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CHAP T E R
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7
Page 278
Conservation of
Energy and Momentum
CHAPTER CONTENTS
Multi-Lab
Energy
Transformations
in Springs
279
7.1 Energy
Transformations
280
Investigation 7-A
Conservation of
Energy of a
Falling Mass
288
Investigation 7-B
Energy
Transformations
of a Pendulum
289
7.2 Conservation of
Total Energy
301
Investigation 7-C
Mechanical and
Thermal Energy
304
7.3 Conservation of
Momentum
310
Investigation 7-D
Examining
Collisions
318
7.4 Technological
Implications
323
W
ith a “whump,” the fireworks shell is lofted upward into
the darkness. As the shell rises, it slows; kinetic energy
transforms into gravitational potential energy. Then, the shell
explodes and chemical potential energy rapidly converts into
heat, light, and sound. The darkness gives way to brilliant colour
and loud bangs startle the crowd below.
There is a balance in all of these transformations and effects.
Energy gained in one form comes at the expense of another.
This is the law of conservation of energy.
In this chapter, you will examine these energy transformations
and the laws that govern these transformations.
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TARGET SKILLS
M ULT I
L A B
Energy Transformations
in Springs
Performing and recording
Analyzing and interpreting
Communicating results
3. When the spring was stretched to a greater
A Spring Pendulum
Hang a spring (at least 50 cm long) from a
rigid support and attach a mass to the lower
end of the spring. Ensure that the mass is
heavy enough to extend the spring noticeably
without overstretching it. The hanging mass
should not come close to the desktop. Pull
the mass to the side and release it, allowing
the spring to swing from side to side. Observe
the motion of the mass and the spring.
Analyze and Conclude
1. What types of periodic (repeating) motion
did you observe?
length than it was when the mass was
at rest, was the mass moving rapidly
or slowly?
4. What was being transferred between
(a) the different types of vibration, and
(b) the mass and the spring?
5. The law of conservation of energy states
that the total energy of an isolated system
remains constant. How was that law illustrated by the action of the spring and the
mass in this case?
6. What eventually happened to the motion
2. When the amplitude of one type of vibra-
tion was at a maximum, what happened
to the amplitude of the other type of
vibration?
of the spring and the mass? Suggest why
this occurred.
7. If the system regularly switched back
and forth between the two patterns of
motion, was the time taken for the change
consistent?
3. At what stage or stages in the action of the
Slinky Motion
Place a Slinky™ toy several steps up from the
bottom of a set of stairs, with the axis of the
spring vertical. Take the top coil and arc it
over and down to touch the next lower step.
Release the Slinky™ and observe its motion.
Analyze and Conclude
1. What is the condition of the coils of the
spring when energy is stored in the spring?
2. At what stage or stages in the action of the
spring is kinetic energy being converted
into elastic potential energy in the spring?
spring is elastic potential energy in the
spring being converted into kinetic
energy?
4. Is there any instant during the motion of
the spring when both the kinetic energy
and the elastic potential energy are at a
maximum? Would you expect this to be
possible? Give a reason for your answer.
5. Any system loses energy due to friction,
which converts mechanical energy into
thermal energy. Why then does the spring
continue going down the stairs? From
where is it getting its energy?
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7.1
SECTION
OUTCOMES
• Describe quantitatively
mechanical energy as the
sum of kinetic and potential
energies.
Page 280
Energy Transformations
In Chapter 6, you learned that by doing work on an object you can
change its position or condition and thus give it energy. The three
forms of energy that you studied — kinetic energy, gravitational
potential energy, and elastic potential energy — are classified as
mechanical energy. In this chapter, you will focus on what
happens to the energy after you have done work on an object.
• Analyze quantitatively problems
related to kinematics and
dynamics using the mechanical
energy concept.
• Analyze common energy
transformation situations using
the closed system work-energy
theorem.
KEY
TERMS
• mechanical energy
• conservative force
• nonconservative force
• amplitude
Figure 7.1 Although this large pendulum might oscillate for days, it will
eventually slow down and stop.
A hockey stick does work on a puck, and the puck slides across
the ice with kinetic energy. However, if the ice surface was long
enough, the puck would eventually come to a stop. You place a
book on the shelf, giving it gravitational potential energy. Then
you bump into the shelf and the book falls off, hits the floor, and
stops. At first, the gravitational potential energy was transformed
into kinetic energy, but soon all of the energy seemed to disappear.
You do work on a mass on the end of a spring and give the spring
elastic potential energy. When you release the mass, it oscillates
up and down for a while, but eventually comes to a stop. You do
work on a pendulum by pulling the bob up and to the side. After
you release it, the pendulum bob will swing back and forth for
minutes, hours, or possibly even days, such as the pendulum in
Figure 7.1. Eventually, however, it too will stop. In every case, the
object lost the mechanical energy that you gave it. In some cases,
however, the object keeps the mechanical energy for extremely
long periods of time. If a system such as a pendulum or a mass
on a spring was “perfect,” the mechanical energy of the system
would remain constant. You could say that the mechanical energy
was conserved. Is there any specific characteristic of a system or
process that you can use to predict whether the mechanical energy
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will be conserved over a long period of time? To answer this question, you need to learn about two classes of forces that do work on
a system.
Conservative and Non-Conservative Forces
If you lift your book one metre above a table and release it, it will
drop back onto the table, gaining kinetic energy as it falls. If you
push your book across the table, will it automatically return to its
original spot, gaining kinetic energy as it moves? Of course not. In
the first case, you were doing work against the gravitational force.
In the second case, you were doing work against a frictional force.
These two forces, gravity and friction, represent two important
classes of forces. Before defining these classes of forces, consider
another property of doing work against them. If you lift your book
to a certain height, then carry it across the room, you have done
the same amount of work that you would do if you simply lifted it
straight up. However, if you push your book from side to side as
you move it from one end of the table to the other, you have done
more work than you would if you pushed it in a straight line. The
amount of work you do against friction depends on the path
through which you push the object.
A conservative force is one that does work on an object in such
a way that the amount of work done is independent of the path
taken. The force of gravity is an example of a conservative force
because it takes the same amount of work to lift a mass to height
Δh, regardless of the path. The work done, and therefore the
gravitational potential energy, depend only on the height Δh.
Friction, on the other hand, is not a conservative force. The
work done against a frictional force when a crate is pushed across
a rough floor depends on whether the path is straight, or curved,
or zigzagged. The work done is path-dependent, and therefore the
force of friction is nonconservative.
Work done by conservative forces results in energy changes that
are independent of the path and are therefore reversible. Work
done by non-conservative forces results in energy changes that
are dependent on the path, and therefore may not be reversed. The
force of gravity and the force that perfectly elastic materials exert
on an object are conservative forces. No real material is perfectly
elastic, but some materials are nearly so. Therefore, in many
situations they can be considered to be perfectly elastic.
Conservation of Mechanical Energy
When all of the work done throughout an energy transformation
process is done by conservative forces, mechanical energy is
conserved. This property of energy transformation processes can
be stated as a law and expressed mathematically as shown in the
box on the next page.
B
E
Δh
A
F
D
C
Figure 7.2 Regardless of
the path taken, the work done
to lift the box to height Δh is
identical. Work done against a
conservative force is independent
of the path.
PROBEWARE
www.mcgrawhill.ca/links/
atlphysics
If your school has probeware
equipment, visit the Internet site
above and follow the links for
several laboratory activities.
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THE LAW OF CONSERVATION OF MECHANICAL ENERGY
When all of the work done throughout a process is done by
conservative forces, the total mechanical energy of the system
after the process is equal to the total mechanical energy of
the system before the process.
E′k + E′g + E′e = Ek + Eg + Ee
Quantity
Symbol
E′k
kinetic energy after process
E′g
gravitational potential energy after process
E′e
elastic potential energy after process
Ek
kinetic energy before process
Eg
gravitational potential energy before process
Ee
elastic potential energy before process
SI unit
J (joule)
J (joule)
J (joule)
J (joule)
J (joule)
J (joule)
Unit Analysis
All of the units are joules.
Note: The symbol that looks like an apostrophe is called a
prime. The prime is used to represent the condition of the
system after the process has been completed.
Gravitational Potential Energy and Kinetic Energy
Concentrate first on energy transformation processes that involve
only gravitational potential energy and kinetic energy.
Conceptual Problem
• A marble oscillates back and forth in a U-shaped track, repeatedly transferring gravitational potential energy to kinetic energy,
and back to gravitational potential energy. Does the law of
conservation of mechanical energy apply? Defend your reasoning by discussing conservative and non-conservative forces.
The example of a falling rock in Figure 7.3 illustrates such a
process. The rock, initially at rest, falls from a height, h, above
the ground. The rock’s gravitational potential energy is being
transformed into kinetic energy as it falls farther and faster. Its
mechanical energy, however, is being conserved until the instant
before it strikes the ground.
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A
total
energy
potential
C
energy
B
kinetic
D
time
rock strikes
ground
Gravitational potential energy
is being transformed into
kinetic energy.
The graph shows the amount of
potential and kinetic energy of
the rock at different positions,
up to the moment before it hits
the ground.
Figure 7.3 Although gravitational potential energy is being transformed
into kinetic energy, the total mechanical energy of the rock remains the
same as it falls.
Conceptual Problems
• Imagine an amusement park that has rides like those illustrated
here. The masses of all of the cars are identical and the same
four people go on each ride. The wheels and track are effectively
frictionless. Each car starts from rest at level A. What are the
relative speeds of the cars when they reach level B?
A
B
• While playing catch by yourself, you throw a ball straight up as
hard as you can throw it. Neglecting air friction, how does the
speed of the ball, when it returns to your hand, compare to the
speed with which it left your hand?
You can say, intuitively, that the kinetic energy of an object,
the instant before it hits a solid surface, is the same as the gravitational potential energy at the point from which it fell from rest.
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However, physicists try to show such relationships mathematically.
Then, they can make calculations based on the principles. Start
with the general expression for kinetic energy, Ek = 12 mv 2 , and
show how the kinetic energy at the end of a free fall relates to the
gravitational potential energy, Eg = mgΔh, at a distance Δh above
the surface.
The first step in deriving this relationship is to find the velocity
of an object after it falls a distance, Δh.
■
Recall from Chapter 3 the equation
of motion for uniform acceleration.
v22 = v12 + 2aΔd
■
The object was initially at rest.
v1 = 0
v22 = 0 + 2aΔd
■
Substitute Δh, representing height, for Δd.
Δd = Δh
v22 = 0 + 2aΔh
■
Since the only force acting on the object
is gravity, the object’s acceleration is g,
the acceleration due to gravity.
a=g
This is the square of the velocity of
the object, just before hitting a surface,
after falling through height Δh.
v22 = 2gΔh
Substitute the previous expression for
v 2 into the general expression for
kinetic energy.
Ek =
1
mv 2
2
The kinetic energy of the object just
before it hits the surface becomes
Ek =
1
m(2gΔh)
2
Notice that the kinetic energy just
before the object hits the solid
surface is identical to
Ek = mgΔh
The equation for gravitational potential
energy a distance Δh above the surface is
Eg = mgΔh
■
■
■
■
■
The total energy of the object is not only the same at the top
and at the bottom, but throughout the entire trip. The mathematical process used above could be applied to any point throughout
the fall. You could show that, for any point within the fall, the
gravitational potential energy that was lost up to that point has
been transformed into kinetic energy. Therefore, the sum of the
kinetic and potential energies remains constant at all times
throughout the fall. Obviously, when an object hits a solid surface
and stops, the mechanical energy is no longer conserved. Also,
any effects of air friction were neglected in the discussion.
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MODEL PROBLEM
Conservation of Energy on the Ski Slopes
2.00
A skier is gliding along with a speed of 2.00 m/s
at the top of a ski hill, 40.0 m high, as shown
in the diagram. The skier then begins to slide
down the icy (frictionless) hill.
m
s
v
40.0 m
(a) What will be the skier’s speed at a height
25.0 m
of 25.0 m?
(b) At what height will the skier have a speed
of 10.0 m/s?
Frame the Problem
■
■
Sketch the two parts of the problem separately.
Label the initial conditions (top of the hill) “1.” Label
the position when h = 25 m as “2.” Label the position
at which the skier is travelling at 10.0 m/s as “3.”
➀ v = 2.00
40.0 m
Assume that the slope is frictionless.
■
The law of conservation of mechanical energy can
be applied.
➁ v =?
25.0 m
Use subscripts 1 and 2 to indicate the initial and final
conditions in part (a) and use subscripts 1 and 3 to
indicate the initial and final conditions in part (b).
■
m
s
➀ v = 2.00
40.0 m
m
s
➂ v = 10.0
m
s
?
Identify the Goal
(a) the speed, v2 , at a height of 25.0 m
(b) the height, h3 , at which the skier’s speed will be 10.0 m/s
Variables and Constants
Known
m
v1 = 2.00
s
h1 = 40.0 m
m
v3 = 10.0
s
h2 = 25.0 m
Implied
m
g = 9.81 2
s
Unknown
v2
h3
Strategy
Calculations
State the law of conservation of mechanical
energy without elastic potential energy.
E′k + E′p = Ek + Ep
Expand by replacing E with the expression
that defines the type of energy.
1
mv22
2
+ mgh2 =
Divide through by m.
1 2
v
2 2
1 2
v
2 1
=
1
mv12
2
+ mgh1
+ gh1 − gh2
continued
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continued from previous page
Simplify and rearrange the equation to
solve for v2 .
Substitute and solve.
