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PHYSICS
BRIDGING COURSE 2014
SCHOOL OF
PHYSICS
The School of Physics would like to acknowledge and pay respect to t he t raditional owners of t he land– t he Gadigal people of t he
Eora Nation. It is upon t heir ancestral lands t hat t he Universit y of Sydney is built. As we share our own knowledge, teaching, learning,
and research practices wit hin t his Universit y may we also pay respect to t he knowledge embedded f orever within t he Aboriginal
Custodianship of Country.
For more information head to
ht t p:/ / sydney.edu.au/ science/ physics/
Physics Bridging Course 2014
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© School of Physics, University of Sydney
PHYSICS BRIDGING COURSE
This course is designed for students entering any first year Physics course at the
University of Sydney who have not undertaken any study of Physics in Years 11 and
12 of High School. It should also provide a sound preparation for students in a
similar situation who are attending any other tertiary institution. The course is an
intensive introduction to basic concepts in Physics. It consists of 14 sessions, each of
which consists of a 1 hour lecture followed by a 2 hour small group tutorial.
The purpose of the course is not to cover the entire HSC Physics syllabus, but instead
to provide a good general introduction to the subject so that students can understand
the lecture and laboratory material from the beginning of the year. A good way of
defining the course objectives is:
 to become familiar with problem solving techniques
 to start to think in a way that Physics students need to
 to cover some basic content
Students are NOT expected to obtain a deep understanding of the subject but rather to
obtain an appreciation for Physics problems may be solved. This course is NOT about
content but rather the development of skills!
TEXT BOOKS
Most of the material covered in the lectures that you will require is included in these
notes but a suitable textbook is always useful. It will contain more details that cannot
be included here and many valuable worked examples and further problems and
exercises. The books recommended for first year Physics at the University of Sydney
depend on your course. They are:

PHYS1001 Regular, PHYS1003 Technological (second semester), PHYS1901
and PHYS1902 Advanced - University Physics (12th edition, with Modern
Physics) by Young and Freedman. (Publishers: Pearson/Addison Wesley)

PHYS1002 Fundamental and PHYS1004 Environmental - College Physics: A
Strategic Approach 2007 (2nd Edition) by Knight, Jones, and Field.
(Publishers: Pearson/Addison Wesley)
Copies of these texts are kept in the Physics Student Support Office (Room 203).
Depending on your choice of units, the textbook you need may be different in first
and second semesters. We will facilitate exchange of textbooks between students to
ensure you only need buy one textbook for the year.
One place to buy texts is the Co-op Bookshop on campus next to the Aquatic Centre.
Copies of this workbook are also available from the School of Physics.
© School of Physics, University of Sydney
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Physics Bridging Course 2014
ACKNOWLEDGEMENTS
Dr George Braoudakis wrote the original notes in 1998 and Dr Pal Fekete and Owen
Shepherd added questions and provided edits prior to 2006. Dr Phil Dooley made
changes after 2006 and Tom Gordon and George Pinniger were responsible for edits
after 2011.
Experiments were added in 2009, and in 2010 Waves were added to Chapter 7 and the
Chapter on Atomic and EM was added by Dr Alex Argyros. Thanks to Chris Stewart
and Gil Vella for their input on these chapters and for providing the material from the
Cumberland Campus Sciences Bridging Course (written by Gil Vella and Ian Cathers)
The School of Physics would like to acknowledge and pay respect to the traditional
owners of the land– the Gadigal people of the Eora Nation. It is upon their ancestral
lands that the University of Sydney is built. As we share our own knowledge,
teaching, learning, and research practices within this University may we also pay
respect to the knowledge embedded forever within the Aboriginal Custodianship of
Country.
4
© School of Physics, University of Sydney
Contents
Chapter 1: Measurement and Units ......................................................................... 6
Chapter 2: Motion in a straight line....................................................................... 13
Chapter 3: Vectors & Motion in 2 & 3 Dimensions.............................................. 28
Chapter 4: Forces and Motion ................................................................................ 37
Chapter 5: Work and Energy ................................................................................. 53
Chapter 6: Momentum ............................................................................................ 71
Chapter 7: Oscillations and Waves ........................................................................ 81
Chapter 8: Atomic Theory and Electromagnetism .............................................. 98
Chapter 9: Electric Charge ................................................................................... 105
Chapter 10: Current and Circuits ........................................................................ 111
Experiments ............................................................................................................ 119
© School of Physics, University of Sydney
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CHAPTER 1
MEASUREMENT AND UNITS
What is Physics?
Physics is the most fundamental of all sciences. Physics is also the foundation of
much modern technology, from maglev trains to hip replacements, from CDs to
sunglasses, from light bulbs to carbon-dating. Physics ranges from the very large (the
universe) to the very small (the quantum world) and everything in between. Physics
looks to understand the behaviour of objects under certain conditions and how these
observations fit into our world.
Models and theories
Physics is based on experiment, i.e. on observing some event and taking careful
measurements of what happens and when it happens. It then tries to explain these
observations. Why does a stone fall when dropped from a height? Why is it safer
inside a car when there is danger of a lightning strike? Does a black hole spin?
Scientists start by making a hypothesis – an educated guess of what they will
observe. In order to explain their observations physicists develop theories. The theory
is usually based on a mathematical model that can be used to predict the outcome of
even more experiments and observations. If these observations agree with the theory
then the theory is accepted. The test of any theory is experiment.
Most naturally occurring events are so complicated (e.g. a fish swimming through
water, the movement of the earth’s tectonic plates, you remembering a funny joke)
that to understand them our theories must make several approximations or
assumptions. That does not mean our theory is wrong. It just makes it simpler than
to do all the complicated mathematics and obtain a fuller answer. As long as we are
aware of our assumptions our predictions should be close to our observations.
In order to study Physics we must be familiar with one of the tools physicists use.
This is mathematics. We will need mathematics to do simple algebraic manipulations
and sometimes some very subtle approximations. The more you study Physics the
more mathematics you will need.
Language and Physics
While studying Physics you will realise that sometimes you must learn new words
and terms and other times you will learn new meanings to everyday familiar words.
Words such as velocity, acceleration, heat, work and energy have very specific and
precise meanings in Physics. When physicists use such words they choose them
carefully so that accurate and correct information is communicated. The everyday use
of such words is less precise.
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© School of Physics, University of Sydney
Chapter 1: Measurement and Units
Measuring things
In order to make observations we must measure quantities. There are many different
quantities we can measure. How long does it take to get to work? What is the
temperature now? What electrical current is drawn by a toaster? The list goes on. In
order to describe a physical quantity we must first define a unit i.e. a measure of the
quantity that is exactly 1. Then we must define a standard which is a reference to
which all other examples of the quantity are compared. The definition of the standard
should be sensible, accurate and practical for physicists all around the world. There
are many different quantities but a standard does not exist for all. Instead there are
standards for the base quantities and all other quantities can be defined in terms of
these.
SI units
Definitions of units are agreed upon by an international committee and in 1971 the
14th General Conference on Weights and Measures decided on a system of seven
base quantities. These make up the International System of Units (S.I. Units).
Physical quantity
Length
Mass
Time
Electric current
Thermodynamic temperature
Luminous intensity
Amount of substance
Table 1.1 SI base units
Unit name
metre
kilogram
second
ampere
kelvin
candela
mole
Unit symbol
m
kg
s
A
K
cd
mol
For each unit there is also a symbol (abbreviation) which is commonly used to
describe a quantity. For example when we say the mass of an object is 15.2
kilograms we write it as 15.2 kg. Derived quantities such as velocity are written as
25 m.s-1 (note the full stop used to separate consecutive symbols). Some derived
quantities are so important they have special names and units assigned. Work, force
and pressure (Joules, Newtons and Pascals respectively) are such quantities. Note
that the units generally begin with lower case but if the unit is named after a person
then the symbol begins with a capital.
Most measurements are much larger or smaller than the units so we use either
scientific notation or prefixes when writing such quantities. For example light takes
0.0000000033 s to travel 1 m. We can write this using scientific notation as 3.3 x 10-9
s. Alternatively we can use the prefix “nano” – meaning 10-9 – to express it as 3.3
nanoseconds (in symbols: 3.3 ns).
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Factor
Prefix
Symbol
12
10
teraT
109
gigaG
106
megaM
103
kilok
2
10
hectoh
101
dekada
Table 1.2 Prefixes for SI units.
Factor
10-12
10-9
10-6
10-3
10-2
10-1
Prefix
piconanomicromillicentideci-
Symbol
p
n

m
c
d
Significant figures
When we make measurements we usually quote all the figures the instrument is
capable of delivering. These are termed the significant figures. When making
calculations using such quantities we must take care to only quote as many
significant figures as justified by the original data. When we multiply or divide
quantities the result should be rounded to the same number of significant figures as in
the least precise piece of data. When we add or subtract quantities the result is
rounded to the decimal place that matches the decimal place of the least precise piece
of data.
A significant figure is one that is non-zero. For example, 123.45 has 5 significant
figures. Zeroes that appear between non-zero digits are significant, eg 501.6 has 4
significant figures. All zeroes after a decimal place are significant, eg 20.00 has 4
significant figures. However, leading zeroes after a decimal place are not significant,
eg 0.002300 has 4 significant figures – the two zeroes before the ‘2’ are not
significant. All zeroes appearing in a number to the right of non-zero digits, but
without a decimal point, are not significant, eg 100 has only 1 significant figure – the
digit ‘1’.
Q.
Determine the speed of a train that travels 1000m in 44.1 s. The speed is
distance
given by
speed =
time
A.
1000 m
1 significant figure up top, 3 on the bottom
44.1 s
= 22.6757... m.s-1 = 20 m.s-1 or 2 x 101 m.s-1
speed =
The final answer should have the same number of significant figures as the least
precise original data, which is 1.
Q.
If a train travels 60.1 km and then a further 4.213 km find the total distance
travelled.
A.
