Module 6: Motion in Two Dimensions
Performance Expectations
Students will explore content and develop skills related to the following Performance Expectations.
Mastery can be assessed using the associated online Applying Practices activities.
Build to Performance Expectations
HS-ESS1-4. Use mathematical or computational representations to predict the motion of orbiting
objects in the solar system. (Mastery in Module 7)
Expand on Performance Expectations
HS-PS2-1. Analyze data to support the claim that Newton’s second law of motion describes the
mathematical relationship among the net force on a macroscopic object, its mass, and its
acceleration. (Mastered in Module 4)
Module 6 • Motion in Two Dimensions
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Module Planner
GO ONLINE to curate your presentations, interactive content, additional resources, and
media library, and find answer keys, materials lists, rubrics, differentiated instruction, and more.
Module Resources
Pacing
(min)
Module
Launch
Lesson
1
Lesson
2
Lesson
3
Module
Close
45
90
45
45
45
Encounter the
Phenomenon
CER
Collect Evidence
Collect Evidence
Collect Evidence
Claim, Evidence,
Make Your Claim
Reasoning
Labs and
Investigations
LL: Projectile
Motion
Go Further: Data
Analysis Lab
QI: Over the Edge;
Projectile Path
PhysicsLAB:
Centripetal Force
PhysicsLAB: Moving
Reference Frame
PhysicsLAB: On
Target
PT: Flight of a Ball
PhET Simulation:
Projectile Motion
Media and OER
Lesson Check
PhET Simulation:
Motion in 2D
Lesson Check
Lesson Check
Assess
KEY:
Revisit the
Phenomenon
Module Vocabulary
Practice
Module Test
LL: Launch Lab
QI: Quick Investigation
VI: Virtual Investigation
PT: Personal Tutor
Three-Course Model
GO ONLINE If teaching a 3-course model, go online to find associated Earth and Space
Science Content.
EARTH AND SPACE SCIENCE
Module: Earth Processes
• Lesson: Weathering and Erosion can be integrated after Lesson 3 of this Module.
• Lesson: Shoreline and Seafloor Features can be integrated after Lesson 3 of this Module.
• Lesson: Plate Tectonics can be integrated after Lesson 3 of this Module.
• Lesson: Volcanoes and Mountains can be integrated after Lesson 3 of this Module.
.
137B
Module 6 • Motion in Two Dimensions
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Module 6: Motion in Two Dimensions
Module Storyline
MODULE 6
In this module, students will seek to answer the Encounter
the Phenomenon Question “Why do thrown basketballs
travel in arcs?” The lessons in this module each provide
part of the answer to this question.
Thomas Barwick/Iconica/Getty Images
MOTION IN TWO DIMENSIONS
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PRESENTATION
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Motion in Two Dimensions
INTERACTIVE CONTENT
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INTERACTIVE CONTENT
Encounter the Phenomenon:
Motion in Two Dimensions
Lesson 1: Projectile Motion
Students will explore how the path of a project is
determined by its launch conditions, gravity, and air
resistance. This will lead them to understand why
basketballs travel in arcs.
•
Lesson 2: Circular Motion
Students will explore centripetal force, centripetal
acceleration, and circular motion, and will use
Newton’s second law for circular motion. This will lead
them to understand the differences between projectile
and circular motion, while recognizing that Newton’s
second law applies in both cases.
•
Lesson 3: Relative Velocity
Students will explore classical relative motion in one
and two dimensions. This will lead them to understand
that motion will appear different in different reference
frames.
ENCOUNTER THE PHENOMENON
Have students read the Encounter the Phenomenon question and study the module opener photo.
Then, have students watch the video either as a class,
in groups, or individually.
Ask Questions
Asking Questions and Defining Problems
Have students revisit the Driving Question Board and
review the Unit question. Then, add the Module title and
the Encounter the Phenomenon Question. Have students
identify the sticky note questions they think will be answered in this Module and place them under the Module
Encounter the Phenomenon Question. Students may also
have additional questions about the Module Phenomenon
to add to the board.
Video Supplied by BBC Worldwide Learning
CER: Motion in Two
Dimensions
137
•
Module 6 • Motion in Two Dimensions
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Module 6: Encounter the Phenomenon
Claim, Evidence, Reasoning
O
btaining, Evaluating, and
Communicating Information
Make Your Claim
A scientific claim answers a question or offers a solution
to a problem. Give students time to reflect and brainstorm,
then have each student take a clear stand and write a
claim in their CER charts.
MODULE 6
MOTION IN TWO DIMENSIONS
ENCOUNTER THE PHENOMENON
Why do thrown basketballs
travel in arcs?
GO ONLINE to play a video about
the path of a projectile.
Collect Evidence
The online Launch Lab Projectile Motion can be used
at this time as a starting point for investigation.
At the end of each reading or activity, students should
record their evidence in the class Summary Table.
Ask Questions
Do you have other questions about the phenomenon? If so, add them to the driving
question board.
CER
Claim, Evidence, Reasoning
Make Your Claim Use your
CER chart to make a claim
about why thrown basketballs
travel in arcs. Explain your
reasoning.
Collect Evidence Use the
lessons in this module to
collect evidence to support
your claim. Record your
evidence as you move
through the module.
Explain Your Reasoning You
will revisit your claim and
explain your reasoning at the
end of the module.
GO ONLINE to access your CER chart and explore resources that can
help you collect evidence.
LESSON 1: Explore & Explain:
Angled Launches
138
LESSON 2: Explore & Explain:
Centripetal Acceleration
Additional Resources
(t)Video Supplied by BBC Worldwide Learning, (bl)David Madison/Digital Vision/Getty Images, (br)Exroy/Shutterstock
After students have made a claim, they are ready to collect
evidence. Research, experimentation, or data interpretation are common sources of scientific evidence.
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Teacher Toolbox
Identifying Preconceptions
The following preconceptions will be addressed at point
of use in the lessons.
Lesson 2
• An object in circular motion can only be accelerating
if its speed is changing.
Lesson 3
• An object continues to move in a circular path if
the net force supplying a centripetal acceleration is
removed.
• An object moves in a curved path if it moves on a
second object and perpendicular to the motion of the
second object.
138
Module 6 • Encounter the Phenomenon
Video Supplied by BBC Worldwide Learning
Lesson 1
• At the top of its trajectory, a projectile has zero
acceleration and no force acting on it.
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Lesson 1: Projectile Motion
Engage
LESSON 1
Launch the Lesson Interactive Content can be assigned the
night before class as a lesson preview, during class to spark
discussion, as a resource during inquiry, or as homework.
PROJECTILE MOTION
FOCUS QUESTION
What forces affect a basketball’s trajectory?
Driving Question Board
Path of a Projectile
Have students revisit the DQB to remind themselves of the
Unit and Module questions. Have them identify the sticky
note questions they think will be answered in this lesson.
Then, have students read the Focus Question and add it
to the DQB. Students will revisit the Focus Question at the
end of the lesson.
A hopping frog, a tossed snowball, and an arrow shot from a bow all move along similar paths.
Each path rises and then falls, always curving downward along a parabolic path. An object shot
through the air is called a projectile. Its path through space is called its trajectory.
You can determine a projectile’s trajectory if you know its initial velocity. Figure 1 shows water
that is launched as a projectile horizontally and at an angle. In both cases, gravity curves the path
downward along a parabolic path.
You can draw a free-body diagram of a launched projectile and identify the forces acting on it. If
you ignore air resistance, after an initial force launches a projectile, the only force on it as it
moves through the air is gravity. Gravity causes the object to curve downward.
Science Journal Remind students to keep records of
their investigations in their Science Journals. Additionally,
be sure that each reading or activity is added to the
class Summary table.
Three-Dimensional Thinking The activities called out
in the Student Edition will allow students to practice threedimensional thinking. Worksheets for these activities can
be found online.
I
Disciplinary Core Ideas
Crosscutting Concepts
Science & Engineering Practices
CCC
INVESTIGATE
GO ONLINE to find these activities and more resources.
PhysicsLAB: On Target
Plan and carry out an investigation to determine what factors affect projectile motion.
Revisit the Encounter the Phenomenon Question
What information from this lesson can help you answer the Unit and Module questions?
Lesson 1 • Projectile Motion
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PRESENTATION
Teacher Presentation:
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Chapter: 06
I
3D THINKING
DC
DC
CCC
COLLECT EVIDENCE
Use your Science Journal to
record the evidence you collect as
you complete the readings and
activities in this lesson.
SEP
Figure 1 A projectile launched horizontally immediately curves downward, but if it is launched upward at an
angle, it rises and then falls, always curving downward.
SEP
Hutchings Photography/Digital Light Source
Explore and Explain
07:22PM
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INTERACTIVE CONTENT
Launch the Lesson:
Projectile Motion
Visual Literacy
Have students study Figure 1.
Draw students’ attention to the difference between the
shapes of the paths in the two photos. Emphasize that
the initial direction is the same as the direction of the
force that launches a projectile, but the gravitational force
changes the shape of the path. Ask students to name other
examples of both types of motion.
EL Support
ELD PII.11/12.2a
Guide students in applying knowledge of familiar language
resources for referring to make texts more cohesive.
IMPLEMENTATION OPTIONS
Jupiterimages/Jupiterfootage/Getty Images
Presentation: Teacher-Facilitated Pathway
Use the Teacher Presentation to support classroom
instruction and spark discourse. Obtain data to
inform your instruction by assigning the Interactive
Content, Additional Resources, and Assessment.
Interactive Content: Student-Led Pathway
Students can use the online Interactive Content,
along with the Student Edition, Science Notebook,
projects, and labs, to collect evidence to support
their claim. They can record their evidence in their
Science Journals and the class Summary Table.
EMERGING LEVEL Draw students’ attention to it in the
caption of Figure 1. Support students in recognizing
that it refers to projectile. Demonstrate moving your
finger from it back to projectile.
EXPANDING LEVEL Draw students’ attention to it in the
caption of Figure 1. Ask: Do you know what it is? Guide
students to recognize what it refers to. projectile With
students, restate the sentence, replacing it with what it
refers to. Support their understanding that the pronoun
it makes the sentence easier to read.
BRIDGING LEVEL Draw students’ attention to it in the
caption of Figure 1. Elicit what it refers to. projectile
Have students look for other uses of it or its on the page
and discuss what they refer to.
Lesson 1 • Projectile Motion
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Lesson 1: Projectile Motion
Quick Demo
Initial Horizontal Velocity or Not
Materials two identical marbles, table
Procedure To show the independent nature of simultaneous horizontal and vertical motions, hold one marble next
to the edge of a tabletop and place an identical marble
on the edge of the table. Drop the first marble and at the
same time give the second marble a horizontal push.
Repeat the experiment with different horizontal velocities
for the first marble. For all experiments, students should
hear and see the two marbles both hit the floor at the
same time.
Est. time: 5 min
Reinforcement
Independence of Velocities Emphasize again that a projectile’s horizontal motion is constant in the absence of air
resistance. The projectile’s vertical velocity changes as the
force of gravity accelerates the projectile.
Caption Question Fig. 2: –9.8 m/s
Get It?