Speed does not involve direction, so choose
the positive root since speed can never be
negative.
v22 = v12 + 2g(h1 − h2)
v2 = v12 + 2g(h1 − h2)
m
m 2
v2 =
+ 2 9.81 2 (40.0 m − 25.0 m)
2.00
s
s
m2
v2 = 298.3 2
s
m
v2 = ±17.271
s
m
v2 ≅ 17.3
s
(a) The speed is 17.3 m/s at a height of 25.0 m.
Write the expanded version of the
conservation of mechanical energy.
Rearrange and solve for height.
1
mv32
2
+ mgh3 =
gh3 =
1 2
v
2 1
h3 =
Substitute numerical values and solve.
h3 =
1 2
v
2 1
1
2
+
1
mv12
2
gh1 − 12 v32
+ mgh1
+ gh1 − 12 v32
g
2.00
m
s
2
+ 9.81
m
s2
(40.0 m) − 12 10.0
9.81
m
s
2
m
s2
h3 = 35.107 m
h3 ≅ 35.1 m
(b) The height must be 35.1 m when the speed is 10.0 m/s.
Validate
(a) The units for the right-hand side of the equation for v2 are
PROBLEM TIP
(b) The units on the right-hand side of the equation for h3 are
Students are often tempted to
apply the equations for linear
motion to the solution of these
problems. However, the paths are
not always linear, so the equations
might not apply. One of the great
advantages of using conservation
of energy is that you generally do
not need to know the exact path
between two points or vector
directions. You need to know the
conditions at only those two points.
12
m 2
m
m
m =
+
.
s
s2
s
These are the correct units for the speed. In addition, since
the skier is going downhill, the final speed must be larger
than the initial speed of 2.00 m/s.
2
m
s
+
2
m
(m) − ms
s2
m
s2
= m.
Since the speed in part (b) is less than the speed in part (a),
the skier should be higher on the hill than in part (a).
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PRACTICE PROBLEMS
Use the following diagram for practice
problems 1, 2, and 3.
4.0
Neglecting air friction, what is the kinetic
energy of each object just before it hits the
ground? What is the velocity of each object
just before it hits the ground?
m
s
7. A 32 kg crate slides down a frictionless
ramp. Its initial velocity at the top of the
ramp was 3.2 m/s. Its velocity when it
reached the bottom of the ramp was 9.7 m/s.
The ramp makes an angle of 25˚ with the
horizontal. How long is the ramp?
B
12.0 m
A
y
4.0 m
8. A worker on the roof of a building that is
1. A car on a roller coaster is moving along a
level section 12.0 m high at 4.0 m/s when it
begins to roll down a slope, as shown in the
diagram. Determine the speed of the car at
point A.
2. What is the height of point B in the roller
coaster track if the speed of the car at that
point is 10.0 m/s?
3. What is the height of y if the speed of the
under construction dropped a 0.125 kg
wrench over the edge. Another workman
on the eighth floor saw the wrench go by
and determined that its speed at that level
was 33.1 m/s. The first floor of the building
is 12.0 m high and each successive floor is
8.00 m high. Neglecting air friction, how
many floors does the building have? How fast
was the wrench falling just before it hit the
ground? What was its kinetic energy just
before it hit the ground?
roller coaster at B is 12.5 m/s?
4. You throw a ball directly upward, giving it
an initial velocity of 10.0 m/s. Neglecting
friction, what would be the maximum height
of the ball? (Explain why you do not need to
know the mass of the ball.)
30˚
5. A wrecking ball, with a mass of 315 kg,
hangs from a crane on 10.0 m of cable. If the
crane swings the wrecking ball so that the
angle that the cable makes with the vertical
is 30.0˚, what is the potential energy of the
wrecking ball in relation to its lowest position? What will be the kinetic energy of the
wrecking ball when it falls back to the vertical position? What will be the speed of the
wrecking ball?
6. A 2.5 kg lead ball and a 55 g piece of lead
shot are both dropped from a height of 25 m.
ELECTRONIC
LEARNING PARTNER
A video clip of a wrecking ball in action can be found
on the Electronic Learning Partner.
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I N V E S T I G AT I O N
7-A
TARGET SKILLS
Predictingand interpreting
Analyzing
Performingconcepts
and recording
Modelling
Analyzing and interpreting
Communicating
results
Conservation of Energy
of a Falling Mass
When scientists set out to test an hypothesis or
challenge a law, they often use the hypothesis
or law to make a prediction and then test it. In
this investigation, you will make a prediction
based on the law of conservation of mechanical
energy.
Problem
To perform a test of the law of conservation
of mechanical energy
■
■
■
■
■
■
■
■
energy, calculate the expected speed of the
cart and the hanging mass just before the
mass strikes the board. If the cart is to be
released from rest, determine the average
speed of the system and then the predicted
time interval for the hanging mass to drop
to the board.
5. With the entire apparatus in place, hold the
Equipment
■
4. Using the law of conservation of mechanical
dynamics cart
balance capable of measuring a mass of 1 to 2 kg
2 pulleys
retort stand and clamps for the pulleys
board (or similar material) to protect the floor from
dropped masses
metric measuring tape
selection of masses, including several that can be
suspended on a string
stopwatch or photogate timers
string about 4 m long
cart still. Release the cart and measure the
time taken for the hanging mass to reach the
board. Be sure to catch the cart before it hits
the lower pulley.
6. Repeat the measurement several times and
average the results.
7. Perform several trial runs with a different pair
of masses for the cart and the hanging mass.
Analyze and Conclude
1. Prepare a table to show all of your data, as
well as your calculations for the final speed,
average speed, and the time interval.
2. What is the percent difference between the
Procedure
1. Determine the mass of the dynamics cart.
2. Select a mass to be the rider and a mass to
be the hanging mass. Decide on the height of
the hanging mass above the board.
3. Set up the apparatus on a long desktop, as
shown in the diagram.
rider
hanging
mass
dynamics
cart
time as predicted by the law of conservation
of energy and the measured average time?
3. Based on the precision of the timing devices,
what range of experimental error would you
expect in this investigation? How does this
range compare with the percent difference
determined in question 2?
4. Do the results of this investigation support the
law of conservation of mechanical energy?
Apply and Extend
5. Which part of this investigation caused the
h
board
288 MHR • Unit 3 Momentum and Energy
greatest difficulty? Provide suggestions for
overcoming this difficulty.
6. List some of the sources of error in this
investigation and suggest how these errors
might be reduced.
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7-B
I N V E S T I G AT I O N
TARGET SKILLS
Energy Transformations
of a Pendulum
Initiating and planning
Predicting
Performing and recording
Analyzing and interpreting
You have learned about mechanical energy and worked problems
based on the conservation of energy. Now you will design an
experiment that will test the law of conservation of mechanical
energy. A well-designed pendulum has such a small amount of
friction that it can be neglected.
Problem
How can you verify the conservation of
mechanical energy for a system in which all
of the work is done by conservative forces?
positions at which the mechanical energy of the
pendulum will be (a) all gravitational potential,
(b) all kinetic, and (c) half gravitational potential and half kinetic.
Prediction
Analyze and Conclude
Make predictions about the type of energy
transformations that will occur and when,
during the cycle, they will occur.
1. Which variables did you control in the
Equipment
3. How well did your results verify the law of
■
■
■
experiment?
2. How might you improve your experiment?
conservation of energy? Was any discrepancy
that you found within limits that are
acceptable to you?
parts for assembling a pendulum
motion sensor
metre stick or other measuring device
4. If mechanical energy was lost, explain how
and why it might have been lost.
Procedure
Write your own procedure for testing the law
of conservation of energy using a pendulum.
Clearly describe the required materials and the
proposed procedure.
Verify, with the teacher, that the experiment is
safe before proceeding.
Make detailed predictions identifying the
Apply and Extend
5. The transformation of energy from one
form to another is never completely efficient.
For example, predict what will happen to the
chemical potential energy of a car’s fuel as it
is transferred to the kinetic energy of the
motion of the car.
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Elastic Potential Energy and Kinetic Energy
In Chapter 6, you learned that the amount of elastic potential
energy stored in a stretched or compressed spring could be
expressed as Ee = 12 kx 2 where x is the amount of stretch of
compression of the spring. In this section, you will follow the
energy transformation processes that occur when the elastic potential energy stored in a spring or other elastic object is transformed
into kinetic energy.
Figure 7.4 An archer does work on the bow by pulling the string and
bending the bow. The bow stores elastic energy that can then be transformed into the kinetic energy of an arrow.
Hooke’s law and the concept of elastic potential energy can
apply to elastic objects other than springs. A practical example of
the transformation of elastic potential energy into kinetic energy
is shown in Figure 7.4 in which an archer is shooting an arrow.
Some other examples of elastic devices are the elastic strips in
sling shots that are sometimes used in paint ball games, and the
bungee cords that are used in some current amusement park rides.
In fact, to some extent, chemical bonds in molecules can act as
springs with the atoms vibrating on the ends of the “springs.”
You can apply the principles that you learn using springs to all
of these devices and more.
A mass oscillating on the end of a spring is an excellent model
from which to develop energy transformation principles. Start
with an ideal spring and mass sliding on a horizontal, frictionless
surface to eliminate the effects of both gravitational potential
energy and of nonconservative forces. As shown in Figure 7.5,
work is done on the mass to stretch it to a distance x = A. The
symbol A is used to represent the maximum extension of the
spring and is called the amplitude. When the mass is released,
the energy stored in the spring will remain in the spring mass
system as either elastic potential energy of the spring or kinetic
energy of the mass. You can express this statement mathematically
as shown.
ET = Ee + Ek
290 MHR • Unit 3 Momentum and Energy
or
ET =
1
mv 2
2
+
1
kx 2
2
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all elastic
potential
energy
Page 291
all elastic
potential
energy
all kinetic
energy
–A
0
x
A
Figure 7.5 When the mass is released from point A, the mass will
oscillate between points A and –A indefinitely.
The point x = 0 is the equilibrium point at which the spring is
neither stretched nor compressed and therefore has no elastic
potential energy. The following points describe, in words and
equations, the events that occur when the mass is released.
■
When the mass is at point x = A, the
total energy of the system is elastic
potential energy.
ET =
1
kA2
2
■
As the mass moves toward x = 0, elastic
potential energy is transformed into
kinetic energy and the total energy
is a combination of both.
ET =
1
kx 2
2
■
When the mass is at x = 0, there is no
elastic potential energy and the total
energy of the system is kinetic energy.
At this point, the speed of the mass is
at its maximum value, vmax.
ET =
1
2
mvmax
2
■
As the mass passes through x = 0, kinetic
energy is transformed back into elastic
potential energy of the spring because
the spring is becoming compressed. The
system contains both elastic potential
energy and kinetic energy. Notice that
the sign of position x and speed v does
not affect the equation because both
quantities are squared, making the
values positive.
ET =
1
kx 2
2
■
When the mass reaches x = −A, all of
the kinetic energy has been transformed
into elastic potential energy and the total
energy is once again all elastic potential
energy. Notice that (−A)2 = A2 .
ET =
1
kA2
2
+
+
1
mv 2
2
1
mv 2
2
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The Energy of a Spring
Bend
a Wall
and Mass
System
This quick lab will help you visualize the distribution of the energy of a spring/mass system at
every point in its oscillatory cycle. Imagine that
you have a perfect spring and mass system such
as the one shown in Figure 7.5. The object has
a mass of 0.500 kg and the spring constant is
600.0 N/m. You pull the mass an amplitude
of 20.0 cm from its rest position and release it.
Make a table with the headings Position,
Elastic Potential Energy, and Kinetic Energy.
In the column under “Position,” write the
following values, one on each row: –20.0 cm,
–15.0 cm, –10.0 cm, –5.0 cm, 0.0 cm, 5.0 cm,
10.0 cm, 15.0 cm, and 20.0 cm. For each position, calculate the elastic potential energy and
place the value in the second column. Using
ET = Ee + Ek , calculate the kinetic energy for
each position. On graph paper, make a plot
of energy versus position. On the same graph,
using different colours, plot elastic potential
TARGET SKILLS
Identifying variables
Analyzing and interpreting
energy versus position and kinetic energy
versus position. Add a line representing total
mechanical energy.
Analyze and Conclude
1. Describe the shape of the curves representing
the different types of energy.
2. How does your graph display the law of
conservation of mechanical energy?
3. Compare your graph with the graph of the
distribution of energy for a falling stone in
Figure 7.3. How does your graph of kinetic
and elastic potential energies differ from the
graph of kinetic and gravitational potential
energies?
4. Examine the equations for gravitational
potential and elastic potential energies and
explain why the graphs differ.
MODEL PROBLEMS
Horizontal Elastic Collisions
1. A low-friction cart with a mass of 0.25 kg
travels along a horizontal track and collides
head on with a spring that has a spring
constant of 155 N/m. If the spring was
compressed by 6.0 cm, how fast was the
cart initially travelling?
N
k = 155 m
m = 0.25 kg
cart
v
Frame the Problem
■
The cart is moving so it has kinetic energy.
■
The spring does negative work on the cart, lowering its kinetic energy.
■
The cart does work on the spring, giving it elastic potential energy.
■
The height of the cart does not change, so there is no change in
gravitational potential energy.
292 MHR • Unit 3 Momentum and Energy
spring
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■
The term low friction tells you to neglect the energy lost to
work done by friction.
■
The law of conservation of mechanical energy applies to
this problem.
Identify the Goal
The initial speed, v, of the cart
Variables and Constants
Known
m = 0.25 kg
N
k = 155
m
Unknown
x = 0.060 m
v
Strategy
Calculations
Write the law of conservation of mechanical
energy, including the energy quantities
associated with the interaction.
E′k + E′e = Ek + Ee
Initially, the spring was not compressed, so
the initial elastic potential energy was zero.
