Chapter 1-7
MASTER NOTES
MASTER CLASS NOTES FOR
Chapter 1,2,3,5,6 & 7
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Chapter 1
The Science of Physics
Table of Contents
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 7
The Science of Physics
Motion in One Dimension
Two- Dimensional Motion and Vectors
Forces and the Laws of Motion
Rotational Motion and the Law of Gravity
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Chapter 1
Section 1 What is Physics
The Topics of Physics
• Physics is simply the study of the physical world.
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Chapter 1
Section 1 What is Physics
The areas of Physics
1. Mechanics - The study of motion and its causes.
– Falling objects, friction, weight, spinning
objects.
2. Thermodynamics – The study of heat and
temperature.
– Melting and Freezing processes, engines,
refrigerators.
3. Vibration and Wave Phenomena – The study of
specific types of repetitive motion.
– Springs, pendulums, sound
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Chapter 1
Section 1 What is Physics
The areas of Physics (cont)
4. Optics – The study of light.
– Mirrors, lenses, color, astronomy
5. Electromagnetism – The study of electricity, magnetism,
and light.
– Electrical charge, circuitry, permanent magnets,
electromagnets.
6. Relativity – The study of particles moving at any speed,
including very high speed.
– Particle collisions, particle accelerators, nuclear energy.
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Chapter 1
Section 1 What is Physics
The areas of Physics (cont.)
7. Quantum Mechanics – The study of
submicroscopic particles.
– The atom and its parts
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Chapter 1
Types of observations
• Qualitative- descriptive, but not true measurements
– Hot
– Large
• Quantitative- describe with numbers and units
– 100C
– 15 meters
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Chapter 1
Section 1 What is Physics
The Scientific Method
• The scientific method is a way to ask and answer scientific
questions by making observations and doing experiments.
• Steps of the scientific :
– Observation (Ask a Question)
– Collect Data (Do Background Research)
– Construct a Hypothesis (Educated guess)
– Test Your Hypothesis by Doing Experiments
– Analyze Your Data and Draw a Conclusion
• The conclusion is only valid if it can be verified by
other people.
– Communicate Your Results
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Chapter 1
Section 1 What is Physics?
The Scientific Method
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Chapter 1
Section 1 What is Physics
The Scientific Method (cont)
• System – A set of items or interactions considered a
distinct physical entity for the purpose of study.
– Decide what to study and eliminate everything else that
has minimal or no effect on the problem.
– Draw a diagram of what remains (Model)
• Models – A replica or description designed to show the
structure or workings of an object, system, or concept.
– Models help guide experimental design
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Chapter 1
Section 1 What is Physics?
The System
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Chapter 1
Section 1 What is Physics?
The Scientific Model
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Chapter 1
Section 1 What is Physics
The Scientific Method (cont)
• Hypothesis – A reasonable explanation for
observations, one that can be tested with
additional experiments.
– The hypothesis must be tested in a controlled
experiment.
• Controlled Experiment- Only one variable
at a time is changed to determine what
influences the phenomenon you are
observing.
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Chapter 1
Section 2 Measurements in
Experiments
Numbers As Measurements
• Numerical measurements in science contain the
value (number) and Dimension.
• Dimension is the physical quantity being measured
(length, mass, time, temperature, electric current)
• Each dimension is measured using units and prefixes
from the SI system.
• The dimension must match the unit. (ex. If you are
measuring length, use the meter(m), not the
kilogram(kg)
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Chapter 1
Section 2 Measurements in
Experiments
• SI is the standard measurement system for science.
• Used so that scientists can communicate with the
same language.
• There are seven base units. They are:
– Meter(m) – length
– kilogram(kg) – Mass
– Second(s) – Time
– Kelvin(K) – Temperature
– Ampere(A) – current
– Mole(mol) – amount of substance
– Candela(cd) – luminous intensity
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Chapter 1
Section 2 Measurements in
Experiments
• Common Metric Prefixes:
-See handout or visit reference section of website
-Be able to convert between any prefix and another.
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How good are the measurements?
• Scientists use two word to describe how good the
measurements are:
• Accuracy- how close the measurement is to the actual
value.
• Precision- how well can the measurement be repeated.
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Differences
• Accuracy can be true of an individual measurement
or the average of several.
– Problems with accuracy are due to error
• Precision requires several measurements before
anything can be said about it.
– Precision describes the limitation of the measuring
instrument.
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Percent Error
• Percent error =
(Experimental Value – Accepted value) x 100
Accepted Value
• Percent error can be negative.
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Significant Figures
Scientific Notation
Accuracy and Precision
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Pacific
Atlantic
Present
Absent
If the decimal point is absent, start at the
Atlantic (right), find the first non zero, and
count all the rest of the digits
230000
1750
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Pacific
Atlantic
Present
Absent
If the decimal point is PRESENT, start at the
Pacific (left), find the first non zero, and
count all the rest of the digits
0.045
1.2300
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Sig Figs for Addition
27.93 + 6.4

+
First line up the decimal places
27.93 Then do the adding..
Find the estimated
6.4
numbers in the problem.
34.33 This answer must be
rounded to the tenths place.
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Multiplication and Division
• Rule is simpler
• Same number of sig figs in the answer as the least in
the question
• 3.6 x 653
• 2350.8
• 3.6 has 2 s.f. 653 has 3 s.f.
• answer can only have 2 s.f.
• 2400
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Multiplication and Division
•
•
•
•
•
•
•
Same rules for division.
practice
4.5 / 6.245
4.5 x 6.245
9.8764 x .043
3.876 / 1980
16547 / 710
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The Metric System
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The Metric System
•
•
•
•
•
Easier to use because it is a decimal system.
Every conversion is by some power of 10.
A metric unit has two parts.
A prefix and a base unit.
prefix tells you how many times to divide or multiply by
10.
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Base Units
•
•
•
•
•
•
•
Length - meter - m
Mass - grams - g
Time - second - s
Temperature - Kelvin K
Energy - Joules- J
Volume - Liter - L
Amount of substance - mole - mol
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Prefixes
•
•
•
•
•
•
kilo k
deci d
centi c
milli m
micro μ
nano n
1000 times
1/10
1/100
1/1000
1/1000000
1/1000000000
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Volume
•
•
•
•
•
•
•
calculated by multiplying L x W x H
Liter the volume of a cube 1 dm (10 cm) on a side
1L = 1 dm3
so 1 L = 10 cm x 10 cm x 10 cm
1 L = 1000 cm3
1/1000 L = 1 cm3
1 mL = 1 cm3
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Mass
• 1 gram is defined as the mass of 1 cm3 of water at 4
ºC.
• 1000 g = 1000 cm3 of water
• 1 kg = 1 L of water
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Converting
k h D
d c m
• how far you have to move on this chart, tells you how
far, and which direction to move the decimal place.
• The box is the base unit, meters, Liters, grams, etc.
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Conversions
k h D
d c m
• Change 5.6 m to millimeters
starts
at the base unit and move three to
the right.
move the decimal point three to the right
56 00
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Conversions
k h D
•
•
•
•
d c m
convert 25 mg to grams
convert 0.45 km to mm
convert 35 mL to liters
It works because the math works, we are dividing or
multiplying by 10 the correct number of times.
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What about micro- and nano-?
k h D
d c m μ n
3
3
• The jump in between is 3 places
• Convert 15000 μm to m
• Convert 0.00035 cm to nm
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Chapter 2
Motion in One Dimension
Table of Contents
Section 1 Displacement and Velocity
Section 2 Acceleration
Section 3 Falling Objects
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Chapter 2
Section 1 Displacement and
Velocity
One Dimensional Motion
• To simplify the concept of motion, we will first
consider motion that takes place in one
direction.
• One example is the motion of a commuter train
on a straight track.
• To measure motion, you must choose a frame of
reference. A frame of reference is a system for
specifying the precise location of objects in space
and time.
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Chapter 2
Section 1 Displacement and
Velocity
Displacement
• Displacement is a change in position.
• Displacement is not always equal to the distance
traveled.
• The SI unit of displacement is the meter, m.
Dx = xf – xi
displacement = final position – initial position
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Chapter 2
Section 1 Displacement and
Velocity
Positive and Negative Displacements
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Chapter 2
Section 1 Displacement and
Velocity
Average Velocity
• Average velocity is the total displacement
divided by the time interval during which the
displacement occurred.
vavg
Dx x f  xi


Dt
t f  ti
change in position
displacement
average velocity =
=
change in time
time interval
• In SI, the unit of velocity is meters per second,
abbreviated as m/s.
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Chapter 2
Section 1 Displacement and
Velocity
Velocity and Speed
• Velocity describes motion with both a direction
and a numerical value (a magnitude).
• Speed has no direction, only magnitude.
• Average speed is equal to the total distance
traveled divided by the time interval.
distance traveled
average speed =
time of travel
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Chapter 2
Section 1 Displacement and
Velocity
Interpreting Velocity Graphically
• For any position-time graph, we can determine
the average velocity by drawing a straight line
between any two points on the graph.
• If the velocity is constant, the graph
of position versus time is a straight
line. The slope indicates the velocity.
– Object 1: positive slope = positive
velocity
– Object 2: zero slope= zero velocity
– Object 3: negative slope = negative
velocity
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Chapter 2
Section 1 Displacement and
Velocity
Interpreting Velocity Graphically, continued
The instantaneous velocity is the velocity of
an object at some instant or at a specific point
in the object’s path.
The instantaneous
velocity at a given time
can be determined by
measuring the slope of
the line that is tangent
to that point on the
position-versus-time
graph.
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Chapter 2
Section 2 Acceleration
Changes in Velocity
• Acceleration is the rate at which velocity changes
over time.
aavg
Dv v f  vi