60.1 km + 4.213 km = 64.313 km which is rounded to 64.3 km (1 decimal
place)
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© School of Physics, University of Sydney
Chapter 1: Measurement and Units
Calculating with units
Notice in the above examples that the units calculate in the same way as the
quantities: the initial units of metres divided by seconds gives the final units of metres
per second.
It is good practice to always check that the units of your answer correspond with those
of the calculation. That way you can always double check the calculation by looking
at the resulting units and also change units if necessary.
Q.
Calculate the average speed of a car that travels 56 km in 48 min.
1000 m
60 s
A.
To convert to SI we use the ratios
and
.
1 km
1 min
The calculation becomes
1000 m
56 km ´
1 km
average speed =
60 s
48 min ´
1 min
=
56 ´ 1000 m
48 ´ 60 s
Note the units of km and min
have cancelled out.
= 19 m.s-1
© School of Physics, University of Sydney
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Questions and Problems
1. In your own words how would you define Physics?
2. What do we mean by the scientific method?
Try to draw a concept map that might describe the scientific method using the
following words:
experiment, law, model, observation, theory
3. Physics is a language in its own right.
a. In what ways is Physics similar/different to other languages (for example
German, Mandarin, English)?
b. What role does mathematics and the use of equations play in this language?
c. What other things make up this language?
4. What do we need to do before we can take a measurement of something?
Consider the differences and similarities between the temperature scales:
Celsius, Fahrenheit and Kelvin.
5.
a.
b.
c.
d.
Define the following terms in your own words:
unit
standard
base quantity
derived quantity
6.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Express each of the following in both scientific notation and using prefixes:
2 300 m
54 000 g
0.00014 s
3 520 000 J
4 700 V
0.000045 A
0.025 m
0.000000067 s
0.0082 V
3 100 000 000 
7. Calculate each of the following quantities to the correct number of
significant figures. Also describe these quantities in terms of their correct SI
units (you may need to convert quantities in some cases).
a. The speed of a train if it travels 1001 m in 44.1s.
b. The total distance travelled by a train if it travels 60.137 km and then another
4.21 km.
c. The average speed of a car when it travels 56 km in 48 min.
d. The time it takes a starter’s gun to be heard if it is fired 1km from the listener.
Assume you see the effect on the gun instantly.
(NB speed of sound is 343 ms-1 at 20o C).
e. The average mass of an egg if 12 eggs weigh 0.75 kg.
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© School of Physics, University of Sydney
Chapter 1: Measurement and Units
The following three questions are a little more difficult.
Do not be surprised if you cannot do them all.
8.
a.
b.
c.
d.
e.
f.
g.
h.
On the sheet of graph paper below draw the following functions.
y=3
x=2
y = -5
y = 3x - 4
y = 5 - 2x
2
y=x
2
y = 3x
2
y=1-x
© School of Physics, University of Sydney
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9. A bowling ball rolls along the bowling alley. A table of distance travelled
versus time taken is given in the table below.
time (s)
distance (m)
0
0
1
3
2
5.5
3
8.7
4
12.4
5
15.1
a. On the graph paper below plot the points from this data.
b. Draw a line of best fit for this data. This is the straight line that passes as close
to as many as the points as possible.
c. The speed is given by the distance travelled divided by the time taken.
Calculate the speed of this ball using the line of best fit.
Go to Experiment 1 in the experiment section in the back of the book.
Student Learning
What is the most important thing you believe you learnt in this topic?
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© School of Physics, University of Sydney
CHAPTER 2
MOTION IN A STRAIGHT LINE
Motion
Everywhere we look things are in motion. Pens scribble across pages, cars drive
along Parramatta Road, students are running late for lectures. The study of such
motions is called kinematics. We will use new terms with exact meanings – do not
confuse them with the meanings of everyday conversation.
Position and displacement
In order to talk about the motion of an object we need to specify its position. This
requires a reference point or origin (zero point). Once we choose our origin, we
define motion in one direction to be positive and motion in the other direction to be
negative. We usually give our reference system a name, such as the x-axis. Here is
an example.
positive direction
negative direction
x1
x2
-5
-4
-3
-2
-1
1
0
origin
2
3
4
x
5
Figure 2.1: Position along a straight line
The position of x1 above is 4 and that of x2 is 2. Note that the sign of the number
(plus or minus) indicates the direction with respect to the origin.
A change from one position, say, x1 to another, say, x2 is called the displacement x
It is the difference in their values and is given by
x = x2  x1
displacement
The symbol  (delta) is used often in Physics to represent a change of some quantity.
When we substitute the values for the positions x1 and x2 we obtain a negative answer
since the object moves towards the left, ie in the negative direction. Displacement is
called a vector quantity because it not only has a magnitude (in the case of x this is
|x| = 6) but also a direction (negative). We will look more carefully at vectors when
we consider motion in 2 and 3 dimensions in Chapter 3.
© School of Physics, University of Sydney
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Average velocity and average speed
Now that we have defined our 1-dimensional axis and position let us consider an
object in motion i.e. one whose position is changing with time. In this case it is
helpful to plot a 2-dimensional graph to represent the object’s motion. We usually
plot the position (x) on the vertical axis and the time (t) on the horizontal axis.
Figure 2.2: Graph of the motion of an object.
Let us consider the motion of an object moving in a straight line. The position of the
object is shown in Fig. 2.2 as a function of time. Graphs such as these are very useful
because they contain so much information that can be quickly absorbed. The object
begins at position -3 m, travels in the positive direction up to position 4 m (at time 7
s) and then turns back to position 1 m (at time 10 s).
We can also obtain other information from the graph, such as how fast the object is
moving. The average velocity, v , is the ratio of the displacement x to the time
interval t over which it occurs. We write this as
average velocity
where (t1, x1) and (t2, x2) are the two points on the graph representing the beginning
and ending of the segment of motion. This is also the slope of the straight line joining
these two points on the graph. The steeper the slope the faster the object is moving.
We can also define another average quantity called the average speed. This is given
by
average speed
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
By total distance we mean the actual path length travelled. The displacement on the
other hand only depends on the initial and final positions not on what happens in
between.
As an example let us calculate v and s of the above object for the whole journey.
(t1, x1) = (0 s, -3 m)
v=
and
(t2, x2) = (10 s, 1 m)
x 2 - x1 1- (-3)m
=
= 0.4m.s-1
t 2 - t1
10 - 0s
and
As we can see the two are completely different and in general they are unrelated.
Both, however, give us very valuable information concerning the motion of the
object.
Instantaneous velocity and speed
So far we can calculate the average velocity of an object but we may also like to know
the velocity at a particular time or instant. This is called the instantaneous velocity,
v, of an object. The velocity at any instant can be found by calculating the average
velocity and letting the time interval t approach 0. As t approaches 0, v
approaches v.
instantaneous velocity
Mathematically, this is differentiation. Velocity is the time derivative of displacement.
Note that the instantaneous velocity has a direction as well as a magnitude so it is a
vector. In terms of the graph of motion, the instantaneous velocity is the slope of the
tangent at the instant (time) of interest. It represents the rate at which the object’s
position is changing with time. Sometimes you will come across the term velocity this usually refers to the instantaneous velocity.
Speed is the magnitude of the velocity i.e. the numerical value with no regard to the
direction. This is different to the case of average velocity and average speed.
Acceleration
When the velocity of an object is changing the object is said to undergo acceleration
(or to accelerate). The average acceleration of an object over a time interval t is
given by
average acceleration
© School of Physics, University of Sydney
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Physics Bridging Course 2013
The instantaneous acceleration or simply the acceleration is given by
instantaneous acceleration
Mathematically, acceleration is the derivative of velocity, which is in turn the
derivative of displacement. Acceleration is the rate at which the velocity of an object
is changing with time. Like velocity it too is a vector quantity and so has both a
magnitude and direction.
Graphs of motion
So far we have seen graphs of position-time but we can now also draw graphs of
velocity-time and of acceleration-time. Let us begin with a graph of position-time for
a train (or any object) starting from rest, accelerating, travelling at constant velocity
and then coming to rest again. This is shown below.
Figure 2.3: Position-time graph for a train.
Using our definition for the velocity we can find the slope of the tangent at every
point on the position-time graph and obtain the velocity-time graph, shown below. It
is not always easy to use the graphical information to obtain accurate results (because
of the curvy segments in the plot) but it is possible in principle.
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
Figure 2.4: Velocity-time graph for a train.
Now using our definition for acceleration we can also find the slope of the tangent at
every point on the velocity-time graph and obtain the acceleration-time graph, as
shown below. Note that a negative acceleration does not necessarily mean the object
is moving in the negative direction (as is true for velocity). A negative acceleration
means the velocity is decreasing; the train is slowing down, but still going forwards.
© School of Physics, University of Sydney
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Physics Bridging Course 2013
Figure 2.5: Acceleration-time graph for train.
Areas under graphs
If we had a velocity-time graph for an object what information apart from the
acceleration could we deduce? It turns out that we can also calculate the
displacement between any 2 points in time. This is given by the area under the
velocity-time graph. In the language of calculus the area corresponds to the integral
(the opposite of a derivative) so we can write
Similarly for an acceleration-time graph the area under the curve between any 2
points is equal to the change in velocity,
Constant acceleration
Motion with constant acceleration is not easily produced, although it is often a good
approximation in many everyday situations. There is a special set of equations that
describe motion under constant acceleration. If we let the initial position and velocity
be x0 and v0 respectively, then we have the following equations of motion
v = v0 + at
x  x0 = v0t + 1 at2
2
equations of motion for constant acceleration
v = v + 2a(x  x0)
where x and v are the position and velocity at time t later.
2
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2
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
Like all algebraic equations we can rearrange the above equations to solve for any
unknown variable (physical quantity).
If you want to solve Physics problems, you need a good acronym.
How about DDESC:
Diagram — draw one
Data — write it all down, and identify what you need to find out
Equation — what’s the best equation(s) to use?
Solve —make the unknown the subject of the equation
Calculate — now sub in the numbers and calculate the answer
Now you should check your answer, try this acronym, SUS.