Ee = 0 J
After the interaction, the cart stopped, so the
final kinetic energy was zero.
E′k = 0 J
Substitute the values for energy listed above.
0 J + E′e = Ek + 0 J
Ek = E′e
Expand by substituting the expressions for
the energies.
1
mv 2
2
Solve for the initial velocity.
v=
=
1
kx 2
2
kx 2
m
155
(0.060 m)2
v=
0.25 kg
m
v = 1.493 99
s
m
v ≅ 1.5
s
Substitute numerical values and solve.
N
m
The cart was travelling at approximately 1.5 m/s before the collision.
Validate
Unit analysis of the equation v =
the standard units for velocity.
kx 2
shows that it is equivalent to m/s,
m
kg · m
N
2
m
m
m2
m
2
m
s
=
=
=
.
kg
kg
s2
s
A velocity of 1.5 m/s is reasonable for a lab cart.
continued
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2. A 225 g mass oscillates on a frictionless surface while attached
to a spring that has a spring constant of 145 N/m. The amplitude
of the oscillations is 80.0 cm.
(a) What is the maximum velocity of the mass?
(b) What is the position of the mass when its velocity is exactly
half of its maximum velocity?
Frame the Problem
■
The total mechanical energy of the mass is the same as the elastic
potential energy when the mass is at its maximum distance from
equilibrium position because its speed and therefore kinetic
energy is zero.
■
There is sufficient data to determine the elastic potential energy
at A.
■
The speed of the mass is at its maximum value when it is at its
equilibrium position.
■
The total mechanical energy of the mass is the same as the kinetic
energy when the mass is at its equilibrium position.
■
At any point in its motion, the total mechanical energy is the
sum of the elastic potential energy and the kinetic energy.
Identify the Goal
(a) The maximum speed, vmax, of the mass
(b) The position, x, at half maximum speed
Variables and Constants
Known
m = 225 g
N
k = 145
m
A = 80.0 cm
Unknown
vmax
x at 12 vmax
ET
Strategy
Calculations
Write the equation for total mechanical
energy at maximum amplitude.
ET =
All the values are known so substitute
and solve.
ET =
294 MHR • Unit 3 Momentum and Energy
1
kA2
2
N
(0.80 m)2
145
m
N·m2
ET = 46.4
m
ET = 46.4 J
1
2
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Write the equation for total mechanical
energy at the equilibrium position.
ET =
1
2
mvmax
2
2ET
2
= vmax
m
2ET
vmax =
m
2(46.4 J)
vmax =
0.225 kg
Solve for speed, v.
vmax =
vmax
vmax
412.444
m
20.3087
s
m
≅ 10.154
s
v1/2(max) =
v1/2(max)
Write the equation for total mechanical
energy an any position then solve for
position, x.
kg
m
= 20.3087
s
m
≅ 20.3
s
(a) The maximum speed of the mass is 20.3 m/s.
Find half maximum speed.
kg·m2
s2
ET =
1
kx 2
2
=
x2 =
x=
x=
x=
x=
x=
1
2
1
mv 2 + 12 kx 2
2
ET − 12 mv 2
2
ET − 12 mv 2
k
2
ET − 12 mv 2
k
m 2
2
1
46.4
J
−
(0.225
kg)
10.154
N
2
s
145 m
m
0.0137931
(46.4 J − 11.5991 J)
N
0.0137931 m
(34.8008 N·m)
N
√
0.48001 m2
x = ±0.69282 m
x ≅ ±0.693 m
(b) The mass has half its maximum speed at 69.3 cm on either side
of its equilibrium position.
Validate
Recalling that a J is equivalent to a N·m and to a kg·m2/s2 , the units
cancel correctly to give m/s for speed and m for position. Also, by
examining the curve for kinetic energy versus position (Quick Lab:
The Energy of a Spring and Mass System) you see that the mass
has travelled well over half the distance from equilibrium before
the speed has dropped to half its maximum speed.
continued
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PRACTICE PROBLEMS
9. A 1.2 kg dynamics cart is rolling to the right
along a horizontal lab desk at 3.6 m/s, when
it collides head on with a spring bumper that
has a spring constant of 2.00 × 102 N/m .
(a) Determine the maximum compression
of the spring.
(b) Determine the speed of the cart at the
moment that the spring was compressed
by 0.10 m.
(c) Determine the acceleration of the cart
at the moment that the spring was
compressed 0.10 m.
10. A circus car with a clown has a total mass
of 150 kg. It is coasting at 6.0 m/s, when it
hits a large spring head on. If it is brought to
a stop by the time the spring is compressed
2.0 m, what is the spring constant of the
spring?
11. An archery bow has an effective spring
constant of 485 N/m. The archer draws the
bow a distance of 45.0 cm. If the arrow has
a mass of 0.030 kg, what will be its speed at
the moment it leaves the bow?
12. A 0.0250 kg mass on a frictionless surface
is attached to a horizontal spring having a
spring constant of 124 N/m. The spring is
stretched to an amplitude of 9.00 cm and
released. Find:
(a) the maximum speed of the mass
(b) the speed of the mass at 3.00 cm on either
side of its equilibrium position.
13. A 0.150 kg mass on a frictionless surface is
attached to a spring having a spring constant
of 215 N/m. A motion detector determines
that the mass is travelling at 15.0 m/s when
it passes the equilibrium position.
(a) What is the amplitude of the motion of
the mass?
(b) How much work was done to stretch the
spring to its maximum amplitude?
14. An object on a frictionless surface is attached
to a spring having a spring constant of
235 N/m. 50.0 J of work were done on the
spring to stretch it to its maximum amplitude. As the mass passed through its
equilibrium position, a motion detector
determined that its speed was 14.6 m/s.
(a) What was the mass of the object?
(b) What was its amplitude?
(c) What was the object’s position when
its speed was 5.00 m/s?
Total Mechanical Energy
When solving problems involving gravitational and elastic potential energy and kinetic energy, the method is the same as you used
for just two of the three forms of mechanical energy. However, you
must take care in choosing your reference level for gravitational
potential energy. Analyze the position that was chosen for zero
potential energy in the following model problem until you
understand the reasoning behind the choice.
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MODEL PROBLEM
Vertical Elastic Collisions
A freight elevator car with a total mass of 100.0 kg is
moving downward at 3.00 m/s, when the cable snaps.
The car falls 4.00 m onto a huge spring with a spring
constant of 8.000 × 103 N/m. By how much will the spring
be compressed when the car reaches zero velocity?
elevator car
m = 100 kg
3.00
m
s
4.00 m
Frame the Problem
■
Initially, the car is in motion and therefore has kinetic energy.
It also has gravitational potential energy.
■
As the car begins to fall, the gravitational potential energy
transforms into kinetic energy. When the elevator hits the
spring, the elevator slows, losing kinetic energy, and the spring
compresses, gaining elastic potential energy.
■
When the elevator comes to a complete stop, it has no kinetic
or gravitational potential energy. All of the energy is now
stored in the spring in the form of elastic potential energy.
■
Since all of the motion is in a downward direction, define
“down” as the positive direction for this problem.
■
Choose the lowest level reached by the elevator as the
reference level for gravitational potential energy
x
spring
Eg = 0
Identify the Goal
The compression of the spring, x, when the car comes to rest
Variables and Constants
Known
mcar = 100.0 kg
m
v = 3.00
[down]
s
N
k = 8.000 × 103
m
h(above spring) = 4.00 m
Implied
m
g = 9.81 2
s
Unknown
x
Strategy
Calculations
Write the law of conservation of energy
for the forms of energy involved in
the problem.
E′g + E′e + E′k = Eg + Ee + Ek
The final gravitational potential energy
of the car is zero.
E′g = 0 J
The car comes to a rest at the lowest
point.
E′k = 0 J
continued
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Initially, the spring is not compressed.
Ee = 0 J
Substitute these initial and final conditions into the equation for conservation
of energy and simplify.
0 J + E′e + 0 J = Eg + 0 J + Ek
E′e = Eg + Ek
Expand by substituting the expressions
for the various forms of energy.
1
kx 2
2
= mgΔh +
1
kx 2
2
= mg(4.00 + x) +
The change in height for the gravitational potential is 4.00 m, plus the
compression of the spring, x. Substitute
this expression into the equation.
1
mv 2
2
1
mv 2
2
Since the equation yields a quadratic
equation, you cannot solve for x.
Substitute in the numerical values
and rearrange so that the right-hand
side is zero.
N 2
m
8.00 × 103
x = (100.0 kg) 9.81 2 (4.00 + x) m
m
s
m 2
+ 12 (100.0 kg) 3.00
s
4.00 × 103x 2 − 981x − 3924 − 450 = 0
Use the quadratic formula to find the
value of x.
4.00 × 103x 2 − 981x − 4374 = 0
√
981 ± (−981)2 − 4(4.00 × 103)(−4374)
x=
2(4.00 × 103)
x = 1.1756 m or − 0.930 25 m
1
2
x ≅ 1.18 m
Compression cannot be negative (or the spring would be stretching),
so choose the positive value. The spring was compressed 1.18 m.
Validate
kg · m
kg · m2
N
2
· m2 = s · m2 =
.
m
m
s2
kg · m2
m
On the right-hand side of the equation, the units are kg · 2 · m =
.
s
s2
Both sides of the equation have the same units, so you can have
confidence in the equation. The answer is also in a range that would
be expected with actual springs.
The units on the left-hand side of the final equation are
Note: The negative root in this problem is interesting in that it does
have meaning. If the car had somehow latched onto the spring during
the collision, the negative value would represent the maximum
extension of the spring if the car had bounced up from the bottom
due to the upward push of the spring.
PRACTICE PROBLEMS
15. A 70.0 kg person steps through the window
of a burning building and drops to a rescue
net held 8.00 m below. If the surface of the
net is 1.40 m above the ground, what must be
298 MHR • Unit 3 Momentum and Energy
the value of the spring constant for the net so
that the person just touches the ground when
the net stretches downward?
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16. A 6.0 kg block is falling toward a spring
located 1.80 m below. If it has a speed of
4.0 m/s at that instant, what will be the
maximum compression of the spring?
The spring constant is 2.000 × 103 N/m.
17. In a “head dip” bungee jump from a bridge
over a river, the bungee cord is fastened to
the jumper’s ankles. The jumper then steps
off and falls toward the river until the cord
becomes taut. At that point, the cord begins
to slow the jumper’s descent, until his head
just touches the water. The bridge is 22.0 m
above the river. The unstretched length of
the cord is 12.2 m. The jumper is 1.80 m tall
and has a mass of 60.0 kg. Determine the
(a) required value of the spring constant for
this jump to be successful
(b) acceleration of the jumper at the bottom
of the descent
PHYSICS & SOCIETY
TARGET SKILLS
Pole-Vaulting: The Art of Energy Transfer
Athlete Stacy Dragila won a gold medal for polevaulting at the 2000 Olympic Games in Sydney, the
first games to feature women’s vaulting. During
her winning vaults she demonstrated a fundamental principle of physics — the conservation of
energy. Dragila has speed, power, and near-perfect
technique. When she vaults, she uses and transfers
energy as efficiently as possible.
Before a pole-vaulter begins to sprint toward
the bar, she is storing chemical energy in her
muscles. If the vaulter is well trained, she will be
able to use the chemical energy as efficiently as
possible. This chemical energy is converted into
kinetic energy as she sprints toward the bar. Near
the end of the sprint, the vaulter places her pole in
a socket. Kinetic energy from the sprint bends the
pole and the pole gains elastic potential energy.
When the pole begins to extend from its bend,
it converts the elastic energy into gravitational
potential energy, lifting the vaulter into the air.
Ideally, a vaulter would be able to convert all of
her kinetic energy into elastic energy. But not even
Dragila has perfect technique. Even if she did, some
of the energy would still be converted to forms that
do not perform useful work. Some kinetic energy is
converted to thermal and sound energy during the
sprint, and some elastic energy is absorbed by the
pole as thermal energy during the lift.
The important thing to remember is that the
amount of energy involved in the pole vault
remains constant. The type of energy and the
efficiency with which the athlete uses the energy
make the difference between champions such as
Dragila and other pole-vaulters.
Analyzing and interpreting
Olympic gold medallist Stacy Dragila in action
Analyze
1. Sporting events are sometimes held at high
altitudes, such as the 1968 Olympic Games in
Mexico City. The thinner air poses challenges
for some athletes, but pole-vaulters, and also
long-jumpers and high-jumpers, tend to benefit.
Why?
2. Pole-vaulting began as an inventive way to get
across rivers and other barriers. The first poles
were made of bamboo. Today’s poles are much
more sophisticated. What principles need to be
considered when designing a pole?
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Section Review
1.
K/U Which types of energy are generally
referred to as mechanical energy?
2.
C Using examples not found in the textbook, describe and explain an example in
which the forces are
5.
K/U A skier starts from rest at the top of a
ski jump, skis down, and leaves the bottom
of the jump with a velocity v. Assuming
that the ski jump is frictionless, how many
times higher would the jump have to be for
the skier to jump with twice the velocity?
6.
MC Some Olympic sports transform kinetic
energy into gravitational potential energy
of the athlete’s body to aid performance.
These sports include:
(a) conservative
(b) non-conservative
3.
I A student is sliding down a frictionless
water slide at an amusement park.
(a) Sketch a graph of gravitational potential
energy against height for the descent.
(No numbers are required on the axes.)
(b) On the same axes, sketch a graph of the
total energy of the student against
height for the descent.
(c) On the same axes, sketch a graph of the
kinetic energy of the student against
height for the descent.
4.