Dt t f  ti
change in velocity
average acceleration =
time required for change
• An object accelerates if its speed, direction, or both
change.
• Acceleration has direction and magnitude. Thus,
acceleration is a vector quantity.
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Chapter 2
Section 2 Acceleration
Changes in Velocity, continued
• Consider a train moving to the right, so that the
displacement and the velocity are positive.
• The slope of the velocity-time graph is the average
acceleration.
– When the velocity in the positive
direction is increasing, the
acceleration is positive, as at A.
– When the velocity is constant, there is
no acceleration, as at B.
– When the velocity in the positive
direction is decreasing, the
acceleration is negative, as at C.
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Chapter 2
Section 2 Acceleration
Velocity and Acceleration
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Chapter 2
Section 2 Acceleration
Motion with Constant Acceleration
• When velocity changes by the same amount during
each time interval, acceleration is constant.
• The relationships between displacement, time,
velocity, and constant acceleration are expressed
by the equations shown on the next slide. These
equations apply to any object moving with constant
acceleration.
• These equations use the following symbols:
Dx = displacement
vi = initial velocity
vf = final velocity
Dt = time interval
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Chapter 2
Section 2 Acceleration
Equations for Constantly Accelerated
Straight-Line Motion
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Chapter 2
Section 2 Acceleration
Sample Problem
Final Velocity After Any Displacement
A person pushing a stroller starts from rest, uniformly
accelerating at a rate of 0.500 m/s2. What is the
velocity of the stroller after it has traveled 4.75 m?
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Chapter 2
Section 2 Acceleration
Sample Problem, continued
1. Define
Given:
vi = 0 m/s
a = 0.500 m/s2
Dx = 4.75 m
Unknown:
vf = ?
Diagram: Choose a coordinate system. The most
convenient one has an origin at the initial location
of the stroller, as shown above. The positive
direction is to the right.
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Chapter 2
Section 2 Acceleration
Sample Problem, continued
2. Plan
Choose an equation or situation: Because the initial
velocity, acceleration, and displacement are known,
the final velocity can be found using the following
equation:
v f 2  vi 2  2aDx
Rearrange the equation to isolate the unknown:
Take the square root of both sides to isolate vf .
v f   vi 2  2aDx
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Chapter 2
Section 2 Acceleration
Sample Problem, continued
3. Calculate
Substitute the values into the equation and solve:
v f   (0 m/s)2  2(0.500 m/s2 )(4.75 m)
v f  2.18 m/s
4. Evaluate
Tip: Think about the physical situation to
determine whether to keep the positive or
negative answer from the square root. In this
case, the stroller starts from rest and ends
with a speed of 2.18 m/s. An object that is
speeding up and has a positive acceleration
must have a positive velocity. So, the final
velocity must be positive.
The stroller’s velocity
after accelerating for 4.75 m is 2.18 m/s to the right.
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Chapter 2
Section 3 Falling Objects
Free Fall
• Free fall is the motion of a body when only the force
due to gravity is acting on the body.
• The acceleration on an object in free fall is called the
acceleration due to gravity, or free-fall
acceleration.
• Free-fall acceleration is denoted with the symbols ag
(generally) or g (on Earth’s surface).
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Chapter 2
Section 3 Falling Objects
Free-Fall Acceleration
• Free-fall acceleration is the same for all objects,
regardless of mass.
• This book will use the value g = 9.81 m/s2.
• Free-fall acceleration on Earth’s surface is –9.81 m/s2
at all points in the object’s motion.
• Consider a ball thrown up into the air.
– Moving upward: velocity is decreasing, acceleration is –
9.81 m/s2
– Top of path: velocity is zero, acceleration is –9.81 m/s2
– Moving downward: velocity is increasing, acceleration is –
9.81 m/s2
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Chapter 2
Section 3 Falling Objects
Sample Problem
Falling Object
Jason hits a volleyball so that it moves with an initial
velocity of 6.0 m/s straight upward. If the volleyball
starts from 2.0 m above the floor, how long will it be
in the air before it strikes the floor?
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Chapter 2
Section 3 Falling Objects
Sample Problem, continued
1. Define
Given:
vi = +6.0 m/s
a = –g = –9.81 m/s2
Dy = –2.0 m
Unknown:
Dt = ?
Diagram:
Place the origin at the
Starting point of the ball
(yi = 0 at ti = 0).
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Chapter 2
Section 3 Falling Objects
Sample Problem, continued
2. Plan
Choose an equation or situation:
Both ∆t and vf are unknown. Therefore, first solve for vf using
the equation that does not require time. Then, the equation for
vf that does involve time can be used to solve for ∆t.
v f 2  vi 2  2aDy
v f  vi  a Dt
Rearrange the equation to isolate the unknown:
Take the square root of the first equation to isolate vf. The second
equation must be rearranged to solve for ∆t.
v f   vi  2aDy
2
Dt 
v f  vi
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Chapter 2
Section 3 Falling Objects
Sample Problem, continued
3. Calculate
Substitute the values into the equation and solve:
First find the velocity of the ball at the moment that it hits the floor.
v f   vi 2  2aDy   (6.0 m/s)2  2(–9.81 m/s2 )(–2.0 m)
v f   36 m2 /s2  39 m2 /s2   75 m2 /s2  –8.7 m/s
Tip: When you take the square root to find vf , select the
negative answer because the ball will be moving toward the
floor, in the negative direction.
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Chapter 2
Section 3 Falling Objects
Sample Problem, continued
Next, use this value of vf in the second equation to solve for ∆t.
Dt 
v f  vi
a

–8.7 m/s  6.0 m/s –14.7 m/s

2
–9.81 m/s
–9.81 m/s 2
Dt  1.50 s
4. Evaluate
The solution, 1.50 s, is a reasonable amount of time for the ball
to be in the air.
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Chapter 2
Standardized Test Prep
Multiple Choice
Use the graphs to answer questions 1–3.
1. Which graph
represents an
object moving
with a constant
positive velocity?
A. I
B. II
C. III
D. IV
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Chapter 2
Standardized Test Prep
Multiple Choice
Use the graphs to answer questions 1–3.
1. Which graph
represents an
object moving
with a constant
positive velocity?
A. I
B. II
C. III
D. IV
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the graphs to answer questions 1–3.
2. Which graph
represents an
object at rest?
F. I
G. II
H. III
J. IV
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the graphs to answer questions 1–3.
2. Which graph
represents an
object at rest?
F. I
G. II
H. III
J. IV
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the graphs to answer questions 1–3.
3. Which graph
represents an
object moving
with a constant
positive
acceleration?
A. I
B. II
C. III
D. IV
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the graphs to answer questions 1–3.
3. Which graph
represents an
object moving
with a constant
positive
acceleration?
A. I
B. II
C. III
D. IV
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
4. A bus travels from El Paso, Texas, to
Chihuahua, Mexico, in 5.2 h with an average
velocity of 73 km/h to the south.What is the
bus’s displacement?
F. 73 km to the south
G. 370 km to the south
H. 380 km to the south
J. 14 km/h to the south
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
4. A bus travels from El Paso, Texas, to
Chihuahua, Mexico, in 5.2 h with an average
velocity of 73 km/h to the south.What is the
bus’s displacement?
F. 73 km to the south
G. 370 km to the south
H. 380 km to the south
J. 14 km/h to the south
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the position-time graph of a squirrel
running along a clothesline to answer questions 5–6.
5. What is the squirrel’s
displacement at time
t = 3.0 s?
A. –6.0 m
B. –2.0 m
C. +0.8 m
D. +2.0 m
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the position-time graph of a squirrel
running along a clothesline to answer questions 5–6.
5. What is the squirrel’s
displacement at time
t = 3.0 s?
A. –6.0 m
B. –2.0 m
C. +0.8 m
D. +2.0 m
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the position-time graph of a squirrel
running along a clothesline to answer questions 5–6.
6. What is the squirrel’s
average velocity
during the time
interval between 0.0 s
and 3.0 s?
F. –2.0 m/s
G. –0.67 m/s
H. 0.0 m/s
J. +0.53 m/s
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
Use the position-time graph of a squirrel
running along a clothesline to answer questions 5–6.
6. What is the squirrel’s
average velocity
during the time
interval between 0.0 s
and 3.0 s?
F. –2.0 m/s
G. –0.67 m/s
H. 0.0 m/s
J. +0.53 m/s
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
7. Which of the following statements is true of
acceleration?
A. Acceleration always has the same sign as
displacement.
B. Acceleration always has the same sign as
velocity.
C. The sign of acceleration depends on both
the direction of motion and how the velocity
is changing.
D. Acceleration always has a positive sign.
•
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
7. Which of the following statements is true of
acceleration?
A. Acceleration always has the same sign as
displacement.
B. Acceleration always has the same sign as
velocity.
C. The sign of acceleration depends on both
the direction of motion and how the velocity
is changing.
D. Acceleration always has a positive sign.
•
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
8. A ball initially at rest rolls down a hill and has an
acceleration of 3.3 m/s2. If it accelerates for 7.5 s,
how far will it move during this time?
F. 12 m
G. 93 m
H. 120 m
J. 190 m
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
8. A ball initially at rest rolls down a hill and has an
acceleration of 3.3 m/s2. If it accelerates for 7.5 s,
how far will it move during this time?
F. 12 m
G. 93 m
H. 120 m
J. 190 m
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
9. Which of the following statements is true for a ball
thrown vertically upward?
A. The ball has a negative acceleration on the way
up and a positive acceleration on the way down.
B. The ball has a positive acceleration on the way
up and a negative acceleration on the way down.
C. The ball has zero acceleration on the way up and
a positive acceleration on the way down.
D. The ball has a constant acceleration throughout
its flight.
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Chapter 2
Standardized Test Prep
Multiple Choice, continued
9. Which of the following statements is true for a ball
thrown vertically upward?
A. The ball has a negative acceleration on the way
up and a positive acceleration on the way down.
B. The ball has a positive acceleration on the way
up and a negative acceleration on the way down.
C. The ball has zero acceleration on the way up and
a positive acceleration on the way down.
D. The ball has a constant acceleration throughout
its flight.
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Chapter 2
Standardized Test Prep
Short Response
10. In one or two sentences, explain the difference
between displacement and distance traveled.
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Chapter 2
Standardized Test Prep
Short Response
10. In one or two sentences, explain the difference
between displacement and distance traveled.
Answer:
Displacement measures only the net change in
position from starting point to end point. The
distance traveled is the total length of the path
followed from starting point to end point and may be
greater than or equal to the displacement.
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Chapter 2
Standardized Test Prep
Short Response, continued
11. The graph shows the position of a runner at
different times during a run. Use the graph to
determine the runner’s displacement and average
velocity:
a. for the time interval from
t = 0.0 min to t = 10.0 min
b. for the time interval from
t = 10.0 min to t = 20.0 min
c. for the time interval from
t = 20.0 min to t = 30.0 min
d. for the entire run
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Chapter 2
Standardized Test Prep
Short Response, continued
11. The graph shows the position of a runner at different times
during a run. Use the graph to determine the runner’s
displacement and average velocity. Answers will vary but
should be approximately as follows:
a. for t = 0.0 min to t = 10.0 min
Answer: +2400 m, +4.0 m/s
b. for t = 10.0 min to t = 20.0 min
Answer: +1500 m, +2.5 m/s
c. for t = 20.0 min to t = 30.0 min
Answer: +900 m, +2 m/s
d. for the entire run
Answer: +4800 m, +2.7 m/s
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Chapter 2
Standardized Test Prep
Short Response, continued
12. For an object moving with constant negative
acceleration, draw the following:
a. a graph of position vs. time
b. a graph of velocity vs. time
For both graphs, assume the object starts with a
positive velocity and a positive displacement from
the origin.
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Chapter 2
Standardized Test Prep
Short Response, continued
12. For an object moving with constant negative
acceleration, draw the following:
a. a graph of position vs. time
b. a graph of velocity vs. time
For both graphs, assume the object starts with a
positive velocity and a positive displacement from
the origin.
Answers:
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Chapter 2
Standardized Test Prep
Short Response, continued
13. A snowmobile travels in a straight line. The
snowmobile’s initial velocity is +3.0 m/s.
a. If the snowmobile accelerates at a rate of
+0.50 m/s2 for 7.0 s, what is its final velocity?
b. If the snowmobile accelerates at the rate of
–0.60 m/s2 from its initial velocity of +3.0 m/s,
how long will it take to reach a complete stop?
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Chapter 2
Standardized Test Prep
Short Response, continued
13. A snowmobile travels in a straight line. The
snowmobile’s initial velocity is +3.0 m/s.
a. If the snowmobile accelerates at a rate of
+0.50 m/s2 for 7.0 s, what is its final velocity?
b. If the snowmobile accelerates at the rate of
–0.60 m/s2 from its initial velocity of +3.0 m/s,
how long will it take to reach a complete stop?
Answers: a. +6.5 m/s
b. 5.0 s
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Chapter 2
Standardized Test Prep
Extended Response
14. A car moving eastward along a straight road
increases its speed uniformly from 16 m/s to 32 m/s
in 10.0 s.
a. What is the car’s average acceleration?
b. What is the car’s average velocity?
c. How far did the car move while accelerating?
Show all of your work for these calculations.
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Chapter 2
Standardized Test Prep
Extended Response
14. A car moving eastward along a straight road
increases its speed uniformly from 16 m/s to 32 m/s
in 10.0 s.
a. What is the car’s average acceleration?
b. What is the car’s average velocity?
c. How far did the car move while accelerating?
Answers: a. 1.6 m/s2 eastward
b. 24 m/s
c. 240 m
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Chapter 2
Standardized Test Prep
Extended Response, continued
15. A ball is thrown vertically upward with a speed of
25.0 m/s from a height of 2.0 m.
a. How long does it take the ball to reach its highest
point?
b. How long is the ball in the air?
Show all of your work for these calculations.
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Chapter 2
Standardized Test Prep
Extended Response, continued
15. A ball is thrown vertically upward with a speed of
25.0 m/s from a height of 2.0 m.
a. How long does it take the ball to reach its highest
point?
b. How long is the ball in the air?
Show all of your work for these calculations.
Answers: a. 2.55 s
b. 5.18 s
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Chapter 3
Two-Dimensional Motion and Vectors
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Chapter 3
Two-Dimensional Motion and Vectors
Table of Contents
Section 1 Introduction to Vectors
Section 2 Vector Operations
Section 3 Projectile Motion
Section 4 Relative Motion
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Chapter 3
Section 1 Introduction to Vectors
Scalars and Vectors
• A scalar is a physical quantity that has magnitude
but no direction.
– Examples: speed, volume, the number of pages
in your textbook
• A vector is a physical quantity that has both
magnitude and direction.
– Examples: displacement, velocity, acceleration
• In this book, scalar quantities are in italics. Vectors
are represented by boldface symbols.
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Chapter 3
Section 1 Introduction to Vectors
Graphical Addition of Vectors
• A resultant vector represents the sum of two or
more vectors.
• Vectors can be added graphically.
A student walks from his
house to his friend’s house
(a), then from his friend’s
house to the school (b).
The student’s resultant
displacement (c) can be
found by using a ruler and a
protractor.
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Chapter 3
Section 1 Introduction to Vectors
Triangle Method of Addition
• Vectors can be moved parallel to themselves in a
diagram.
• Thus, you can draw one vector with its tail starting at
the tip of the other as long as the size and direction
of each vector do not change.
• The resultant vector can then be drawn from the tail
of the first vector to the tip of the last vector.
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Chapter 3
Section 1 Introduction to Vectors
Properties of Vectors
• Vectors can be added in any order.
• To subtract a vector, add its opposite.
• Multiplying or dividing vectors by scalars results in
vectors.
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Chapter 3
Section 2 Vector Operations
Coordinate Systems in Two Dimensions
• One method for diagraming
the motion of an object
employs vectors and the use
of the x- and y-axes.
• Axes are often designated
using fixed directions.
• In the figure shown here, the
positive y-axis points north
and the positive x-axis points
east.
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Chapter 3
Section 2 Vector Operations
Determining Resultant Magnitude and
Direction
• In Section 1, the magnitude and direction of a
resultant were found graphically.
• With this approach, the accuracy of the answer
depends on how carefully the diagram is drawn
and measured.
• A simpler method uses the Pythagorean theorem
and the tangent function.
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Section 2 Vector Operations
Chapter 3
Determining Resultant Magnitude and
Direction, continued
The Pythagorean Theorem
• Use the Pythagorean theorem to find the magnitude of the
resultant vector.
• The Pythagorean theorem states that for any right triangle,
the square of the hypotenuse—the side opposite the right
angle—equals the sum of the squares of the other two
sides, or legs.
c  a b
2
2
2
(hypotenuse)2  (leg 1)2  (leg 2)2
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Chapter 3
Section 2 Vector Operations
Determining Resultant Magnitude and
Direction, continued
The Tangent Function
• Use the tangent function to find the direction of the
resultant vector.
• For any right triangle, the tangent of an angle is defined as
the ratio of the opposite and adjacent legs with respect to
a specified acute angle of a right triangle.
opposite leg
tangent of angle  =
adjacent leg
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Chapter 3
Section 2 Vector Operations
Sample Problem
Finding Resultant Magnitude and Direction
An archaeologist climbs the Great Pyramid in
Giza, Egypt. The pyramid’s height is 136 m and its
width is 2.30  102 m. What is the magnitude and
the direction of the displacement of the
archaeologist after she has climbed from the
bottom of the pyramid to the top?
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
1. Define
Given:
Dy = 136 m
Dx = 1/2(width) = 115 m
Unknown:
d= ?
=?
Diagram:
Choose the archaeologist’s starting
position as the origin of the coordinate
system, as shown above.
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
2. Plan
Choose an equation or situation: The
Pythagorean theorem can be used to find the
magnitude of the archaeologist’s displacement.
The direction of the displacement can be found by
using the inverse tangent function.
Dy
2
2
2
d  Dx  Dy
tan  
Dx
Rearrange the equations to isolate the unknowns:
2
2
–1  Dy 
d  Dx  Dy
  tan  
 Dx 
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
3. Calculate
d  Dx  Dy
2
2
d  (115 m) 2  (136 m) 2
d  178 m
 Dy 
  tan  
 Dx 
–1  136 m 
  tan 