Significant figures – don’t use more than your data has
Units – make sure you write them in
Sense – check that the answer makes sense, isn’t faster than the speed of
light, or smaller than an atom, for example
Q.
A car accelerates from rest with uniform acceleration of 2.0 m.s-2. It
stops accelerating once the velocity has reached 10 m.s-1. How far has
it travelled during this time?
A.
Diagram:
x0 = 0; + dir
a = 2.0 m.s-2,
v0 = 0 m.s-1
v = 10 m.s-1
x0 = 0 m (defined as zero to make calculation easy)
Unknown:
x
Data:
Appropriate equation is
v 2 = v 02 + 2a(x - x 0 )
Solve to make x the subject
v 2 - v 02
x=
+ x0
2a
Calculate
(10 m.s-1) 2 - (0 m.s-1 ) 2
x=
+0
2 x 2 m.s-2
= 25 m
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Physics Bridging Course 2013
Free-fall acceleration
Galileo was the first to show that under the ideal conditions of no air resistance all
objects fall with the same constant acceleration. So in these cases we can treat objects
thrown vertically into the air as motion in a straight line under constant acceleration.
This acceleration is called the acceleration due to gravity and is taken as 9.8 m.s-2
and represented by g. The actual value varies slightly from place to place on the
Earth’s surface and changes with height above the surface.
We can use our equations of motion for constant acceleration when solving problems
with objects falling due to gravity. We normally take the vertically upward direction
as positive (even if the object is falling down) and this means the acceleration has a
negative value. But if you choose positive as down (so g is positive) everything will
still work out correctly, so long as you are consistent.
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
Q.
A boy throws a tennis ball vertically upwards with an initial speed of 12 m.s-1.
How long does it take to reach the maximum height and how high does it rise?
A.
Diagram:
Ball O
^
+ dir : up. Displacement (y direction) = 0
Data:
v0 = 12 m.s-1
a = g = -9.8 m.s-2
Unknowns: t, y (call it y instead of s) for vertical
To find 2 unknowns we will need 2 equations
PART 1
For constant acceleration:
v = v 0 + at
Rearranging for t:
v - v0
t=
a
At the highest point the ball will stop, i.e the velocity will be 0. Substituting v = 0 we
obtain:
0 - 12 m.s-1
t=
-9.8 m.s-2
t = 1.2 s
PART 2
To find the height (displacement) reached use our first answer and a different
equation:
v 0 = v - 2as (where s = y). Rearranging
v 2 - v 02
y=
2a
It is good practice to use y for variables in the vertical direction since we will be using
this for motion in 2 and 3-dimensions later. Substituting we obtain
(0) 2 - (12 m.s-1) 2
y=
2(-9.8 m.s-2 )
y = 7.3 m
© School of Physics, University of Sydney
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Physics Bridging Course 2013
Questions and Problems
1. Define the terms distance and displacement.
2. Define the terms average velocity and average speed and explain how they are
related.
3. Can the speed of a particle ever be negative? If so, give an example; if not
explain why?
4. For each of these questions give an example if your answer is yes; explain why if
your answer is no.
a. Can an object have zero velocity and still be accelerating?
b. Can an object have a constant velocity and still have a varying speed? Discuss
your answer.
5. Can the velocity of an object reverse direction when its acceleration is constant?
If so, give an example; if not, explain why?
6. In the 2008 Olympics Usain Bolt won the 100 m sprint in 9.69 s and Samuel
Wanjiru won the marathon of about 42 km in 2 h 6 min 32 s.
a. What is each of their average speeds?
b. If Bolt had maintained his sprint speed during a marathon, how long would he
have taken to finish?
7. A car travels on a straight road for 40 km at 30 km.h-1. It then continues in the
-1
same direction for another 40 km at 60 km.h .
a. What is the average speed of the car during this 80 km trip?
b. Graph displacement x versus time t and indicate how the average speed is
found on the graph.
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
8. For the position-time graph shown below find the following (in each case explain
how you arrived at the result):
a.
b.
c.
d.
the total distance travelled for the journey
the displacement for the journey
the average speed and average velocity for the journey.
indicate on the graph regions where the velocity is positive and where it is
negative.
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Physics Bridging Course 2013
9. The velocity-time graph for a model train is shown below. Find the following
(show all working):
a.
b.
c.
d.
the maximum positive velocity
the maximum negative velocity
the maximum speed
displacement for the time interval between t = 2 s and t = 4.5 s. What is the
displacement for the entire journey
e. indicate on the various regions of the graph whether the acceleration is
positive, negative or zero
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© School of Physics, University of Sydney
Chapter 2: Motion in a Straight Line
10. A tomato falls from a table top and lands on the floor below. The velocity-time
graph for the motion is shown below.
a. Draw position-time and acceleration-time graphs for the motion.
b. How far does the tomato fall?
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11.Sketch position-time, velocity-time and acceleration-time graphs for an object
moving with:
a. constant velocity
b. constant acceleration
12. A car moving at 20 m.s-1 takes 6 s to come to a complete stop in a straight line.
Find
a. the acceleration of the car
b. the distance travelled during the braking
13. A jumbo jet must reach a speed of 360 km.h-1 to take off. What is the least
acceleration required if the runway is 1.8 km long?
14. If the maximum acceleration that people can comfortably tolerate in a train is 1.34
m.s-2, what is the maximum speed a train can attain between two stations that are 806
m apart? (assume that the train must stop at the second station)
15. Consider the first experiment you performed. What equation was needed to
calculate the ball’s falling time?
16. Two trains, one travelling at 72 km.h-1 and the other at 144 km.h-1 are headed
toward each other along a straight level track. The drivers apply the brakes when the
trains are 950 m apart. If the acceleration during braking is 1.0 m.s-2 determine if the
trains collide or not.
If experimental equipment is free, go to the back of the book and carry out
Experiment 2.
17. At a construction site a wrench hits the ground with a speed of 24 m.s-1.
a. From what height did it fall?
b. How long did it take to reach the ground?
18. A tennis ball is hit vertically upwards with a speed of 65 m.s-1. If it was initially
1.2 m off the ground
a. how high does it travel?
b. How long does it take to hit the ground?
3
2
19. The position of an object moving in a straight line is given by x = t - 4t + 3t
where x is in metres and t in seconds.
c. What is the position of the object at t = 1, 2, 3 and 4 s?
d. What is the object’s displacement between t = 0 and t = 4 s?
e. What is the velocity of the object at t = 2 s?
20. A balloon moving vertically downwards at a constant speed of 49 ms-1 drops a
sandbag at an elevation of 980 m
a. How long until the sandbag hits the ground?
b. What will be the speed of the sandbag on impact?
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Chapter 2: Motion in a Straight Line
21. A car accelerates from rest at 5 m.s-2 for 4s, then accelerates at 2 m.s-2 for a
further 5 s. The car then maintains constant velocity for 3 s and then brakes uniformly
stopping in a further 10 s. The motion is on a straight line on a level surface.
a. Draw the velocity time graph for the motion of the object over the 22s
interval.
b. Determine the acceleration during the section when the brakes were applied.
c. Find the displacement at times 4 s, 9 s, 12 s and 22 s respectively.
Student Learning
What was the most difficult area in this topic?
© School of Physics, University of Sydney
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CHAPTER 3
VECTORS & MOTION IN 2 & 3 DIMENSIONS
Vectors and scalars
So far we have considered objects moving in a straight line. Such objects can move
in only two directions – forwards or backwards. We specify these using either a plus
or minus sign. For us everyday, moving in 3 dimensions, however, a plus or minus
sign is not enough. The object could be moving in any direction and as such we need
an arrow to indicate its direction of travel. This arrow is called a vector.
A vector has both a magnitude (the size of the vector which is always positive) and
a direction. A vector quantity is any quantity that has both a magnitude and
direction associated with it. We have already come across several vector quantities:
displacement, velocity and acceleration. Others include: force, electric field and
magnetic field.
Not all physical quantities have a direction, for example: time, mass, temperature and
energy have no direction. Such quantities are called scalars. They only have
magnitude.
The simplest and most familiar vector is displacement. When an object moves from
a position A to a position B we say it undergoes a displacement from A to B. We
represent vectors graphically using an arrow with a triangle (the head) at the end
pointing in the direction of the vector. All the vectors below are identical since they
represent the same change of position and have the same magnitude and direction.
B’
B
A’
B”
P
A”
A
O
Figure 3.1: Displacement vectors
A displacement vector represents the difference in the initial and final positions and
it doesn’t matter how the object got there. In the above diagram, the curved path O
 P has the same displacement as the straight line.
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Chapter 3: Vectors and Motion in 2 & 3 Dimensions
Vectors and their components
Working with vectors may seem awkward because they can point in various
directions but it is greatly simplified if we resolve them into their components.
Once we have done this we deal with each component as though it is 1-dimensional.
As an example let us consider motion in 2 dimensions. This includes any motion that
is restricted to a plane. We can split the position vector of any object into two parts:
its horizontal component and vertical component. These are usually referred to as
the x and y components. This is shown below.
y
A
a
ay

ax
x
Figure 3.2: A vector and its components.
The x and y axes form a rectangular system of co-ordinates (they are at right
angles) and as such we can form right angle triangles and resolve the components
using simple trigonometric rules. We have
ax = a cos 
and
ay = a sin 
where  is the angle the vector a makes with the direction of increasing x and a is
the magnitude of a. Note that we use bold to indicate a vector quantity. In
handwriting it is usual to place an arrow over the symbol to indicate it is a vector e.g.
a. The magnitude of a and the angle  are given by
ay
2
2
and
tan  =
a = a = ax + ay
ax
The above can be extended to 3 dimensions by introducing a third component az and
a new set of equations relating the components, angles and magnitude of the vector.
Adding and subtracting vectors
Imagine an object that first moves from the origin to position A and then to position
B. We want to find the displacement for the whole journey, i.e. the vector sum a + b.
Graphically this is found by placing the tail of one vector on the head of the other.