An extreme sport in-line skating competition often involves aerial tricks at the
top of a curved ramp that has the shape
of a cylinder cut in half. Describe how
this activity demonstrates the concept
of conservation of mechanical energy.
■
bob sledding
■
high diving off a diving board
■
pole vaulting
■
snow ski racing
■
snow ski jumping
■
trampoline jumping
(a) In which of these sports can you
increase your gravitational potential
energy by using better sports equipment? (for example, a better diving
board will allow you to jump higher).
C
(b) Suppose you can use better materials to
produce equipment with greater elastic
potential energy. Which sports will be
affected?
UNIT PROJECT PREP
Athletic activities from walking to
skydiving involve work being done
by conservative and non-conservative
forces.
300 MHR • Unit 3 Momentum and Energy
■
Identify conservative and non-conservative forces involved with your
project topic.
■
How are non-conservative forces used
in the design of sports equipment to
increase safety?
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Conservation of
Total Energy
In the real world, there are no frictionless surfaces, ideal springs, or
vacuums in which fluid friction is absent. Some type of frictional
force influences nearly every real process, and frictional forces are
nonconservative forces. Nevertheless, the law of conservation of
energy is a fundamental law of nature even in the real world. To
develop the law of conservation of energy and apply it to real
processes, it is helpful to define and use the concept of a system.
Any object or group of objects can be defined as a system. Once
a system is defined, forces are classified as internal or external
forces. An internal force is any force exerted on any object in the
system due to another object in the system. An external force is
any force exerted by an object that is not part of the system on an
object within the system.
Scientists classify systems according to their interaction with
their surroundings, as illustrated in Figure 4.6. An open system
can exchange both matter and energy with its surroundings.
Matter does not enter or leave a closed system, but energy can
enter or leave. Neither matter nor energy can enter or leave an
isolated system.
SECTION
OUTCOMES
• Analyze quantitatively the relationships among mass, speed,
and thermal energy, using the
law of conservation of energy.
• Compare empirical and theoretical values of total energy and
account for discrepancies.
• Analyze common energy
transformation situations using
the closed system work-energy
theorem.
KEY
TERMS
• internal force
• external force
• open system
• closed system
• isolated system
open system
closed system
isolated system
An open pot of potatoes boiling on the stove represents an
open system, because heat is entering the pot and water vapour is leaving
the system. A pressure cooker prevents any matter from escaping but heat
is entering, so the pressure cooker represents a closed system. If the pot is
placed inside a perfect insulator, neither heat nor water can enter or leave
the system, making it an isolated system.
Figure 7.6
An external force can do work on a closed system, thus increasing the energy of the system. Conversely, a closed system can
do work on its surroundings, thus decreasing the energy of the
system. An isolated system, however, does not interact in any way
with its surroundings. The law of conservation of (total) energy is
based on an isolated system and stated in the following box.
MISCONCEPTION
A Closed System Is Not Isolated
Many people confuse the terms
“closed” and “isolated” as they
apply to systems. Although it
might sound as though closed
systems would not exchange
anything with their surroundings,
they do allow energy to enter or
leave. Only isolated systems
prevent the exchange of energy
with the surroundings.
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PHYSICS FILE
The first law of thermodynamics
is another way of stating the law
of conservation of energy. It
states that the change in the
energy of a system is equal to the
amount of work done on or by the
system plus the amount of heat
that enters or leaves the system.
The first law of thermodynamics
can be stated mathematically as
follows, where Q symbolizes heat.
ΔE = W + Q
LAW OF CONSERVATION OF ENERGY
Energy can neither be created nor destroyed, but it can be
transformed from one form to another or transferred from one
object to another. The total energy of an isolated system,
including all forms of energy, always remains constant.
To bring the law of conservation of energy to a practical level,
consider the example of skiers on a hill.
Friction is a non-conservative force. The amount of work done
by a non-conservative force depends on the path taken by the
force and the object. For example, the amount of energy transferred to the snow in Figure 7.7 depends on the path taken by
the skier. The skier going straight down the slope should reach
the bottom with a greater speed than the skier who is tracking
back and forth across the slope.
Figure 7.7 Although friction between the skis and the snow is small,
friction nevertheless does some work on the skiers, slowing their velocity
a little. The work done by friction is greater along the longer of the paths.
ELECTRONIC
LEARNING PARTNER
To enhance your understanding of
energy transformation, refer to
your Electronic Learning Partner.
Friction causes the skier to do work on the environment. The
snow heats up slightly and is moved around. For the skier, this is
negative work — the skier is losing energy and cannot regain it as
useful kinetic or potential energy. The sum of the skier’s kinetic
and gravitational potential energy at the end of the run will be
less than it was at the start.
Wind pressure is another example of a nonconservative force.
If the skier had the wind coming from behind, the wind (the
environment) could be doing work on the skier. This would be
positive work. The sum of the skier’s kinetic and gravitational
potential energies could increase beyond the initial total. However,
the amount of energy transferred by the wind would depend to
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a large extent on the path of the skier, so the wind would be a
nonconservative force.
When dealing with nonconservative forces, the law of conservation of energy still applies. However, you must account for the
energy exchanged between the moving object and its environment.
One approach to this type of situation is to define the system as
the skier and the local environment; that is, the skier, wind,
and snow become the system.
All real mechanical processes involve some type of frictional
forces that do work on the system, causing a reduction of the
amount of mechanical energy in the system at the end of the
process. You will often read that the system “lost” energy.
Since energy cannot be created or destroyed, what happens
to the energy? Work done by nonconservative forces transforms
mechanical energy into thermal energy, which is the random
kinetic energy of the molecules in a substance. Heat is the transfer
of thermal energy during a process. You can now state the law
of conservation of energy mathematically as shown in the
following box.
PHYSICS FILE
Quite often, you might not want
your forces to be conservative.
Without friction, many of your
clothes would simply fall apart
into strands as you moved. In
addition, keep-fit programs would
have to be greatly modified. A
person who rides an exercise
bicycle to lose mass (through
chemical reactions that provide
the energy) does not want the
energy back. It simply is dissipated as sound and heat. Likewise,
the weight lifter who does work
to lift a bar bell does not expect
to receive that energy back when
the bar bell is lowered.
CONSERVATION OF TOTAL ENERGY
The work done by nonconservative forces is the difference of
the final mechanical energy and the initial mechanical energy
of a system.
PHYSICS FILE
Wnc = Efinal − Einitial
Quantity
work done by nonconservative forces
Symbol
Wnc
SI unit
J (joule)
mechanical energy of the system
after the process
Efinal
J (joule)
mechanical energy of the system
before the process
Einitial
J (joule)
Unit Analysis
All of the units are joules.
In many processes, the temperature change in the objects
involved is so small that it is almost imperceptible, making the
calculations seem artificial. The following investigation, however,
will make temperature changes resulting from a loss of mechanical
energy seem real.
One of James Joule’s scientific
goals was to determine the
“mechanical equivalent of heat.”
A story is often told that, while
Joule was on his honeymoon in
Switzerland, he spent his time
measuring the temperature of
water at the top and at the bottom of waterfalls. By determining
the difference in the water
temperature, he could determine
how much thermal energy the
water would gain by falling a
certain distance.
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I N V E S T I G AT I O N
7-C
Mechanical and Thermal Energy
Before 1800, physicists and chemists did not
know that a relationship existed between
mechanical energy and heat. Count Rumford
(Benjamin Thompson: 1753–1814) was the first
to observe such a relationship, followed by
Julius Robert Mayer (1814–1878). Rumford and
Mayer made some very important discoveries.
Mayer was unable to express himself clearly
in writing, however, so his discoveries were
overlooked. Eventually, James Prescott Joule
(1818–1889) was credited with the determination of the mechanical equivalence of heat. In
this investigation you will perform experiments
similar to those of Mayer and Joule.
Problem
How much heat is produced when a mass of
lead pellets is repeatedly lifted and dropped
through a known distance?
Equipment
■
■
■
■
■
■
balance
thermometer (˚C)
lead shot
cardboard or plastic tube with a small hole in the
side, close to one end; the ends must be able to
be closed
metre stick
small amount of masking or duct tape
Procedure
1. Determine the mass of the lead shot.
2. Place the lead shot into the tube and close up
the tube. Let the tube sit upright on a desk
for several minutes to allow the tube and its
contents to come to room temperature. Make
sure that the hole is close to the bottom of
the tube.
3. Insert the thermometer or temperature probe
through the hole in the tube and nestle the
end in the lead shot. Measure and record
the temperature.
304 MHR • Unit 3 Momentum and Energy
TARGET SKILLS
Performing and recording
Analyzing and interpreting
Identifying variables
4. Close the hole.
5. Measure the length of the tube.
6. Repeatedly invert the tube for several
minutes, waiting only to allow the lead
shot to fall to the bottom on each inversion.
Keep track of the number of inversions.
7. Finish the inversions with the hole near the
bottom of the tube. Remove the tape and
measure the temperature of the lead shot.
Record the final temperature.
Analyze and Conclude
1. What were the initial and final temperatures
of the lead shot? What was the total mass of
the lead shot?
2. Determine the quantity of heat gained by
the lead shot using the formula, Q = mcΔT,
where Q is the quantity of heat, m is the
mass of the lead shot, c is the specific heat
capacity of lead (128 J/kg·C˚), and ΔT is the
temperature change.
3. Determine the total distance through which
the lead shot was lifted by the inversions
and calculate the total gain in gravitational
potential energy of the lead.
4. Determine the percentage of the gravitational
potential energy that was converted into heat.
5. If the conversion into heat does not account
for all of the gravitational potential energy
gained by the lead shot, where else might
some of the energy have gone?
Apply and Extend
6. How could this investigation be improved?
Try to design a better apparatus and, if possible, carry out the investigation again.
7. Do research and write a summary of the
work of Rumford, Mayer, and Joule on the
mechanical equivalence of heat.
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MODEL PROBLEMS
Energy Conversions on a Roller Coaster
mcar = 200 kg
1. A roller-coaster car with a mass of 200.0 kg
(including the riders) is moving to the right
at a speed of 4.00 m/s at point A in the diagram. This point is 15.00 m above the
ground. The car then heads down the slope
toward point B, which is 6.00 m above the
ground. If 3.40 × 103 J of work are done by
friction between points A and B, determine
the speed of the car at point B.
4.00
A
m
s
15.0 m
B
6.0 m
Frame the Problem
■
As the roller-coaster car moves down the track, most of the
gravitational potential energy is converted into kinetic energy,
but some is lost as heat due to friction.
■
The law of conservation of total energy applies.
■
The work done by friction is negative because it removed energy
from the system.
Identify the Goal
The speed of the car at point B, vB
Variables and Constants
Known
hA = 15.00 m
hB = 6.00 m
m
vA = 4.00
s
Wnc = −3.40 × 103 J
m = 200.0 kg
Implied
m
g = 9.81 2
s
Unknown
vB
Strategy
Calculations
Write the law of conservation of energy.
Rearrange to solve for E′k .
Wnc = Efinal + Einitial
Wnc = E′k + E′g − Ek − Eg
E′k = Wnc − E′g + Ek + Eg
Expand by substituting the expressions
for the forms of energy. Solve for the
speed of the car at point B
vB =
= Wnc − mghB +
1
mvA2
2
+ mghA
2(Wnc − mghB +
+ mghA)
m
2(Wnc − mghB + 12 m(vA)2 + mghA)
vB =
m
2 −(3.40 × 103 J) − 1.1772 × 104 kg ·2m2 + 1.6 × 103 kg m22 + 2.943 × 104
s
s
vB =
1
mvB2
2
1
mvA2
2
vB2 =
kg · m2 s2
200.0 kg
3.1716 × 104 kgs·2m
200.0 kg
2
continued
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continued from previous page
Since the problem only requires
magnitude, choose the positive value.
m2
s2
m
vB = ±1.2593 × 101
s
m
vB ≅ 12.6
s
vB =
1.5858 × 102
The speed of the car at point B will be 12.6 m/s.
Validate
The speed at point B is expected to be larger than its speed at point A.
2. Work Done by Air Friction
A 65.0 kg skydiver steps out from a hot air balloon that
is 5.00 × 102 m above the ground. After free-falling a
short distance, she deploys her parachute, finally
reaching the ground with a velocity of 8.00 m/s
(approximately the speed with which you would hit
the ground after having fallen a distance of 3.00 m).
(a) Find the gravitational potential energy of the skydiver,
relative to the ground, before she jumps.
(b) Find the kinetic energy of the skydiver just before
she lands on the ground.
(c) How much work did the nonconservative frictional
force do?
Einitial
5.00 × 102 m
Frame the Problem
■
Make a sketch of the problem and label it.
■
The hot air balloon can be considered to be at rest, so the vertical velocity
of the skydiver initially is zero.
■
As she starts to fall, the force of air friction does negative work on her,
converting much of her gravitational potential energy into thermal energy.
■
The rest of her gravitational potential energy was converted into kinetic energy.
Identify the Goal
Initial gravitational potential energy, Eg , of the skydiver relative to the ground
Final kinetic energy, Ek , of the skydiver just before touchdown
Work done by non-conservative force, Wnc
Variables and Constants
Known
m = 65.0 kg
m
g = 9.81 2
s
Δh = 5.00 × 102 m
m
v = 8.00 2
s
Unknown
Einitial
Eg
Wnc
Efinal
Ek
306 MHR • Unit 3 Momentum and Energy
Ffriction
Efinal
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Strategy
Calculations
Use the equation for gravitational potential energy.
All of the needed variables are known, so substitute
into the equation.
Eg = mgΔh
m Eg = 65.0 kg 9.81 2 5.00 × 102 m
s
m
Eg = 3.188 × 105 kg 2 m
s
Multiply.