115


  49.8
–1
4. Evaluate
Because d is the hypotenuse, the archaeologist’s
displacement should be less than the sum of the height and
half of the width. The angle is expected to be more than 45
because the height is greater than half of the width.
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Chapter 3
Section 2 Vector Operations
Resolving Vectors into Components
• You can often describe an object’s motion more
conveniently by breaking a single vector into two
components, or resolving the vector.
• The components of a vector are the projections
of the vector along the axes of a coordinate
system.
• Resolving a vector allows you to analyze the
motion in each direction.
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Chapter 3
Section 2 Vector Operations
Resolving Vectors into Components, continued
Consider an airplane flying at 95 km/h.
• The hypotenuse (vplane) is the resultant vector
that describes the airplane’s total velocity.
• The adjacent leg represents the x component
(vx), which describes the airplane’s horizontal
speed.
•
The opposite leg represents
the y component (vy),
which describes the
airplane’s vertical speed.
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Chapter 3
Section 2 Vector Operations
Resolving Vectors into Components, continued
• The sine and cosine functions can be used to
find the components of a vector.
• The sine and cosine functions are defined in terms
of the lengths of the sides of right triangles.
opposite leg
sine of angle  =
hypotenuse
adjacent leg
cosine of angle  =
hypotenuse
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Section 2 Vector Operations
Chapter 3
Adding Vectors That Are Not Perpendicular
• Suppose that a plane travels first 5 km at an angle
of 35°, then climbs at 10° for 22 km, as shown
below. How can you find the total displacement?
• Because the original displacement vectors do not
form a right triangle, you can not directly apply the
tangent function or the Pythagorean theorem.
d2
d1
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Chapter 3
Section 2 Vector Operations
Adding Vectors That Are Not Perpendicular,
continued
• You can find the magnitude and the direction of
the resultant by resolving each of the plane’s
displacement vectors into its x and y components.
• Then the components along each axis can be
added together.
As shown in the figure, these sums will
be the two perpendicular components
of the resultant, d. The resultant’s
magnitude can then be found by using
the Pythagorean theorem, and its
direction can be found by using the
inverse tangent function.
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Chapter 3
Section 2 Vector Operations
Sample Problem
Adding Vectors Algebraically
A hiker walks 27.0 km from her base camp at 35°
south of east. The next day, she walks 41.0 km in
a direction 65° north of east and discovers a forest
ranger’s tower. Find the magnitude and direction
of her resultant displacement
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
1 . Select a coordinate system. Then sketch and
label each vector.
Given:
d1 = 27.0 km
d2 = 41.0 km
1 = –35°
2 = 65°
Tip: 1 is negative, because clockwise
movement from the positive x-axis
is negative by convention.
Unknown:
d=?
=?
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
2 . Find the x and y components of all vectors.
Make a separate sketch of the
displacements for each day. Use the cosine
and sine functions to find the components.
For day 1 :
Dx1  d1 cos1  (27.0 km)(cos –35) = 22 km
Dy1  d1 sin 1  (27.0 km)(sin –35) = –15 km
For day 2 :
Dx2  d2 cos 2  (41.0 km)(cos 65) = 17 km
Dy2  d2 sin  2  (41.0 km)(sin 65) = 37 km
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
3 . Find the x and y components of the total
displacement.
Dxtot  Dx1  Dx2  22 km + 17 km = 39 km
Dytot  Dy1  Dy2  –15 km + 37 km = 22 km
4 . Use the Pythagorean theorem to find the
magnitude of the resultant vector.
d 2  (Dxtot )2  (Dytot )2
d  (Dxtot )2  (Dytot )2  (39 km)2  (22 km)2
d  45 km
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Chapter 3
Section 2 Vector Operations
Sample Problem, continued
5 . Use a suitable trigonometric function to find
the angle.
 Dy 
–1  22 km 
  tan   = tan 

D
x
39
km
 


  29 north of east
–1
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Chapter 3
Section 3 Projectile Motion
Projectiles
• Objects that are thrown or launched into the air
and are subject to gravity are called projectiles.
• Projectile motion is the curved path that an
object follows when thrown, launched,or otherwise
projected near the surface of Earth.
• If air resistance is disregarded, projectiles follow
parabolic trajectories.
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Chapter 3
Section 3 Projectile Motion
Projectiles, continued
• Projectile motion is free fall
with an initial horizontal
velocity.
• The yellow ball is given an
initial horizontal velocity and
the red ball is dropped. Both
balls fall at the same rate.
– In this book, the horizontal
velocity of a projectile will be
considered constant.
– This would not be the case if we
accounted for air resistance.
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Chapter 3
Section 3 Projectile Motion
Kinematic Equations for Projectiles
• How can you know the displacement, velocity, and
acceleration of a projectile at any point in time
during its flight?
• One method is to resolve vectors into components,
then apply the simpler one-dimensional forms of
the equations for each component.
• Finally, you can recombine the components to
determine the resultant.
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Chapter 3
Section 3 Projectile Motion
Kinematic Equations for Projectiles, continued
• To solve projectile problems, apply the
kinematic equations in the horizontal and
vertical directions.
• In the vertical direction, the acceleration ay will
equal –g (–9.81 m/s2) because the only vertical
component of acceleration is free-fall
acceleration.
• In the horizontal direction, the acceleration is
zero, so the velocity is constant.
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Chapter 3
Section 3 Projectile Motion
Kinematic Equations for Projectiles, continued
• Projectiles Launched Horizontally
– The initial vertical velocity is 0.
– The initial horizontal velocity is the initial velocity.
• Projectiles Launched At An Angle
– Resolve the initial velocity into x
and y components.
– The initial vertical velocity is the y
component.
– The initial horizontal velocity is
the x component.
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Chapter 3
Section 3 Projectile Motion
Sample Problem
Projectiles Launched At An Angle
A zookeeper finds an escaped monkey hanging
from a light pole. Aiming her tranquilizer gun at the
monkey, she kneels 10.0 m from the light
pole,which is 5.00 m high. The tip of her gun is
1.00 m above the ground. At the same moment
that the monkey drops a banana, the zookeeper
shoots. If the dart travels at 50.0 m/s,will the dart
hit the monkey, the banana, or neither one?
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Chapter 3
Section 3 Projectile Motion
Sample Problem, continued
1 . Select a coordinate system.
The positive y-axis points up, and the positive xaxis points along the ground toward the pole.
Because the dart leaves the gun at a height of
1.00 m, the vertical distance is 4.00 m.
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Chapter 3
Section 3 Projectile Motion
Sample Problem, continued
2 . Use the inverse tangent function to find the
angle that the initial velocity makes with the xaxis.
 Dy 
1  4.00 m 

tan


  21.8
 Dx 
 10.0 m 
  tan 1 
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Chapter 3
Section 3 Projectile Motion
Sample Problem, continued
3 . Choose a kinematic equation to solve for time.
Rearrange the equation for motion along the xaxis to isolate the unknown Dt, which is the time
the dart takes to travel the horizontal distance.
Dx  (vi cos  )Dt
Dt 
Dx
10.0 m