The sum is simply the new vector formed by joining the unconnected tail and
unconnected head, shown as the vector s. Trying to obtain numerical values from
graphs is messy but once we resolve the vectors a and b into their components we
simply add the horizontal components together and then the vertical ones. We can
extend this for more than 2 vectors by placing consecutive vectors tail to head.
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y
b
B
A
s
a
x
Figure 3.3: The addition of vectors. s = a + b . The vectors
corresponding to positions A and B are bold, lower case;
In order to subtract a vector we simply add the negative of that vector. The negative
of a vector points in the opposite direction to that of the vector. For the above
vectors the resultant of a - b is given by
y
B
A
a – b = a + (-b)
-b
a
a-b
x
Figure 3.4: The subtraction of vectors.
Unit vectors
When writing vector quantities it is useful to write the components in terms of unit
vectors. A unit vector is a vector that has a magnitude of exactly 1 and points in a
particular direction. It has no units (m.s-1, m or anything like that) just a direction.
We define unit vectors in the directions along the positive x, y and z axes and label
them as xˆ , yˆ and zˆ . The vector a above would then be written as
a = a cosq xˆ + a sinq yˆ
Projectile motion
Let’s look at motion in a plane i.e. 2-dimensions. This includes throwing a ball at
some angle or dropping objects whilst moving along. Again we will neglect air
resistance to simplify the mathematics. Any object travelling through the air under
free-fall is called a projectile.
We will use the equations of motion for constant acceleration as we did for free-fall.
We take components in the horizontal and vertical directions and solve for each
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Chapter 3: Vectors and Motion in 2 & 3 Dimensions
direction independent of the other. Below is a diagram showing the components of
velocity along the path of a projectile.
y
vy=0
v
vx
vy
v
vx
vx
vy
v0
v0y
v
0
v0x
vx
x
vy
v
vx
vy
v
Figure 3.5: The path of a projectile.
Let us consider the motion of a projectile for the general case. We launch the
projectile from a position (x0, y0) with an initial velocity v. We can write v in terms
of the components as
v = v0 cos 0 xˆ + v0 sin 0 yˆ
where 0 is the angle v0 makes with the positive x axis.
In the vertical direction the motion experiences acceleration due to gravity, however
in the horizontal direction there is no acceleration so our equations of motion reduce
to:
vx = v0 cos 0
x  x0 = v0 cos 0 t
Note that in the equations the velocity is the horizontal component of the initial
velocity. These equations tell us that the horizontal component of the velocity does
not change throughout the motion. This can be seen in the figure above.
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Q.
A rescue plane is flying at constant elevation of 1200 m with a speed of
430 km.h-1 toward a point directly above a person struggling in the
water. At what distance should the pilot release a rescue capsule if it is
to strike close to the person in the water?
A.
The initial velocity of the capsule is the same as the plane. It therefore
has a horizontal component of 430 km.h-1 and a vertical component of 0
m.s-1. We can place our origin at the position of the plane when the
capsule is dropped. Using our equations of motion we can solve for the
time to fall
y  y0 = v0 sin 0 t  1 gt2
2
y - y0 = 1200 m since the capsule moves down and 0 = 0 since the capsule
(in the plane) is initially moving horizontally. Substituting and solving for t
yields
1200 m = 0  1 (9.8 m.s-2)t2
2
(2)(1200 m)
t=
= 15.65 s
9.8 m.s-2
The horizontal distance covered by the capsule (and plane) during this time is
given by
x  x0 = v0 cos 0 t
æ1000 m öæ 1 h ö
= 430 km.h-1 ç
֍
÷cos  (15.65 s)
è 1 km øè 3600 s ø
= 1870 m
The pilot must therefore release the capsule 1870 m before reaching the
person.
So far we have neglected the effects of air resistance and for some projectiles there
are even more complicated aerodynamic effects that influence the motion of the
projectile – for example top spin on a tennis ball or slice on a golf ball. Such
examples can still be analysed using Physics but require some new concepts and
ideas.
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Chapter 3: Vectors and Motion in 2 & 3 Dimensions
Questions and Problems
1. Explain the difference between a vector and a scalar.
2. If two vectors add to give zero what can you say about their magnitudes and their
directions?
3. Can the magnitude of the sum of two vectors ever be greater than the sum of the
magnitudes of these vectors?
4. Compare the distance travelled and the displacement for the following paths, the
arrows indicate direction of travel.
C
A
B
5. For the points indicated below give the position vector and the distance from the
origin.
6. A small town is situated 156 km West and 65 km North of Sydney. How far is it
from Sydney?
7. In a car rally the contestants begin in Melbourne and complete the following
journeys:
56 km North, 30 km West
then 105 km North, 78 km East
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a. Sketch a plot showing the journeys.
b. How far are they from Melbourne?
c. If Sydney is 710 km North and 250 km East of Melbourne, how far are the
contestants from Sydney?
8. A boy sitting in a train moving at constant velocity throws a ball straight up into
the air. Will the ball land in front of him, behind him or in his hands? Explain.
9. You are driving behind a truck, going at the same speed as the truck. A crate
falls from the truck onto the road. Will your car hit the crate before the crate
reaches the ground (you neither swerve nor brake)? Explain.
10. A dart is thrown horizontally towards a bull’s-eye at an initial speed of 10 m.s-1.
If the dartboard is 3 m away where will the dart land relative to the bull’s-eye?
11. A ball rolls off the edge of a table that is 0.9 m high. It strikes the floor 1.5m
horizontally away from the edge.
a. How long was the ball in the air?
b. What was the initial speed of the ball?
12. A boy kicks a soccer ball with an initial angle of 30° to the horizontal and an
initial speed of 25 m.s-1.
a. What height does the ball reach?
b. How far does it travel in the horizontal direction before it lands?
13. A cannon has a muzzle velocity of 85 m.s-1. If it is aimed at 40° to the horizontal
what is the range of the cannon?
14. A volcano that is 3300 m above sea level erupts and sends rock fragments
hurtling into the sea 9.4 km away. If the fragments were ejected at an angle of
35° what was their initial speed?
15. A boat is travelling upstream at 14 km.h-1 with respect to the water of a river.
The water itself is flowing at 9 km.h-1 with respect to the ground.
a. What is the velocity of boat with respect to the ground?
b. If a child on the boat walks from the front to the rear at 6 km.h-1 with respect
to the boat, what is the child’s velocity with respect to the ground?
16. Snow is falling vertically at a constant speed of 8.0 m.s-1. A man is driving at a
constant speed of 20 m.s-1. At what angle do the snowflakes appear to be falling?
17. We saw that you can resolve a vector in a 2-dimensional system of co-ordinates
as:
a = a cosq xˆ + a sinq yˆ
How would you write a vector in a 3-dimensional system of co-ordinates?
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Chapter 3: Vectors and Motion in 2 & 3 Dimensions
Student Learning
How has your grasp of these concepts changed since the previous tutorial?
© School of Physics, University of Sydney
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Physics Bridging Course 2013
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© School of Physics, University of Sydney
CHAPTER 4
FORCES AND MOTION
Mechanics
We have seen how to describe the motion of objects. Now we will see what makes
them move the way they do. As part of our daily experiences we all know that to
make an object move in a particular way you must apply the appropriate force i.e. a
push or pull in the direction you wish the object to go. The harder you push or pull
the more it moves. These simple observations can be explained by Newton’s laws of
motion. This is the field of Physics known as mechanics. They allow us to predict
the motions of the planets, the path of Shane Warne’s googly and many more
everyday events.
Newton’s first law of motion
Before Newton formulated his laws of motion most people held the view of the
ancient Greek philosopher Aristotle that an object remains at rest unless a force
makes it move. In other words the ‘natural’ state of an object was that of rest, not
motion. This seems reasonable when we consider some everyday examples. If the
engine of a car doesn’t provide a force the car soon comes to a stop. If you push a
book across the table it too will come to a stop. So it seems if you want to keep an
object moving you need to keep applying a force to it.
But let us perform a few more careful experiments. What if we push the book over a
highly polished table? We find that it travels further.
Now try it over a skating rink or an air track. It travels even further. We now
imagine that by using even smoother surfaces the book will travel further and further.
In the limit if we have an perfectly smooth (frictionless) surface the book will
continue moving without slowing down.
We now conclude that you do not need a force to keep an object moving. We see
that it is the presence of other forces (friction, air resistance etc.) that slows them
down. This leads us to a formulation of Newton’s first law of motion
NEWTON’S FIRST LAW: Consider an object on which no
net force acts. If the body is at rest, it will remain at rest. If
the body is moving with constant velocity it will continue to
do so.
Note that forces may still act on the object but the net (resultant) force must be zero.
Newton’s first law is sometimes called the law of inertia and it holds only in a
special set of reference frames. A reference frame is quite simply a set of coordinate axes we use to make our measurements and observations. For example we
can measure position relative to the Earth’s surface. We say the Earth is our
reference frame.
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If we are moving along in a train and making measurements we can use the train as
our reference frame. If the train is moving at a constant velocity we will find that
Newton’s first law of motion is true. It is an inertial reference frame.
If we tried to do the same from a merry-go-round that was turning we would
probably find that Newton’s first law was not observed to be true. This is because
objects on the merry-go-round are moving in a circle, so will have a force acting on
them (provided by friction with the floor). Is the Earth’s surface an inertial reference
frame?
Newton’s second law of motion
Newton’s second law deals with the effect of forces on an object. It’s fairly intuitive
that the harder you push, the faster it goes. Newton’s law says that the acceleration
(the “go”) is proportional to the sum of the forces (the push), and the proportionality
depends on the mass. Mathematically this is written
NEWTON’S SECOND LAW:
åF
= ma
In other words the sum of all the forces acting on the object is equal to the mass of
that object times the acceleration. Note that this is a vector equation and the direction
of the acceleration is the same as the net force. You can see this equation in
everyday life. If we slide a book across a table we can do it quite easily and achieve
a reasonable acceleration. Try pushing on a whole bookcase and the result is quite
different. The larger the mass of the object the smaller the acceleration, with the
same amount of force.