1 kg
m
s2
Eg = 3.188 × 105 N · m
is equivalent to 1 N.
Eg = 3.188 × 105 J
1 N · m is equivalent to 1 J.
(a) The skydiver’s initial gravitational energy, before starting the
descent, was 3.19 × 105 J .
Use the skydiver’s velocity just before touchdown to
find the kinetic energy before touchdown. All of the
needed values are known, so substitute into the
equation for kinetic energy, then multiply.
1 kg
m2
s2
Ek =
1
mv 2
2
1
65.0
2
m 2
kg 8.00
s
m2
3
Ek = 2.080 × 10 kg 2
s
Ek =
Ek = 2.080 × 103 J
is equivalent to 1 J.
(b) The skydiver’s kinetic energy just before landing was 2.08 × 103 J .
Use the equation for the conservation of total energy
to find the work done by air friction. The final
mechanical energy is entirely kinetic energy and the
initial mechanical energy was entirely gravitational
potential energy. These values were just calculated.
Wnc = Efinal − Einitial
Wnc = 2.080 × 103 J − 3.188 × 105 J
Wnc = −3.167 × 105 J
(c) Air friction did 3.17 × 105 J of negative work on the skydiver.
Validate
The 3.17 × 105 J of work done by air resistance
is very large. This is fortunate for the skydiver,
as it allows her to land softly on the ground,
having lost 99.3 percent of her mechanical
energy. This large loss of mechanical energy
is what a parachute is designed to do.
How fast would the skydiver in the model problem
be travelling just before hitting the ground if the
work done by air resistance was ignored?
continued
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PRACTICE PROBLEMS
18. Determine the speed of the roller-coaster
(b) Ignoring frictional effects, determine the
car in the sample problem at point C if
point C is 8.0 m above the ground and another
4.00 × 102 J of heat energy are dissipated by
friction between points B and C.
mcar = 200 kg
4.00
A
speed of the ball as it strikes the floor,
assuming that it fell from rest.
(c) How fast is the ball moving just before
you catch it?
22. A 2.0 g bullet initially moving with a
m
s
C
15.0 m
B
6.0 m
8.0 m
velocity of 87 m/s passes through a block of
wood. On exiting the block of wood, the
bullet’s velocity is 12 m/s. How much work
did the force of friction do on the bullet as it
passed through the wood? If the wood block
was 4.0 cm thick, what was the average force
that the wood exerted on the block?
23. The Millennium Force, the tallest roller
19. A sled at the top of a snowy hill is moving
forward at 8.0 m/s, as shown in the diagram.
The height of the hill is 12.0 m. The total
mass of the sled and rider is 70.0 kg.
Determine the speed of the sled at point X,
which is 3.0 m above the base of the hill,
if the sled does 1.22 × 103 J of work on the
snow on the way to point X.
8.0
m
s
coaster in North America, is 94.5 m high at
its highest point. What is the maximum possible speed of the roller coaster? The roller
coaster’s actual maximum speed is 41.1 m/s.
What percentage of its total mechanical energy is lost to thermal energy due to friction?
24. A 15 kg child slides, from rest, down a play-
ground slide that is 4.0 m long, as shown in
the figure. The slide makes a 40˚ angle with
the horizontal. The child’s speed at the bottom is 3.2 m/s. What was the force of friction
that the slide was exerting on the child?
12.0 m
X
3.0 m
4.0 m
Y
20. If the sled in the previous question reaches
the base of the hill with a speed of 15.6 m/s,
how much work was done by the snow on
the sled between points X and Y?
21. A 0.50 kg basketball falls from a 2.3 m shelf
onto the floor, then bounces up to a height
of 1.4 m before you catch it.
(a) Calculate the gravitational potential
energy of the ball before it falls.
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40˚
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Concept Organizer
Work done
W = F Δd cos θ
Mechanical
kinetic energy
Ek =
1
mv 2
2
Gravitational
potential energy
Eg = mg Δh
yes
Conservation of
mechanical energy
Et′otal = Etotal
Work
done by
conservative
force
no
Conservation of
total energy
E ′final
mechanical
energy
− E initial
mechanical
energy
= W nc
Elastic
potential energy
Ee =
1
kx 2
2
Figure 7.8 How many energy transformations are
taking place in the photograph?
7.2
Section Review
1.
K/U How are conservation of mechanical
energy and conservation of energy
different?
2.
C How is the conservation of energy
demonstrated if an in-line skater falls
during a competition?
3.
A bullet is fired into a block of wood
and penetrates to a distance Δd. An identical bullet is fired into a block of wood with
a velocity three times that of the first.
How much farther, in terms of Δd, does
the second bullet penetrate the wood?
K/U
4.
MC
(a) In an amusement park there is a ride
on which children sit in a simulated log
while it slides rapidly down a watercovered slope. At the bottom, the log
slams into a trough of water, which
slows it down. Why did the ride designers not simply have the log slam into a
large spring?
(b) Steel or plastic barrels are located along
highways to cushion the impact if a car
skids into a bridge abutment. These
barrels are often filled with energyabsorbing material. Why are these
barrels used instead of large springs to
bring the cars to a stop?
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7.3
SECTION
OUTCOMES
• apply qualitatively the law of
conservation of momentum
to one-dimensional collisions
and recoil.
• Determine experimentally
whether a collision is elastic
or inelastic
KEY
TERMS
• elastic collisions
• inelastic collisions
• conservation of momentum
Page 310
Conservation of Momentum
When the cue ball hits the eight ball in billiards, the eight ball hits
the cue ball. When a rock hits the ground, the ground hits the
rock. In any collision, two objects exert forces on each other.
Collisions are another type of process in which conservation laws
play an essential role. As is the case with all processes, total
energy is conserved in a collision. If the materials involved in the
collision are perfectly elastic, mechanical energy — more specifically kinetic energy — is conserved. This condition divides
collisions into two categories, elastic collisions in which kinetic
energy is conserved and inelastic collisions in which kinetic
energy is not conserved.
Another important conservation law is also extremely useful in
analyzing collisions — the law of conservation of momentum. In
Chapter 5, you discovered that rearranging Newton’s second law
led to the concepts of impulse and momentum. You learned about
the impulse-momentum theorem which states that the impulse
exerted on an object is equal to the change in the momentum of
the object during a collision. The impulse-momentum theorem
can be stated mathematically as FΔt = mv2 − mv1 . In this chapter,
you will take the process one step further and apply Newton’s
third law to the impulse-momentum theorem.
Figure 7.9
The game of billiards offers many excellent examples of
collisions.
Newton’s Third Law and Momentum
Newton’s third law states that “For every action force on object B
due to object A, there is a reaction force, equal in magnitude but
opposite in direction, acting on object A due to object B.” Unlike
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Newton’s second law focusses on the motion of one specific object
but his third law deals with the interaction between two objects.
When you apply Newton’s third law to collisions, you discover one
of the most important laws of physics — the law of conservation
of momentum. The following steps, along with the diagram in
Figure 7.10, show you how to derive the law of conservation of
momentum by applying Newton’s third law to a collision between
two objects.
F BΔt
mA
v A′
mA
mA
vA
mB
v B′
vA
mB
FAΔt
mB
Figure 7.10 Object A
exerts a force on object B,
causing a change in B’s
momentum. At the same
time, object B exerts a
force equal in magnitude
and opposite in direction
on object A, changing A’s
momentum.
■
Write the impulse-momentum
theorem for each of two objects, A
and B, that collide with each other.
v A2 − mA
v A1
F AΔt = mA
F BΔt = mBv B2 − mBv B1
■
Apply Newton’s third law to the
forces that A and B exert on each
other.
F A = −F B
The duration of the collision is the
same for both objects. Therefore,
you can multiply both sides of the
equation above by Δt.
F AΔt = −F BΔt
Substitute the expressions for
change in momentum in the first
step into the equation in the third
step and then simplify.
mA
v A2 − mA
v A1 = −(mB
v B2 − mB
v B1)
■
■
■
Algebraically rearrange the last
equation so that (1) the terms representing the before-collision conditions precede the equals sign and
(2) the terms for the after-collision
conditions follow the equals sign.
mA
v A2 − mA
v A1 = −mB
v B2 + mB
v B1
v A1 + mB
v B1 = mA
v A2 + mB
v B2
mA
The last equation is a mathematical expression of the law of
conservation of momentum, which states that the total momentum
of two objects before a collision is the same as the total momentum of the same two objects after they collide.
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PHYSICS FILE
LAW OF CONSERVATION OF MOMENTUM
When working with collisions,
instead of using subscripts such
as “2,” physicists often use
a superscript symbol called a
“prime,” which looks like an
apostrophe, to represent the
variables after a collision. The
variable is said to be “primed.”
Look for this notation in the box
on the right.
The sum of the momenta of two objects before a collision is
equal to the sum of their momenta after they collide.
v A + mB
v B = mA
v A′ + mB
v B′
mA
Quantity
Symbol
SI unit
mass of object A
mA
kg (kilograms)
mass of object B
mB
kg (kilograms)
velocity of object A
before the collision
vA
m
(metres per second)
s
velocity of object B
before the collision
vB
m
(metres per second)
s
velocity of object A
after the collision
v A′
m
(metres per second)
s
velocity of object B
after the collision
v B′
m
(metres per second)
s
The law of conservation of momentum can be broadened to
more than two objects by selecting all of the objects in an isolated
system. To demonstrate that the momentum of an isolated system
is conserved, start with the impulse-momentum theorem, where
psys represents the total momentum of all of the objects within
the system.
■
■
■
COURSE CHALLENGE:
SCANNING TECHNOLOGIES
Momentum
A moving planet, bowling ball, and
an electron share the property of
momentum. Learn more about
momentum conservation as it
relates to the Course Challenge
in your e-book.
An impulse on a system due to an
external force causes a change in the
momentum of the system.
F extΔt = Δ
psys
If a system is isolated, the net
external force acting on the system
is zero.
(0.0 N) Δt = Δ
psys
If the impulse is zero, the change in
momentum must be zero.
kg · m
Δ
psys = 0.0
s
The last expression is an alternative form of the equation for the
conservation of momentum. The equation states that the change in
momentum of an isolated system is zero. The particles or objects
within the system might interact with each other and exchange
momentum, but the total momentum of the isolated system does
not change.
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In reality, systems are rarely perfectly isolated. In nearly all real
situations, immediately after a collision, frictional forces and interactions with other objects change the momentum of the objects
involved in the collision. Therefore, it might appear that the law
of conservation of momentum is not very useful. However, the law
always applies to a system from the instant before to the instant
after a collision. If you know the conditions just before a collision,
you can always use conservation of momentum to determine the
momentum and, thus, velocity of an object at the instant after a
collision. Often, these values are all that you need to know.
Collisions in One Dimension
Since momentum is a vector quantity, both the magnitude and the
direction of the momentum must be conserved. Therefore, momentum is conserved in each dimension, independently. For complex
situations, it is often convenient to separate the momentum into
its components and work with each dimension separately. Then
you can combine the results and find the resultant momentum of
the objects in question. Solving problems that involve only one
dimension is good practice for tackling more complex problems.
Figure 7.11 A moving car and
its occupants can be defined as
being a system. The children in
the car might be exerting forces
on each other or on objects that
they are handling. Although they
are exchanging momentum
between themselves and the
objects, these changes have no
effect on the total momentum
of the system.
MODEL PROBLEM
v = 5.45
Analyzing a Collision between Boxcars
A 1.75 × 104 kg boxcar is rolling down
a track toward a stationary boxcar that
has a mass of 2.00 × 104 kg. Just before
the collision, the first boxcar is moving
east at 5.45 m/s. When the boxcars
collide, they lock together and continue
down the track. What is the velocity
of the two boxcars immediately after
the collision?
v =0
m
s
A
B
A
B
A
B
Frame the Problem
■
Make a sketch of the momentum vectors
representing conditions just before and just
after the collision.
■
Before the collision, only one boxcar (A) is
moving and therefore has momentum.
■
At the instant of the collision, momentum
is conserved.
before
m
v A = 5.45
s
mA = 1.75 × 104 kg
after
m
v B = 0.00
s
mB = 2.00 × 104 kg
v AB
′ =?
continued
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continued from previous page
■
After the collision, the two boxcars (A and B) move as one mass,
with the same velocity.
Identify the Goal
v A′ B , of the combined boxcars immediately after the collision
The velocity, Variables and Constants
Known
mA = 1.75 × 104 kg
mB = 2.00 × 104 kg
m
v A = 5.45
[E]
s
Implied
m
v B = 0.00
s
Unknown
v A′ B
Strategy
Calculations
Apply the law of conservation of
momentum.
mA
v A + mB
v B = mA
v A′ + mB
v B′
After the collision, the two masses
act as one, with one velocity. Rewrite
the equation to show this condition.
v A + mB
v B = (mA + mB)
mA
v A′ B
v A′ B .
Solve for m
v + mB
vB
v A′ B = A A
(mA + mB)
Substitute values and solve.
(1.75 × 104 kg)(5.45 ms [E]) + (2.00 × 104 kg)(0.00
v A′ B =
(1.75 × 104 kg + 2.00 × 104 kg)
9.53 75 × 104 kgs· m [E]
v A′ B =
3.75 × 104 kg
m
v A′ B = 2.543
[E]
s
m
[E]
v A′ B ≅ 2.54
s
The locked boxcars were rolling east down the track at 2.54 m/s.
Validate
The combined mass of the boxcars was nearly double the mass
of the boxcar that was moving before the collision. Since the exponents of mass and velocity are always one, making the relationships
linear, you would expect that the velocity of the combined boxcars
would be just under half of the velocity of the single boxcar before
the collision. Half of 5.45 m/s is approximately 2.7 m/s. The
calculated value of 2.54 m/s is very close to what you would expect.