Independence of Motion in
Two Dimensions
Think about two softball players warming up for a game, tossing
high fly balls back and forth. What does the path of the ball through
the air look like? Because the ball is a projectile, it follows a parabolic path.
Imagine you are standing directly behind one of the players and you
are watching the softball as it is being tossed. What would the
motion of the ball look like? You would see it go up and back down,
just like any object that is tossed straight up in the air.
If you were watching the softball from a hot-air balloon high above
the field, what motion would you see? You would see the ball move
from one player to the other at a constant speed, just like any object
that is given an initial horizontal velocity, such as a hockey puck
sliding across ice. The motion of projectiles is a combination of
these two motions.
Get It?
Illustrate Draw sketches of a drone’s-eye view of the
person at the beginning of the module shooting a basketball. The sketches should show the view from directly
behind the person, from directly above the court, and
from the sideline.
Figure 2 The ball on the left was dropped with no
initial velocity. The ball on the right was given an initial
horizontal velocity. The balls have the same vertical
motion as they fall.
Why do projectiles behave in this way? After a softball leaves a
Identify What is the vertical velocity of the balls after
falling for 1 s?
player’s hand, what forces are exerted on the ball? If you ignore air
resistance, there are no contact forces on the ball. There is only the
field force of gravity in the downward direction. How does this affect
the ball’s motion? Gravity causes the ball to have a downward
acceleration.
Comparing motion diagrams
The trajectories of two balls
are shown in Figure 2. The red ball was dropped, and the blue ball
was given an initial horizontal velocity of 2.0 m/s. What is similar
about the two paths?
Look at their vertical positions. The horizontal lines indicate equal
vertical distances. At each moment that a picture was taken, the
heights of the two balls were the same. Because the change in vertical
position was the same for both, their average vertical velocities during
each interval were also the same. The increasingly large distance
traveled vertically by the balls, from one time interval to the next,
shows that they were accelerating downward due to the force of
gravity.
McGraw-Hill Education
Quick Practice
Developing and Using Models Throughout the lesson,
have students use the models they developed in previous
modules to analyze constant-velocity motion and constant-acceleration motion to analyze a ­projectile’s horizontal and vertical motions.
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INTERACTIVE CONTENT
INTERACTIVE CONTENT
Explore and Explain: What is
a projectile?
Explore and Explain:
Independence of Motion in
Two Dimensions
Differentiated Instruction
AL Struggling Students Provide students with a copy of
a picture similar to Figure 2. Using a ruler, have students
draw a horizontal line between each side-by-side pair
of objects. Using a square, have students verify that the
line is straight with respect to the left or right edge of the
image. Reinforce for students that the vertical position of
the two objects is the same at each interval and, therefore, the objects are falling at the same rate regardless of
the horizontal component of motion.
140
Module 6 • Motion in Two Dimensions
(l)Richard Hutchings/Digital Light Source, (r)McGraw-Hill Education
The sketch from behind the person is a straight vertical
line. The sketch from above the court is a straight horizontal line. The sketch from the sideline is a parabola.
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Get It?
The horizontal and vertical velocities of the dropped
object are independent, so the vertical velocity does not
depend on the initial horizontal velocity.
Notice that the horizontal motion of the launched ball does not affect its vertical motion.
A projectile launched horizontally has initial horizontal velocity, but it has no initial vertical
velocity. Therefore, its vertical motion is like that of an object dropped from rest. Just like the red
ball, the blue ball has a downward velocity that increases regularly because of the acceleration
due to gravity.
Get It?
Activity
Explain why a dropped object has the same vertical velocity as an object launched
horizontally.
Constant Horizontal Velocity Place a large bottle with
about a 5-cm-wide mouth on the floor where students
can walk past it. Give each student a ball that will easily fit
through the mouth of the bottle. Have students walk past
the bottle at a constant speed, holding the ball at their
side, and drop the ball into the bottle. After the activity, ask
students at what point they should drop the ball so that it
falls into the container. Students should ­release the ball
before it is above the opening.
Horizontally Launched Projectiles
Imagine a person standing near the edge of a cliff and kicking a pebble horizontally. Like all
horizontally launched projectiles, the pebble will have an initial horizontal velocity and no initial
vertical velocity. What will happen to the pebble as it falls from the cliff?
Separate motion diagrams
Recall that the horizontal motion of a projectile does not
affect its vertical motion. It is therefore easier to analyze the horizontal motion and the vertical
motion separately. Separate motion diagrams for the x-components and y-components of a
horizontally launched projectile, such as a pebble kicked off a cliff, are shown on the left in
Figure 3.
Horizontal motion Notice the horizontal vectors in the diagram on the left. Each of the
velocity vectors is the same length, which indicates that the object’s horizontal velocity is not
changing. The pebble is not accelerating horizontally. This constant velocity in the horizontal
direction is exactly what should be expected, because after the initial kick, there is no horizontal
force acting on the pebble. (In reality, the pebble’s speed would decrease slightly because of air
resistance, but remember that we are ignoring air resistance in this module.)
Get It?
Get It?
Explain why the horizontal motion of a projectile is constant.
If we ignore air resistance, there are no forces acting
in the horizontal direction. Therefore, there is no acceleration in the horizontal direction, and the horizontal
velocity is constant.
Vectors in Two Dimensions
Begin
Begin
vx
ax = 0
vy
ay
+y
Caption Question Fig. 3: –9.8 m/s2
+x
Figure 3 To describe the motion of a horizontally launched projectile, the x- and y-components can be treated
independently. The resultant vectors of the projectile are tangent to a parabola.
Decide What is the value of ay?
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INTERACTIVE CONTENT
Sutham/Shutterstock
Explore and Explain:
Horizontally Launched
Projectiles
Lesson 1 • Projectile Motion
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Lesson 1: Projectile Motion
Reinforcement
Projectiles, Kinematics, and Vectors Have students
make lists of the concepts from previous modules needed to ­analyze projectile motion. Such lists should include
velocity, acceleration, free fall, vector ­resolution, and the
­independence of perpendicular vectors.
Get It?
There is a net force in the vertical direction, so the
projectile accelerates in the vertical direction and the
vertical component of the velocity vector changes.
There is no net force in the horizontal direction, so the
projectile cannot accelerate in the horizontal direction
and the horizontal component of the velocity vector
cannot change.
Patterns
Students can make use of technology by recording video
of each launch to view the trajectory. Encourage students
to conduct launches in front of a grid to make evidence
of the pattern of parabolic motion easier to identify.
What patterns do you observe in the data presented in
the output?
What does the pattern of data you see allow you to
­conclude from the experiment?
Vertical motion Now look at the vertical velocity vectors in the diagram on the left. Each
velocity vector has a slightly longer length than the one above it. The changing length shows that
the object’s vertical velocity is increasing and the object is accelerating downward. Again, this is
what should be expected, because in this case the force of gravity is acting on the pebble.
Get It?
Describe why the vertical velocity vectors change length but the horizontal velocity
vectors do not for a projectile.
Parabolic path
When the x- and y-components of the object’s motion are treated
independently, each path is a straight line. The diagram on the right in Figure 3 shows the actual
parabolic path. The horizontal and vertical components at each moment are added to form the
total velocity vector at that moment. You can see how the combination of constant horizontal
velocity and uniform vertical acceleration produces a trajectory that has a parabolic shape.
PROBLEM-SOLVING STRATEGIES
Motion in Two Dimensions
When solving projectile problems, use the following strategies.
1. Draw a motion diagram with vectors for the projectile at its initial position
Motion Equations
and its final position. If the projectile is launched at an angle, also show its
maximum height and the initial angle.
Horizontal (constant speed)
x f = v tf + x i
2. Consider vertical and horizontal motion independently. List known and
unknown variables.
Vertical (constant acceleration)
3. For horizontal motion, the acceleration is ax = 0.0 m/s2. If the projectile is
vf = v i + a t f
launched at an angle, its initial vertical velocity and its vertical velocity when
1
xf = xi + vitf + __
at 2
it falls back to that same height have the same magnitude but different
2 f
direction: v yi = -v yf.
vf2 = vi2 + 2a(xf – xi)
4. For vertical motion, ay = -9.8 m/s2 (if you choose up as positive). If the
projectile is launched at an angle, its vertical velocity at its highest point
is zero: v y, max = 0.0 m/s.
5. Choose the motion equations that will enable you to find the unknown variables. Apply them to vertical and
horizontal motion separately. Remember that time is the same for horizontal and vertical motion. Solving for
time in one of the dimensions identifies the time for the other dimension.
6. Sometimes it is useful to apply the motion equations to part of the projectile’s path. You can choose any initial
and final points to use in the equations.
CROSSCUTTING CONCEPTS
Patterns Gather empirical evidence to identify the pattern of motion that
projectiles follow. Launch an object at various angles and record the motion it
follows. Create a presentation that illustrates your findings.
142 Module 6 • Motion in Two Dimensions
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Quick Practice
Developing and Using Models To help students with
visual impairments get an idea of the shape of a trajectory,
attach pieces of string at equal intervals along a meterstick.
The intervals represent equal time intervals. The string
lengths should have a ratio of 1:4:9:16:25, and so on. The
lengths of the pieces of string represent vertical distances
traveled. Attach a small ­washer to the end of each string.
Students can then “feel” what the trajectory “looks” like. By
holding the stick at different angles, students can simulate
the trajectories of projectiles with different launch angles.
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ADDITIONAL RESOURCE
Virtual Investigation:
Soccer Kick
LAB
Reteach
Independence of Velocities Give students the horizontal
and vertical components of the initial velocity of a projectile. Ask them to compute the velocity components at
later times in the projectile’s flight. Stress that the vertical
­velocity is continuously changing whereas the horizontal
velocity remains constant.
142
Module 6 • Motion in Two Dimensions
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PRACTICE Problems
1. a. 4.0 s
b. 2.0×101 m
c. vx = 5.0 m/s, vy = −39 m/s
2. 1.1 m/s
3. 0.60 cm
EXAMPLE Problem 1
A SLIDING PLATE You are preparing breakfast and slide a plate on the countertop. Unfortunately, you
slide it too fast, and it flies off the end of the countertop. If the countertop is 1.05 m above the floor
and the plate leaves the top at 0.74 m/s, how long does it take to fall, and how far from the end of the
counter does it land?
1 ANALYZE AND SKETCH THE PROBLEM
Draw horizontal and vertical motion diagrams. Choose the coordinate system
so that the origin is at the top of the countertop. Choose the positive x direction
in the direction of horizontal velocity and the positive y direction up.
Known
xi = y i = 0 m
ax = 0 m/s
v yi = 0 m/s
yf = −1.05 m
v xi = 0.74 m/s
Begin
vx
ax = 0
Unknown
t=?
2
ay = −9.8 m/s2
xf = ?
ay
vy
2 SOLVE FOR THE UNKNOWN
Use the equation of motion in the y direction to find the time of fall.
+y
Clarify a Preconception
+x
Apex Acceleration Activity Have students draw a freebody diagram of a projectile at the apex of its trajectory.