We can now use Newton’s second law to carefully define force. Let us take the
standard kilogram as our object. We place the object on a frictionless table and pull
it along until, by trial and error, it has an acceleration of 1 m.s-2. This force is equal
to 1 kg.m.s-2 and is given a special name called the Newton (abbreviated N). It is the
SI unit of force.
Force is a vector quantity and so obeys the laws of addition and subtraction for
vectors. If 2 or more forces act on a body we can find the net force by placing the
tail of one force vector on the head of the next and so on. That forces behave as
vectors is a fact that was verified by experiment. This means that, as with the 2dimensional motion studied in Chapter 3, we can also take components of forces.
Using components we can write Newton’s second law as
åF
x
= max,
åF
y
= may,
åF
z
= maz
Newton’s second law also allows us to define the mass of an object. It is the ratio of
the force and the corresponding acceleration. This means if the force produces an
acceleration of 10 m.s-2 for the standard 1 kg mass, and an acceleration of 5 m.s-2 for
some other mass, e.g. a pink brick, then the mass of the pink brick is
F = (1 kg) (10 m.s-2) = m (5 m.s-2)
m = 2 kg
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Chapter 4: Forces and Motion
This is also referred to as the inertial mass of an object. It is a characteristic of the
object that relates the force on the object to the resulting acceleration. Less
precisely, we can think of it as a measure of the amount of matter in the object. This
is different to the weight of an object, which is the force due to gravity on that object.
Newton’s second law shows that weight and mass are related but not equal.
The easy way to use Newton’s second law to solve problems is to apply it to only
one object at a time. We must find all the forces acting on that body and then we
draw them in a free-body diagram. In it, the object is represented by a dot (make
sure it is clearly visible) and each force by a vector with its tail on the dot. We only
consider external forces, i.e. forces exerted on the object by other objects. Internal
forces in which parts of the object exert forces on other parts of itself are not
important here (we will see later that they cancel out).
Q.
A student pushes a loaded sled of mass m = 240 kg for a distance d of 2.3 m
over the frictionless surface of a frozen lake. In doing so he exerts a
constant horizontal force F with magnitude F = 130 N. If the sled starts
from rest, what is its final velocity?
A.
Step 1 is to draw a free-body diagram. This is shown below. The sled is
represented by the dot and the force by an arrow. We also draw the relevant
co-ordinate axes. In this case it is only a 1-dimensional problem so only the
x-axis is drawn. We choose the force to be along the positive x-axis (we are
free to do this since we are not told otherwise). Using Newton’s second law
of motion we can find the acceleration of the sled.
m
F
x
Consider the x-component of F and the acceleration along the x-axis. We get from
Newton’s second law (written in component form)
Fx = max
F 130 N
ax = x =
m 240 kg
= 0.542 m.s-2
From our equations of motion for constant acceleration we have
v2 = v 02 + 2a(x - x0)
Since the sled starts from rest v0 = 0 and we identify that x - x0 = d. Our equation
simplifies to
v = 2ad
= (2)(0.542 m.s-2 )(2.3 m)
= 1.6 m.s-1
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Physics Bridging Course 2013
Q.
In a 2-dimensional tug-of-war, Alex, Betty and Charles pull on a car tyre as
shown in the diagram below. Alex pulls with a force FA (220 N) and Charles
with a force FC (170 N). With what force FB does Betty pull if the tyre remains
stationary?
y
Alex
Charles
FA
FC
47 o
137

x
o
Betty
FB
A.
In this example we need to use 2-dimensions to deal with the forces. First we
draw our free-body diagram as above. We are free to choose any x-y coordinate system with respect to the actual forces. It is usually a good idea to
align at least one force with one of the axes. Now we use Newton’s second
law. Since the tyre is stationary, the acceleration is zero, so we can write
åF = F
A
+ FB + FC = 0
You will find that the component form of this vector equation is more useful when
solving problems. In terms of the x and y components we have
åF
= FC cos FA cos 47.0˚ = 0
åF
= FC sin FA sin 47.0˚FB = 0
x
and
y
Note the use of minus signs to indicate a component is in the direction opposite to
the positive x or y axes. Substituting values into the first equation we get
(170 N)( cos  = (220 N) (cos 47.0˚)
é(220 N) (0.682) ù
cos-1
= 28.0˚
ë
170 N
û
Substituting into the other equation we get
FB = FC sin FA sin 47.0˚
= (170 N) (sin 28.0˚) + (220 N)(sin 47.0˚)
= 241 N
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Chapter 4: Forces and Motion
Some particular forces
There are many forces around us but there are some particular types that are very
important and very common. We will now take a closer look at these.
Weight
The weight of an object is the force on that object due to the local gravitational field.
If that object is near the Earth then the weight is due to the Earth’s gravitational field.
From Newton’s second law we can write the weight of an object near the Earth’s
surface as
W = mg = m (9.8 m.s-2)
Since weight is a force it has a direction. The direction is down towards the Earth’s
centre. On the moon the same object would weigh less because the magnitude of the
gravitational field on the moon is not as large as the Earth’s. The mass of the object,
however, will be the same. It is a property of the object. In our everyday speech we
tend to interchange the word weight and mass but in Physics they have completely
different meanings.
Normal force
Consider a book resting flat on a table. There is a weight force acting on the book.
This, however, cannot be the only force since the net force must be zero, the book is
stationary. There is a force exerted by the table on the book. This force is normal
(perpendicular) to the surface of the table and is called the normal force (usually
indicated by N). This is shown in the diagram below.
y
Normal force N
N
x
W
Weight W
We must remember that the normal force is always perpendicular to the surface and
that the normal force is not always equal to the weight force.
Friction
If we try to slide the book over the table (or any surface) we find that the motion is
resisted by a force. This force is called friction (usually indicated by f). It is parallel
to the surface and opposite in direction to the intended motion. Friction is due to
attraction (bonding) between the surface and the object. Sometimes we simplify a
problem by assuming there is no (or very little) friction. Such a surface is called
frictionless.
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Tension
When an object is being pulled along by a rope, the rope is under tension. There is a
force exerted by the rope on the object. This force is called tension (usually
indicated by T). The direction of the tension is along the rope, away from the object
and toward the centre of the rope. This is indicated in the figure below.
T
T
T
T
The rope is usually assumed to be massless. This means its mass is much less than
the mass of the object. Any pulleys are also assumed to be massless and frictionless.
This means its mass is also much smaller than the object and that the friction on the
pulley axle is very small.
Fundamental forces
All the forces we have discussed so far and all forces so far known to physicists are
manifestations of the four fundamental forces of nature. They are
 gravitational
 electromagnetic
 weak nuclear
 strong nuclear
Most everyday forces (such as friction, tension, normal force) are due to
electromagnetic forces.
Newton’s third law of motion
The most familiar of Newton’s laws of motion is probably his third and final law of
motion. We are all aware that when we push on an object it pushes back. For
instance to get out of a chair you may push down on that chair (arm rests or
somewhere) and it pushes back on you. Newton’s third law talks about the pair of
forces acting between 2 bodies.
NEWTON’S THIRD LAW: If a body A exerts a force FBA
on a body B then that body exerts and equal and opposite
force FAB on body A.
Since force is a vector and has direction opposite means in the opposite direction. So
the two forces have the same magnitude but are in the opposite direction.
Mathematically we can write Newton’s third law as
FAB = FBA
Here the notation FAB means the force exerted on body A by body B. Sometimes you
will hear people talk about action and reaction forces and they state Newton’s third
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Chapter 4: Forces and Motion
law as “the reaction force is equal and opposite to the action force”. There is nothing
wrong with this but it is not always clear which force is the action and which is the
reaction.
You may now think that if the two forces are equal and opposite they should cancel
out and the net force will always be zero (so then nothing will ever accelerate!). The
point to keep in mind here is that the 2 forces are acting on different objects. In this
case the forces do not cancel out.
Let’s take a closer look at a few examples of action-reaction pairs. Imagine a
satellite orbiting around the Earth. The only force acting on the satellite is FSE the
force exerted on the satellite by the gravitational attraction of the Earth. By
Newton’s third law the satellite should exert an equal and opposite force on the
Earth. This is the gravitational pull of the satellite on the Earth. Gravity is the
mutual attraction of 2 objects. The satellite’s pull on Earth has the same magnitude,
but because the mass of the Earth is so large Newton’s second law tells us that the
acceleration will be very small.
FSE
FES
Now consider a book resting on a table. We can draw the forces acting on the book.
There is the gravitational pull of the Earth on the book FBE (weight). Since the book
is not accelerating this force must be cancelled by the normal force of the table on
the book FBT. However, these 2 forces do not form an action-reaction pair. They are
equal and opposite but do not act on the same pair of objects. The ‘reaction’ to FBE
is the gravitational pull on the Earth due to the book FEB. The ‘reaction’ of the
normal force due to the table is FTB, the force of the book on the table.
FBT (normal from
FBE
table)
FBT
FEB
FBE (weight of
book)
FTB
Friction
We are all familiar with the effects of friction: from a car’s brakes to air flow over a
cyclist’s helmet, from carpet burns to a car doing burnouts. However, the effects are
not always to our disadvantage. Without friction we would not be able to walk, or
write and objects would slide around. Friction provides the force that makes all these
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Physics Bridging Course 2013
possible. There are different types of friction but in our case we will consider the
friction between two dry solid surfaces (usually an object and some surface).
Let us begin our investigation of friction with a few simple experiments. First we
slide a book over the surface of a table. We observe that the book gradually slows
down and eventually comes to rest. This is the first property of friction, it opposes
motion. If we want to keep the book moving at a constant velocity we have to push
it with a force. This force is equal and opposite to the frictional force.
In our second experiment we gently push on the book. Even though we exert a force
on the book it does not move. As we push even harder the book still does not move.