 0.215 s
vi cos  (50.0 m/s)( cos 21.8)
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Chapter 3
Section 3 Projectile Motion
Sample Problem, continued
4 . Find out how far each object will fall during
this time. Use the free-fall kinematic equation in
both cases.
For the banana, vi = 0. Thus:
Dyb = ½ay(Dt)2 = ½(–9.81 m/s2)(0.215 s)2 = –0.227 m
The dart has an initial vertical component of velocity equal to vi
sin , so:
Dyd = (vi sin )(Dt) + ½ay(Dt)2
Dyd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81 m/s2)(0.215 s)2
Dyd = 3.99 m – 0.227 m = 3.76 m
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Chapter 3
Section 3 Projectile Motion
Sample Problem, continued
5 . Analyze the results.
Find the final height of both the banana and the dart.
ybanana, f = yb,i+ Dyb = 5.00 m + (–0.227 m)
ybanana, f = 4.77 m above the ground
ydart, f = yd,i+ Dyd = 1.00 m + 3.76 m
ydart, f = 4.76 m above the ground
The dart hits the banana. The slight difference is due to
rounding.
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Chapter 3
Section 4 Relative Motion
Objectives
• Describe situations in terms of frame of reference.
• Solve problems involving relative velocity.
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Chapter 3
Section 4 Relative Motion
Frames of Reference
• If you are moving at 80 km/h north and a car
passes you going 90 km/h, to you the faster car
seems to be moving north at 10 km/h.
• Someone standing on the side of the road would
measure the velocity of the faster car as 90 km/h
toward the north.
• This simple example demonstrates that velocity
measurements depend on the frame of reference
of the observer.
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Chapter 3
Section 4 Relative Motion
Frames of Reference, continued
Consider a stunt dummy dropped from a plane.
(a) When viewed from the plane, the stunt dummy falls straight
down.
(b) When viewed from a stationary position on the ground, the
stunt dummy follows a parabolic projectile path.
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Chapter 3
Section 4 Relative Motion
Relative Velocity
• When solving relative velocity problems, write down the
information in the form of velocities with subscripts.
• Using our earlier example, we have:
• vse = +80 km/h north (se = slower car with respect to
Earth)
• vfe = +90 km/h north (fe = fast car with respect to Earth)
• unknown = vfs (fs = fast car with respect to slower car)
• Write an equation for vfs in terms of the other velocities. The
subscripts start with f and end with s. The other subscripts
start with the letter that ended the preceding velocity:
• vfs = vfe + ves
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Chapter 3
Section 4 Relative Motion
Relative Velocity, continued
• An observer in the slow car perceives Earth as moving south
at a velocity of 80 km/h while a stationary observer on the
ground (Earth) views the car as moving north at a velocity of
80 km/h. In equation form:
• ves = –vse
• Thus, this problem can be solved as follows:
• vfs = vfe + ves = vfe – vse
• vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n
• A general form of the relative velocity equation is:
• vac = vab + vbc
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Chapter 3
Standardized Test Prep
Multiple Choice
1. Vector A has a magnitude of 30 units. Vector B is
perpendicular to vector A and has a magnitude of 40
units. What would the magnitude of the resultant
vector A + B be?
A. 10 units
B. 50 units
C. 70 units
D. zero
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Chapter 3
Standardized Test Prep
Multiple Choice
1. Vector A has a magnitude of 30 units. Vector B is
perpendicular to vector A and has a magnitude of 40
units. What would the magnitude of the resultant
vector A + B be?
A. 10 units
B. 50 units
C. 70 units
D. zero
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
2. What term represents the magnitude of a velocity
vector?
F. acceleration
G. momentum
H. speed
J. velocity
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
2. What term represents the magnitude of a velocity
vector?
F. acceleration
G. momentum
H. speed
J. velocity
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer questions 3–4.
3. What is the direction of the
resultant vector A + B?
A. 15º above the x-axis
B. 75º above the x-axis
C. 15º below the x-axis
D. 75º below the x-axis
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer questions 3–4.
3. What is the direction of the
resultant vector A + B?
A. 15º above the x-axis
B. 75º above the x-axis
C. 15º below the x-axis
D. 75º below the x-axis
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer questions 3–4.
4. What is the direction of the
resultant vector A – B?
F. 15º above the x-axis
G. 75º above the x-axis
H. 15º below the x-axis
J. 75º below the x-axis
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer questions 3–4.
4. What is the direction of the
resultant vector A – B?
F. 15º above the x-axis
G. 75º above the x-axis
H. 15º below the x-axis
J. 75º below the x-axis
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A motorboat heads due east at 5.0 m/s across a river
that flows toward the south at a speed of 5.0 m/s.
5. What is the resultant velocity relative to an observer
on the shore ?
A. 3.2 m/s to the southeast
B. 5.0 m/s to the southeast
C. 7.1 m/s to the southeast
D. 10.0 m/s to the southeast
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A motorboat heads due east at 5.0 m/s across a river
that flows toward the south at a speed of 5.0 m/s.
5. What is the resultant velocity relative to an observer
on the shore ?
A. 3.2 m/s to the southeast
B. 5.0 m/s to the southeast
C. 7.1 m/s to the southeast
D. 10.0 m/s to the southeast
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A motorboat heads due east at 5.0 m/s across a river
that flows toward the south at a speed of 5.0 m/s.
6. If the river is 125 m wide, how long does the boat
take to cross the river?
F. 39 s
G. 25 s
H. 17 s
J. 12 s
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A motorboat heads due east at 5.0 m/s across a river
that flows toward the south at a speed of 5.0 m/s.
6. If the river is 125 m wide, how long does the boat
take to cross the river?
F. 39 s
G. 25 s
H. 17 s
J. 12 s
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
7. The pilot of a plane measures an air velocity of 165
km/h south relative to the plane. An observer on the
ground sees the plane pass overhead at a velocity of
145 km/h toward the north.What is the velocity of the
wind that is affecting the plane relative to the
observer?
A. 20 km/h to the north
B. 20 km/h to the south
C. 165 km/h to the north
D. 310 km/h to the south
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
7. The pilot of a plane measures an air velocity of 165
km/h south relative to the plane. An observer on the
ground sees the plane pass overhead at a velocity of
145 km/h toward the north.What is the velocity of the
wind that is affecting the plane relative to the
observer?
A. 20 km/h to the north
B. 20 km/h to the south
C. 165 km/h to the north
D. 310 km/h to the south
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
8. A golfer takes two putts to sink his ball in the hole
once he is on the green. The first putt displaces the
ball 6.00 m east, and the second putt displaces the
ball 5.40 m south. What displacement would put the
ball in the hole in one putt?
F. 11.40 m southeast
G. 8.07 m at 48.0º south of east
H. 3.32 m at 42.0º south of east
J. 8.07 m at 42.0º south of east
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
8. A golfer takes two putts to sink his ball in the hole
once he is on the green. The first putt displaces the
ball 6.00 m east, and the second putt displaces the
ball 5.40 m south. What displacement would put the
ball in the hole in one putt?
F. 11.40 m southeast
G. 8.07 m at 48.0º south of east
H. 3.32 m at 42.0º south of east
J. 8.07 m at 42.0º south of east
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
9. What is the initial speed of the girl’s ball relative to
the boy?
A. 1.0 m/s
C. 2.0 m/s
B. 1.5 m/s
D. 3.0 m/s
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
9. What is the initial speed of the girl’s ball relative to
the boy?
A. 1.0 m/s
C. 2.0 m/s
B. 1.5 m/s
D. 3.0 m/s
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
10. If air resistance is disregarded, which ball will hit the
ground first?
F. the boy’s ball
H. neither
G. the girl’s ball
J. cannot be determined
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
10. If air resistance is disregarded, which ball will hit the
ground first?
F. the boy’s ball
H. neither
G. the girl’s ball
J. cannot be determined
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
11. If air resistance is disregarded, which ball will have
a greater speed (relative to the ground) when it hits
the ground?
A. the boy’s ball
C. neither
B. the girl’s ball
D. cannot be determined
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
11. If air resistance is disregarded, which ball will have
a greater speed (relative to the ground) when it hits
the ground?
A. the boy’s ball
C. neither
B. the girl’s ball
D. cannot be determined
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
12. What is the speed of the girl’s ball when it hits the
ground?
F. 1.0 m/s
H. 6.2 m/s
G. 3.0 m/s
J. 8.4 m/s
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Chapter 3
Standardized Test Prep
Multiple Choice, continued
Use the passage to answer questions 9–12.
A girl riding a bicycle at 2.0 m/s throws a tennis ball
horizontally forward at a speed of 1.0 m/s from a
height of 1.5 m. At the same moment, a boy standing
on the sidewalk drops a tennis ball straight down
from a height of 1.5 m.
12. What is the speed of the girl’s ball when it hits the
ground?
F. 1.0 m/s
H. 6.2 m/s
G. 3.0 m/s
J. 8.4 m/s
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Chapter 3
Standardized Test Prep
Short Response
13. If one of the components of one vector along the
direction of another vector is zero, what can you
conclude about these two vectors?
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Chapter 3
Standardized Test Prep
Short Response
13. If one of the components of one vector along the
direction of another vector is zero, what can you
conclude about these two vectors?
Answer: They are perpendicular.
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Chapter 3
Standardized Test Prep
Short Response, continued
14. A roller coaster travels 41.1 m at an angle of 40.0°
above the horizontal. How far does it move
horizontally and vertically?
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Chapter 3
Standardized Test Prep
Short Response, continued
14. A roller coaster travels 41.1 m at an angle of 40.0°
above the horizontal. How far does it move
horizontally and vertically?
Answer: 31.5 m horizontally, 26.4 m vertically
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Chapter 3
Standardized Test Prep
Short Response, continued
15. A ball is thrown straight upward and returns to the
thrower’s hand after 3.00 s in the air. A second ball
is thrown at an angle of 30.0° with the horizontal. At
what speed must the second ball be thrown to reach
the same height as the one thrown vertically?
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Chapter 3
Standardized Test Prep
Short Response, continued
15. A ball is thrown straight upward and returns to the
thrower’s hand after 3.00 s in the air. A second ball
is thrown at an angle of 30.0° with the horizontal. At
what speed must the second ball be thrown to reach
the same height as the one thrown vertically?
Answer: 29.4 m/s
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Chapter 3
Standardized Test Prep
Extended Response
16. A human cannonball is shot out of a cannon at 45.0°
to the horizontal with an initial speed of 25.0 m/s. A
net is positioned at a horizontal distance of 50.0 m
from the cannon. At what height above the cannon
should the net be placed in order to catch the
human cannonball? Show your work.
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Chapter 3
Standardized Test Prep
Extended Response
16. A human cannonball is shot out of a cannon at 45.0°
to the horizontal with an initial speed of 25.0 m/s. A
net is positioned at a horizontal distance of 50.0 m
from the cannon. At what height above the cannon
should the net be placed in order to catch the
human cannonball? Show your work.
Answer: 10.8 m
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Chapter 3
Standardized Test Prep
Extended Response, continued
Read the following passage to answer question 17.
Three airline executives are discussing ideas for
developing flights that are more energy efficient.
Executive A: Because the Earth rotates from west to
east, we could operate “static flights”—a helicopter or
airship could begin by rising straight up from New
York City and then descend straight down four hours
later when San Francisco arrives below.
Executive B: This approach could work for one-way
flights, but the return trip would take 20 hours.
continued on the next slide
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Chapter 3
Standardized Test Prep
Extended Response, continued
Executive C: That approach will never work. Think
about it.When you throw a ball straight up in the air, it
comes straight back down to the same point.
Executive A: The ball returns to the same point
because Earth’s motion is not significant during such
a short time.
17. State which of the executives is correct, and explain
why.
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Chapter 3
Standardized Test Prep
Extended Response, continued
17. State which of the executives is correct, and explain
why.
Answer: Executive C is correct. Explanations should
include the concept of relative velocity—when a
helicopter lifts off straight up from the ground, it is
already moving horizontally with Earth’s horizontal
velocity. (We assume that Earth’s motion is constant
for the purposes of this scenario and does not
depend on time.)
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Chapter 3
Section 3 Projectile Motion
Projectiles
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Chapter 5
Work and Energy
Table of Contents
Section 1 Work
Section 2 Energy
Section 3 Conservation of Energy
Section 4 Power
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Chapter 5
Section 1 Work
Definition of Work
• Work is done on an object when a force causes a
displacement of the object.
• Work is done only when components of a force are
parallel to a displacement.
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Chapter 5
Section 1 Work
Definition of Work
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Chapter 5
Section 2 Energy
Kinetic Energy
• Kinetic Energy
The energy of an object that is due to the object’s
motion is called kinetic energy.
• Kinetic energy depends on speed and mass.
1
KE  mv 2
2
1
2
kinetic energy =  mass   speed 
2
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Chapter 5
Section 2 Energy
Kinetic Energy, continued
• Work-Kinetic Energy Theorem
– The net work done by all the forces acting on an
object is equal to the change in the object’s kinetic
energy.
• The net work done on a body equals its change in
kinetic energy.
Wnet = ∆KE
net work = change in kinetic energy
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Chapter 5
Section 2 Energy
Potential Energy
• Potential Energy is the energy associated with an
object because of the position, shape, or condition of
the object.
• Gravitational potential energy is the potential
energy stored in the gravitational fields of interacting
bodies.
• Gravitational potential energy depends on height
from a zero level.
PEg = mgh
gravitational PE = mass  free-fall acceleration  height
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Chapter 5
Section 2 Energy
Potential Energy, continued
•
Elastic potential energy is the energy available for
use when a deformed elastic object returns to its
original configuration.
1 2
PEelastic  kx
2
elastic PE =
1
 spring constant  (distance compressed or stretched)
2
•
The symbol k is called the spring constant, a
parameter that measures the spring’s resistance to
being compressed or stretched.
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2
Chapter 5
Section 2 Energy
Elastic Potential Energy
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Chapter 5
Section 3 Conservation of
Energy
Conserved Quantities
• When we say that something is conserved, we mean
that it remains constant.
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Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy
• Mechanical energy is the sum of kinetic energy and
all forms of potential energy associated with an object
or group of objects.
ME = KE + ∑PE
• Mechanical energy is often conserved.
MEi = MEf
initial mechanical energy = final mechanical energy
(in the absence of friction)
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem
Conservation of Mechanical Energy
Starting from rest, a child zooms down a frictionless
slide from an initial height of 3.00 m. What is her
speed at the bottom of the slide? Assume she has a
mass of 25.0 kg.
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
1. Define
Given:
h = hi = 3.00 m
m = 25.0 kg
vi = 0.0 m/s
hf = 0 m
Unknown:
vf = ?
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan
Choose an equation or situation: The slide is
frictionless, so mechanical energy is conserved.
Kinetic energy and gravitational potential energy are
the only forms of energy present.
1
KE 
mv 2
2
PE  mgh
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan, continued
The zero level chosen for gravitational potential
energy is the bottom of the slide. Because the child
ends at the zero level, the final gravitational potential
energy is zero.
PEg,f = 0
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan, continued
The initial gravitational potential energy at the top of
the slide is
PEg,i = mghi = mgh
Because the child starts at rest, the initial kinetic
energy at the top is zero.
KEi = 0
Therefore, the final kinetic energy is as follows:
1
KEf  mv f2
2
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
3. Calculate
Substitute values into the equations:
PEg,i = (25.0 kg)(9.81 m/s2)(3.00 m) = 736 J
KEf = (1/2)(25.0 kg)vf2
Now use the calculated quantities to evaluate the
final velocity.
MEi = MEf
PEi + KEi = PEf + KEf
736 J + 0 J = 0 J + (0.500)(25.0 kg)vf2
vf = 7.67 m/s
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
4. Evaluate
The expression for the square of the final speed can
be written as follows:
2mgh
2
vf 
 2gh
m
Notice that the masses cancel, so the final speed
does not depend on the mass of the child. This
result makes sense because the acceleration of an
object due to gravity does not depend on the mass
of the object.
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Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy, continued
•
Mechanical Energy is
not conserved in the
presence of friction.
•
As a sanding block
slides on a piece of
wood, energy (in the
form of heat) is
dissipated into the
block and surface.
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Chapter 5
Section 4 Power
Rate of Energy Transfer
• Power is a quantity that measures the rate at which
work is done or energy is transformed.
P = W/∆t
power = work ÷ time interval
• An alternate equation for power in terms of force and
speed is
P = Fv
power = force  speed
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Chapter 5
Standardized Test Prep
Multiple Choice
1. In which of the following situations is work not being
done?
A. A chair is lifted vertically with respect to the floor.
B. A bookcase is slid across carpeting.
C. A table is dropped onto the ground.
D. A stack of books is carried at waist level across a
room.
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
1. In which of the following situations is work not being
done?
A. A chair is lifted vertically with respect to the floor.
B. A bookcase is slid across carpeting.
C. A table is dropped onto the ground.
D. A stack of books is carried at waist level across a
room.
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
2. Which of the following equations correctly describes
the relation between power,work, and time?
P
F. W 
t
t
G. W 
P
W
H. P 
t
t
J. P 
W
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
2. Which of the following equations correctly describes
the relation between power,work, and time?
P
F. W 
t
t
G. W 
P
W
H. P 
t
t
J. P 
W
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the graph below to answer questions 3–5. The
graph shows the energy of a 75 g yo-yo at different
times as the yo-yo moves up and down on its string.
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
3. By what amount does the mechanical energy of the
yo-yo change after 6.0 s?
A. 500 mJ
B. 0 mJ
C. –100 mJ
D. –600 mJ
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
3. By what amount does the mechanical energy of the
yo-yo change after 6.0 s?
A. 500 mJ
B. 0 mJ
C. –100 mJ
D. –600 mJ
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
4. What is the speed of the yo-yo after 4.5 s?
F. 3.1 m/s
G. 2.3 m/s
H. 3.6 m/s
J. 1.6 m/s
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
4. What is the speed of the yo-yo after 4.5 s?
F. 3.1 m/s
G. 2.3 m/s
H. 3.6 m/s
J. 1.6 m/s
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
5. What is the maximum height of the yo-yo?
A. 0.27 m
B. 0.54 m
C. 0.75 m
D. 0.82 m
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
5. What is the maximum height of the yo-yo?
A. 0.27 m
B. 0.54 m
C. 0.75 m
D. 0.82 m
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
6. A car with mass m requires 5.0 kJ of work to move
from rest to a final speed v. If this same amount of
work is performed during the same amount of time
on a car with a mass of 2m, what is the final speed
of the second car?
F. 2v
G.
2v
v
H.
2
v
J.
2
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
6. A car with mass m requires 5.0 kJ of work to move
from rest to a final speed v. If this same amount of
work is performed during the same amount of time
on a car with a mass of 2m, what is the final speed
of the second car?
F. 2v
G.
2v
v
H.
2
v
J.
2
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 7–8.
A 70.0 kg base runner moving at a speed of 4.0 m/s
begins his slide into second base. The coefficient of
friction between his clothes and Earth is 0.70. His
slide lowers his speed to zero just as he reaches the
base.
7. How much mechanical energy is lost because of
friction acting on the runner?
A. 1100 J
B. 560 J
C. 140 J
D. 0 J
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 7–8.
A 70.0 kg base runner moving at a speed of 4.0 m/s
begins his slide into second base. The coefficient of
friction between his clothes and Earth is 0.70. His
slide lowers his speed to zero just as he reaches the
base.
7. How much mechanical energy is lost because of
friction acting on the runner?
A. 1100 J
B. 560 J
C. 140 J
D. 0 J
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 7–8.
A 70.0 kg base runner moving at a speed of 4.0 m/s
begins his slide into second base. The coefficient of
friction between his clothes and Earth is 0.70. His
slide lowers his speed to zero just as he reaches the
base.
8. How far does the runner slide?
F. 0.29 m
G. 0.57 m
H. 0.86 m
J. 1.2 m
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 7–8.
A 70.0 kg base runner moving at a speed of 4.0 m/s
begins his slide into second base. The coefficient of
friction between his clothes and Earth is 0.70. His
slide lowers his speed to zero just as he reaches the
base.
8. How far does the runner slide?
F. 0.29 m
G. 0.57 m
H. 0.86 m
J. 1.2 m
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 9–10.
A spring scale has a spring with a force constant of
250 N/m and a weighing pan with a mass of 0.075
kg. During one weighing, the spring is stretched a
distance of 12 cm from equilibrium. During a second
weighing, the spring is stretched a distance of 18 cm.
How far does the runner slide?
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
9. How much greater is the elastic potential energy of
the stretched spring during the second weighing than
during the first weighing?
9
A.
4
3
B.
2
2
C.
3
4
D.
9
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
9. How much greater is the elastic potential energy of
the stretched spring during the second weighing than
during the first weighing?
9
A.
4
3
B.
2
2
C.
3
4
D.
9
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
10. If the spring is suddenly released after each
weighing, the weighing pan moves back and forth
through the equilibrium position. What is the ratio of
the pan’s maximum speed after the second weighing
to the pan’s maximum speed after the first weighing?
Consider the force of gravity on the pan to be
negligible.
9
2
F.
H.
4
3
3
4
G.
J.
2
9
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Chapter 5
Standardized Test Prep
Multiple Choice, continued
10. If the spring is suddenly released after each
weighing, the weighing pan moves back and forth
through the equilibrium position. What is the ratio of
the pan’s maximum speed after the second weighing
to the pan’s maximum speed after the first weighing?
Consider the force of gravity on the pan to be
negligible.
9
2
F.
H.
4
3
3
4
G.
J.
2
9
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Chapter 5
Standardized Test Prep
Short Response
11. A student with a mass of 66.0 kg climbs a staircase
in 44.0 s. If the distance between the base and the
top of the staircase is 14.0 m, how much power will
the student deliver by climbing the stairs?
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Chapter 5
Standardized Test Prep
Short Response, continued
11. A student with a mass of 66.0 kg climbs a staircase
in 44.0 s. If the distance between the base and the
top of the staircase is 14.0 m, how much power will
the student deliver by climbing the stairs?
Answer: 206 W
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Chapter 5
Standardized Test Prep
Short Response, continued
Base your answers to questions 12–13 on the
information below.
A 75.0 kg man jumps from a window that is 1.00 m
above a sidewalk.
12. Write the equation for the man’s speed when he
strikes the ground.
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Chapter 5
Standardized Test Prep
Short Response, continued
Base your answers to questions 12–13 on the
information below.
A 75.0 kg man jumps from a window that is 1.00 m
above a sidewalk.
12. Write the equation for the man’s speed when he
strikes the ground.
Answer: v  2gh
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Chapter 5
Standardized Test Prep
Short Response, continued
Base your answers to questions 12–13 on the
information below.
A 75.0 kg man jumps from a window that is 1.00 m
above a sidewalk.
13. Calculate the man’s speed when he strikes the
ground.
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Chapter 5
Standardized Test Prep
Short Response, continued
Base your answers to questions 12–13 on the
information below.
A 75.0 kg man jumps from a window that is 1.00 m
above a sidewalk.
13. Calculate the man’s speed when he strikes the
ground.
Answer: 4.4 m/s
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Chapter 5
Standardized Test Prep
Extended Response
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
14. The work done on the projectile by gravity.
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Chapter 5
Standardized Test Prep
Extended Response, continued
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
14. The work done on the projectile by gravity.
Answer: 1200 J
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Chapter 5
Standardized Test Prep
Extended Response, continued
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
15. The change in kinetic energy since the projectile
was fired.
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Chapter 5
Standardized Test Prep
Extended Response, continued
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
15. The change in kinetic energy since the projectile
was fired.
Answer: 1200 J
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Chapter 5
Standardized Test Prep
Extended Response, continued
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
16. The final kinetic energy of the projectile.
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Chapter 5
Standardized Test Prep
Extended Response, continued
Base your answers to questions 14–16 on the
information below.
A projectile with a mass of 5.0 kg is shot horizontally
from a height of 25.0 m above a flat desert surface.
The projectile’s initial speed is 17 m/s. Calculate the
following for the instant before the projectile hits the
surface:
16. The final kinetic energy of the projectile.
Answer: 1900 J
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Chapter 5
Standardized Test Prep
Extended Response, continued
17. A skier starts from rest at the top of a hill that is
inclined at 10.5° with the horizontal. The hillside is
200.0 m long, and the coefficient of friction between
the snow and the skis is 0.075. At the bottom of the
hill, the snow is level and the coefficient of friction is
unchanged. How far does the skier move along the
horizontal portion of the snow before coming to rest?
Show all of your work.
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Chapter 5
Standardized Test Prep
Extended Response, continued
17. A skier starts from rest at the top of a hill that is
inclined at 10.5° with the horizontal. The hillside is
200.0 m long, and the coefficient of friction between
the snow and the skis is 0.075. At the bottom of the
hill, the snow is level and the coefficient of friction is
unchanged. How far does the skier move along the
horizontal portion of the snow before coming to rest?
Show all of your work.
Answer: 290 m
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Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy
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Chapter 6
Momentum and Collisions
Chapter 6
Momentum and Collision
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Chapter 6
• Momentum
– Momentum (p) is the quantity of motion in a body.
Momentum is a vector. It has a size and a
direction. Only moving object have momentum.
– Calculation: The size of the momentum is equal
to the mass of the object(kg) multiplied by the size
of the object's velocity (m/s).
– Equation: p=mv
Units: kgm/s
– The direction of the momentum is the same as the
direction of the object's velocity..
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• The Law of Conservation of Momentum
– Often physics problems deal with momentum
before and after a collision. In such cases the total
momentum of the bodies before collision is taken
as equal to the total momentum of the bodies after
collision. That is to say: momentum is conserved.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
total initial momentum = total final momentum
– A collision is an event where momentum or
kinetic energy is transferred from one object to
another. The two types are elastic and inelastic
collisions.
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• An inelastic collisions occurs when
two objects collide and do not bounce
away from each other
–Momentum is conserved. Calculated
by using:
m1v1,i + m2v2,i = (m1 + m2)vf
–Kinetic energy is not conserved.
Calculated by using:
∆KE = KEf - KEi
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A completely inelastic collision:
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• An elastic collision occurs when the two
objects "bounce" apart when they collide.
Two rubber balls are a good example.
• Momentum and kinetic energy are
conserved and can be calculated
using:
m1v1,i  m2v 2,i  m1v1,f  m2v 2,f
1
1
1
1
2
2
2
2
m1v1,i  m2v 2,i  m1v1,f  m2v 2,f
2
2
2
2
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Elastic Collisions
In elastic collisions, both kinetic energy and
momentum are conserved.
One-dimensional elastic collision:
Newton’s Laws and Momentum
• Momentum refers to inertia in motion.
• Momentum is a measure of how difficult it is to
stop an object; a measure of “how much motion”
an object has.
• Inertia is the tendency of a body at rest to remain at
rest or of a body in straight line motion to stay in
motion in a straight line unless acted on by an
outside force.
• Force, on the other hand, is the push or pull that is
applied to an object to CHANGE its momentum
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• Newton's second law of motion defines force as the
product of mass times ACCELERATION (vs.
velocity). F=ma
• Since acceleration is the change in velocity divided
by time, (a= ∆V/∆t ) ,you can connect the two
concepts with the following relationship:
– The impulse-momentum theorem states that
when a net force is applied to an object over a
certain time interval, the force will cause a change
in the object’s momentum.
F∆t = ∆p = mvf – mvi
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Momentum and Impulse
• Momentum is a measure of how difficult it is to
stop an object. p=mv
• change in momentum is given by: Δp = mΔv
• Impulse occurs as force is applied to an object over
a period of time: I = FΔt
• Newton's Second Law of Motion is the rate of change
of momentum:
• F=ma so F= mΔv/Δt so FΔt=mΔv so I= Δp and
F = Δp/t
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Chapter 6
SAMPLE PROBLEMS
Sample Problems
Calculating Momentum
Calculate the momentum of a 11.35kg wagon rolling
down a hill at 12m/s.
p = mv
p = (11.35kg) (12m/s)
p = 136.2 kgm/s down the hill
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Chapter 6
SAMPLE PROBLEMS
Sample Problems
Conservation of Momentum
A 76 kg boater, initially at rest in a stationary 45 kg
boat, steps out of the boat and onto the dock. If the
boater moves out of the boat with a velocity of 2.5
m/s to the right,what is the final velocity of the boat?
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Chapter 6
Sample Problem, continued
Conservation of Momentum
1. Define
Given:
m1 = 76 kg m2 = 45 kg
v1,i = 0
v2,i = 0
v1,f = 2.5 m/s to the right
Unknown:
v2,f = ?
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Chapter 6
Sample Problem, continued
Conservation of Momentum
2. Plan
Choose an equation or situation: Because the total
momentum of an isolated system remains constant,
the total initial momentum of the boater and the boat
will be equal to the total final momentum of the boater
and the boat.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
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Chapter 6
Sample Problem, continued
Conservation of Momentum
2. Plan, continued
Because the boater and the boat are initially at rest,
the total initial momentum of the system is equal to
zero. Therefore, the final momentum of the system
must also be equal to zero.
m1v1,f + m2v2,f = 0
Rearrange the equation to solve for the final velocity
of the boat.
m2 v 2,f  – m1v1,f
v 2,f
 m1 
 –
 v1,f
 m2 
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Chapter 6
Sample Problem, continued
Conservation of Momentum
3. Calculate
Substitute the values into the equation and solve:
v 2,f
v 2,f
 76 kg 
 –
 2.5 m/s to the right 

