level sidewalk. The child
does this by applying a 22-newton force to
the wagon handle, which is inclined at 35°
to the sidewalk as shown below.
Date _____________
39. The constant velocity means acceleration
a=0, so Fnet=0. If Fnet=0 that means the
component of Fnet along the direction of
motion (x-axis) will be zero as well:
Fnetx=0
The forces along the x-direction are:
Fnetx=0=Fappcos35o-Ffk
Ffk=Fappcos35o=22Ncos35o=18N
Therefore the answer is (3).
What is the magnitude of the force of
friction on the wagon?
(1) 11 N
(2) 13 N
(3) 18 N
(4) 22 N
40. The diagram below shows the
arrangement of three small spheres, A, B,
and C, having charges of 3q, q, and q,
respectively. Spheres A and C are located
distance r from sphere B.
40. The magnitude of the electrostatic force
between two charges q1 and q2 placed at a
distance r away from each other, is given by
Coulomb’s law, which can be found on the
Reference Table:
kq q
Fe  12 2
r
The magnitude of electrostatic force
between charges B and C will be:
FBC 
Compared to the magnitude of the
electrostatic force exerted by sphere B on
sphere C, the magnitude of the electrostatic
force exerted by sphere A on sphere C is
(1) the same
(2) twice as great
(3) 3/4 as great
(4) 3/2 as great
kqq kq 2
 2
r2
r
The magnitude of electrostatic force
between charges A and C will be:
FAC 
k 3qq 3kq 2 3

 FBC
4
( 2r ) 2
4r 2
Therefore the answer is (3).
Student _________________
Class ___________
41. A space probe is launched into space
from Earth’s surface. Which graph
represents the relationship between the
magnitude of the gravitational force exerted
on Earth by the space probe and the distance
between the space probe and the center of
Earth?
Date _____________
41. The magnitude of the gravitational force
between two objects with masses m1 and m2
placed at a distance r away from each other
(in this problem the two objects interacting
are the space probe and Earth), is given by
Newton’s law of universal gravitation,
which can be found on the Reference Table:
Fg 
Gm1m2
r2
1
, which is an inverse
r2
square relationship represented by a
hyperbola. Therefore the answer is (2).
As we can see Fg ~
42. Which graph represents the relationship
between the gravitational potential energy
(GPE) of an object near the surface of Earth
and its height above the surface of Earth?
42. The formula for the gravitational
potential energy can be found on the
Reference Table:
PEg=mgh
As we can see PEg~h, which is a direct
relationship represented by a straight line.
Therefore the answer is (2).
Student _________________
Class ___________
43. Two parallel metal plates are connected
to a variable source of potential difference.
When the potential difference of the source
is increased, the magnitude of the electric
field strength between the plates increases.
The diagram below shows an electron
located between the plates.
Which graph represents the relationship
between the magnitude of the electrostatic
force on the electron and the magnitude of
the electric field strength between the
plates?
Date _____________
43. The formula of the electric force acting
on charge placed in an electric field is found
on the Reference Table:
Fe=qE
For an electron Fe=eE (this is not important
for this question anyway).
As we can see Fe~E, which is a direct
relationship represented by a straight line.
Therefore the answer is (4). Again, this
question is designed to look complicated on
purpose. You should be able to see beyond
unnecessary information and get to the point
directly.
Student _________________
Class ___________
44. The diagram below represents a circuit
consisting of two resistors connected to a
source of potential difference. What is the
current through the 20.-ohm resistor?
Date _____________
44. This is a series connection of two
resistors (one loop only), so from the
Reference Table for the series connection
we get:
Req=R1+R2=10.+20. =30. 
To find the current we use Ohm’s law found
on the Reference Table:
I=I1=I2=V/Req = (120V)/( 30. ) = 4.0A
Therefore the answer is (4).
(1) 0.25 A
(2) 6.0 A
(3) 12 A
(4) 4.0 A
45. A small electric motor is used to lift a
0.50-kilogram mass at constant speed. If the
mass is lifted a vertical distance of 1.5
meters in 5.0 seconds, the average power
developed by the motor is
(1) 0.15 W
(2) 1.5 W
(3) 3.8 W
(4) 7.5 W
46. A ball is dropped from the top of a cliff.
Which graph best represents the relationship
between the ball’s total energy and elapsed
time as the ball falls to the ground? [Neglect
friction.]