Some students might believe that a projectile’s acceleration at the top of its trajectory is zero and will indicate no
force acting on the projectile at that time. Point out that
a single force acts on the projectile; namely, the force of
gravity. Because gravity acts downward, the projectile must
always have a downward acceleration until another force
acts to counter gravity.
1 a t2
yf = yi + __
2 y
√
_______
t=
√
2(y
- y i)
f
______
ay
Rearrange the equation to solve for time.
2(-1.05 m - 0 m)
____________
= 0.46 s
-9.8 m/s2
Substitute yf = −1.05 m, yi = 0 m, ay = −9.8 m/s2.
_____________
=
Use the equation of motion in the x direction to find where the plate hits the floor.
xf = vxt = (0.74 m/s to the right)(0.46 s) = 0.34 m to the right of the counter
3 EVALUATE THE ANSWER
• Are the units correct? Time is measured in seconds. Position in measured in meters.
• Do the signs make sense? Both are positive. The position sign agrees with the coordinate
choice.
• Are the magnitudes realistic? A fall of about a meter takes about 0.5 s. During this time, the
horizontal displacement of the plate would be about 0.5 s × 0.74 m/s.
PRACTICE Problems
ADDITIONAL PRACTICE
1. You throw a stone horizontally at a speed of
5.0 m/s from the top of a cliff that is 78.4 m high.
a. How long does it take the stone to reach the
bottom of the cliff?
b. How far from the base of the cliff does the stone
hit the ground?
c. What are the horizontal and vertical components
of the stone’s velocity just before it hits the
ground?
2. Lucy and her friend are working at an assembly
plant making wooden toy giraffes. At the end of
the line, the giraffes go horizontally off the edge of
a conveyor belt and fall into a box below. If the box
is 0.60 m below the level of the conveyor belt and
0.40 m away from it, what must be the horizontal
velocity of giraffes as they leave the conveyor belt?
3. CHALLENGE You are visiting a friend
from elementary school who now lives in
a small town. One local amusement is the
ice-cream parlor, where Stan, the shortorder cook, slides his completed ice-cream
sundaes down the counter at a constant
speed of 2.0 m/s to the servers. (The
counter is kept very well polished for this
purpose.) If the servers catch the sundaes
7.0 cm from the edge of the counter,
how far do they fall from the edge of the
counter to the point at which the servers
catch them?
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ADDITIONAL RESOURCE
PhET: Projectile Motion
Lesson 1 • Projectile Motion
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Lesson 1: Projectile Motion
vx
ax = 0
ay
vy
+y
+x
Get It?
Its vertical velocity will be 1.2 m/s down. The ball rises
and falls the same amount of time, so it slows down
from 1.2 m/s up to 0 m/s, then speeds up from 0 m/s to
1.2 m/s down.
vx
ax = 0
Get It?
at the top
ay
ay
Maximum
height
vy
+y
Critical Thinking
Projectile Motion Explain to students that three objects
of equal mass are shot straight upward with the same
initial velocity. One of the objects is on the Moon, one is
on Earth’s surface, and one is on Earth but submerged
in a pool of water so deep that the projectile does not
leave the water. Have students compare and contrast the
shapes of the objects’ trajectories. Each trajectory will have
a range of zero. The trajectories, ranked from highest to
lowest, are Moon, Earth, and underwater.
Parabolic Trajectory
Vertical and Horizontal Components
θ
Begin
+x
Range
Figure 4 When a projectile is launched at an upward angle, its parabolic path is upward and then downward. The
up-and-down motion is clearly represented in the vertical component of the vector diagram.
a
Angled Launches
y
When a projectile is launched at an angle, the initial velocity has a vertical component as well as a
Maximum
horizontal component.
If the object is launched upward, like a ball tossed straight up in the air, it
height
rises with slowing speed, reaches the top of its path where its speed is momentarily zero, and
θ with increasing speed.
descends
Begin
Separate motion diagrams The left diagram of Figure 4 shows the separate verticaland horizontal-motion
diagrams for the trajectory. In the coordinate system, the x-axis is
Range
horizontal and the y-axis is vertical. Note the symmetry. At each point in the vertical direction,
the velocity of the object as it is moving upward has the same magnitude as when it is moving
downward. The only difference is that the directions of the two velocities are opposite. When
solving problems, it is sometimes useful to consider symmetry to determine unknown
quantities.
Get It?
Analyze When a ball on the ground is kicked, the ball moves with an initial vertical
velocity of 1.2 m/s up. When the ball hits the ground, what will its vertical velocity be?
Explain your reasoning.
Parabolic path
The right diagram of Figure 4 defines two quantities associated with the
trajectory. One is the maximum height, which is the height of the projectile when the vertical
velocity is zero and the projectile has only its horizontal-velocity component. The other quantity
depicted is the range (R), which is the horizontal distance the projectile travels when the initial
and final heights are the same. Not shown is the flight time, which is how much time the
projectile is in the air. For football punts, flight time often is called hang time.
Get It?
Describe At what point of a projectile’s trajectory is its vertical velocity zero?
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INTERACTIVE CONTENT
Explore and Explain: Angled
Launches
BL Physics Challenge Activity Have students watch a
video of a football kickoff. Have them use video analysis
software to measure the time of flight and the distance
that the ball is kicked. From these measurements, have
them calculate the initial horizontal and vertical velocities,
the initial velocity (magnitude and angle), and the maximum height. Have students select another sport in which
projectile motion is evident, such as volleyball or soccer.
They should analyze the motion of the projectile and
compare the motion of the two objects.
144
Module 6 • Motion in Two Dimensions
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Differentiated Instruction
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PRACTICE Problems
4. a. 2.8 s
b. 9.3 m
c. 65 m
5.hang time = 4.8 s; distance = 65 m;
maximum height = 28 m
6. 32 m/s at 83° below horizontal
EXAMPLE Problem 2
THE FLIGHT OF A BALL A ball is launched at 4.5 m/s at 66° above the horizontal.
It starts and lands at the same distance from the ground. What are the
maximum height above its launch level and the flight time of the ball?
1 ANALYZE AND SKETCH THE PROBLEM
• Establish a coordinate system with the initial position of the ball at the origin.
• Show the positions of the ball at the beginning, at the maximum
height, and at the end of the flight. Show the direction of Fnet.
• Draw a motion diagram showing v and a.
Known
yi = 0.0 m
θi = 66°
vi = 4.5 m/s
ay = -9.8 m/s2
v y, max = 0.0 m/s
+y
ymax
vi
Begin
Unknown
ymax = ?
Fg =Fnet
vyi = vi(sin θi)
= (4.5 m/s)(sin 66°) = 4.1 m/s
Use symmetry to find the y-component of vf.
+x
v
vyi
t=?
2 SOLVE FOR THE UNKNOWN
Find the y-component of vi.
R
vi
a
vxi
ADDITIONAL IN-CLASS Example
Substitute vi = 4.5 m/s, θi = 66°.
Use with Example Problem 2.
v yf = -vyi = -4.1 m/s
Solve for the maximum height.
v y, max 2 = v yi 2 + 2ay( ymax - yi)
(0.0 m/s)2 = v yi 2 + 2ay( ymax - 0.0 m)
Problem Courtney kicks a soccer ball that is at rest on
level ground and gives it an initial velocity of 7.8 m/s at an
angle of 32° above the ground. Assume that forces due to
air drag on the ball are insignificant.
v yi2
ymax = - ___
2a
y
(4.1 m/s)
2
= - __________
= 0.86 m
2(-9.8 m/s2)
Substitute vyi = -4.1 m/s; ay = -9.8 m/s2.
Solve for the time to return to the launching height.
vyf = vyi + ayt
vyf - vyi
t = _____
a
y
-4.1 m/s - 4.1 m/s
= ____________
= 0.84 s
-9.8 m/s2
a. How long will the ball be in the air?
b. How high will the ball go?
c. What will be its range?
Substitute vyf = -4.1 m/s; vyi = 4.1 m/s; ay = -9.8 m/s2.
3 EVALUATE THE ANSWER
• Are the magnitudes realistic? For an object that rises less than 1 m, a time of less than 1 s is reasonable.
PRACTICE Problems
ADDITIONAL PRACTICE
Trajectory
Response
​ ​= ​v i​​ sin θ = (7.8 m/s) sin 32°
​v yi
60.0°
y
4. A player kicks a football from ground level with an
initial velocity of 27.0 m/s, 30.0° above the horizontal,
as shown in Figure 5. Find each of the following.
Assume that forces from the air on the ball are
negligible.
a. the ball’s hang time
b. the ball’s maximum height
c. the horizontal distance the ball travels before
hitting the ground
5. The player in the previous problem then kicks the ball
with the same speed but at 60.0° from the horizontal.
What is the ball’s hang time, horizontal distance
traveled, and maximum height?
30.0°
x
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​v ​yi​= 4.1 m/s
Figure 5
6. CHALLENGE A rock is thrown from a 50.0-m-high
cliff with an initial velocity of 7.0 m/s at an angle of
53.0° above the horizontal. Find its velocity when it
hits the ground below.
03:08PM
v​ xi​ ​= 6.6 m/s
a. At landing, y = 0
​ ​t - _
​  1 ​g​t ​2​
0 = 0 + ​v yi
2
t = 2​v yi​ ​/g = 2(4.1 m/s)/(9.8 m/​s ​2​) = 0.84 s
2
​ ​​ _
b.​
y ​max​ = ​v yi
​  1 t​ ​ - _
​  1 ​g ​​ _
​  1 t​ ​ ​ ​
(2 )
2
(2 )
= (4.1 m/s)(0.42 s) - _
​  1 ​(9.8 m/​s ​2​)(0.42 s​) ​2​
2
= 0.86 m
c. R = ​v ​xi​t = (6.6 m/s)(0.84 s) = 5.5 m
Visual Literacy
Have students study Figure 6 on page 146.
Have students compare the paths of the water in the two
lower photos with the path of the water in the upper ­photo.
Remind them that it is the net force on an object that
changes the object’s velocity.
Elaborate
Return to the DQB and have students determine what
questions they can answer. At this point, they should be
able to answer the Focus Question.
Lesson 1 • Projectile Motion
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Lesson 1: Projectile Motion
No Effect from Air
Evaluate
No Effect from Air
Formative Assessment Check
The effect of forces due to air has been ignored so far in this chapter, but
think about why a kite stays in the air or why a parachute helps a skydiver
fall safely to the ground. Forces from the air can significantly change the
motion of an object.
Ask: If you could switch gravity off just at the moment a
punter kicks a football, how will the football’s trajectory
change? Gravity acts downward, so only the vertical component of the trajectory will change. Because no forces
(neglecting air resistance) act on the football in the vertical
direction, it will not accelerate vertically and so continue
forever upward with its initial vertical velocity. Its horizontal
motion is not affected by gravity, so the football will move
horizontally as it does when gravity is on.
What if the direction of wind is at an angle relative to a moving object? The
horizontal component of the wind affects only the horizontal motion of an
object. The vertical component of the wind affects only the vertical motion
of the object. In the case of the water, for example, a strong updraft could
decrease the downward speed of the water.