The frictional force seems to adjust so as to be equal but opposite to the force we
exert. What if we push really hard? The book does begin to move. At first it
accelerates but as we reduce our push it moves with a constant velocity. So there is a
maximum force that friction can exert but once the object moves then the frictional
force is less.
N
N
N
fs
F
W
W
Book resting on table Frictional force equal
but opposite to push F
a
fk
F
W
Frictional force equal to
maximum value
v
N
W
Book accelerates initially as
F is greater than fk
fs
F
N
F
fk
W
F is reduced until it is
equal to fk
By performing such experiments carefully physicists have been able to formulate 2
expressions to account for the above behaviour. In the case where the object is not
moving we call such friction the static frictional force and denote it by fs. The
maximum value (magnitude) of this force is given by
fs,max = sN
where s is the static coefficient of friction (it has no units, it is dimensionless) and
N is the normal force acting on the object. If the external force that is pushing is less
than this maximum frictional force then the frictional force is equal to the external
force i.e.
fs  sN
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Chapter 4: Forces and Motion
Once the object is moving then the frictional force decreases. It is called the kinetic
frictional force and is denoted by fk . The magnitude of this force is given by
fk = kN
where k is the coefficient of kinetic friction.
Note that neither of the expressions for the frictional force say anything about
direction. They are scalar expressions. The directions of fs and fk are always parallel
to the surface and tend to oppose any other forces parallel to the surface or oppose
any motion. These expressions are empirical (based on experimental data) and aren’t
fundamental equations like Newton’s second and third laws. Under some situations
our expressions for frictional forces may not be accurate, at high speed for instance.
They are approximations of what actually happens.
The coefficients of friction depend on the 2 surfaces in contact. We usually state
these as the coefficient of static friction between a rubber tyre and asphalt. The
coefficient of static friction between a rubber tyre and wood might be very different.
Tables of such useful values exist. Friction is basically due to the attractive forces
between the 2 surfaces.
Q
A 100 kg chest standing on a floor has a coefficient of static friction of 0.6. A
boy approaches the chest and leans horizontally against it.
(a) What is the frictional force exerted by the floor on the chest before the boy
comes into contact with it?
(b) What is the maximum force the boy can exert before the chest begins to
slip?
(c) What is the frictional force exerted by the floor when the exerts boy a force
of 100 N on the chest?
A
Drawing a free-body diagram we have
N
W
(a) Initially the only forces acting on the chest are the weight and normal force.
The chest is at rest so there is no friction initially (friction would result in an
acceleration and force).
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Physics Bridging Course 2013
(b) Now we add the push of the boy and the frictional force in the free body
diagram.
Taking components in the horizontal direction we get
åF
x
N
= fs  F = 0
fs
F
and in the vertical direction we have
åF
y
W
= N  W = 0
From the first equation we get
fs  F
so F is a maximum when fs is a maximum i.e. F = fs,max
From the last equation we have N = W.
Our equation for the static frictional force is
fs,max = sN
So we get
F = fs,max = sW
Substituting we get
F = 0.6 x (100 kg) (9.8 m.s-2)
= 588 N
(c) If the boy exerts a force of 100 N this is less than the maximum frictional
force (588 N) so the block remains at rest and the frictional force is equal and
opposite to the force the boy exerts.
fs = 100 N
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Chapter 4: Forces and Motion
Q
A block is at rest on an inclined plane as shown below. The coefficient of
static friction between the block and incline is s. What is the maximum possible
angle of inclination max of the surface for which the block will not slip?
N
fs


W
A
In this problem we choose our co-ordinate axes to be parallel and
perpendicular to the inclined surface. This will make the mathematics easier, the
Physics is always going to be the same.
Taking components parallel to the surface and applying Newton’s second law
fs  W sin  = 0
(taking up the incline as positive)
taking components perpendicular to the incline we get
N  W cos  = 0 (taking upwards as positive)
The block begins to slip when fs = fs,max . Our expression for fs,max is
fs,max = sN
Substituting for fs and N we get
fs,max = W sin s W cos 
rearranging we get
sin q
= tan 
cos q
= tan-1 s
s =
So max
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Questions and Problems
1. Describe Newton’s first law of motion in your own words. Give an original
example from everyday life where you can see it. Compare it with a classmate’s.
2. When you are standing in a bus if the bus starts to move forward you tend to jolt
backwards. When the bus comes to a stop you tend to jolt forward. Explain this
in terms of Newton’s first law.
3. What do we mean by reference frame? What is an inertial reference frame?
4. Describe Newton’s second law of motion in your own words. Give an original
example from everyday life where you can see it. Compare it with a classmate’s.
5. If two forces act on a body can the body move with a constant velocity? Can it
remain at rest? Give examples to support your answers.
6. What is the difference between the weight of an object and its mass?
7. A student pulls a sled of mass 60kg across a frictionless surface with a force of
100 N directed at 30° to the horizontal. Draw a free-body diagram showing all
the forces acting on the sled and calculate the horizontal acceleration.
F
30
8. Two masses are connected by a rope and pulled across a frictionless surface by a
constant force of 50 N. Draw a free-body diagram for each mass and calculate
the acceleration of the masses and the tension in the rope.
m1 = 5.0 kg
m2 = 2.0
kg
50 N
9. Calculate the tensions in each of the segments of rope in the following diagram.
The masses are at rest. Draw a free-body diagram for each mass.
T1
m1 = 4.0
kg
T2
m2 = 2.0
kg
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Chapter 4: Forces and Motion
10. A mass is suspended from a ceiling by two cords. If the mass is 5.0 kg find the
tension in each cord. Draw a free-body diagram indicating all the forces on the
mass.
60
T2
30
T1
11. A mass, 2.0 kg, is sliding down a frictionless inclined plane of angle 30°. What
is the acceleration of the mass? Draw a free-body diagram indicating all the
forces acting on the mass. (Hint choose the co-ordinate axes to be parallel and
perpendicular to the surface).
30
12. Consider the second experiment you performed, rolling a ball down a slope. How
does your measured acceleration compare with the component of gravity along
the rail?
13. Two masses on a frictionless surface are connected by a pulley as shown in the
diagram below. Draw free-body diagrams for the masses and find the tension in
the rope and the acceleration of the masses.
m2 = 4.0 kg
m1 = 3.0 kg
14. A man is standing in an lift. Draw a free-body indicating the forces acting on the
man and calculate the normal force acting for the following situations
a. lift is stationary
b. lift is accelerating upwards at 2.0 m.s-2
c. lift is accelerating downwards at 1.5 ms-2.
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15. Describe Newton’s third law of motion in your own words. Give an original
example from everyday life where you can see it. Compare it with a classmate’s.
16. Consider the two masses shown below, sitting on a frictionless surface. A force
of 10 N acts on mass m1 (3.0 kg). It applies a force on m2 (which has a mass of
2.0 kg). According to Newton’s third law, m2 must apply an equal and opposite
force on m1. Show that these forces are equal and opposite (calculate them).
What is the acceleration of the two masses? Why don’t these forces cancel to
produce no acceleration?
10 N
m1 = 3.0 kg
m2 = 2.0
kg
17. A boy walks along pulling a cart on a rope. Draw all the ‘action-reaction’ pairs.
(Hint you must consider friction if he is walking)
18. Explain how friction helps you walk.
19. If there was a person at rest on a frictionless surface of ice, say an ice-skating
rink, could they get off the ice by walking, rolling, kicking or any other way?
20. What is meant by the coefficient of static friction? The coefficient of kinetic
friction?
21. A crate heavier than you rests on the ground. If the coefficient of static friction
between you and the ground is the same as between the crate and the ground, can
you push the crate across the ground?
22. Find the x and y-components of the following force.
y
50 N
60
x
23. A table of mass 45kg is resting on the floor. If the coefficient of static friction
between the table and floor is 0.45 what is the minimum force required to just
start the table moving?
24. In order to just move a bookcase across the floor a student must exert a force of
150 N. If the mass of the bookcase and books is 110 kg what is the coefficient of
static friction between the floor and the bookcase?
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Chapter 4: Forces and Motion
25. A worker pushes horizontally on a 35 kg crate with a force of 110 N. The
coefficient of static friction between the crate and the floor is 0.37.
a. What is the maximum frictional force, fs,max that may be exerted by the floor
on the crate?
b. What is the frictional force exerted by the floor on the crate?
c. Does the crate move? If not, with what minimum force must the worker push
to make it move?
26. The coefficient of static friction between the tyres on a car and the road surface is
0.6. If the mass of the car is 1200 kg, what is the maximum frictional force on
the car? What is the maximum acceleration of the car?
27. The coefficient of static friction between teflon and scrambled eggs is about 0.04.
What is the maximum angle that you can tilt a teflon coated pan without the eggs
sliding?
28. A block is sliding down an incline. If the coefficient of kinetic friction between
the surface and block is 0.20, what is the acceleration of the block? (Hint: draw a
free-body diagram).
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Physics Bridging Course 2013
Student Learning
Summarise the three most important points from this topic.
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© School of Physics, University of Sydney
CHAPTER 5
WORK AND ENERGY
So far we have been looking at problems and analysing them using Newton’s laws of
motion. We now consider another method of solving problems, that we will find
useful when dealing with objects moving at speeds close to that of light or for
particles that are very small (like the electrons and protons that make up atoms). In
such cases Newton’s laws of motion need to be modified before they are valid. Our
new methods are always valid. They involve terms that you are already familiar with
but as always physicists give them very precise meanings.
Work: motion in 1-dimension with a constant force
Let us assume that a student pushes a toaster along a frictionless surface. The force
she applies F is parallel to the surface as is the resulting straight-line motion. She
pushes the toaster through a displacement d. We define a quantity work, W, as given
by
W = Fd
where F is the magnitude of F and d the magnitude of the displacement d. We say
that W is the work done by F in moving the toaster through a displacement d. The
work is done by a force. Note that work is a scalar quantity although force and
displacement are vectors.