314 MHR • Unit 3 Momentum and Energy
m
)
s
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PRACTICE PROBLEMS
25. Claude and Heather are practising pairs
26. Two amusement park “wrecker cars” are
skating for a competition. Heather (47 kg) is
skating with a velocity of 2.2 m/s. Claude
(72 kg) is directly behind her, skating with a
velocity of 3.1 m/s. When he reaches her, he
holds her waist and they skate together. At
the instant after he takes hold of her waist,
what is their velocity?
heading directly toward each other. The combined mass of car A plus driver is 375 kg and
it is moving with a velocity of +1.8 m/s. The
combined mass of car B plus driver is 422 kg
and it is moving with a velocity of −1.4 m/s.
When they collide, they attach and continue
moving along the same straight line. What is
their velocity immediately after they collide?
Recoil
Imagine yourself in the situation illustrated in Figure 7.12. You are
in a small canoe with a friend and you decide to change places.
Assume that the friction between the canoe and the water is
negligible. While the canoe is not moving in the water, you very
carefully stand up and start to take a step. You suddenly have the
sense that the boat is moving under your feet. Why?
Figure 7.12 If you start to step forward in a canoe, the canoe recoils
under your feet.
When you stepped forward, your foot pushed against the
bottom of the canoe and you started to move. You gained momentum due to your velocity. Momentum of the system — you, your
friend, and the canoe — must be conserved, so the canoe started to
move in the opposite direction. The interaction that occurs when
two stationary objects push against each other and then move
apart is called recoil. You can use the equation for conservation
of momentum to solve recoil problems, as the following problem
illustrates.
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MODEL PROBLEM
Recoil of a Canoe
For the case described in the text, find the velocity of the canoe
and your friend at the instant that you start to take a step, if your
velocity is 0.75 m/s[forward]. Assume that your mass is 65 kg
and the combined mass of the canoe and your friend is 115 kg.
Frame the Problem
■
Make a simple sketch of the conditions before and after you took a step.
before
after
mA = 65 kg
mB = 115 kg
mA = 65 kg
mB = 115 kg
m
v A = 0.00
s
m
v B = 0.00
s
m
v A′ = 0.75
s
v B′ = ?
■
The canoe was not moving when you started to take a step.
■
You gained momentum when you started to move. Label yourself “A”
and consider the direction of your motion to be positive.
■
The canoe had to move in a negative direction in order to conserve
momentum. Label the canoe and your friend “B.”
Identify the Goal
v B′ of the canoe and your friend
The initial velocity, Variables and Constants
Known
mA = 65 kg
m
v A′ = 0.75
s
mB = 115 kg
Implied
m
v A = 0.00
s
m
v B = 0.00
s
Unknown
v B′
Strategy
Calculations
Apply conservation of momentum.
mA
v A + mB
v B = mA
v A′ + mB
v B′
Velocities before the interaction were zero;
therefore, the total momentum before the
interaction was zero. Set these values
equal to zero and solve for the velocity
of B after the reaction.
kg · m
= mA
v A′ + mB
v B′
s
v A′ = −mB
v B′
mA
m v′
v B′ = − A A
mB
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0.0
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(65 kg)(0.75
v B′ = −
115 kg
m
v B′ = −0.4239
s
m
v B′ ≅ −0.42
s
Substitute values and solve.
m
)
s
The velocity of the canoe and your friend, immediately after you
started moving, was –0.42 m/s.
Validate
Since the mass of the canoe plus your friend was larger than your
mass, you would expect that the magnitude of their velocity would
be smaller, which it was. Also, the direction of the velocity of the
canoe plus your friend must be negative, that is, in a direction
opposite to your direction. Again, it was.
PRACTICE PROBLEMS
27. A 1385 kg cannon containing a 58.5 kg
28. While you are wearing in-line skates, you are
cannon ball is on wheels. The cannon fires
the cannon ball, giving it a velocity of
49.8 m/s north. What is the initial velocity
of the cannon the instant after it fires the
cannon ball?
vA
mA
mB
vB
standing still and holding a 1.7 kg rock.
Assume that your mass is 57 kg. If you throw
the rock directly west with a velocity of
3.8 m/s, what will be your recoil velocity?
29. The mass of a uranium-238 atom is
3.95 × 10−25 kg. A stationary uranium atom
emits an alpha particle with a mass of
6.64 × 10−27 kg. If the alpha particle has a
velocity of 1.42 × 104 m/s, what is the recoil
velocity of the uranium atom?
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I N V E S T I G AT I O N
7-D
TARGET SKILLS
Predicting
Performing and recording
Analyzing and interpreting
Examining Collisions
Are there characteristics that allow you to
predict whether a collision will be elastic? In
this investigation, you will observe and analyze
several collisions and draw conclusions regarding
whether a type of collision will be elastic or
inelastic. If you do not have the necessary
equipment, you can find excellent simulations
on the Internet.
photogates so that the velocity flags will pass
through the gates smoothly but will trigger
the gates.
photogate
timers
Problem
What are the characteristics of elastic and
inelastic collisions?
air pump
4. Label one glider “A” and the other glider
Equipment
■
■
■
■
■
■
■
■
air track (with source of compressed air)
2 gliders (identical, either middle- or
large-sized)
2 photogate timers
laboratory balance
4 glider bumper springs
2 Velcro™ bumpers (or a needle and a piece
of wax)
2 velocity flags (10 cm) (or file cards cut to a
10 cm length)
modelling clay
Procedure
1. Set up the air track and adjust the levelling
screw to ensure that the track is horizontal.
You can test whether the track is level by
turning on the air pressure and placing a
glider on the track. Hold the glider still and
then release it. If the track is level, the glider
will remain in place. If the glider gradually
starts moving, the air track is not level.
2. Attach a velocity flag (or 10 cm card) and
two bumper springs to each glider. If only
one bumper spring is attached, the glider
might not be properly balanced.
3. Position the photogates about one fourth the
length of the track from each end, as shown
in the diagram. Adjust the height of the
318 MHR • Unit 3 Momentum and Energy
“B.” Use the laboratory balance to determine
accurately the mass of each glider.
5. With the air flowing, place glider A on the left
end of the air track and glider B in the centre.
6. Perform a test run by pushing glider A so
that it collides with glider B. Ensure that the
photogates are placed properly so that the
flags are not inside the gates when the gliders
are in contact. Adjust the positions of the
photogates, if necessary.
7. The first set of trials will be like the test run,
with glider A on the left end of the track and
glider B in the centre. Turn on the photogates
and press the reset button. Push glider A and
allow it to collide with glider B. Allow both
gliders to pass through a photogate after the
collision, then catch them before they
bounce back and pass through a photogate
again. Record the data in a table similar to
the one shown on the next page. Since all of
the motion will be in one dimension, only
positive and negative signs will be needed to
indicate direction. Vector notations will not
be necessary.
The displacement, Δd, is the distance
that the gliders travelled while passing
through the photogates. This displacement is
the length of the flag. Time Δti is the time
that a glider spent in the photogate before the
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collision, while Δtf is the time the glider
took to pass through the photogate after the
collision. Calculate velocity, v, from the
displacement and the time interval. Be sure
to include positive and negative signs.
Trial
Glider A (mass = ?)
Δti
vi
Δd
vf
Δt f
1
2
3
Trial
Glider B (mass = ?)
Δt f
Δd
vf
1
2
3
8. Increase the mass of glider B by attaching
some modelling clay to it. Be sure that the
clay is evenly distributed along the glider.
If the glider tips to the side or to the front
or back, the motion will not be smooth.
Determine the mass of glider B. Repeat step 7
for the two gliders, which are now of
unequal mass.
9. Exchange the gliders and their labels. That is,
the glider with the extra mass is on the left
and becomes glider “A.” The glider with no
extra mass should be in the centre and
labelled “B.” Repeat step 7 for the new
arrangement of gliders.
10. Remove the clay from the glider. Place one
glider at each end of the track. Practise starting both gliders at the same time, so that
they collide near the middle of the track.
The collision must not take place while
either glider is in a photogate. When you
have demonstrated that you can carry out
the collision correctly, perform three trials
and record the data in a table similar to the
ones shown here. This table will need two
additional columns — one for the initial time
for glider B and a second for the initial
velocity of glider B.
11. Remove the bumper springs from one end of
each glider and attach the Velcro™ bumpers.
(If you do not have Velcro™ bumpers, you
can attach a large needle to one glider and a
piece of wax to the other. Test to ensure that
the needle will hit the wax when the gliders
collide.)
12. With the Velcro™ bumpers attached, perform
three sets of trials similar to those in steps 7,
8, and 9. You might need to perform trial
runs and adjust the position of the photogates so that both gliders can pass through
the photogate before reaching the right-hand
end of the air track. Record the data in tables
similar to those you used previously.
Analyze and Conclude
1. For each glider in each trial, calculate the
initial momentum (before the collision) and
the final momentum (after the collision).
2. For each trial, calculate the total momentum
of both gliders before the collision and the
total momentum of both after the collision.
3. For each trial, compare the momentum
before and after the collision. Describe how
well the collisions demonstrated conservation of momentum.
4. In any case for which momentum did not
seem to be conserved, provide possible
explanations for errors.
5. Calculate the kinetic energy of each glider in
each trial. Then calculate the total kinetic
energy of both gliders before and after the
collision for each trial.
continued
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continued from previous page
6. Compare the kinetic energies before and after
7. Examine the nature of the collisions that you
the collisions and decide which collisions
were elastic and which were inelastic. Due
to measurement errors, do not expect the
kinetic energies to be identical before and
after a collision. Decide if the values appear
to be close enough that the differences could
be attributed to measurement errors.
considered to be elastic and those that you
classed as inelastic. Look for a trend that
would permit you to predict whether a collision would be elastic or inelastic. Discuss
your conclusions with the rest of class. How
well did your conclusions agree with those
of other class members?
MODEL PROBLEM
Classifying a Collision
A 0.250 kg billiard ball moving at 5.00 m/s collides head on with a
stationary, 0.800 kg steel ball. If the billiard ball bounces directly
backwards at 2.62 m/s, was the collision elastic?
Frame the Problem
■
Momentum is always conserved in a collision.
■
If the collision is elastic, kinetic energy must also be conserved.
■
The motion is in one dimension so only positive and negative
signs will be used to indicate direction.
■
If the billiard ball bounced directly backwards, its velocity after
the collision must be negative.
■
Let subscript b represent the billiard ball and subscript s represent
the steel ball.
Identify the Goal
Is the collision elastic? If it is elastic, then Ekb = E k′ b + Eks
Variables and Constants
Known
mb = 0.250 kg
ms = 0.800 kg
m
vb = 5.00
s
m
v b′ = −2.62
s
320 MHR • Unit 3 Momentum and Energy
Implied
Unknown
m
vs = 0.0
s
v s′
E′kb
Ekb
E′ks
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Strategy
Calculations
Since momentum is conserved in all collisions,
use the law of conservation of momentum to
find the velocity of the steel ball after the
collision.
mbvb + msvs = mbv′b + msv′s
mbvb + 0.0 − mbv′b = msv′s
v′s =
v′s =
v′s =
mbvb − mbv′b
ms
(0.250 kg)(5.00
1.25
m
s
− (0.250 kg)(−2.62
0.800 kg
m
)
s
− (−0.655
0.800
m
)
s
m
)
s
1.905 ms
0.800
m
v′s = 2.38125
s
v′s =
Calculate the total kinetic energy before the
collision. This is just the kinetic energy of
the billiard ball since the steel ball was
not moving.
Ekb =
1
m v2
2 b b
Ekb =
1
(0.250
2
Ekb =
1
2
m 2
kg) 5.00
s
2
kg·m
6.25
s2
Ekb = 3.125 J
Calculate the sum of the kinetic energies of
the billiard ball and the steel ball. This sum
is the total kinetic energy after the collision.
E′kb =
1
m v′ 2
2 b b
1
(0.250
2
m 2
kg) 2.62
s
kg·m2
1
E′kb = 2 1.7161
s2
E′kb = 0.85805 J
E′kb =
E′ks =
1
m v′2
2 s s
1
(0.800
2
m 2
kg) 2.38125
s
kg·m2
1
E′ks = 2 4.536281
s2
E′ks = 2.26814 J
E′ks =
E′kb + E′ks = 0.85805 J + 2.26814 J
E′kb + E′ks = 3.12619 J
The total kinetic energies before and after the collision are the same
to the third decimal place. Therefore, the collision was elastic.
Validate
Although the kinetic energies before and after the collision differ in
the fourth decimal place, the difference is less than 1%. Since the data
contained only three significant digits, this difference could easily be
due to the precision of the measurement. Therefore, it is fair to say
that the collision was elastic.
continued
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PRACTICE PROBLEMS
30. A billiard ball of mass 0.155 kg moves with
31. Car A, with a mass of 1735 kg, was travelling
a velocity of 12.5 m/s toward a stationary
billiard ball of identical mass and strikes it
in a head on collision. The first billiard ball
comes to a complete stop. Determine
whether the collision was elastic.
north at 55.5 km/h and Car B, with a mass of
2540 kg, was travelling north at 37.7 km/h
when Car A struck Car B from behind. If the
cars stuck together after the collision and
continued in a straight line, what was their
combined momentum after the collision?
Was the collision elastic or inelastic?
7.3
1.
Section Review
C Explain qualitatively how Newton’s
third law is related to the law of conservation of momentum.
2.
What is the difference between an
internal force and an external force?
3.
How does a closed system differ from
an isolated system?
4.