Fg
Fg
9. 27 m
Fg
Force of Air That Decreases Velocity
Figure 6 Forces from air can increase or
decrease the velocity of a moving object.
9. Projectile Motion A ball is thrown out a
window 28 m high at 15.0 m/s and 20.0° below
the horizontal. How far does the ball move
horizontally before it hits the ground?
10. Projectile Motion What is the maximum height
of a softball thrown at 11.0 m/s and 50.0° above
the horizontal?
11. Critical Thinking Suppose an object is thrown
with the same initial velocity and direction on
Earth and on the Moon, where the acceleration
due to gravity is one-sixth its value on Earth.
How will vertical velocity, time of flight, maximum height, and horizontal distance change?
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Motion Diagrams
On the table
In the air
Chapter: 06
INTERACTIVE CONTENT
a
Explore and Explain: Forces
from Air
03:08PM
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ADDITIONAL RESOURCE
Vocabulary Flashcards:
Projectile Motion
Fg
C06-01A-865893_A
10. 3.6 m
11.no change in horizontal velocity, time of flight is longer
on the Moon, maximum height is larger on the Moon,
horizontal distance is larger on the Moon
Formative Assessment: Lesson Check
GO ONLINE You might want to assign from the
Additional Resources the pre-made Lesson Check based
on key concepts and disciplinary core ideas, or you can
customize your own using the customization tool.
146
Force of Air that Decreases Velocity
Richard Hutchings/Digital Light Source
8. Free-Body Diagram An ice cube slides
without friction across a table at a constant
velocity. It slides off the table and lands on the
floor. Draw free-body and motion diagrams of
the ice cube at two points on the table and at
two points in the air.
146
a=0
Force
Velocity
Forceof
of Air
Air that
That Decreases
Increases Velocity
Module 6 • Motion in Two Dimensions
ADDITIONAL RESOURCE
Lesson Check: Projectile
Motion
ADDITIONAL RESOURCE
Inspire Physics LearnSmart
Richard Hutchings/Digital Light Source
FN
No Effect from Air
Force of Air that Increases Velocity
Check Your Progress
7. Initial Velocity Two baseballs are pitched
horizontally from the same height but at
different speeds. The faster ball crosses home
plate within the strike zone, but the slower ball
is below the batter’s knees. Why do the balls
pass the batter at different heights?
7. The faster ball is in the air a shorter time and thus
gains a smaller vertical velocity.
FN
Force of Air that Increases Velocity
Force of Air that Decreases Velocity
The effects shown in Figure 6 occur because the air is moving enough to
significantly change the motion of the water. Even air that is not moving,
however, can have a significant effect on some moving objects. A piece of
paper held horizontally and dropped, for example, falls slowly because of air
resistance. The air resistance increases as the surface area of the object that
faces the moving air increases.
Check Your Progress
Free-Body Diagrams
On the table
In the air
Force of Air that Increases Velocity
What happens if there is wind? Moving air can change the motion of a
projectile. Consider the three cases shown in Figure 6. In the top photo, water
is flowing from the hose pipe with almost no effect from air. In the middle
photo, wind is blowing in the same direction as the water’s initial movement.
The path of the water changes because the air exerts a force on the water in
the same direction as its motion. The horizontal distance the water travels
increases because the force increases the water’s horizontal speed. The
direction of the wind changes in the bottom photo. The horizontal distance
the water travels decreases because the air exerts a force in the direction
opposite the water’s motion.
Remediation Throw a ball vertically upward and
ask students the following questions. How does the
velocity change as the ball rises? The upward velocity
decreases. What is the velocity of the ball at its highest
point? zero How does the velocity change as the
ball falls? The downward velocity increases. What is
the acceleration of the ball as it rises? The free-fall
acceleration is 9.8 m/s2 downward. What is the ball’s
acceleration at its highest point? 9.8 m/s2 downward
What is the ball’s acceleration as the ball falls? 9.8 m/s2
downward
8.
No Effect from Air
Forces from Air
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Lesson 2: Circular Motion
Forces and Motion
LESSON 2
PS2.A Newton’s second law accurately predicts
changes in the motion of macroscopic objects.
CIRCULAR MOTION
FOCUS QUESTION
*Bold font indicates the part of the DCI covered in this
lesson.
What force causes an object to move in a circle?
Describing Circular Motion
Are objects moving in a circle at a constant speed, such as a fixed horse on a carousel, accelerating? At first, you might think they are not because their speeds do not change. But remember
that acceleration is related to the change in velocity, not just the change in speed. Because their
directions are changing, the objects must be accelerating.
Engage
Uniform circular motion is the movement of an object at a constant speed around a circle with a
fixed radius. The position of an object in uniform circular motion, relative to the center of the
circle, is given by the position vector r. Remember that a position vector is a displacement vector
with its tail at the origin. Two position vectors, r1 and r2, at the beginning and end of a time
interval are shown on the left in Figure 7.
Position and
Velocity Vectors
v1
Displacement Vector
r1
r2
Launch the Lesson Interactive Content can be assigned
the night before class as a lesson preview, during class
to spark discussion, as a resource during inquiry, or as
homework.
v2
∆r
r1
Driving Question Board
r2
Have students revisit the DQB to remind themselves of the
Unit and Module questions. Have them identify the sticky
note questions they think will be answered in this lesson.
Then, have students read the Focus Question and add it
to the DQB. Students will revisit the Focus Question at the
end of the lesson.
Figure 7 For an object in uniform circular motion, the velocity is tangent to the circle. It is in the same direction as
the displacement.
I
CCC
3D THINKING
Disciplinary Core Ideas
COLLECT EVIDENCE
Use your Science Journal to
record the evidence you collect as
you complete the readings and
activities in this lesson.
Crosscutting Concepts
Science & Engineering Practices
INVESTIGATE
GO ONLINE to find these activities and more resources.
PhysicsLAB: Centripetal Force
Analyze data about the relationship between centripetal force and centripetal acceleration.
Review the News
Obtain information from a current news story about circular motion. Evaluate your source and
communicate your findings to your class.
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I
INTERACTIVE CONTENT
Launch the Lesson: Circular
Motion
Science Journal Remind students to keep records of
their investigations in their Science Journals. Additionally,
be sure that each reading or activity is added to the
class Summary table.
Three-Dimensional Thinking The activities called out
in the Student Edition will allow students to practice threedimensional thinking. Worksheets for these activities can
be found online.
DC
PRESENTATION
Teacher Presentation:
Circular Motion
Chapter: 06
03:08PM
Explore and Explain
SEP
DC
SEP
Analyze How can you tell from the diagram that the motion is uniform?
CCC
Activity
IMPLEMENTATION OPTIONS
Presentation: Teacher-Facilitated Pathway
Use the Teacher Presentation to support classroom
instruction and spark discourse. Obtain data to
inform your instruction by assigning the Interactive
Content, Additional Resources, and Assessment.
flun/123RF
Interactive Content: Student-Led Pathway
Students can use the online Interactive Content,
along with the Student Edition, Science Notebook,
projects, and labs, to collect evidence to support
their claim. They can record their evidence in their
Science Journals and the class Summary Table.
Centripetal Force Draw a large circle (with a diameter of
at least 50 cm) on a large piece of paper. Have students
observe as you have one or more volunteers try to make
a ball roll along the circumference of the circle by only
tapping the ball. Have students identify each tap as a force.
Ask them what they noticed about the direction of each
force. Each force should be directed toward the center of
the circle.
Tie to Prior Knowledge
Vector Quantities Remind students that velocity and
acceleration are vector quantities because they have both
magnitude and direction.
Caption Question Fig. 7: The velocity vectors have the
same length.
Lesson 2 • Circular Motion
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Lesson 2: Circular Motion
Visual Literacy
Use the following with Figure 8. Explain to students that
the diagram at the bottom of ­Figure 8 makes use of the
definition of Δv = vf − vi by drawing it in the form
vi + Δv = vf.
Discussion
Ask: In circular motion, what aspects of the velocity and
acceleration vectors are similar? What aspects are different? Both have constant magnitude and changing direction. The direction of the velocity is tangential to the orbit,
whereas the direction of the acceleration is radially inward.
Quick Demo
Centripetal Acceleration
Materials rotating stool, accelerometer
Procedure
1.Have a student sit on the stool and slowly spin while
holding an accelerometer at arm’s length.
2.Have the student hold the accelerometer tangentially
to the circular path and observe the accelerometer.
3. Repeat step 2 but with the accelerometer held radially.
4.Have students compare the two situations.
The accelerometer will indicate no acceleration in
step 2 and an acceleration directed toward the center
in step 3.
As the object moves around the circle, the length of the position vector does not
change, but the direction does. The diagram also shows two instantaneous
velocity vectors. Notice that each velocity vector is tangent to the circular path, at
a right angle to the corresponding position vector. To determine the object’s
velocity, you first need to find its displacement vector over a time interval. You
Δx
know that a moving object’s average velocity is defined as ___
Δt , so for an object in
Δr .
circular motion, v = ___
Δt The right side of Figure 7 shows Δr drawn as the
displacement from r1 to r2 during a time interval. The velocity for this time
interval has the same direction as the displacement, but its length would be
different because it is divided by Δt.
v2
a
r1
r2
a
a
a
Centripetal Acceleration
A velocity vector of an object in uniform circular motion is tangent to the circle.
What is the direction of the acceleration? Figure 8 shows the velocity vectors v1
and v2 at the beginning and end of a time interval. The difference in the two
vectors (Δv) is found by subtracting the vectors, as shown at the bottom of the
Δv
figure. The average acceleration (a = ___
Δt ) for this time interval is in the same
direction as Δv. For a very small time interval, Δv is so small that a points
toward the center of the circle. As the object moves around the circle, the direction of the acceleration vector changes, but it always points toward the center of
the circle. For this reason, the acceleration of an object in uniform circular
motion is called center-seeking or centripetal acceleration.
v1
v2
∆v
Figure 8 The acceleration of an object in
uniform circular motion is the change in
velocity divided by the time interval. The
direction of centripetal acceleration is
always toward the center of the circle.
Magnitude of acceleration What is the magnitude of an object’s centripetal
acceleration? Look at the starting points of the velocity vectors in the top of Figure 8. Notice the
triangle the position vectors at those points make with the center of the circle. An identical
triangle is formed by the velocity vectors in the bottom of Figure 8. The angle between r1 and r2
is the same as that between v1 and v2. Therefore, similar triangles are formed by subtracting the
two sets of vectors, and the ratios of the lengths of two corresponding sides are equal.
Δr
Δv
___
Thus, ___
r = v . The equation is not changed if both sides are divided by Δt.
Δr
Δv
Δr
___
___
rΔt = vΔt
Δv
___
But, v = ___
Δt and a = Δt .
__ ___
(__r )(___
Δt )=( v )( Δt )
1
Δr
Δr
1
Δv
Δv
___
Substituting v = ___
Δt in the left-hand side and a = Δt in the right-hand side gives the following
equation:
v __a
__
r=v
Solve for acceleration, and use the symbol ac for centripetal acceleration.