 45 kg 
 –4.2 m/s to the right
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Chapter 6
Sample Problem, continued
Conservation of Momentum
4. Evaluate
The negative sign for v2,f indicates that the boat is
moving to the left, in the direction opposite the motion
of the boater. Therefore,
v2,f = 4.2 m/s to the left
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Chapter 6
Sample Problem
Kinetic Energy in Perfectly Inelastic Collisions
Two clay balls collide head-on in a perfectly inelastic
collision. The first ball has a mass of 0.500 kg and an
initial velocity of 4.00 m/s to the right. The second ball
has a mass of 0.250 kg and an initial velocity of 3.00
m/s to the left. What is the decrease in kinetic energy
during the collision?
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
1. Define
Given:
m1= 0.500 kg
m2 = 0.250 kg
v1,i = 4.00 m/s to the right, v1,i = +4.00 m/s
v2,i = 3.00 m/s to the left, v2,i = –3.00 m/s
Unknown:
∆KE = ?
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
2. Plan
Choose an equation or situation: The change in
kinetic energy is simply the initial kinetic energy
subtracted from the final kinetic energy.
∆KE = KEf - KEi
Determine both the initial and final kinetic energy.
1
1
2
Initial: KEi  KE1,i  KE2,i  m1v1,i  m2v 2,2 i
2
2
1
Final: KEf  KE1,f  KE2,f   m1  m2  v f2
2
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
2. Plan, continued
Use the equation for a perfectly inelastic collision to
calculate the final velocity.
vf 
m1v1,i  m2v 2,i
m1  m2
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate
Substitute the values into the equation and
solve: First, calculate the final velocity, which will be
used in the final kinetic energy equation.
(0.500 kg)(4.00 m/s)  (0.250 kg)(–3.00 m/s)
vf 
0.500 kg  0.250 kg
v f  1.67 m/s to the right
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate, continued
Next calculate the initial and final kinetic energy.
1
1
2
2
0.500
kg
4.00
m/s