45. The power formula from the Reference
Table is:
P=Fv
Substituting we get:
P=mg(d/t) = (0.50kg9.81m/s2)(1.5m/5.0s)
P= 1.5 W
Therefore the answer is (2).
46. If we neglect friction that means the total
mechanical energy of the ball remains
constant. Therefore the answer is (4).
Student _________________
Class ___________
47. A child, starting from rest at the top of a
playground slide, reaches a speed of 7.0
meters per second at the bottom of the slide.
What is the vertical height of the slide?
[Neglect friction.]
(1) 0.71 m
(2) 1.4 m
(3) 2.5 m
(4) 3.5 m
48. The graph below represents the
relationship between the current in a
metallic conductor and the potential
difference across the conductor at constant
temperature.
Date _____________
47. This question is about the law of
conservation of energy. If we neglect
friction that means the total mechanical
energy of the child remains constant. The
child starts from rest at the top (KEtop=0)
and has zero potential energy at the bottom
of the slide (PEgbottom=0). So we can write:
MEtop=MEbottom
PEgtop+KEtop=PEgbottom+KEbottom
mgh+0=0+(1/2)mv2
m’s cancel out and we get:
h=v2/(2g)=(7.0m/s)2/(29.81m/s2)=2.5m
Therefore the answer is (3).
48. The Ohm’s law is a formla you can find
on the Reference Table:
R=V/I
or, we can write: 1/R=I/V
The given graph is of current I vs potential
difference V, therefore: slope = I/V=1/R
Slope = I/V = (0.5-0)A/(1.0-0)V =
0.5A/V
R=1/slope = 1/(0.5) V/A = 2.0 
The resistance of the conductor is
(1) 1.0 
(3) 0.50 
(2) 2.0 
(4) 4.0 
Therefore the answer is (2).
Student _________________
Class ___________
49. A student throws a baseball vertically
upward and then catches it. If vertically
upward is considered to be the positive
direction, which graph best represents the
relationship between velocity and time for
the baseball? [Neglect friction.]
50. A 15.0-kilogram mass is moving at 7.50
meters per second on a horizontal,
frictionless surface. What is the total work
that must be done on the mass to increase its
speed to 11.5 meters per second?
(1) 120. J
(2) 422 J
(3) 570. J
(4) 992 J
Date _____________
49. Velocity is a vector, which means it
could be either positive or negative. If a ball
is thrown upward, the ball goes from a
positive value to zero (at the top when the
ball stops for a split of a second). On the
way back the velocity becomes negative and
the baseball speeds up until it reaches the
value it had when it was initially thrown up
(speed is always positive, while the velocity
on the way down is negative). Therefore the
answer is (4).
50. This is horizontal motion, for which the
gravitational potential energy does not
change. To solve this question we will use
the formula you have on the Reference
Table:
W=ET
Since we do not consider PEg at all, the
formula becomes:
W=KE=KEf-KEi
which we called the work-change of kinetic
energy theorem during class (this formula is
not on Reference Table as such).
W=(1/2)mvf2-(1/2)mvi2 =
=0.5(15.0kg)[(11.5m/s)2-(7.50m/s)2]
= 570. J
Therefore the answer is (3).
Student _________________
Class ___________
FREE-RESPONSE QUESTIONS
1. A 0.50-kilogram frog is at rest on the
bank surrounding a pond of water. As the
frog leaps from the bank, the magnitude of
the acceleration of the frog is 3.0 meters per
second2. Calculate the magnitude of the net
force exerted on the frog as it leaps. [Show
all work, including the equation and
substitution with units.]
Base your answers to questions 2 and 3 on
the information below.
A student and the waxed skis he is wearing
have a combined weight of 850 newtons.
The skier travels down a snow-covered hill
and then glides to the east across a snowcovered, horizontal surface.
Date _____________
SHOW ALL WORK
1.