Centripetal Acceleration
Centripetal acceleration always points to the center of the circle. lts magnitude is equal to
the square of the speed divided by the radius of motion.
v
ac = __
r
2
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Describing Circular Motion
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INTERACTIVE CONTENT
Explore and Explain:
Centripetal Acceleration
EL Support
ELD PI.11/12.6a
Guide students in applying knowledge of language
devices used to make text more cohesive.
EMERGING LEVEL Draw students’ attention to the use
of by + gerund (-ing form) in the second paragraph of
page 148. Support students in recognizing that by + gerund tells us how. Ask: How do you find the difference in
two vectors? by subtracting
EXPANDING LEVEL Draw students’ attention to the use
BRIDGING LEVEL Ask students which language
structure in the second paragraph of page 148 tells
us how to find the difference between two vectors.
by + gerund (Ex: by subtracting) Help students use
other gerunds in sentences to tell how. by running, by
watching, by helping, by showing
148
Module 6 • Motion in Two Dimensions
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of by + gerund (by + subtracting) in the second paragraph
of page 148. Ask: What does by subtracting tell us? how
to find the difference in two vectors
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Systems and System Models
Models should indicate that during a turn, there is no force
acting on the object away from the center of the turn.
Students might use an accelerometer to indicate the lack
of a force directed outward.
Period of revolution One way to describe the speed of an object moving in a circle is to
measure its period (T), the time needed for the object to make one complete revolution. During
this time, the object travels a distance equal to the circumference of the circle (2πr).
2π r
The speed, then, is represented by v = ___
T . If you substitute for v in the equation for centripetal
acceleration, you obtain the following equation:
(2πr/T)2
Centripetal force
4π2r
ac = ______
= ____
.
r
T2
Because the acceleration of an object moving in a circle is always in the
direction of the net force acting on it, there must be a net force toward the center of the circle.
This force can be provided by any number of agents. For Earth circling the Sun, the force is the
Sun’s gravitational force on Earth. When a hammer thrower swings the hammer, as in Figure 9,
the force is the tension in the chain attached to the massive ball. When an object moves in a
circle, the net force toward the center of the circle is called the centripetal force. To accurately
analyze centripetal acceleration situations, you must identify the agent of the force that causes
the acceleration. Then you can apply Newton’s second law for the component in the direction of
the acceleration in the following way.
What are the key parts of the system?
How do the different components of the system ­interact?
Clarify a Preconception
Newton's Second Law for Circular Motion
The net centripetal force on an object moving in a circle is equal to
the object's mass times the centripetal acceleration.
dpa picture alliance archive/Alamy Stock Photo
Fnet = mac
Direction of acceleration When solving problems, you have
found it useful to choose a coordinate system with one axis in the
direction of the acceleration. For circular motion, the direction of the
acceleration is always toward the center of the circle. Rather than labeling
this axis x or y, call it c, for centripetal acceleration. The other axis is in
the direction of the velocity, tangent to the circle. It is labeled tang for
tangential. You will apply Newton’s second law in these directions, just
as you did in the two-dimensional problems you have solved before.
Remember that centripetal force is just another name for the net force in
the centripetal direction. It is the sum of all the real forces, those for
which you can identify agents that act along the centripetal axis.
Figure 9 As the hammer thrower swings the ball
around, tension in the chain is the force that causes
the ball to have an inward acceleration.
Predict Neglecting air resistance, how would the
horizontal acceleration and velocity of the hammer
change if the thrower released the chain?
In the case of the hammer thrower in Figure 9, in what direction does the hammer fly when the
chain is released? Once the contact force of the chain is gone, there is no force accelerating the
hammer toward the center of the circle, so the hammer flies off in the direction of its velocity, which
is tangent to the circle. Remember, if you cannot identify the agent of a force, then it does not exist.
CROSSCUTTING CONCEPTS
Systems and System Models Provide evidence to help overcome the
misconception of centrifugal "force" that is held by many people by developing a
model to demonstrate what occurs as a car makes a turn.
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03:08PM
Acceleration in Circular Motion Students might think that
an object in circular motion can only be accelerating if its
speed is changing. Explain that the word acceleration is
often used in this way in everyday language, but in science
an object also accelerates if the direction of its motion
changes, even if its speed remains constant.
Circular Motion Students might believe that, if the net
force supplying a centripetal acceleration is removed from
an object, the object will continue to move in a circular
path. Whirl a foam ball on the end of a string above your
head then release the string. The ball moves in a straight
line at a tangent to its original circular path. Explain that the
ball moves in a circle because the string provides a centripetal force to cause the centripetal acceleration. Once
the string is released, the ball continues in a straight line
because there are no horizontal forces acting on it.
Caption Question Fig. 9: The horizontal acceleration
would be zero; the horizontal velocity would remain
­constant.
Reinforcement
Net Force Activity Glue some tubing in the shape of a
semicircle onto a piece of wood or heavy cardboard. Roll a
marble around the inside of the channel and have students
observe that, when it emerges from the channel, it follows
a straight line tangent to the channel at the exit point.
Quick Practice
Using Mathematics and Computational Thinking Have
students view one of the scenes from the movie 2001:
A Space Odyssey that shows the space station rotating.
According to the film’s scientific consultant, the ­diameter of
the station was supposed to be 305 m (1000 ft). Using the
video, have students measure the station’s ­period of rotation and calculate the centripetal acceleration of a person
in the station. Have students compare this centripetal acceleration to the free-fall acceleration on Earth. To achieve
the free-fall acceleration of Earth, the station’s period of
rotation would have _______________
to be

m)/(9.8 m/s2) ​​ = 25 s.
T = 2π​  √r /g ​ = 2π​ ​​ (152.5
√
Lesson 2 • Circular Motion
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Lesson 2: Circular Motion
PRACTICE Problems
12.
13.
14.
15.
16.
3.1 m/s2, friction
8.1 km
1.2×102 N
0.74 N
8.6 m/s2, 0.88
EXAMPLE Problem 3
UNIFORM CIRCULAR MOTION A 13-g rubber stopper is attached to a 0.93-m string. The stopper is
swung in a horizontal circle, making one revolution in 1.18 s. Find the magnitude of the tension force
exerted by the string on the stopper.
Known
Unknowns
m = 13 g
FT = ?
r = 0.93 m
v2
a
T = 1.18 s
v1
FT
+ tang
+c
2 SOLVE FOR THE UNKNOWN
Find the magnitude of the centripetal acceleration.
Problem Jay is on a fishing trip and is having such a good
time that he decides to swing a fishing-line weight around
on the end of a piece of fishing line. The weight has a mass
of 0.028 kg, the length of the fishing line between his hand
and the weight is 0.75 m, and the weight makes one revolution in 1.2 s. What is the magnitude of the force exerted
by the string on the weight?
Response
0.58 N
4π2r
ac = ____
T2
4π2(0.93 m)
= ________
(1.18 s)2
Substitute r = 0.93 m, T = 1.18 s.
= 26 m/s2
Use Newton’s second law to find the magnitude of the tension in the string.
FT = mac
= (0.013 kg)(26 m/s2)
Substitute m = 0.013 kg, ac = 26 m/s2.
= 0.34 N
3 EVALUATE THE ANSWER
• Are the units correct? Dimensional analysis verifies that ac is in meters per second squared and
F T is in newtons.
• Do the signs make sense? The signs should all be positive.
• Are the magnitudes realistic? The force is almost three times the weight of the stopper, and the
acceleration is almost three times that of gravity, which is reasonable for such a light object.
PRACTICE Problems
​a ​c​ = 4​π ​2​r/​T ​2​
F
​ T​ ​ = m​a c​ ​= 4​π ​2​mr /​T ​2​
kg)(0.75 m)/(1.2
m
• Establish a coordinate system labeled tang and c. The directions of ac and FT
are parallel to c.
Use with Example Problem 3.
=
r
• Include the radius and the direction of motion.
ADDITIONAL IN-CLASS Example
4​π ​2​(0.028
v
1 ANALYZE AND SKETCH THE PROBLEM
• Draw a free-body diagram for the swinging stopper.
s​) ​2​
= 0.58 N
Critical Thinking
Banked Curves Ask students why curves on highways
are banked. The inclined roadway produces a horizontal
component of the normal force. This component is in the
same direction as the desired centripetal acceleration and
so contributes to accelerating the car to make it round the
curve.
ADDITIONAL PRACTICE
12. A runner moving at a speed of 8.8 m/s rounds a
bend with a radius of 25 m. What is the centripetal
acceleration of the runner, and what agent exerts the
centripetal force on the runner?
13. An airplane traveling at 201 m/s makes a turn.
What is the smallest radius of the circular path
(in kilometers) the pilot can make and keep the
centripetal acceleration under 5.0 m/s2?
14. A 45-kg merry-go-round worker stands on the
ride’s platform 6.3 m from the center, as shown in
Figure 10. If her speed (vworker) as she goes around
the circle is 4.1 m/s, what is the force of friction (Ff)
necessary to keep her from falling off the platform?
15. A 16-g ball at the end of a 1.4-m string is swung in a
horizontal circle. It revolves once every 1.09 s. What
is the magnitude of the string's tension?
16. CHALLENGE A car racing on a flat track travels at
22 m/s around a curve with a 56-m radius. Find
the car’s centripetal acceleration. What minimum
coefficient of static friction between the tires and
the road is necessary for the car to round the curve
without slipping?
vworker
Ff
Figure 10
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Real-World Physics
g Forces Pilots who fly combat and aerobatic aircraft
undergo the strain of what is called a “g force,” which is a
measure of an apparent increase in gravitational force due
to the force ­exerted by the pilot’s seat. In a tight 4- or 5-g
turn, the ­pilot’s apparent body weight is 4 or 5 times its
normal weight. At about 9 g, eyesight ­begins to fail and the
pilot begins to experience tunnel vision. The pilot momentarily experiences a “grayout,” in which everything appears
to be black and white.
Elaborate
Return to the DQB and have students determine what
questions they can answer. At this point, they should be
able to answer the Focus Question.
150
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Evaluate
Path of car
Centrifugal “Force”
Path of
passenger
without
car
If a car you are in stops suddenly, you are thrown forward into your
safety belt. Is there a forward force on you? No, because according to
Newton’s first law, you will continue moving with the same velocity
unless there is a net force acting on you. The safety belt applies that
force.
Figure 11 shows a car turning left as viewed from above. A passenger
in the car would continue to move straight ahead if it were not for the
force of the door acting in the direction of the acceleration. As the car
goes around the curve, the car and the passenger are in circular
motion, and the passenger experiences centripetal acceleration.
Recall that centripetal acceleration is always directed toward the
center of the circle. There is no outward force on the passenger. You
feel as if you are being pushed only because you are accelerating
relative to your surroundings. The so-called centrifugal, or outward,
force is a fictitious, nonexistent force. There is no real force because
there is no agent exerting a force.
Figure 11 When a car moves around a curve, a rider
feels a fictitious centrifugal force directed outward.
In fact, the force on the person is the centripetal force,
which is directed toward the center of the circle and is
exerted by the seat on which the person is sitting.
Check Your Progress
17. Circular Motion A ball on a rope swung at a
constant speed in a circle above your head is
in uniform circular motion. In which direction
does it accelerate? What force causes this?