0.250
kg
–3.00
m/s





  5.12 J
2
2
1
2
KEf   0.500 kg  0.250 kg1.67 m/s   1.05 J
2
KEi 
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Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate, continued
Finally, calculate the change in kinetic energy.
DKE  KEf – KEi  1.05 J – 5.12 J
DKE  –4.07 J
4. Evaluate The negative sign indicates that kinetic
energy is lost.
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Chapter 6
Sample Problem, continued
Elastic Collisions
A 0.015 kg marble moving to the right at 0.225 m/s
makes an elastic head-on collision with a 0.030 kg
shooter marble moving to the left at 0.180 m/s. After
the collision, the smaller marble moves to the left at
0.315 m/s. Assume that neither marble rotates before
or after the collision and that both marbles are
moving on a frictionless surface. What is the velocity
of the 0.030 kg marble after the collision?
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Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Elastic Collisions
1. Define
Given: m1 = 0.015 kg m2 = 0.030 kg
v1,i = 0.225 m/s to the right, v1,i = +0.225 m/s
v2,i = 0.180 m/s to the left, v2,i = –0.180 m/s
v1,f = 0.315 m/s to the left, v1,i = –0.315 m/s
Unknown:
v2,f = ?
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Chapter 6
Sample Problem, continued
Elastic Collisions
2. Plan
Choose an equation or situation: Use the equation for
the conservation of momentum to find the final velocity
of m2, the 0.030 kg marble.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Rearrange the equation to isolate the final velocity of m2.
m2 v2,f  m1v1i,  m2 v2,i – m1v1,f
v 2,f 
m1v1,i  m2 v2,i – m1v1,f
m2
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Chapter 6
Sample Problem, continued
Elastic Collisions
3. Calculate
Substitute the values into the equation and solve: The
rearranged conservation-of-momentum equation will
allow you to isolate and solve for the final velocity.
 0.015 kg 0.225 m/s   0.030 kg  –0.180 m/s  – 0.015 kg –0.315 m/s 
v 
2,f
v 2,f
v 2,f
0.030 kg
3.4  10