2. As the skier glides across the horizontal
surface there is no motion along the vertical
direction, therefore the net force in the vertical
direction must be zero:
Fnety=0=FN-Fg
FN=Fg=850N
2. Determine the magnitude of the normal
force exerted by the snow on the skis as the
skier glides across the horizontal surface.
3.
3. Calculate the magnitude of the force of
friction acting on the skis as the skier glides
across the snow-covered, horizontal surface.
[Show all work, including the equation and
substitution with units.]
4.
4. Calculate the kinetic energy of a particle
with a mass of 3.34 × 10–27 kilogram and a
speed of 2.89 × 105 meters per second.
[Show all work, including the equation and
substitution with units.]
Student _________________
Class ___________
5. The graph below represents the
relationship between the work done by a
person and time.
Date _____________
5.
From the Reference Table we get the
formula:
P=W/t
The slope of the given graph is the rate at
which work is being done, which is power.
6.
Identify the physical quantity represented by
the slope of the graph.
6. The heating element in an automobile
window has a resistance of 1.2 ohms when
operated at 12 volts. Calculate the power
dissipated in the heating element. [Show all
work, including the equation and
substitution with units.]
7. Ohm’s law for an electric circuit can be
found on the Reference Table as:
Base your answers to questions 7 and 8 on
the information below.
R=V/I
= (1.50V)/(24.0A) =0.0625  =6.25x10-2 
A 3.50-meter length of wire with a cross
sectional area of 3.14 × 10–6 meter2 is at
20° Celsius. The current in the wire is
24.0 amperes when connected to a 1.50-volt
source of potential difference.
8.
7. Determine the resistance of the wire.
8. Calculate the resistivity of the wire.
[Show all work, including the equation and
substitution with units.]
Student _________________
Class ___________
Date _____________
Base your answers to questions 9 and 10 on
the information below.
In an experiment, a 0.028-kilogram rubber
stopper is attached to one end of a string. A
student whirls the stopper overhead in a
horizontal circle with a radius of 1.0 meter.
The stopper completes 10. revolutions in 10.
seconds.
9. The period of rotations is:
T=10.s/10 = 1.0s
The speed of rotations is:
v=d/t = (2r)/T
= (23.141.0m)/(1.0s) = 6.28m/s  6.3 m/s
9. Determine the speed of the whirling
stopper.
10. Calculate the magnitude of the
centripetal force on the whirling stopper.
[Show all work, including the equation and
substitution with units.]
10.
Student _________________
Class ___________
Base your answers to questions 11 through
13 on the information below.
11.
A kicked soccer ball has an initial velocity
of 25 meters per second at an angle of 40.°
above the horizontal, level ground. [Neglect
friction.]
11. Calculate the magnitude of the vertical
component of the ball’s initial velocity.
[Show all work, including the equation and
substitution with units.]
12.
12. Calculate the maximum height the ball
reaches above its initial position. [Show all
work, including the equation and
substitution with units.]
13. On the diagram on the right, sketch the
path of the ball’s flight from its initial
position at point P until it returns to level
ground.
Base your answers to questions 65 through
67 on the information and diagram below.
A 15-ohm resistor, R1, and a 30.-ohm
resistor, R2, are to be connected in parallel
between points A and B in a circuit
containing a 90.-volt battery.
13.
Date _____________
Student _________________
Class ___________
14. Complete the diagram on your right to
show the two resistors connected in parallel
between points A and B.
Date _____________
14.
15. Determine the potential difference across
resistor R1.
15. Since we have a parallel connection of
two resistors, the potential difference across
both resistors is the same as the voltage
provided by the battery, which is 90. V.
16. Calculate the current in resistor R1.
[Show all work, including the equation and
substitution with units.]
The spring in a dart launcher has a spring
constant of 140 newtons per meter. The
launcher has six power settings, 0 through 5,
with each successive setting having a spring
compression 0.020 meter beyond the
previous setting. During testing, the
launcher is aligned to the vertical, the spring
is compressed, and a dart is fired upward.
The maximum vertical displacement of the
dart in each test trial is measured. The
results of the testing are shown in the table
below.
16.