21. Centripetal Acceleration An object swings in
a horizontal circle, supported by a 1.8-m string.
It completes a revolution in 2.2 s. What is the
object's centripetal acceleration?
18. Circular Motion What is the direction of the
force that acts on clothes spinning in a top-load
washing machine? What exerts the force?
22. Centripetal Force The 40.0-g stone in
Figure 12 is whirled horizontally at a speed of
19. Centripetal Acceleration A newspaper article
states that when turning a corner, a driver
must be careful to balance the centripetal and
centrifugal forces to keep from skidding. Write
a letter to the editor that describes physics
errors in this article.
20. Free-Body Diagram You are sitting in the
back seat of a car going around a curve to the
right. Sketch motion and free-body diagrams
to answer these questions:
a. What is the direction of your acceleration?
b. What is the direction of the net force on
you?
c. What exerts this force?
2.2 m/s. What is the tension in the string?
23. Amusement-Park Ride A ride park has people
stand around a 4.0-m radius circle with their
backs to a wall. The ride then spins them with a
1.7-s period of revolution. What are the centripetal acceleration and velocity of the riders?
24. Critical Thinking You are always in uniform
circular motion due to Earth’s rotation. What
agent supplies the force? If you are on a scale,
how does the motion affect your weight?
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Explore and Explain:
Centrifugal “Force”
Remediation Ask students to describe an object’s speed,
velocity, and acceleration as it moves in uniform circular
motion. Its speed is constant, but the direction of the
velocity changes constantly to follow a circular path.
The magnitude of the acceleration is constant, but the
direction of the acceleration changes so that it is always
oriented toward the center of the circle.
Check Your Progress
Figure 12
Lesson 2 • Circular Motion
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Ask: Why do airplanes bank to turn? By banking, a
component of the lift force of the wings becomes horizontal
and perpendicular to the plane’s velocity. This force causes
the plane to accelerate centripetally (i.e., turn).
0.60 m
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and reinforce your understanding.
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ADDITIONAL RESOURCE
17.The ball accelerates toward the center of the circle
because of the centripetal force.
18.The force is toward the center of the tub. The walls
exert the force on the clothes.
19.There is an acceleration because the direction of the
velocity is changing. There must be a net force toward
the center of the circle. The road supplies that force,
and the friction between the road and the tires allows
the force to be exerted on the tires. The seat exerts
the force on the driver towards the center of the circle.
The note should also make it clear that centrifugal
force is not a real force.
20.
Vocabulary Flashcards:
Circular Motion
a
v
ADDITIONAL RESOURCE
Lesson Check: Circular
Motion
ADDITIONAL RESOURCE
Inspire Physics LearnSmart
Fnet
a.C06-02A-865893_A
right
b. right
c. the car’s seat
21. 15 m/s2
22. 0.20 N
23. 55 m/s2, 15 m/s
24.Earth’s gravity supplies the force that accelerates you.
The scale will record a lower weight if you are in uniform circular motion.
Formative Assessment: Lesson Check
GO ONLINE You might want to assign from the
Additional Resources the pre-made Lesson Check based
on key concepts and disciplinary core ideas, or you can
customize your own using the customization tool.
Lesson 2 • Circular Motion
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Lesson 3: Relative Velocity
Engage
Launch the Lesson Interactive Content can be assigned
the night before class as a lesson preview, during class
to spark discussion, as a resource during inquiry, or as
homework.
Driving Question Board
Have students revisit the DQB to remind themselves of the
Unit and Module questions. Have them identify the sticky
note questions they think will be answered in this lesson.
Then, have students read the Focus Question and add it
to the DQB. Students will revisit the Focus Question at the
end of the lesson.
LESSON 3
RELATIVE VELOCITY
FOCUS QUESTION
Does your description of motion depend on your frame of reference?
Relative Motion in One Dimension
Suppose you are in a school bus that is traveling at a velocity of 8 m/s in a positive direction. You
walk with a velocity of 1 m/s toward the front of the bus. If a friend is standing on the side of the
road watching the bus go by, how fast would your friend say you are moving? If the bus is traveling at 8 m/s, its speed as measured by your friend in a coordinate system fixed to the road is
8 m/s. When you are standing still on the bus, your speed relative to the road is also 8 m/s, but
your speed relative to the bus is zero. How can your speed be different?
Different reference frames
In this example, your motion is viewed from different
coordinate systems. A coordinate system from which motion is viewed is a reference frame.
Walking at 1 m/s toward the front of the bus means your velocity is measured in the reference
frame of the bus. Your velocity in the road’s reference frame is different. You can rephrase the
problem as follows: given the velocity of the bus relative to the road and your velocity relative to the
bus, what is your velocity relative to the road? A vector representation of this problem is shown in
Figure 13. If right is positive, your speed relative to the road is 9 m/s, the sum of 8 m/s and 1 m/s.
Same Direction
vbus relative to street
Explore and Explain
SEP
I
Activity
Moving Sidewalks Place a constant-speed car on a long
board on a tabletop and start it. Ask students to describe
two situations in which the moving car has zero velocity.
(1) Pull the board while walking with the same speed as the
car but in the opposite direction (so that the velocity of the
car relative to the tabletop is zero). (2) Walk with the same
speed and in the same direction as the car (so that the
velocity of the car relative to the walker is zero).
vyou relative to bus
vyou relative to street
vyou relative to street
SEP
Figure 13 When an object moves in a moving reference frame, you add the velocities if they are in the same
direction. You subtract one velocity from the other if they are in opposite directions.
Recall What do the lengths of the velocity vectors indicate?
I
DC
CCC
vyou relative to bus
DC
Science Journal Remind students to keep records of
their investigations in their Science Journals. Additionally,
be sure that each reading or activity is added to the
class Summary table.
Three-Dimensional Thinking The activities called out
in the Student Edition will allow students to practice threedimensional thinking. Worksheets for these activities can
be found online.
Opposite Direction
vbus relative to street
CCC
3D THINKING
Disciplinary Core Ideas
COLLECT EVIDENCE
Use your Science Journal to
record the evidence you collect as
you complete the readings and
activities in this lesson.
Crosscutting Concepts
Science & Engineering Practices
INVESTIGATE
GO ONLINE to find these activities and more resources.
PhysicsLAB: Moving Reference Frame
Carry out an investigation to examine the motion of a system in different reference frames.
Identify Crosscutting Concepts
Create a table of the crosscutting concepts and fill in examples you find as you read.
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PRESENTATION
Teacher Presentation:
Relative Velocity
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03:08PM
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INTERACTIVE CONTENT
Launch the Lesson: Relative
Velocity
Tie to Prior Knowledge
Caption Question Fig. 13: speed
Use an Analogy
Net Motion and Net Wage Draw students’ attention to the
vector showing their net motion relative to the street in the
diagram on the right in Figure 13. To help them understand
this vector, use the following analogy: If your hourly wage
is $12/h and you must pay $2/h for parking while you work,
your net hourly wage is $10/h.
152
Module 6 • Motion in Two Dimensions
IMPLEMENTATION OPTIONS
Presentation: Teacher-Facilitated Pathway
Use the Teacher Presentation to support classroom
instruction and spark discourse. Obtain data to
inform your instruction by assigning the Interactive
Content, Additional Resources, and Assessment.
Interactive Content: Student-Led Pathway
Students can use the online Interactive Content,
along with the Student Edition, Science Notebook,
projects, and labs, to collect evidence to support
their claim. They can record their evidence in their
Science Journals and the class Summary Table.
Tom Merton/age fotostock
Velocity and Vector Addition Students extend their understanding of velocity to include relative velocity. Students
apply vector addition to velocity vectors.
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Quick Practice
Suppose that you now walk at the same speed toward the rear of the bus. What would be your
velocity relative to the road? Figure 13 shows that because the two velocities are in opposite
directions, the resultant speed is 7 m/s, the difference between 8 m/s and 1 m/s. You can see that
when the velocities are along the same line, simple addition or subtraction can be used to determine the relative velocity.
Using Mathematics and Computational Thinking
Explain that va/b is the velocity of object a as measured
in observer b’s reference frame, and vb/c is the motion of
observer b’s reference frame as measured in observer c’s
reference frame. Thus, va/c is the velocity of the object as
measured in observer c’s reference frame. Note that, by
canceling the repeated subscripts, the correct order of
addition is assured (e.g., va/b + vb/c = va/c).
Get It?
Identify When would you use subtraction to determine the relative velocity?
PHYSICS Challenge
TENSION IN A ROPE Phillipe whirls a stone of mass m on a rope in a horizontal circle above
his head such that the stone is at a height h above the ground. The circle has a radius of r,
and the magnitude of the tension in the rope is T. Suddenly the rope breaks, and the stone
falls to the ground. The stone travels a horizontal distance x from the time the rope breaks
until it impacts the ground. Find a mathematical expression for x in terms of T, r, m, and h.
Does your expression change if Phillipe is walking 0.50 m/s relative to the ground?
Combining velocity vectors Take a closer look at how the relative velocities in Figure 13
were obtained. Can you find a mathematical rule to describe how velocities are combined when
the motion is in a moving reference frame? For the situation in which you are walking in a bus,
you can designate the velocity of the bus relative to the road as v b/r. You can designate your
velocity relative to the bus as vy/b and the velocity of you relative to the road as vy/r. To find the
velocity of you relative to the road in both cases, you added the velocity vectors of you relative to
the bus and the bus relative to the road. Mathematically, this is represented as v y/b + vb/r = vy/r.
The more general form of this equation is as follows.
Reinforcement
Adding Vectors The sum of two vectors acting in one
dimension is their algebraic sum and not their arithmetic
sum.
Relative Velocity
The relative velocity of object a to object c is the vector sum of object a’s velocity relative to
object b and object b’s velocity relative to object c.
Get It?
va/b + vb/c = va/c
when the velocities are in opposite directions
Get It?
Explain A flea is running on the back of a dog that is running along a road. Write an
equation to determine the velocity of the flea relative to the road.
Physics Challenge
√
_____
STEM CAREER Connection
Commercial Pilot
A person wanting to be a commercial pilot must have an understanding of relative
motion to get the plane or helicopter to its intended destination. Commercial
pilots who drop fire-retardant gel to combat forest fires must also understand
projectile motion to hit their targets.
2Trh
x = ​​ ​ _____
m​a​ ​ ​​ ​​​
Lesson 3 • Relative Velocity
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Relative Motion in One
Dimension
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grav
Yes, the expression changes if Phillipe is walking 0.50 m/s
relative to the ground. If the stone travels in the same
direction as Phillipe is walking, the velocity of the stone
relative to the ground will be greater, resulting in a larger
value of x.
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Get It?
vf/d + vd/r = vf/r
STEM Connections
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selection of people and groups that have made important
contributions to society through science and technology.