Under what circumstances is the
change in momentum of a system equal
to zero?
K/U
5.
322 MHR • Unit 3 Momentum and Energy
Define and give an example of recoil.
6.
What is the difference between an
elastic collision and an inelastic collision?
7.
Describe an example of an elastic
collision and an example of an inelastic
collision that were not discussed in
the text.
8.
Given a set of data for a collision,
describe a step-by-step procedure that
you could use to determine whether the
collision was elastic.
K/U
K/U
K/U
K/U
C
C
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Technological Implications
Compare the safety equipment worn by the two football players in
Figure 7.13. Although the game has probably become much more
physical during the same period of time, the chances of a serious
injury have been significantly reduced by applying the principles
of energy and momentum. Millions of dollars of basic research
have gone into studying and improving sports equipment. In
football equipment alone, imagine how many types of pads are
designed on the principle of increasing the time interval of the
collision in order to decrease the force on one part of the body.
In addition to safety, new designs in sports equipment have
resulted in materials that help athletes run faster, hit golf balls
farther, and jump higher.
SECTION
OUTCOMES
• Analyze and describe examples
where energy- and momentumrelated technologies were
developed and improved
over time.
• Describe and evaluate the
design of technological
solutions and the way they
function using principles of
energy and momentum.
Figure 7.13 Sports equipment
has come a long way in the last
fifty years.
Athletes choose to participate in sports but all citizens must
avail themselves of transportation. As you learned in Chapter 5
when you read about and designed crumple zones for vehicles,
a knowledge of energy and momentum is critical in auto safety.
Go back to page 195 and look at Figure 5.18 of the race car accident. Do you have a better understanding of the reasons that the
driver was able to walk away from the accident now than when
you first saw the photograph? The research that goes into race
cars is eventually applied to passenger vehicles. Modern vehicles
have many safety features built into them today.
Safety in industry is also improving, from hard hats to steel
toes of shoes for construction workers. The Canadian Standards
Association (CSA) studies the materials and design of safety
equipment for industry, the general public, and sports and then
sets standards for each piece of equipment. For example, if you are
shopping for a bicycle helmet, be sure that it has been approved
by the CSA. In this section, you will read features, perform an
experiment, and then choose a topic of interest in which to do
in-depth research.
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CAREERS IN PHYSICS
Science of the Sole
When the world’s athletes take a run at the gold in
the Olympics, few people know that a Canadian
physicist is running in spirit right beside them.
Dr. Benno Nigg, a professor at the University of
Calgary and founder and chairperson of its Human
Performance Laboratory, is a leading expert in
biomechanics. Biomechanics is the science of how
living things move. With biomechanics, scientists
can determine how dinosaurs really walked, tens
of millions of years ago. Biomechanics can also
shed light on how humans walk, or, more importantly for athletes, how humans run.
It may sound strange, but there is a lot of
physics in the design of running shoes. Dr. Nigg’s
work demands a precise application of forces,
energy, and thermal physics. For example, the
hardness of the shoe affects which kinds of muscle
fibres are activated. This, in turn, will have an
effect on the runner’s fatigue. So an in-depth
understanding of how the mechanical energy of
the impact of the shoe is transmitted to the runner
is vital. These discoveries are even now being used
by athletics-wear companies to design the next
generation of running shoe.
Dr. Nigg’s interest in this kind of physics has
taken him far. He has won awards from around the
world, and is a member of the Olympic Order and
the International Olympic Committee Medical
Commission. He has consulted for all of the largest
sportswear and equipment companies. If you look
at the qualifications for researchers and technicians
at any of these companies, you will see very
quickly that almost every position requires a
knowledge of biomechanics. This is truly a career
that can bring an interest in sports and an interest
in science together.
Dr. Nigg will tell you that, when he first entered
the field, his main qualifications were a broad educational background and a deep interest in physics.
He says these enabled him to go anywhere:
“When you study physics, you open doors.”
324 MHR • Unit 3 Momentum and Energy
Dr. Benno Nigg
Going Further
According to Dr. Nigg, the best way to learn about
biomechanics and the careers available is to get
hands-on experience. Many students don’t realize
that very often university and corporate laboratories offer summer programs specifically aimed at
high school students. Call a local university or an
athletics-equipment manufacturer and see what
they have to offer.
Web Link
www.mcgrawhill.ca/links/atlphysics
The Internet has a number of excellent resources that can
help to explain some of the more difficult concepts and
considerations involved in determining how runners run
and what we can do to help them run faster. For example,
Dr. Nigg’s Biomechanigg page illustrates a number of these
concepts. Go to the web site shown above to find out
where to go next. Try to create your own method for
measuring the forces involved in walking and running.
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PHYSICS & SOCIETY
TARGET SKILLS
The Physics of Car Safety
When a car stops suddenly, you keep going. This
example of Newton’s first law of motion has been
responsible for many traffic injuries. Countless
drivers and passengers have survived horrible
crashes because they were wearing seat belts,
and air bags have also played a major role.
To understand the physics behind the design
of air bags, imagine that the car you are driving is
suddenly involved in a head-on collision. At the
instant of impact, the car begins to decelerate. Your
head and shoulders jerk forward, and the air bag
pops out of its compartment. The bag must inflate
rapidly, before your head reaches the wheel, and
then start to deflate as your head hits it. This causes
your head to decelerate at a slower rate. In addition,
the force of your impact with the air bag is exerted
over a wider area of your body, instead of being
concentrated at the impact site of your head
with something small, such as the top of the
steering wheel.
Physics is also involved in the design of car tires.
The key consideration is the amount of tire area
that stays in contact with the road during braking
Analyzing and interpreting
Hypothesizing
and turning; the more tire contact, the better your
control of the car. Also important is having tires
that resist “hydroplaning” on wet roads — at slow
speeds, water skiers sink; at high speeds, they glide
over the surface of the water. That’s just what you
do not want your car tires to do in the rain.
Engineers used various physics principles to design
tires with a centre groove that pumps water away
from the surface as the tires roll over wet pavement.
Analyze
1. Air bags have come under increased scrutiny.
Research the reasons for this debate.
2. To keep more rubber in contact with the road,
tires could be made wider. The ultimate would
be a single tire as wide as the car. What would
be the disadvantage of this type of tire? What
might limit the maximum width of a tire?
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L A B
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Safety First
Bend a Wall
TARGET SKILLS
Initiating and planning
Modelling concepts
Conducting research
Part A Egg Drop Contest
How well can you design safety equipment for
a raw egg? Work in pairs for this activity. With
your partner, write down everything involving
forces, impulse, momentum, and energy that
might help you design your equipment.
Brainstorm the shape and type of materials
that will best protect your raw egg based on the
principles of physics that you wrote down.
The only restriction is that the total mass of
your equipment cannot exceed 750 g.
As a class, decide on the height (between 3 m
and 15 m) and a location from which you will
drop your eggs. Obtain your materials, build
your safety equipment for your egg, and carry
out the contest. If there are several winners
whose eggs were unbroken, your class might
choose a higher location and have another
contest among the winners of the first.
Analyze and Conclude
1. What physics principles seemed to be the
most important in the design of your egg
safety equipment?
2. What materials were used by the most
successful contestants? What property of
these materials do you think provided the
most protection for the eggs?
3. Discuss a practical application for the princi-
ples that you learned while designing your
egg safety equipment.
Part B Safety Equipment Poster
Your goal in this activity is to prepare a poster
based on in depth research into one type of safety
equipment. Work in small groups and assign
different areas of research to each member of the
group. Within the group, choose some type of
safety equipment from the categories of sports,
transportation, or industry. Include in your
research and poster presentation information
about the history of the product. For example,
before the equipment was developed, what type
7.4
1.
of accidents or injuries were occurring? What
stimulated the early developers of the equipment to pursue research into this area? Include
scientific principles used by the researchers and
developers of the equipment. How does your
new knowledge of physics help you to understand the research? Finally describe the modern
equipment and its capabilities. Include photographs of the equipment. Make your poster
attractive and interesting for others to read.
Section Review
List at least five different types of
safety equipment that were not mentioned
in this section for which the principles
of energy and momentum were pivotal
in the design.
MC
326 MHR • Unit 3 Momentum and Energy
2.
Describe the safety equipment that most
affects your life today.
C
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7
Review
REFLECTING ON CHAPTER 7
■
■
■
■
■
■
Applied forces are conservative if the work
that they do on an object as it moves between
two points is independent of the path taken by
the object while moving between those points.
Applied forces are non-conservative if the
amount of work that they do on an object as
it moves between two points is dependent on
the path of the object between those points.
Work done by an object on its environment is
negative work and decreases the total energy
of that object. Work done on an object by its
environment is positive work and increases
the total energy of the object.
Mechanical energy of a system is conserved
when work is done by conservative forces.
Total energy is conserved even when work is
done by non-conservative forces. The work
done by non-conservative forces decreases
the mechanical energy of the system.
Wnc = Efinal − Einitial
■
■
■
■
■
■
■
The law of conservation of energy states that,
in an isolated system, the total energy is
conserved, but can be transformed from one
form to another.
The concept of impulse plays a significant role
in the design of safety systems. By extending
the time, Δt, of a collision, you can reduce the
amount of force, F , exerted.
By applying Newton’s third law, you can show
that momentum is conserved in a collision.
The momentum of an isolated system is
conserved.
Recoil is the interaction of two objects that are
in contact with each other and exert a force
on each other. Momentum is conserved
during recoil.
Kinetic energy is conserved in elastic collisions.
Kinetic energy is not conserved in inelastic
collisions.
An isolated system is one that neither gains
energy from its environment nor loses energy
to its environment.
Knowledge/Understanding
1. Explain what happens to the total mechanical
energy over a period of time for open systems,
closed systems, and isolated systems.
2. Explain what it means when you say that work
done on an object is independent of the path
taken by the object?
3. (a) Describe a situation in which kinetic energy
is converted into elastic potential energy.
(b) Describe a situation in which gravitational
energy is converted first into elastic potential energy and then into kinetic energy.
4. In the derivation that showed that the kinetic
energy of an object after falling a distance Δh,
is equal to the gravitational potential energy of
the object before it fell, neglecting any frictional
forces, how did you use the kinematic equation
5.
6.
7.
8.
9.
that relates final velocity, initial velocity, acceleration and displacement?
Explain the difference between a closed system
and an isolated system.
Explain, in words, how Newton’s third law is
used to show that momentum is conserved in
a collision.
How can the principles used in problems
involving conservation of momentum in one
dimension apply to problems involving two
or three dimensions?
You wind up the spring of a toy car and then
release it so that it travels up a ramp. Describe
all of the energy transformations that take place.
When two objects recoil, they start at rest and
then push against each other and begin to
move. Initially, since their velocities are zero,
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they have no momentum. When they begin to
move, they have momentum. Explain how
momentum can be conserved during recoil,
when the objects start with no momentum and
then acquire momentum.
Inquiry
10. Design a small wooden cart, with several raw
eggs as passengers. Incorporate elements into
your design to ensure that the passengers
would suffer no injury if the cart was involved
in a collision while travelling at 5.0 m/s. If
possible, test your design.
11. Design and carry out an experiment in which
an object initially has gravitational potential
energy that is soon converted into kinetic energy.
The object then collides with another object
that is stationary. Include in your design a
method for testing whether mechanical energy
is conserved in the first part of the experiment.
If possible, test your design.
Communication
12. A child descends a slide in the playground.
Write expressions to show the total mechanical
energy of the child at the top, halfway down,
and at the bottom of the slide. Write a mathematical expression that relates the energy total
at the three positions.
13. It is a calm day on a lake and you and a friend
are on a sailboat. Your friend suggests attaching
a fan to the sailboat and blowing air into the
sails to propel the sailboat ahead. Explain
whether this would work.
14. Imagine you are standing, at rest, in the middle
of a pond on perfectly frictionless ice. Explain
what will happen when you try to walk back to
shore. Describe a possible method that you
could use to start moving. Would this method
allow you to reach shore? Explain.
15. You and a friend arrive at the scene of a car
crash. The cars were both severely mangled.
Your friend is appalled at the damage to the
328 MHR • Unit 3 Momentum and Energy
cars and says that cars ought to be made to
be sturdier. Explain to your friend why this
reaction to the crash is unwarranted.
16. Start with expressions that apply the impulsemomentum theorem to two objects and use
Newton’s third law to derive the conservation
of momentum for a collision between the two
objects. Explain and justify every substitution
and mathematical step in detail.
Making Connections
17. Car bumper systems are designed to absorb
the impact of slow-speed collisions in such
a way that the vehicles involved sustain no
permanent damage. Prepare a presentation on
how a bumper system works, including an
explanation of the energy transformations
involved.
18. In many sports such as baseball and golf, “follow
through” is considered to be very important.
Based on your knowledge of elastic collisions,
provide an explanation regarding the reason
that “follow through” will improve the results
of hitting a golf ball or baseball.
19. The floors of some gymnasiums are built so that
they move a little when an athlete lands and
then spring back in place. Explain, based on
impulse and momentum, how this “spring”
in the floor might reduce impact injuries to
athletes’ joints.
Problems for Understanding
20. A 0.500 kg mass resting on a frictionless
surface is attached to a horizontal spring with a
spring constant of 45 N/m. When you are not
looking, your lab partner pulls the mass to one
side and then releases it. When it passes the
equilibrium position, its speed is 3.375 m/s.
How far from the equilibrium position did your
lab partner pull the mass before releasing it?
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21. A 2.00 kg mass is attached to a 3.00 m string
26. A 48.0 kg in-line skater begins with a speed of
and is raised at an angle of 45˚ relative to
the rest position, as shown. Calculate the
gravitational potential energy of the pendulum
relative to its rest position. If the mass is
released, determine its velocity when it
reaches its rest position.