F
F
d
d
What if the force was not parallel to the motion, as shown in the figure below? In this
case we define the work to be

W = Fd cos 
where  is the angle between the force and the displacement vectors (when placed tail
to tail). We can think of it as the component of the displacement in the direction of
the force or the other way around (they are equivalent). Notice that for our initial
situation  = 0° so our more general equation reduces to the special case.
F
F

d
d
Sometimes physicists like to write the above equation in a more compact form using
the scalar (or dot) product. It is another way of saying “include the cosine of the
angle between the 2 vectors”
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Physics Bridging Course 2013
W = F.d
Our more general equation for work has some more special cases. If the force and
displacement vectors are in opposite directions then  = 180˚ so that cos  = -1. Since
F and d are always positive this means the work done in such a situation will be
negative. If the force and displacement vectors are perpendicular to each other then 
= 90˚ so that cos  = 0. In this case there is no work done.
To illustrate these possibilities consider the following situation. You pick up a case of
beer and put it on a shopping trolley, push it with a constant velocity across the car
park and place it on the floor of your van. In the first part you apply a force F to lift
the beer. The force is upwards and so is the displacement. The force (you) does
positive work on the beer. While you are walking across the car park with constant
velocity the trolley still applies an upward force F on the beer (otherwise it would
fall) but the displacement is perpendicular to this. The force (you) does no work in
this case. Finally as you place the beer down you still exert an upward force (as you
place the beer down slowly) but the displacement is downwards so the force does
negative work.
Note that throughout the lifting and carrying you must exert a force but that does not
mean you are always doing work, at least not as far as Physics is concerned. If you
carried the beer instead of using the trolley you would have done no more work. Your
muscles would have strained more to support the weight of the beer and consumed
energy but not necessarily done more work (the body is not a very efficient machine!)
The SI unit for work is the Newton-metre since they are the units for force and
displacement. However, work is such a useful quantity that the unit has been given a
special name, the joule (J). It is the work done by a force of 1 N moving through a
displacement of 1 m.
As we have previously seen there can be several forces acting on an object. How do
we calculate the work done in such a case? There are 2 ways. We can first calculate
the net force acting on the object and then use our general equation for work. Or we
can calculate the work done by each of the forces and simply add them up. Either
way we obtain the same answer.
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Chapter 5: Work and Energy
Q.
A weight lifter raises a mass of 250 kg a distance of 2 m. How much work has
he done?
A.
Draw a free-body diagram
To find the force exerted by the weight lifter we
assume the weight moves at a constant velocity so
the net force must be zero.
d
F
F - mg = 0
OR
F = mg
The work done is
W = Fd cos
= (250 kg) (9.8 m.s-2) (2 m) cos 0˚
= 4900 J.
weights
mg
Work done by a variable force.
We now look at the work done by a variable force (one that is not constant). We will
assume that the direction of the force is the same as the displacement (i.e.  = 0) and
then generalise. The force F varies in some arbitrary way that depends on x, F(x), as
indicated in the figure below. We cannot calculate the work done using our previous
equation because the force is not constant. Instead we divide the motion up into
smaller sections, x1, x2, ... xn, each of displacement x. During each of these sections
the force does not change much and we approximate it as constant F(x). The work
done during each small section is W where
F(x)
F(x)
xi
x
xf
xi
x
xf
x
The total work done is approximately the sum of the work done in each of the smaller
sections.
W =
=
å
F(x) x
We can make this approximation better by making the sections smaller and taking
more of them. In the limit we allow the displacement of the sections to approach
zero, x0, and we obtain the work done by the variable force F(x).
W =
å
F(x)
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Physics Bridging Course 2013
This limit is what we define as the integral under the curve F(x). So the work done by
a variable force is given by
W =
ò
xf
xi
F(x) dx
where xi and xf are the initial and final positions of the object.
Work done by a spring
As a practical example of a situation where the force varies according to the position
we will look at the force exerted by a spring on a toaster. A spring that is not being
stretched or extended is said to be in its relaxed state. Let us fix one end of the spring
(attach it to a wall or hook) and place a toaster on the other (free) end. To measure
the displacement of the toaster we define our origin to be the position when the spring
is in the relaxed state. Furthermore, we consider only 1-dimension, x, along the axis
of the spring.
When we stretch the spring there is a force in the opposite direction to the
displacement trying to restore the spring to its relaxed state. When we compress the
spring the force is again opposite to the displacement trying to restore the spring to
the relaxed state. This force due to the spring is called the restoring force. We
usually define the x-axis such that when the spring is extended x is positive and when
the spring is compressed x is negative.
x=0
F=0
x negative
F positive
x positive
F negative
F
F
x
0
x
x
0
0
It turns out that the force F is proportional to the displacement x from the relaxed
state to a good approximation for most springs. We write the force as:
F = -kx
and this is called Hooke’s law. The constant k is termed the spring constant and is a
property of each individual spring. It is a measure of how stiff (difficult to stretch)
the spring is. The SI units for the spring constant are Newton per metre (N.m-1). The
minus sign in Hooke’s law indicates that the force is in the opposite direction to the
displacement. Although the above is a vector equation you will often see it written in
terms of the components as
F = -kx
Hooke’s law gives us the explicit dependence of the force on the displacement so that
it is possible to calculate the work done by the spring using our integral method. Let
us assume that we move the above toaster (attached to a spring) from an initial
position xi to a final position xf. The work done by the spring is
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Chapter 5: Work and Energy
W =
ò
xf
xi
F(x) dx =
[ ]
= ( 1 k) x 2
2
xf
xi
ò
xf
xi
-kx dx = k
ò
xf
xi
x dx
= ( 1 k)( x 2f - x i2)
2
Note that since F(x) is the force exerted by the spring it is the work done by the
spring. The work done by us is the negative of this (we exert an opposite force).
If the initial position of the toaster (on the spring) is at x i = 0 then the work done in
moving it to a final position of xf is
W = - 1 kx 2f
2
Q.
A toaster is attached to a spring with spring constant 500 N.m-1. If the spring is
stretched by 10 cm from the relaxed state how much work does the spring do? If
it is stretched another 10 cm how much work is done?
A.
Work done by the spring is given by
W = ( 1 k)( x 2f - x i2)
2
where xi = 0 m and xf = 10 cm = 0.10 m. Substituting we obtain
W = ( 1 (500 N.m-1)) ( (0.10 m)2  (0 m)2
2
= 2.5 J
To move a further 10 cm we have xi = 10 cm = 0.10 m and xf = 20 cm = 0.20 m
W = ( 1 (500 N.m-1)) ( (0.20 m)2  (0.10 m)2 )
2
= 7.5 J
(The spring does negative work in each case but the force stretching the spring
does positive work.)
© School of Physics, University of Sydney
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Kinetic energy
A girl picks up a ball and throws it. We neglect the initial lifting and focus on the
actual throw. During the throwing motion the girl exerts a force on the ball. This
force is in the same direction as the motion, so it does positive work. When the ball
eventually leaves the girl’s hand it is moving with a speed v. What is it about the state
of the ball, after the throw, that tells you work has been done on it?
There is a property of any object (including the ball) called the kinetic energy. The
kinetic energy is defined to be
K = 1 mv2
2
where m is the mass of the object and v the speed. The SI units for kinetic energy are
the same as for work, the joule (J). The kinetic energy of an object is always positive
(or zero) it can never be negative.
The kinetic energy of an object is related to the total work done on that object (or the
work done by the net force):
W = K f - KI
where Ki and Kf are the initial and final kinetic energies of the object. This is called
the work-kinetic energy theorem and says that the total work done on an object is
equal to the change in kinetic energy of that object. So when the girl does work on
the ball she also increases the kinetic energy of the ball.
Q.
A pot plant falls from a window sill and drops 10 m to the ground
below. How fast was it falling just before hitting the ground?
A.
We can solve this problem either by using our equations for free-fall
or the work-kinetic energy theorem. Let’s use the latter.
The only force acting during free-fall is due to gravity
F = mg
Since this is constant and in the same direction as motion
W = mgx
The initial speed is 0 so Ki = 0. The work-kinetic energy theorem
gives
mgx = 1 m(v 2f -v i2 ) )
2
v f = 2gx
=
2 (9.8 m.s-2 ) (10 m) = 14 m.s-2
Note using equations of motion would give an identical answer, and
the working would look similar too – compare with Chapter 1 Question 19
b.
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Chapter 5: Work and Energy
Conservation of Energy
So far we have looked at only one form of energy but there are many more. Energy is
a term familiar to all of us but very difficult to explain, even for physicists. It remains
an abstract concept. Energy is a quantity that is associated with the state of an object.
Kinetic energy, for example, is associated with the state of motion of the object.
Despite this abstract definition, energy is very useful concept as we will see.
Potential energy
When we throw a ball vertically into the air the ball gradually slows down until it
comes to rest, momentarily, and then falls gradually speeding up. If we consider the
kinetic energy we find that the kinetic energy decreases, becomes zero and then
increases again. If we assume there is no air resistance then the ball will have the
same kinetic energy when it reaches the same height, on the way up and down (since
the speed is the same). The kinetic energy doesn’t simply disappear and then
reappear, instead it changes form from kinetic to potential and then from potential
back to kinetic. Potential energy is energy associated with a system (at least 2
objects) due to the configuration (positions) of that system. In the case of throwing
the ball the system is the ball and the Earth and the configuration has to do with the
height of the ball and the gravitational field. Sometimes, however, we simply say the
potential energy of the ball but we must remember that we mean the ball-Earth
system.
Figure 5.1: Changes in kinetic energy during motion. Left: A ball thrown into
the air; right: a toaster on a spring
As another example consider a mass at the end of a spring. We can pull the mass,
extending the spring and then let it go. The mass will oscillate, compressing the
spring and then extending again. If the surface is frictionless it will continue this
indefinitely. Again the kinetic energy is increasing and then decreasing (this time
continuously). In this case there is also potential energy associated with the massspring system. Energy is transformed from potential to kinetic and vice versa.