Mint Images Limited/Alamy Stock Photo
Teacher Toolbox
Content Background
Jet Streams and Air Travel Jet streams are fast-flowing, confined air currents found in the atmosphere at an
altitude of around 12 km. In the northern hemisphere,
the streams are most commonly found between latitudes 30°–70° and between latitudes 20°–50°. The wind
speeds vary according to the temperature gradient; they
average 55 km/h in summer and 120 km/h in winter,
although speeds of over 400 km/h do occur. The location of the jet stream is an extremely important datum
for airlines. In the United States, the time needed to fly
from the east coast to the west coast can decrease or
increase by about 30 minutes depending on whether the
airplane flies with or against the jet stream, respectively.
Lesson 3 • Relative Velocity
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Lesson 3: Relative Velocity
Reteach
Relative Velocity Ask each student to write a relative velocity problem on a sheet of paper and to write its solution
on the back of the sheet. (You may want to specify the
number of dimensions.) Have students exchange sheets
and solve the problems. The students may then present
their solutions to the class, and the class can arbitrate any
differences in solutions.
Caption Question Fig. 14: The air vector would be in the
opposite direction, and the placement of the resultant
­vector would change in both length and direction.
yes
Get It?
Get It?
The velocity of the boat relative to the water would have
to be the same as the velocity of the water relative to
the shore but in the opposite direction.
Clarify a Preconception
Relative Velocity Path Activity Ask students to sketch
the displacement with respect to the water of the marble
discussed in Example Problem 4 on page 155. The path is
a straight line in the direction of vm/w. Some students may
draw the path as a curve. Help students model relative
motion using the strategy found on page 155 in Using
Models.
Reference frame
moving relative to ground
y'
vo/m
θo/m
vm/g
x'
Figure 14 Vectors are placed tip-to-tail to find the relative velocity
vector for two-dimensional motion. The subscript o/g refers to an
object relative to ground, o/m refers to an object relative to a moving
reference frame, and m/g refers to the moving frame relative to ground.
Analyze How would the resultant vector change if the ground
reference frame were considered to be the moving reference frame?
vo/g
y
θm/g
θo/g
x
Ground
reference frame
Relative Motion in Two Dimensions
Adding relative velocities also applies to motion in two dimensions. As with one-dimensional
motion, you first draw a vector diagram to describe the motion, and then you solve the problem
mathematically.
Vector diagrams The method of drawing vector diagrams for relative motion in two
dimensions is shown in Figure 14. The velocity vectors are drawn tip-to-tail. The reference frame
from which you are viewing the motion, often called the ground reference frame, is considered
to be at rest. One vector describes the velocity of the second reference frame relative to ground.
The second vector describes the motion in that moving reference frame. The resultant shows the
relative velocity, which is the velocity relative to the ground reference frame.
Get It?
Decide Can a moving car be a reference frame?
An example is the relative motion of an airplane. Airline pilots cannot expect to reach their
destinations by simply aiming their planes along a compass direction. They must take into
account the plane’s speed relative to the air, which is given by their airspeed indicators, and their
direction of flight relative to the air. They also must consider the velocity of the wind at the
altitude they are flying relative to the ground. These two vectors must be combined to obtain the
velocity of the airplane relative to the ground. The resultant vector tells the pilot how fast and in
what direction the plane must travel relative to the ground to reach its destination. A similar
situation occurs for boats that are traveling on water that is flowing.
Get It?
Describe How could a boat on a flowing river stay in one location relative to the shore?
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ELD PII.11/12.2a
Guide students in applying knowledge of familiar language
resources for referring to make texts more cohesive.
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INTERACTIVE CONTENT
Explore and Explain:
Relative Motion in Two
Dimensions
EMERGING LEVEL Draw students’ attention to An example in the third paragraph of page 154. Support students
in recognizing that an example refers to relative motion
in two dimensions. Demonstrate by moving your finger
from An example to the page header, relative motion in
two dimensions.
EXPANDING LEVEL Draw students’ attention to An
BRIDGING LEVEL Draw students’ attention to An
example in the third paragraph on page 154. Elicit what
An example refers to. relative motion in two dimensions
Have students look for other referents in the paragraph
and discuss what they refer to. They = airline pilots;
its = plane’s
154
Module 6 • Motion in Two Dimensions
ITN/Getty Images
example in the third paragraph of page 154. Ask: Do
you know what an example refers to? Guide students to
recognize what an example refers to. relative motion in
two dimensions Ask: What example is used? motion of an
airplane relative to the ground
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Get It?
Combining velocities
Vector Resolution
You can use the equations in
Figure 15 to solve problems for relative motion in two
dimensions. The velocity of a reference frame moving relative to
the ground is labeled vm/g. The velocity of an object in the
moving frame is labeled vo/m. The relative velocity equation gives
the object’s velocity relative to the ground: vo/g = vo/m + vm/g.
Moving reference
Object relative to
frame relative
moving reference frame
to ground
vo/m sinθo/m
vm/g sin θm/g
vo/m
θo/m
vo/m cos θo/m
vo/g
To determine the magnitude of the object’s velocity relative to
the ground (vo/g), first resolve the velocity vectors of the object
and the moving reference frame into x- and y-components. Then
apply the Pythagorean theorem. The general equation is shown
in Figure 15, but for many problems the equation is simpler
because the vectors are along an axis. As shown in the example
problem below, you can find the angle of the object’s velocity
relative to the ground by observing the vector diagram and
applying a trigonometric relationship.
vm/g
Object relative
to ground
θm/g
vm/g cosθm/g
vy
vx
Equation
vo/g2 = vx2 + vy2
Get It?
where vx = vo/m cos θo/m + vm/g cosθm/g
vy = vo/m sinθo/m + vm/g sin θm/g
Figure 15 To find the velocity of an object in a moving
reference frame, resolve the vectors into x- and y-components.
Explain How are vectors used to describe relative
motion in two dimensions?
EXAMPLE Problem 4
RELATIVE VELOCITY OF A MARBLE Ana and Sandra are riding on a ferry boat traveling east at 4.0 m/s.
Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What
is the velocity of the marble relative to the water?
v b/w
1 ANALYZE AND SKETCH THE PROBLEM
Establish a coordinate system. Draw vectors for the velocities.
Known
vb/w = 4.0 m/s
Unknown
vm/b = 0.75 m/s
v m/b
vm/w = ?
2 SOLVE FOR THE UNKNOWN
The velocities are perpendicular, so we can use the
Pythagorean theorem.
vm/w = vb/w + vm/b
_________
vm/w = vb/w2 + vm/b2
__________________
= (4.0 m/s)2 + (0.75 m/s)2 = 4.1 m/s
2
2
v m/b
2
√
√
Find the angle of the marble’s velocity.
v m/w
0.75 m/s
3 EVALUATE THE ANSWER
• Are the units correct? Dimensional analysis verifies units of meters per second for velocity.
• Do the signs make sense? The signs should all be positive.
• Are the magnitudes realistic? The resulting velocity is of the same order of magnitude as the
velocities given in the problem and slightly larger than the larger of the two.
Lesson 3 • Relative Velocity
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6.0 m/s
0.73 m/s
2.0 m/s, against the boat
a. 4.0 m/s
b. 19° south of east
29. 1.7×102 km/h
30. a. 250.0 km/h
b. 150.0 km/h
Problem Lalei places her lunch tray on a cafeteria
­conveyor belt that moves westward at 0.150 m/s. With
respect to the tray, a ladybug on the tray crawls northward
at 0.050 m/s. What is the ladybug’s velocity with respect to
the ground?
= tan-1 (______
4.0 m/s ) = 11° north of east
The marble travels 4.1 m/s at 11° north of east.
152-156_PHYS_CA_S_CH06_L03_674235.indd
25.
26.
27.
28.
Use with Example Problem 4.
v b/w
Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s.
Vm/b
03:08PM
PRACTICE Problems
ADDITIONAL IN-CLASS Example
+y
θ = tan-1 ___
( Vb/w )
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Resolve the velocity vectors into their x- and y-components. Each component gives the speed in the corresponding direction relative to a given reference frame.
03:08PM
Response
0.16 m/s, 18° north of west
​ ​ = ​v l/t
​ ​ + ​v t/g
​ ​
​v l/g
​ ​ ​2​ = ​​v l/t
​ ​ ​2​ + ​​v t/g
​ ​ ​2​
​​v l/g
_________
√
​ ​ ​2​ ​ ​​
​v ​l/g​ = ​​ ​v ​l/t​ ​2​ + ​​v t/g
____________________
√
= ​​ (0.15
m/s)2 + (0.050 m/s)2 ​​
= 0.16 m/s
​ ​/​v t/g
​ ​)
θ = ta​n ​-1​(​v l/t
= ta​n ​-1​((0.050 m/s)/(0.150 m/s))
= 18° north of west
Using Models
Simulating Constant-Velocity Motion To simulate a
constant-velocity boat moving in a river, use a constant-­
velocity toy car and a long sheet of paper. The motion
of the car represents vb/w and the sheet can be pulled at
constant speed to simulate the motion of the current, vw/g.
Have students observe that the velocity of the boat relative
to the ground (vb/g) is a straight line.
Elaborate
Return to the DQB and have students determine what
questions they can answer. At this point, they should be
able to answer the Focus Question.
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Lesson 3: Relative Velocity
Evaluate
Formative Assessment Check
Ask: Consider Figure 2. If the camera were moving with
the same velocity as the initial velocity of the blue ball
with respect to the ground, vb/g, i, what would the picture
look like? Consider the reference frames of the ground (g)
and the camera (c). In g, the camera moves ­horizontally
at the same velocity as the initial velocity of the blue ball
(i.e., vb/g, i = vc/g). Therefore, the blue ball has no horizontal
velocity in c (vb/c, i + vc/g = vb/g, i ⇒ vb/c, i = 0), so it will fall
straight down in the photo. In c, the initial ­velocity of the
red ball will be opposite to the initial ­velocity of the blue
ball in g: vr/c, i + vc/g = vr/g, i ⇒ vr/c, i = –vc/g = –vb/g, i (note that
vr/g, i = 0). Therefore, in c, the initial velocity of the red ball
is opposite that of the blue ball in g, so the red ball will execute the same parabolic trajectory as the blue ball does
in g, except it will be to the left instead of to the right.
Remediation Review the general form of the relative
velocity equation. Then ask students to ­explain why it is
important that the equation involves velocities rather than
speeds. Speed may be used only if the reference frames
and the objects inside of them are all moving in the same
direction (e.g., horizontally or v­ ertically). If any of these
are moving in different directions, you have to consider
the speed in the different dimensions (e.g., horizontal and
vertical) to describe the motion. A quantity that requires
two scalars to describe it is called a vector.
PRACTICE Problems
ADDITIONAL PRACTICE
25. You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at
4.0 m/s relative to the bus. What is your speed relative to the street?
26. Rafi is pulling a toy wagon through a neighborhood at a speed of 0.75 m/s. A caterpillar in the wagon is
crawling toward the rear of the wagon at 2.0 cm/s. What is the caterpillar’s velocity relative to the ground?
27. A boat moves directly upriver at 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving
at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat?
28. A boat is traveling east at a speed of 3.8 m/s. A person walks across the boat
Tailwind
with a velocity of 1.3 m/s south.
a. What is the person’s speed relative to the water?
b. In what direction, relative to the ground, does the person walk?