27.
45˚
3m
Δh
28.
22. A roller coaster at a popular amusement park
has a portion of the track that is similar to the
diagram provided. Assuming that the roller
coaster is frictionless, find its velocity at the
top of the loop.
29.
V1 = 0
70 m
V2 = ?
30.
30 m
23. A simple pendulum swings freely and rises at
the end of its swing to a position 8.5 cm above
its lowest point. What is its speed at its lowest
point?
24. A 50.0 g pen has a retractable tip controlled
by a button on the other end and an internal
spring that has a constant of 1200 N/m.
Suppose you hold the pen vertically on a table
with the tip pointing up. Clicking the button
into the table compresses the spring 0.50 cm.
When the pen is released, how fast will it rise
from the table? To what vertical height will it
rise? (Assume for simplicity that the mass of
the pen is concentrated in the button.)
25. A spring with a spring constant of 950 N/m is
compressed 0.20 m. What speed can it give to
a 1.5 kg ball when it is released?
31.
2.2 m/s. Friction also does –150 J of work on
her. Assume that she did not push on the
ground any more. If her final speed is 5.9 m/s,
(a) determine the change (final − initial)
in her gravitational potential energy.
(b) By how much, and in which direction
(up or down), has her height changed?
A 48.0 kg skateboarder kicks his 7.0 kg board
ahead with a velocity of 2.6 m/s[E]. If he runs
with a velocity of 3.2 m/s[E] and jumps onto
the skateboard, what is the velocity of the
skateboard and skateboarder immediately after
he jumps on the board?
Astrid, who has a mass of 37.0 kg, steps off a
stationary 8.0 kg toboggan onto the snow. If her
forward velocity is 0.50 m/s, what is the recoil
velocity of the toboggan? (Assume that the
snow is level and the friction is negligible.)
A 60.0 t submarine, initially travelling forward
at 1.5 m/s, fires a 5.0 × 102 kg torpedo straight
ahead with a velocity of 21 m/s in relation to
the submarine. What is the velocity of the submarine immediately after it fires the torpedo?
Suppose that a 75.0 kg goalkeeper catches a
0.40 kg ball that is moving at 32 m/s. With
what forward velocity must the goalkeeper
jump when she catches the ball so that the
goalkeeper and the ball have a horizontal
velocity of zero?
You and a colleague are on a spacewalk, repairing your spacecraft that has stalled in deep
space. Your 60.0 kg colleague, initially at rest,
asks you to throw her a hammer, which has a
mass of 3.0 kg. You throw it to her with a
velocity of 4.5 m/s[forward].
(a) What is her velocity after catching the
hammer?
(b) What impulse does the hammer exert on her?
(c) What percentage of kinetic energy is lost in
the collision?
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Review
Knowledge/Understanding
True/False
In your notebook, indicate whether each statement
is true or false. Correct each false statement.
1. Doing work on an object does not change the
object’s energy.
2. Mechanical kinetic energy is stored energy due
to gravity.
3. Work done is proportional to the applied force
and the square of the displacement in the same
direction.
4. Work is done on an object by a force acting
perpendicularly to the displacement.
5. It is possible to do negative work.
6. Gravitational energy is always measured from
the same reference point.
7. Conservation of mechanical energy requires
that a non-conservative force does the work.
8. The force of friction is a non-conservative force.
9. Efficiency is a ratio of useful work input
compared to the amount of work output.
Multiple Choice
In your notebook, write the letter of the best
answer for each of the following questions.
10. Work is not energy itself, but rather
(a) it is a form of kinetic energy.
(b) it is a form of gravitational potential energy.
(c) it is a force.
(d) it is a transfer of mechanical energy.
(e) it is a result of parallel forces.
11. Which of the following are equivalent
to a joule (J)?
(a) N · m2
(d) N m
s
2
(b) kg m
(e)
both
(b) and (c)
2
s
(c) N · m
12. Work done is zero when
(a) an applied force does not result in any
motion.
(b) uniform motion exists in the absence
of a force.
(c) the applied force is perpendicular to the
displacement.
330 MHR • Unit 3 Momentum and Energy
(d) both (a) and (c).
(e) all of the above.
13. A weight lifter lowers a barbell at constant
speed. Down is assigned as positive. In doing
so, the weight lifter
(a) does positive work on the barbell.
(b) allows gravity to do negative work on
the barbell.
(c) does not do any work on the barbell.
(d) does negative work on the barbell.
(e) allows the kinetic energy of the barbell
to increase.
14. You throw a rock straight up into the air. While
it rises and falls, its kinetic energy
(a) remains constant
(b) increases steadily
(c) changes direction only
(d) decreases then increases
(e) increases then decreases
15. Starting from rest at the top of a hill, a bicyclist
pedals furiously on the way down. The kinetic
energy of the bicycle and rider at the bottom
will be equal to
(a) lost potential energy
(b) work done
(c) work done plus lost potential energy
(d) work done plus kinetic energy plus
potential energy
(e) zero
Short Answer
16. Explain how work and a transformation of
energy are related.
17. What type of energy does a wind-up toy
contain after being wound just before release?
18. Does your arm do any work on your textbook as
you carry it down the hall if the book’s vertical
position does not change?
19. Describe the work done by a nail on a hammer
as the nail is driven into a wall. What evidence
is there that the work done is negative?
20. What is the factor by which a javelin’s kinetic
energy is changed, if its velocity is increased to
five times its initial velocity?
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21. Draw a graphical representation showing total
22.
23.
24.
25.
26.
27.
28.
mechanical energy, gravitational energy, and
mechanical kinetic energy versus time for the
following processes.
(a) A block of ice, initially at rest, slides down
a frictionless slope.
(b) A moving block of ice slides up a frictionless slope and instantaneously comes to rest.
Distinguish between an open system, a closed
system, and an isolated system.
Explain why a water hose recoils when the
water is turned on.
Explain why the first hill of a roller-coaster ride
must be the highest hill.
(a) Under what conditions will a marble of
mass m1 and a rock of mass 3m1 have
the same gravitational potential energy?
(b) Under what conditions will a moving
marble of mass m1 and a moving rock of
mass 3m1 have the same kinetic energy?
Write a general equation for the amount of
mechanical energy in a system and include
expressions for as many different forms of
potential energy as you can locate.
Consider two bodies, A and B, moving in the
same direction with the same kinetic energy. A
has a mass twice that of B. If the same retarding
force is applied to each, how will the stopping
distances of the bodies compare?
(a) Under what circumstances does the work
done on a system equal its change in kinetic
energy only?
(b) Under what circumstances does the work
done on a system equal the change in
gravitational potential energy only?
(c) Under what circumstances does the change
in kinetic energy of a system equal the
change in gravitational potential energy?
Inquiry
29. Investigate the energy transformations that
take place when an athlete is pole-vaulting.
30. Design a pogo stick for a child. Designate the
age range of the child you hope will enjoy
the stick and calculate the required spring
constant. Determine other parameters, such as
the length of the stick, the size of the spring,
and the range of distances that the child will be
able to depress the spring. Include a sketch of
your design.
31. Imagine that you found a very unusual spring
that did not obey Hooke’s law. In fact, you
performed experiments on the spring and
discovered that the restoring force was proportional to the square distance that the spring was
stretched or compressed from its equilibrium
or F = −kx 2 .
(a) Describe an experiment that you might have
done to find the expression for the restoring
force.
(b) Describe a method for finding the elastic
potential energy stored in the spring when it
is stretched a distance x.
Communication
32. Can the gravitational potential energy of a golf
ball ever be negative? Explain without using a
formula.
33. Describe how the graph represents conservation
of energy.
potential
kinetic
Energy
Total Mechanical Energy
34. Explain whether it is possible to exert a force
and yet not cause a change in kinetic energy.
35. You blow up a balloon and release the open
end, causing the balloon to fly around the room
as the air is rapidly exhausted. What exerts the
force that causes the balloon to accelerate?
Making Connections
36. Do research to find the efficiency of motion in
various animals. Compare the efficiencies of
walking, running, and swimming for humans.
For example, how much energy does a person
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use to walk a kilometre, run a kilometre, and
swim a kilometre? Compare these efficiencies
for other animals. Include kangaroos, because
a comparison of walking and hopping is quite
interesting.
37. When gymnasts perform routines in the uneven
bars or other events, they strive to land firmly
and stop without allowing their feet to move to
another position. Make some estimates about
their motion just before landing to calculate their
momentum. Since an impulse is required to
change their momentum, find out how much
force their bodies must endure in order to stop
abruptly. How to they spread this force around
to prevent injury?
41. A tow truck pulls a car by a cable that makes
an angle of 21˚ to the horizontal. The tension in
the cable is 6.5 × 103 N .
+
21˚
(a) How large is the force that causes the car
to move horizontally?
(b) How much work has the tow truck done
42.
Problems for Understanding
38. How fast will a 2.55 kg bowling ball be travel-
ling if the 358 J of work done to the ball are
transformed into kinetic energy?
39. A 250 kg roller coaster cart loaded with people
has an initial velocity of 3.0 m/s. Find the
velocity of the cart at A, B, and C. Assume
that friction is negligible.
43.
3.0 m
s
C
B
44.
60 m
44 m
A
10 m
35 m
40. A 45 kg cyclist travelling 15 m/s on a 7.0 kg
bike brakes suddenly and slides to a stop in
3.2 m.
(a) Calculate the work done by friction to stop
the cyclist.
(b) Calculate the coefficient of friction between
the skidding tires and the ground.
(c) Are you able to determine if the tires were
digging into the ground from your answer
in part (b)? Explain.
332 MHR • Unit 3 Momentum and Energy
+
45.
46.
on the car after pulling it 3.0 km?
A 550 kg car travelling at 24.0 m/s[E] collides
head-on with a 680 kg pickup truck. Both
vehicles come to a complete stop on impact.
(a) What is the momentum of the car before the
collision?
(b) What is the change in the car’s momentum?
(c) What is the change in the truck’s momentum?
(d) What is the velocity of the truck before the
collision?
A rocket is travelling 160 m/s[forward] in outer
space. It has a mass of 750 kg, which includes
130 kg of fuel. Burning all of the fuel produces
an impulse of 41 600 N · s . What is the new
velocity of the rocket?
An 8.0 kg stone falls off a 10.0 m cliff.
(a) How much work is done on it by the
gravitational force?
(b) How much gravitational potential energy
does it lose?
You are in a 1400 kg car, coasting down a
25˚ slope. When the car’s speed is 15 m/s, you
apply the brakes. If the car is to stop after
travelling 75 m, what constant force (parallel
to the road) must be applied?
An archery bow has a spring constant of
1.9 × 102 N/m. By how much does its elastic
potential energy increase if it is stretched
(a) 5.0 cm and (b) 71.0 cm?
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47. You exert 72 N to compress a spring with a
spring constant of 225 N/m a certain distance.
(a) What distance is the spring displaced?
(b) What is the elastic potential energy of the
displaced spring?
48. A 2.50 kg mass is attached to one end of a spring
on a horizontal, frictionless surface. The other
end of the spring is attached to one end of a
spring is attached to a solid wall. The spring
has a spring constant of 75.0 N/m. The spring is
stretched to 25.0 cm from its equilibrium point
and released.
(a) What is the total energy of the mass-spring
system?
(b) What is the velocity of the mass when it
passes the equilibrium position?
(c) What is the elastic potential energy stored in
the spring when the mass passes a point that
is 15.0 cm from its equilibrium position?
(d) What is the velocity of the spring when it
passes a point that is 15.0 cm from its equilibrium position?
49. A 275 g ball is resting on top of a spring that is
mounted to the floor. You exert a force of 325 N
on the ball and it compresses the spring 44.5 cm.
If you release the ball from that position, how
high, above the equilibrium position of the
spring-ball system will the ball rise?
50. A 186 kg cart is released at the top of a hill.
(a) How much gravitational potential energy is
lost after it descends through a vertical
height of 8.0 m?
(b) If the amount of friction acting on the cart is
negligible, determine the kinetic energy and
the speed of the cart after it has descended
through a vertical height of 8.0 m.
51. A small 95 g toy consists of a piece of plastic
52.
53.
54.
55.
attached to a spring with a spring constant of
365 N/m. You compress the spring against the
floor through a displacement of 5.5 cm, then
release the toy. How fast is it travelling when it
rises to a height of 10.0 cm?
Suppose a 1.5 kg block of wood is slid along a
floor and it compresses a spring that is attached
horizontally to a wall. The spring constant is
555 N/m and the block of wood is travelling
9.0 m/s when it hits the spring. Assume that
the floor is frictionless and the spring is ideal.
(a) By how much does the block of wood
compress the spring?
A spring with a spring constant of 120 N/m is
stretched 5.0 cm from its rest position.
(a) Calculate the average force applied.
(b) Calculate the work done.
(c) If the spring is then stretched from its 5.0 cm
position to 8.0 cm, calculate the work done.
(d) Sketch a graph of the applied force versus
the spring displacement to show the extension of the spring. Explain how you can
determine the amount of work done by
analyzing the graph.
A 32.0 kg child descends a slide 4.00 m high.
She reaches the bottom with a speed of 2.40 m/s.
Was the mechanical energy conserved? Explain
your reasoning and identify the energy
transformations involved.
A 2.5 kg wooden block slides from rest down
an inclined plane that makes an angle of 30˚
with the horizontal.
(a) If the plane is frictionless, what is the speed
of the block after slipping a distance of 2.0 m?
(b) If the plane has a coefficient of kinetic
friction of 0.20, what is the speed of the
block after slipping a distance of 2.0 m?
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