–3
 
 
kg  m/s  –5.4  10 –3 kg  m/s – –4.7  10 –3 kg m/s

0.030 kg
2.7  10 –3 kg  m/s

3.0  10 –2 kg
v2,f  9.0  10 –2 m/s to the right
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Chapter 6
Sample Problem, continued
Elastic Collisions
4. Evaluate Confirm your answer by making sure kinetic
energy is also conserved using these values.
1
1
1
1
m1v1,2i  m2v 2,2 i  m1v1,2f  m2v 2,2 f
2
2
2
2
1
1
2
2
KEi   0.015 kg  0.225 m/s    0.030 kg  –0.180 m/s 
2
2
 8.7  10 –4 kg  m2 /s2  8.7  10 –4 J
1
1
2
2
KEf   0.015 kg  0.315 m/s    0.030 kg 0.090 m/s 
2
2
 8.7  10 –4 kg  m2 /s2  8.7  10 –4 J
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Standardized Test
Prep
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Chapter 6
Standardized Test Prep
Multiple Choice
1. If a particle’s kinetic energy is zero, what is its
momentum?
A. zero
B. 1 kg • m/s
C. 15 kg • m/s
D. negative
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
1. If a particle’s kinetic energy is zero, what is its
momentum?
A. zero
B. 1 kg • m/s
C. 15 kg • m/s
D. negative
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
2. The vector below represents the momentum of a car
traveling along a road.
The car strikes another car, which is at rest, and the
result is an inelastic collision. Which of the following
vectors represents the momentum of the first car
after the collision?
F.
G.
H.
J.
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
2. The vector below represents the momentum of a car
traveling along a road.
The car strikes another car, which is at rest, and the
result is an inelastic collision. Which of the following
vectors represents the momentum of the first car
after the collision?
F.
G.
H.
J.
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
3. What is the momentum of a 0.148 kg baseball thrown
with a velocity of 35 m/s toward home plate?
A. 5.1 kg • m/s toward home plate
B. 5.1 kg • m/s away from home plate
C. 5.2 kg • m/s toward home plate
D. 5.2 kg • m/s away from home plate
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
3. What is the momentum of a 0.148 kg baseball thrown
with a velocity of 35 m/s toward home plate?
A. 5.1 kg • m/s toward home plate
B. 5.1 kg • m/s away from home plate
C. 5.2 kg • m/s toward home plate
D. 5.2 kg • m/s away from home plate
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 4–5.
After being struck by a bowling ball, a 1.5 kg bowling
pin slides to the right at 3.0 m/s and collides head-on
with another 1.5 kg bowling pin initially at rest.
4. What is the final velocity of the second pin if the first
pin moves to the right at 0.5 m/s after the collision?
F. 2.5 m/s to the left
G. 2.5 m/s to the right
H. 3.0 m/s to the left
J. 3.0 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 4–5.
After being struck by a bowling ball, a 1.5 kg bowling
pin slides to the right at 3.0 m/s and collides head-on
with another 1.5 kg bowling pin initially at rest.
4. What is the final velocity of the second pin if the first
pin moves to the right at 0.5 m/s after the collision?
F. 2.5 m/s to the left
G. 2.5 m/s to the right
H. 3.0 m/s to the left
J. 3.0 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 4–5.
After being struck by a bowling ball, a 1.5 kg bowling
pin slides to the right at 3.0 m/s and collides head-on
with another 1.5 kg bowling pin initially at rest.
5. What is the final velocity of the second pin if the first
pin stops moving when it hits the second pin?
A. 2.5 m/s to the left
B. 2.5 m/s to the right
C. 3.0 m/s to the left
D. 3.0 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 4–5.
After being struck by a bowling ball, a 1.5 kg bowling
pin slides to the right at 3.0 m/s and collides head-on
with another 1.5 kg bowling pin initially at rest.
5. What is the final velocity of the second pin if the first
pin stops moving when it hits the second pin?
A. 2.5 m/s to the left
B. 2.5 m/s to the right
C. 3.0 m/s to the left
D. 3.0 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
6. For a given change in momentum, if the net force that
is applied to an object increases, what happens to
the time interval over which the force is applied?
F. The time interval increases.
G. The time interval decreases.
H. The time interval stays the same.
J. It is impossible to determine the answer from the
given information.
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
6. For a given change in momentum, if the net force that
is applied to an object increases, what happens to
the time interval over which the force is applied?
F. The time interval increases.
G. The time interval decreases.
H. The time interval stays the same.
J. It is impossible to determine the answer from the
given information.
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
7. Which equation expresses the law of conservation of
momentum?
A. p = mv
B. m1v1,i + m2v2,i = m1v1,f + m2v2,f
C. (1/2)m1v1,i2 + m2v2,i2 = (1/2)(m1 + m2)vf2
D. KE = p
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
7. Which equation expresses the law of conservation of
momentum?
A. p = mv
B. m1v1,i + m2v2,i = m1v1,f + m2v2,f
C. (1/2)m1v1,i2 + m2v2,i2 = (1/2)(m1 + m2)vf2
D. KE = p
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
8. Two shuffleboard disks of equal mass, one of which
is orange and one of which is yellow, are involved in
an elastic collision. The yellow disk is initially at rest
and is struck by the orange disk, which is moving
initially to the right at 5.00 m/s. After the collision, the
orange disk is at rest. What is the velocity of the
yellow disk after the collision?
F. zero
G. 5.00 m/s to the left
H. 2.50 m/s to the right
J. 5.00 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
8. Two shuffleboard disks of equal mass, one of which
is orange and one of which is yellow, are involved in
an elastic collision. The yellow disk is initially at rest
and is struck by the orange disk, which is moving
initially to the right at 5.00 m/s. After the collision, the
orange disk is at rest. What is the velocity of the
yellow disk after the collision?
F. zero
G. 5.00 m/s to the left
H. 2.50 m/s to the right
J. 5.00 m/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the information below to answer questions 9–10.
A 0.400 kg bead slides on a straight frictionless wire and moves
with a velocity of 3.50 cm/s to the right, as shown below. The
bead collides elastically with a larger 0.600 kg bead that is
initially at rest. After the collision, the smaller bead moves to the
left with a velocity of 0.70 cm/s.
9. What is the large bead’s velocity after the collision?
A. 1.68 cm/s to the right
B. 1.87 cm/s to the right
C. 2.80 cm/s to the right
D. 3.97 cm/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the information below to answer questions 9–10.
A 0.400 kg bead slides on a straight frictionless wire and moves
with a velocity of 3.50 cm/s to the right, as shown below. The
bead collides elastically with a larger 0.600 kg bead that is
initially at rest. After the collision, the smaller bead moves to the
left with a velocity of 0.70 cm/s.
9. What is the large bead’s velocity after the collision?
A. 1.68 cm/s to the right
B. 1.87 cm/s to the right
C. 2.80 cm/s to the right
D. 3.97 cm/s to the right
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the information below to answer questions 9–10.
A 0.400 kg bead slides on a straight frictionless wire and moves
with a velocity of 3.50 cm/s to the right, as shown below. The
bead collides elastically with a larger 0.600 kg bead that is
initially at rest. After the collision, the smaller bead moves to the
left with a velocity of 0.70 cm/s.
10. What is the total kinetic energy of the system after the collision?
F. 1.40  10–4 J
G. 2.45  10–4 J
H. 4.70  10 –4 J
J. 4.90  10 –4 J
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Chapter 6
Standardized Test Prep
Multiple Choice, continued
Use the information below to answer questions 9–10.
A 0.400 kg bead slides on a straight frictionless wire and moves
with a velocity of 3.50 cm/s to the right, as shown below. The
bead collides elastically with a larger 0.600 kg bead that is
initially at rest. After the collision, the smaller bead moves to the
left with a velocity of 0.70 cm/s.
10. What is the total kinetic energy of the system after the collision?
F. 1.40  10–4 J
G. 2.45  10–4 J
H. 4.70  10 –4 J
J. 4.90  10 –4 J
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Chapter 6
Standardized Test Prep
Short Response
11. Is momentum conserved when two objects with zero
initial momentum push away from each other?
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Chapter 6
Standardized Test Prep
Short Response, continued
11. Is momentum conserved when two objects with zero
initial momentum push away from each other?
Answer: yes
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Chapter 6
Standardized Test Prep
Short Response, continued
12. In which type of collision is kinetic energy
conserved?
What is an example of this type of collision?
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Chapter 6
Standardized Test Prep
Short Response, continued
12. In which type of collision is kinetic energy
conserved?
Answer: elastic collision
What is an example of this type of collision?
Answer: Two billiard balls collide and then move
separately after the collision.
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Chapter 7
Circular Motion and Gravitation
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Chapter 7
Circular Motion and Gravitation
Table of Contents
Section 1 Circular Motion
Section 2 Newton’s Law of Universal Gravitation
Section 3 Motion in Space
Section 4 Torque and Simple Machines
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Chapter 7
Section 1 Circular Motion
Objectives
• Solve problems involving centripetal acceleration.
• Solve problems involving centripetal force.
• Explain how the apparent existence of an outward
force in circular motion can be explained as inertia
resisting the centripetal force.
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Chapter 7
Section 1 Circular Motion
Tangential Speed
• The tangential speed (vt) of an object in circular
motion is the object’s speed along an imaginary line
drawn tangent to the circular path.
• Tangential speed depends on the distance from the
object to the center of the circular path.
• When the tangential speed is constant, the motion is
described as uniform circular motion.
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration
http://www.youtube.com/watch?v=fSfVVz0eIis
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration
• The acceleration of an object moving in a circular
path and at constant speed is due to a change in
direction.
• An acceleration of this nature is called a centripetal
acceleration.
CENTRIPETAL ACCELERATION
vt 2
ac 
r
(tangential speed)2
centripetal acceleration =
radius of circular path
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration, continued
• (a) As the particle moves
from A to B, the direction of
the particle’s velocity vector
changes.
• (b) For short time intervals,
∆v is directed toward the
center of the circle.
• Centripetal acceleration is
always directed toward the
center of a circle.
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration, continued
• You have seen that centripetal acceleration
results from a change in direction.
• In circular motion, an acceleration due to a
change in speed is called tangential
acceleration.
• To understand the difference between centripetal
and tangential acceleration, consider a car
traveling in a circular track.
– Because the car is moving in a circle, the car has a
centripetal component of acceleration.
– If the car’s speed changes, the car also has a tangential
component of acceleration.
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Chapter 7
Section 1 Circular Motion
Centripetal Force
• Consider a ball of mass m that is being whirled in a
horizontal circular path of radius r with constant speed.
• The force exerted by the string has horizontal and vertical
components. The vertical component is equal and
opposite to the gravitational force. Thus, the horizontal
component is the net force.
• This net force, which is is directed toward the center of the
circle, is a centripetal force.
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Chapter 7
Section 1 Circular Motion
Centripetal Force, continued
Newton’s second law can be combined with the
equation for centripetal acceleration to derive an
equation for centripetal force:
vt 2
ac 
r
mvt 2
Fc  mac 
r
mass  (tangential speed)2
centripetal force =
radius of circular path
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Chapter 7
Section 1 Circular Motion
Centripetal Force, continued
• Centripetal force is simply the name given to the
net force on an object in uniform circular motion.
• Any type of force or combination of forces can
provide this net force.
– For example, friction between a race car’s tires
and a circular track is a centripetal force that
keeps the car in a circular path.
– As another example, gravitational force is a
centripetal force that keeps the moon in its
orbit.
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Chapter 7
Section 1 Circular Motion
Centripetal Force, continued
• If the centripetal force vanishes, the object stops
moving in a circular path.
• A ball that is on the end of a
string is whirled in a vertical
circular path.
– If the string breaks at the position
shown in (a), the ball will move
vertically upward in free fall.
– If the string breaks at the top of the
ball’s path, as in (b), the ball will
move along a parabolic path.
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Chapter 7
Section 1 Circular Motion
Describing a Rotating System
• To better understand the motion of a rotating
system, consider a car traveling at high speed and
approaching an exit ramp that curves to the left.
• As the driver makes the sharp left turn, the
passenger slides to the right and hits the door.
• What causes the passenger to move toward the
door?
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Chapter 7
Section 1 Circular Motion
Describing a Rotating System, continued
• As the car enters the ramp and travels along a
curved path, the passenger, because of inertia,
tends to move along the original straight path.
• If a sufficiently large centripetal force acts on the
passenger, the person will move along the same
curved path that the car does. The origin of the
centripetal force is the force of friction between the
passenger and the car seat.
• If this frictional force is not sufficient, the
passenger slides across the seat as the car turns
underneath.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Objectives
• Explain how Newton’s law of universal gravitation
accounts for various phenomena, including satellite
and planetary orbits, falling objects, and the tides.
• Apply Newton’s law of universal gravitation to solve
problems.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force
• Orbiting objects are in free fall.
• To see how this idea is true, we can use a thought
experiment that Newton developed. Consider a
cannon sitting on a high mountaintop.
Each successive cannonball
has a greater initial speed, so
the horizontal distance that
the ball travels increases. If
the initial speed is great
enough, the curvature of
Earth will cause the
cannonball to continue falling
without ever landing.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
• The centripetal force that holds the planets in orbit
is the same force that pulls an apple toward the
ground—gravitational force.
• Gravitational force is the mutual force of attraction
between particles of matter.
• Gravitational force depends on the masses and on
the distance between them.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
• Newton developed the following equation to describe
quantitatively the magnitude of the gravitational force
if distance r separates masses m1 and m2:
Newton's Law of Universal Gravitation
Fg  G
m1m2
r2
gravitational force  constant 
mass 1 mass 2
(distance between masses)2
• The constant G, called the constant of universal
gravitation, equals 6.673  10–11 N•m2/kg.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Newton’s Law of Universal Gravitation
http://www.youtube.com/watch?v=Y50HeIUS4tk
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
• The gravitational forces that two masses exert on
each other are always equal in magnitude and
opposite in direction.
• This is an example of Newton’s third law of motion.
• One example is the Earth-moon system, shown on
the next slide.
• As a result of these forces, the moon and Earth each
orbit the center of mass of the Earth-moon system.
Because Earth has a much greater mass than the
moon, this center of mass lies within Earth.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Newton’s Law of Universal Gravitation
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Applying the Law of Gravitation
• Newton’s law of gravitation accounts for ocean tides.
• High and low tides are partly due to the gravitational
force exerted on Earth by its moon.
• The tides result from the difference between the
gravitational force at Earth’s surface and at Earth’s
center.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Applying the Law of Gravitation, continued
• Cavendish applied Newton’s law of universal
gravitation to find the value of G and Earth’s mass.
• When two masses, the distance between them, and
the gravitational force are known, Newton’s law of
universal gravitation can be used to find G.
• Once the value of G is known, the law can be used
again to find Earth’s mass.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Applying the Law of Gravitation, continued
• Gravity is a field force.
• Gravitational field strength,
g, equals Fg/m.
• The gravitational field, g,
is a vector with magnitude
g that points in the
direction of Fg.
• Gravitational field
strength equals free-fall The gravitational field vectors
represent Earth’s gravitational
acceleration.
field at each point.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Applying the Law of Gravitation, continued
• weight = mass  gravitational field strength
• Because it depends on gravitational field
strength, weight changes with location:
weight = mg
Fg GmmE GmE
g