29. An airplane flies due north at 150 km/h relative to the air. There is a wind
blowing at 75 km/h to the east relative to the ground. What is the plane’s
speed relative to the ground?
30. CHALLENGE The airplane in Figure 16 flies at 200.0 km/h relative to the
air. What is the velocity of the plane relative to the ground if it flies during the
following wind conditions?
a. a 50.0-km/h tailwind
b. a 50.0-km/h headwind
Check Your Progress
Figure 16
Check Your Progress
31. Relative Velocity A plane travels 285 km/h
west relative to the air. A wind blows 25 km/h
east relative to the ground. Find the plane’s
speed and direction relative to the ground.
32. Relative Velocity A fishing boat with a maximum speed of 3 m/s relative to the water is in a
river that is flowing at 2 m/s. What is the maximum speed the boat can obtain relative to the
shore? The minimum speed? Give the direction
of the boat, relative to the river’s current, for the
maximum speed and the minimum speed
relative to the shore.
33. Relative Velocity of a Boat A motorboat heads
due west at 13 m/s relative to a river that flows
due north at 5.0 m/s. What is the velocity (both
magnitude and direction) of the motorboat
relative to the shore?
35. Relative Velocity An airplane flies due south at
175 km/h relative to the air. There is a wind
blowing at 85 km/h to the east relative to the
ground. What are the plane’s speed and direction relative to the ground?
36. A Plane’s Relative Velocity An airplane flies
due north at 235 km/h relative to the air. There
is a wind blowing at 65 km/h to the northeast
relative to the ground. What are the plane’s
speed and direction relative to the ground?
37. Critical Thinking You are piloting a boat across
a fast-moving river. You want to reach a pier
directly opposite your starting point. Describe
how you would navigate the boat in terms of
the components of your velocity relative to the
water.
34. Boating Martin is riding on a ferry boat that is
traveling east at 3.8 m/s. He walks north across
the deck of the boat at 0.62 m/s. What is
Martin’s velocity relative to the water?
Go online to follow your personalized learning path to review, practice,
and reinforce your understanding.
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31. 260 km/h west
32. 5 m/s with the current, 0 m/s against the current
33. 14 m/s, 69° west of north
34. 3.8 m/s, 9.3° north of east
35. 1.9×102 km/h, 64° south of east
36. 2.8×102 km/h, 81° north of east
37.Choose the component of your velocity along the
direction of the river to be equal and opposite to the
velocity of the river.
Headwind
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Vocabulary Flashcards:
Relative Velocity
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Lesson Check: Relative
Velocity
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Formative Assessment: Lesson Check
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Additional Resources the pre-made Lesson Check based
on key concepts and disciplinary core ideas, or you can
customize your own using the customization tool.
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Module 6: STEM at Work
Fighting Fire with Forces
STEM AT WORK
Purpose
Fighting Fire with Forces
Students will understand that smokejumpers are professional wildland firefighters who are first responders to fires
in remote areas. They parachute into an area to try to contain a fire before it gets bigger. When smokejumpers leap
from planes, they are moving in two dimensions. The initial
force that propels them forward and the force of gravity
work together to shape their parabolic trajectory.
Smokejumpers and Parabolic Trajectory
A smokejumper’s leap from a plane is a classic example of the
parabolic path of a projectile moving in two dimensions.
Firefighters accepted into the smokejumper program
must have years of previous experience fighting
fires. Even then, applicants must go through a
rigorous training program before they are ready to
work as smokejumpers. They learn how to properly
exit an airplane, deploy their parachute, and land
safely on the ground. Smokejumpers also receive
first aid training and practice climbing trees and
digging fire lines.
A spotter flies with every smokejumper mission.
Spotters take wind speed and direction into account
as they help plan the trajectories of both the smokejumpers and packages of supplies. The packages—
which contain food, water, first aid kit, and
fire-fighting equipment—ensure that smokejumpers
can work 48 hours or more without other outside
assistance. Spotters also relay information about
how and where the fire is moving.
When smokejumpers skydive out of airplanes to
parachute down to the site of a fire, they are moving
in two dimensions at the same time—along the x and
y axes. They are subject to both the initial force that
launches them and the force of gravity. Gravity
causes their path to curve downward. Smokejumpers must plan their parabolic trajectories so that they
land in a safe place.
Work on the Ground and in the Air
Smokejumpers on the ground build firebreaks and
use water and fire retardant to contain fires. They
build backfires, which burn in the direction of the
existing fire, to eliminate possible fuel. Smokejumpers on planes dump water and fire retardant from
above. Smokejumpers protect vast expanses of
wildlands from destruction every year.
I
SEP
USE MATHEMATICAL
REPRESENTATIONS
CCC
Write a problem about the parabolic trajectory of a
smokejumper. Swap problems with a classmate.
Solve each other’s problem.
DC
Darin Oswald/Idaho Statesman/MCT/Getty Images
Smokejumpers are professional wildland
firefighters who parachute into remote areas
to control and extinguish wildfires. Their job is
incredibly dangerous. They jump out of
planes up to 450 meters above the ground
into blazing fires. How do they land safely?
Parachute jumping from a plane requires
extensive training and an understanding of
two-dimensional parabolic motion.
Module 6 • STEM at Work
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Ask: How can you determine a projectile’s trajectory?
­Possible answer: You must know the projectile’s initial
­velocity in order to determine its trajectory. Remind students that projectiles may be launched vertically or at an
angle, but that in both types of projectile motion, gravity
will cause the projectile’s path to be parabolic.
Background
Smokejumpers toss streamers from the plane before jumping. The streamers show them the direction and strength
of the wind and help them decide exactly when to jump.
But because even the most careful planning of parachute
trajectories can be thrown off by a sudden strong gust of
wind, smokejumpers must practice making rough landings
in difficult terrain. They use a landing simulator with varying
speeds that lifts them up and then drops them. Recruits
learn how to hit the ground in ways that minimize injury.
DC
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Guiding Questions
I
USE MATHEMATICAL
­REPRESENTATIONS
CCC
If students struggle to write a projectile problem,
­encourage them to review the problem-solving strategies,
­example problem, and practice problems in the student
edition.
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Module 6: Study Guide
Review and Connect
Use the following tools to help students review the
content in this Module and to connect to the broader
topics of the Unit.
MODULE 6
STUDY GUIDE
GO ONLINE to study with your Science Notebook.
Lesson 1 PROJECTILE MOTION
Driving Question Board
Revisit the DQB and have students determine what
questions they can answer with their new knowledge.
At this point, they should be able to answer the Module
Encounter the Phenomenon Question.
Summary Table
As a class, review the Summary Table. If you do any endof-module activities, add them to the table.
• The vertical and horizontal motions of a projectile are independent. When there is no air resistance, the horizontal motion
component does not experience an acceleration and has constant
velocity; the vertical motion component of a projectile experiences a constant acceleration under these same conditions.
• The curved flight path a projectile follows is called a trajectory
and is a parabola. The height, time of flight, initial velocity, and
horizontal distance of this path are related by the equations of
motion. The horizontal distance a projectile travels before
returning to its initial height depends on the acceleration due to
gravity and on both components of the initial velocity.
Lesson 2 CIRCULAR MOTION
• An object moving in a circle at a constant speed has an acceleration toward the center of the circle because the direction of its
velocity is constantly changing.
• Acceleration toward the center of the circle is called centripetal
acceleration. It depends directly on the square of the object’s
speed and inversely on the radius of the circle.
Students can use the Module Vocabulary Practice to
review key terms.
LearnSmart® is an adaptive learning tool tailored to
the unique needs of each student. Have students use
LearnSmart® for review, to practice for assessment, and to
monitor the progress of their learning.
v2
ac = __
r
Fnet = mac
• reference frame
• A coordinate system from which you view motion is called a
reference frame. Relative velocity is the velocity of an object
observed in a different, moving reference frame.
• You can use vector addition to solve motion problems of an object
in a moving reference frame.
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• uniform circular motion
• centripetal acceleration
• centripetal force
• A net force must be exerted by external agents toward the circle’s
center to cause centripetal acceleration.
Lesson 3 RELATIVE VELOCITY
Module Vocabulary Practice
• projectile
• trajectory
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SEP
I
Module Wrap-Up
DC
THREE-DIMENSIONAL THINKING
CCC
DC
SEP
REVISIT THE PHENOMENON
I
CCC
THREE-DIMENSIONAL THINKING
Claim, Evidence, Reasoning
Module Wrap-Up
Have students review the evidence they collected in
their Science Journals and the class Summary Table. If
students found evidence that contradicts their claims, their
claims are likely incorrect. Encourage students to use the
evidence they recorded to revise their claims.
REVISIT THE PHENOMENON
Why do thrown basketballs
travel in arcs?
CER
Claim, Evidence, Reasoning
When providing reasoning, students should demonstrate
three-dimensional thinking by applying the disciplinary
core ideas, science and engineering practices, and
­crosscutting concepts learned in the module. They should
describe the scientific principle(s) they used to create
their arguments and explain why their evidence supports
their claims.
Explain Your Reasoning Revisit the claim you made when you encountered the
phenomenon. Summarize the evidence you gathered from your investigations and
research and finalize your Summary Table. Does your evidence support your claim? If not,
revise your claim. Explain why your evidence supports your claim.
STEM UNIT PROJECT
Now that you’ve completed the module, revisit your STEM unit project. You will
summarize your evidence and apply it to the project.
GO FURTHER
Data Analysis Lab
Does the baseball clear the wall?
A baseball player hits a belt-high (1.0 m) fastball down the left-field
line. The player hits the ball with an initial velocity of 42.0 m/s at an
angle of 26° above the horizontal. The left-field wall is 96.0 m from
home plate at the foul pole and is 14 m high.
STEM UNIT PROJECT
Performance Task
CER Analyze and Interpret Data
1. Write the equation for the height of the ball (y) as a function of its
distance from home plate (x).
2. Use a computer or a graphing calculator to plot the path of the
ball. Trace along the path to find how high above the ground the
ball is when it is at the wall.
3. Claim Is the hit a home run?
4. Evidence and Reasoning Justify your claim.
Thomas Barwick/Iconica/Getty Images
Check in with students on their progress with their Unit
Projects. Encourage them to apply what they have learned
in the module to the project.
GO FURTHER
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Go Further: Does the
baseball clear the wall?
Go Further presents another opportunity for students to
practice making claims by analyzing information and data
and supporting their claims with evidence and reasoning.
05:08PM
Data Analysis Lab
Does the baseball clear the wall?
CER Analyze and Interpret Data
1. y = yi + vyi​​ __
​  ​ ​​ + __
​​   ​​a​​ __
​  ​ ​​
( vx ) 2 ( vx )
2. Students should provide the plots they generate.
3. Yes, the ball clears the wall by 2.1 m.
4. Students should use their sketches to explain their
reasoning and calculations to justify their claims. The
minimum speed at which the ball could be hit and clear
the wall is 41 m/s. The range of angles will be 25° to 73°.
x
1
x
2
Summative Assessment: Module Test
GO ONLINE You might want to assign from the
Additional Resources the pre-made Module Test based
on key concepts and disciplinary core ideas, or you can
customize your own using the customization tool.
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