Student _________________
Class ___________
Date _____________
17.
Directions (17–20): Using the information
in the data table, construct a graph on the
grid on your right, following the directions
below.
17. Plot the data points for the dart’s
maximum vertical displacement versus
spring compression.
18. Draw the line or curve of best fit on the
grid on your right. Explain how you drew
the line of best fit.
18. The curve seems to be the best fit in this
case, not the line. We drew the curve of best
fit by going through the middle of as many
points as possible (which doesn’t necessarily
mean connecting all the dots). The curve is a
much better fit than a straight line
19.
19. Using information from your graph,
calculate the energy provided by the
compressed spring that causes the dart to
achieve a maximum vertical displacement of
3.50 meters. [Show all work, including the
equation and substitution with units.]
20. Determine the magnitude of the force, in
newtons, needed to compress the spring
0.040 meter.
20. The formula for Hooke’s law can be
found on the Reference Table:
Fs=kx
Fs=(140 N/s)(0.040m) = 5.6 N
Student _________________
Class ___________
A pop-up toy has a mass of 0.020 kilogram
and a spring constant of 150 newtons per
meter. A force is applied to the toy to
compress the spring 0.050 meter.
21.
21. Calculate the potential energy stored in
the compressed spring. [Show all work,
including the equation and substitution with
units.]
22. The toy is activated and all the
compressed spring’s potential energy is
converted to gravitational potential energy.
Calculate the maximum vertical height to
which the toy is propelled. [Show all work,
including the equation and substitution with
units.]
A horizontal force of 8.0 newtons is used to
pull a 20.-newton wooden box moving
toward the right along a horizontal, wood
surface, as shown.
22.
Date _____________
Student _________________
Class ___________
23. Calculate the magnitude of the frictional
force acting on the box. [Show all work,
including the equation and substitution with
units.]
24. Determine the magnitude of the net force
acting on the box.
Date _____________
23.
24. The motion is along the horizontal
direction, therefore the net force must be
along that direction as well.
Fnetx = F-Ff
= 8.0N – 6.0 N = 2.0N to the right
25.
25. Calculate the magnitude of the
acceleration of the box. [Show all work,
including the equation and substitution with
units.]
Suggested time: 3 hours
Important: I declare that all the work
shown above is solely based on my
individual effort. I have not consulted any
tutors, any friends and/or other materials
regarding this homework.
Total credit: 40 points
Due date: April 2, 2012
Student (please print)
___________________________
Class: Physics Honors
Student (please sign)
___________________________
Teacher: B. Fejzo
Student _________________
June 2011 (Part A)
1. 1
Class ___________
23. 4
24. 1
2. 2
25. 2
3. 1
26. 1
4. 4
27. 4
5. 3
28. 4
6. 1
29. 3
7. 1
30. 3
8. 4
31. 2
9. 3
32. 3
10. 4
33. 3
11. 2
34. 2
12. 4
35. 2
13. 2
14. 3
June 2011 (Part B1)
15. 2
36. 2
16. 1
37. 1
17. 4
38. 3
18. 2
39. 3
40. 3
June 2010 (Part A)
41. 2
19. 2
42. 2
20. 3
43. 4
21. 3
44. 4
22. 4
Date _____________
Student _________________
Class ___________
June 2010 (Part B1)
June 2008 (Part C)
45. 2
11.
Date _____________
46. 4
47. 3
48. 2
49. 4
12.
50. 3
June 2011 (Part B2)
1. Fnet = 1.5 N
2. FN = 850 N
3. Ff = 40 N
4. KE = 1.39x10-16 J
5. slope = power
13. (parabola starts at P, ends at the ground)
14.
6. P = 120 W
June 2010 (Part B2)
7. R = 6.25 x10-2 
8.  = 5.61x10-8  m
15. 90. V
9. v = 6.3 m/s
16.
10. Fc = 1.1 N
Student _________________
Class ___________
17. error: +- .3 grid space
24. Fnet = 2.0 N
18. curve of best fit (parabola)
25.
19.
20. Fs = 5.6 N
June 2007 (Part C)
21.
22.
23.
Date _____________