 2
2
m
mr
r
• On the surface of any planet, the value of g, as
well as your weight, will depend on the planet’s
mass and radius.
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Chapter 7
Section 3 Motion in Space
Objectives
• Describe Kepler’s laws of planetary motion.
• Relate Newton’s mathematical analysis of
gravitational force to the elliptical planetary orbits
proposed by Kepler.
• Solve problems involving orbital speed and period.
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws
Kepler’s laws describe the motion of the planets.
• First Law: Each planet travels in an elliptical orbit
around the sun, and the sun is at one of the focal
points.
• Second Law: An imaginary line drawn from the sun
to any planet sweeps out equal areas in equal time
intervals.
• Third Law: The square of a planet’s orbital period
(T2) is proportional to the cube of the average
distance (r3) between the planet and the sun.
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws, continued
• Kepler’s laws were developed a generation before
Newton’s law of universal gravitation.
• Newton demonstrated that Kepler’s laws are
consistent with the law of universal gravitation.
• The fact that Kepler’s laws closely matched
observations gave additional support for Newton’s
theory of gravitation.
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws, continued
According to Kepler’s second law, if the time a
planet takes to travel the arc on the left (∆t1) is equal
to the time the planet takes to cover the arc on the
right (∆t2), then the area A1 is equal to the area A2.
Thus, the planet
travels faster when it
is closer to the sun
and slower when it is
farther away.
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws, continued
• Kepler’s third law states that T2  r3.
• The constant of proportionality is 4p2/Gm, where m is
the mass of the object being orbited.
• So, Kepler’s third law can also be stated as follows:
2

 3
4
p
2
T 
r
 Gm 
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws, continued
• Kepler’s third law leads to an equation for the period
of an object in a circular orbit. The speed of an object
in a circular orbit depends on the same factors:
r3
T  2p
Gm
m
vt  G
r
• Note that m is the mass of the central object that is
being orbited. The mass of the planet or satellite that is
in orbit does not affect its speed or period.
• The mean radius (r) is the distance between the
centers of the two bodies.
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Chapter 7
Section 3 Motion in Space
Planetary Data
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Chapter 7
Section 3 Motion in Space
Sample Problem
Period and Speed of an Orbiting Object
Magellan was the first planetary spacecraft to be
launched from a space shuttle. During the spacecraft’s
fifth orbit around Venus, Magellan traveled at a mean
altitude of 361km. If the orbit had been circular, what
would Magellan’s period and speed have been?
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
1. Define
Given:
r1 = 361 km = 3.61  105 m
Unknown:
T=?
vt = ?
2. Plan
Choose an equation or situation: Use the equations for
the period and speed of an object in a circular orbit.
r3
T  2p
Gm
vt 
Gm
r
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
Use Table 1 in the textbook to find the values for the
radius (r2) and mass (m) of Venus.
r2 = 6.05  106 m
m = 4.87  1024 kg
Find r by adding the distance between the spacecraft
and Venus’s surface (r1) to Venus’s radius (r2).
r = r1 + r2
r = 3.61  105 m + 6.05  106 m = 6.41  106 m
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
3. Calculate
r3
(6.41  10 6 m)3
T  2p
=2p
Gm
(6.673  10 –11 N•m 2 /kg 2 )(4.87  10 24 kg)
T  5.66  10 3 s
Gm
(6.673  10 –11 N•m 2 /kg 2 )(4.87  10 24 kg)
vt 

r
6.41  10 6 m
vt  7.12  10 3 m/s
4. Evaluate
Magellan takes (5.66  103 s)(1 min/60 s)  94 min to complete
one orbit.
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Chapter 7
Section 3 Motion in Space
Weight and Weightlessness
To learn about apparent weightlessness, imagine that
you are in an elevator:
– When the elevator is at rest, the magnitude of the
normal force acting on you equals your weight.
– If the elevator were to accelerate downward at 9.81
m/s2, you and the elevator would both be in free fall.
You have the same weight, but there is no normal
force acting on you.
– This situation is called apparent weightlessness.
– Astronauts in orbit experience apparent
weightlessness.
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Chapter 7
Section 3 Motion in Space
Weight and Weightlessness
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Chapter 7
Standardized Test Prep
Multiple Choice
1. An object moves in a circle at a constant speed.
Which of the following is not true of the object?
A. Its acceleration is constant.
B. Its tangential speed is constant.
C. Its velocity is constant.
D. A centripetal force acts on the object.
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Chapter 7
Standardized Test Prep
Multiple Choice
1. An object moves in a circle at a constant speed.
Which of the following is not true of the object?
A. Its acceleration is constant.
B. Its tangential speed is constant.
C. Its velocity is constant.
D. A centripetal force acts on the object.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
2. What is the centripetal acceleration of the car?
F. 2.4  10-2 m/s2
G. 0.60 m/s2
H. 9.0 m/s2
J. zero
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
2. What is the centripetal acceleration of the car?
F. 2.4  10-2 m/s2
G. 0.60 m/s2
H. 9.0 m/s2
J. zero
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
3. What is the most direct cause of the car’s centripetal
acceleration?
A. the torque on the steering wheel
B. the torque on the tires of the car
C. the force of friction between the tires and the road
D. the normal force between the tires and the road
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
3. What is the most direct cause of the car’s centripetal
acceleration?
A. the torque on the steering wheel
B. the torque on the tires of the car
C. the force of friction between the tires and the road
D. the normal force between the tires and the road
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Chapter 7
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Multiple Choice, continued
4. Earth (m = 5.97  1024 kg) orbits the sun (m = 1.99 
1030 kg) at a mean distance of 1.50  1011 m. What is
the gravitational force of the sun on Earth? (G =
6.673  10-11 N•m2/kg2)
F. 5.29  1032 N
G. 3.52  1022 N
H. 5.90  10–2 N
J. 1.77  10–8 N
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
4. Earth (m = 5.97  1024 kg) orbits the sun (m = 1.99 
1030 kg) at a mean distance of 1.50  1011 m. What is
the gravitational force of the sun on Earth? (G =
6.673  10-11 N•m2/kg2)
F. 5.29  1032 N
G. 3.52  1022 N
H. 5.90  10–2 N
J. 1.77  10–8 N
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Chapter 7
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Multiple Choice, continued
5. Which of the following is a correct interpretation of
the expression ag  g  G mE ?
r2
A. Gravitational field strength changes with an
object’s distance from Earth.
B. Free-fall acceleration changes with an object’s
distance from Earth.
C. Free-fall acceleration is independent of the falling
object’s mass.
D. All of the above are correct interpretations.
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Chapter 7
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Multiple Choice, continued
5. Which of the following is a correct interpretation of
the expression ag  g  G mE ?
r2
A. Gravitational field strength changes with an
object’s distance from Earth.
B. Free-fall acceleration changes with an object’s
distance from Earth.
C. Free-fall acceleration is independent of the falling
object’s mass.
D. All of the above are correct interpretations.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
6. What data do you need to calculate the orbital speed
of a satellite?
F. mass of satellite, mass of planet, radius of orbit
G. mass of satellite, radius of planet, area of orbit
H. mass of satellite and radius of orbit only
J. mass of planet and radius of orbit only
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
6. What data do you need to calculate the orbital speed
of a satellite?
F. mass of satellite, mass of planet, radius of orbit
G. mass of satellite, radius of planet, area of orbit
H. mass of satellite and radius of orbit only
J. mass of planet and radius of orbit only
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Chapter 7
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Multiple Choice, continued
7. Which of the following choices correctly describes the
orbital relationship between Earth and the sun?
A. The sun orbits Earth in a perfect circle.
B. Earth orbits the sun in a perfect circle.
C. The sun orbits Earth in an ellipse, with Earth
at one focus.
D. Earth orbits the sun in an ellipse, with the sun
at one focus.
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Chapter 7
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Multiple Choice, continued
7. Which of the following choices correctly describes the
orbital relationship between Earth and the sun?
A. The sun orbits Earth in a perfect circle.
B. Earth orbits the sun in a perfect circle.
C. The sun orbits Earth in an ellipse, with Earth
at one focus.
D. Earth orbits the sun in an ellipse, with the sun
at one focus.
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Short Response
14. Explain how it is possible for all the water to remain
in a pail that is whirled in a vertical path, as shown
below.
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Chapter 7
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Short Response
14. Explain how it is possible for all the water to remain
in a pail that is whirled in a vertical path, as shown
below.
Answer: The water
remains in the pail even
when the pail is upside
down because the water
tends to move in a
straight path due to
inertia.
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Short Response, continued
15. Explain why approximately two high tides take place
every day at a given location on Earth.
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Chapter 7
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Short Response, continued
15. Explain why approximately two high tides take place
every day at a given location on Earth.
Answer: The moon’s tidal forces create two bulges on
Earth. As Earth rotates on its axis once per day, any
given point on Earth passes through both bulges.
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Short Response, continued
16. If you used a machine to increase the output force,
what factor would have to be sacrificed? Give an
example.
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Chapter 7
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Short Response, continued
16. If you used a machine to increase the output force,
what factor would have to be sacrificed? Give an
example.
Answer: You would have to apply the input force over
a greater distance. Examples may include any
machines that increase output force at the expense
of input distance.
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Chapter 7
Standardized Test Prep
Extended Response
17. Mars orbits the sun (m = 1.99  1030 kg) at a mean
distance of 2.28  1011 m. Calculate the length of
the Martian year in Earth days. Show all of your
work. (G = 6.673  10–11 N•m2/kg2)
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Chapter 7
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Extended Response
17. Mars orbits the sun (m = 1.99  1030 kg) at a mean
distance of 2.28  1011 m. Calculate the length of
the Martian year in Earth days. Show all of your
work. (G = 6.673  10–11 N•m2/kg2)
Answer: 687 days
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