At any time in both the above situations the sum of the potential energy and the
kinetic energy of the system is a constant. This constant is called the mechanical
energy of the system.
E = U1 + K1 = U2 + K2 = constant
Conservation of
Mechanical Energy
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Physics Bridging Course 2013
where E is the mechanical energy of the system, U1 and K1 the respective potential
and kinetic energies at a given moment. This is called the conservation of
mechanical energy. Another way of writing this is in terms of the changes in kinetic
and potential energies. If the sum is a constant then the change must be zero
K + U = 0
Determining the potential energy
Now we look at the potential energy for different systems. Our equation above tells
us that the change in potential energy is the negative of the change in kinetic energy
U = K
But our work-kinetic energy theorem says that the change in kinetic energy is equal to
the work done by the net force so that
U = W
If we can calculate the work done by the net force then we can find the change in
potential energy of the system. Note that potential energy has the same units as work
and kinetic energy, the joule (J).
Although the kinetic energy of an object is well defined, it depends on the mass and
speed of the object, the potential energy of a system is not so well defined. This is
because we must select a reference point (configuration) at which the system has zero
potential energy. We can then calculate the potential energy at all other points. It is
only differences in potential energy that matter not the actual values. To illustrate this
let’s take a closer look at our previous examples.
Gravitational potential energy
Consider throwing a ball vertically into the air. We choose the y-axis to be increasing
upwards and take y = 0 at the Earth’s surface. When the ball leaves your hand the
only force acting is the gravitational forcemg (negative since it is pointing
downwards).
The change in potential energy between any two points yi and yf is given by
U = W = 
= 
ò
yf
yi
ò
yf
yi
F(y) dy
-mg dy
= mg [ y ] y f
y
i
= mgyf  mgyI
If we let yi = 0 then for any other point yf the change in potential energy is simply
mgyf . So if we define the zero of potential energy at y = 0 then the gravitational
potential energy at any other point y is given by
U = mgy
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© School of Physics, University of Sydney
Chapter 5: Work and Energy
where m is the mass of the object, g the acceleration due to gravity and y the height
above some reference point. The conservation of mechanical energy for a system
where gravity is the only force acting is expressed by
E = mgy + 1 mv2 = constant
2
The above is also true for a swinging pendulum. (Even though there is a force due to
the string, it is always at right angles to the motion, and therefore does no work.)
Elastic potential energy
Let’s go back to our oscillating toaster example from the beginning of the chapter. We
want to compute the change in potential energy from an initial position xi to a final
position xf. We choose our reference to be the x-axis and define our zero to be when
the spring is in the relaxed state. The change in potential energy is given by
U = W = - ò
= -ò
xf
xi
xf
xi
F(x) dx
-kx dx
[ ]
= 1 k x2
2
xf
xi
= 1 kx - 1 kx i2
2
2
2
f
If we let xi = 0 then the change in potential energy for any other point xf is given by
1 kx 2f . So if we set the zero of potential at x = 0 then the elastic potential energy at
2
any point x is given by
U = 1 kx2
2
where k is the spring constant and x the displacement from the relaxed state. Again
we could set the zero of potential energy to be anywhere but that would make the
mathematics more complicated. The conservation of mechanical energy for a massspring system becomes
E = 1 kx2 + 1 mv2 = constant
2
2
The fact that we can choose the zero of potential to be where we like is an advantage;
we should choose it so as to make solving the problem easier.
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Q.
A 2.0 kg branch breaks off a tree and falls 5.0 m to the ground below.
What is the branch’s initial potential energy if we take the zero of potential to
be:
a) at the ground
b) at the height of a nearby balcony, 3 m above the ground
c) at the height of the branch
What is the change in potential for all the reference systems?
What is the speed of the branch just before it hits the ground?
The gravitational potential energy of the branch (branch-Earth system) is given by
U = mgy
If the ground is the reference point, then y = 5.0 m
U = (2.0 kg) (9.8 m.s-2) (5.0 m)
= 98 J
If the balcony is 3 m above the ground then the branch is 2.0 m above it. So
U = (2.0 kg) (9.8 m.s-2) (2.0 m)
= 39 J
Since the reference point is at the height of the branch y = 0
U = (2.0 kg) (9.8 m.s-2) (0.0 m)
= 0J
5m
3m
The change in potential energy is given by
DU = mg(yf - yi)
Since the change in height is the same regardless of the reference system the
change in potential energy will be the same. Writing out explicitly though we obtain
a)
b)
c)
U = (2.0 kg) (9.8 m.s-2) ( (0.0 m) - (5.0 m) ) = -98 J
U = (2.0 kg) (9.8 m.s-2) ( (-3.0 m) - (2.0 m) ) = -98 J
U = (2.0 kg) (9.8 m.s-2) ( (-5.0 m) - (0.0 m) ) = -98 J
The conservation of mechanical energy states that
E = U1 + K1 = U2 + K2 = constant
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Chapter 5: Work and Energy
Initially the kinetic energy is zero. If we choose reference system a) we
obtain
1
98 J + 0 J = 0 J + 2 (2.0 kg) v2
v = 9.9 m.s-1
Q.
The spring of a spring gun is compressed a distance of 3.2 cm from its relaxed
state and a ball of mass 12 g is placed in the barrel. With what speed will the ball
leave the barrel once the gun is fired? The spring constant k is 7.5 N cm-1. Assume
no friction and that the gun is fired horizontally.
A.
The only force we need to consider here is the force exerted by the spring.
There is a gravitational force on the ball but it is perpendicular to the direction of
motion so it does no work (no change in kinetic energy). So the conservation of
mechanical energy gives
E = U1 + K1 = U2 + K2 = constant
We choose our zero reference for the spring-ball system to be when the spring
is in the relaxed state. Initially the spring is compressed and so has potential energy
but the ball is at rest so has no kinetic energy. When the ball is ejected the spring-ball
system has no potential energy. All the initial potential energy is converted into
kinetic energy
1
2 (7.5
1
N 100 cm
) (0.032 m) + (0 J) = (0 J) + 2 (0.012 kg) v2
´
cm
1m
v = 8.0 m.s-1
Kinetic frictional force
What would happen to our mass-spring system if there was friction between the mass
and horizontal surface? Eventually, the mass would slow down and stop. The
mechanical energy of the system will gradually decrease. A force such as friction
(and air resistance) is called dissipative. It is also clearly non-conservative. So what
happens to our conservation of mechanical energy? As we have seen, energy is
transformed into other forms.
Conservation of energy
In all processes the total energy of an isolated system is conserved. In our previous
example with friction the mechanical energy that is lost is transformed into thermal
energy of the mass and surface. In other words the mass and surface get hot. We are
all familiar with this when we rub our hand across the surface of a table. This thermal
energy is usually written as Eint to signify the internal energy of the system.
Sometimes there are other forms of energy that have to be considered, electrical,
magnetic and so on. As long as we consider all of them we find that the total energy
has not changed. We can write the conservation of energy for an isolated system as
K1 + U1 + E1int + (other forms of energy)1 =
K2 + U2 + E2int + (other forms of energy)2
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An isolated system is one in which there is no external force. We can always expand
our system to include all the forces acting. If there are forces acting external to the
system then the work done by those forces is equal to the change in the total energy of
the system
W = K + U + int
Work done by friction
When we considered friction acting on our mass-spring system we noted two effects it
produced. It reduced the mechanical energy of the mass-spring system and it
increased the internal energy of the surface and mass. If we consider our system to be
the mass-spring then the frictional force is external and the work done by friction Wf
is given by
Wf = K + U + int
Surprisingly if the frictional force is given by f and it acts over a displacement d then
the quantity fd is the work done by the frictional force and is only the loss in
mechanical energy
fd = K + U
This is because friction is a complicated force. The quantity f represents the effect is
has on the motion of the object, not the effect it has on the internal energy of the
object.
Q.
A steel ball whose mass is 5.2 g is fired vertically downward from a height
h1 of 18 m with an initial speed of 14 m.s-1. It hits some sand and travels a
further distance h2 of 21 cm before coming to rest.
What is the change in mechanical energy of the ball?
What is the change in the internal energy of the ball-Earth-sand system?
What is the average force exerted by the sand on the ball?
A.
The mechanical energy is given by
E = mgy + 1 mv2
2
We set the reference point for potential energy to be the ground
E = Ef - Ei = mg(y f - y i ) + 1 m(v f -v i )
2
= (0.0052 kg) (9.8 m.s-2) (- 0.21 m - 18 m) +
1 (0.0052 kg)( (0) - (14 m.s-2)2 )
2
= -1.4 J
2
2
The conservation of energy says that
K1 + U1 + E1int = K2 + U2 + E2int
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Chapter 5: Work and Energy
where we neglect other forms of energy since they are the same.
If we rearrange this equation we find that
E = Eint
So the change in internal energy is 1.4 J. All the mechanical energy is
transformed into internal energy.
For friction we have that
f.d = K + U
f .(0.21 m) = 1.4 J
f = 6.8 N
Power
Sometimes we are not interested in the work done by a force but the rate at which the
work is done. For example the top speed of a car is determined by the rate at which
the motor can do work. This quantity is called the power and the average power is
defined to be
where W is the work done in a time t. The instantaneous power is given by
dW
dt
The SI units for power are the joule per second (J.s-1) but it is also used very often and
so has a special name, the watt (W).
P =
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Questions and Problems
1. For the following situations state whether the work done by the force indicated is
positive negative or zero.
F
F
F
d
d
d
2. When you whirl a ball around in a circle at the end of a piece of rope there is a
force on the ball (tension). Does that force do any work? If so, is it positive or
negative?
3. What is the work-kinetic energy theorem?
4. When you lift a book from the floor to the table you apply a force on the book and
do work on it. Yet if the book starts at rest and ends at rest, the change in kinetic
energy is zero. The work kinetic-energy theorem says no work is done on the
book. Explain.
5. You lift a book, originally on the floor, and place it on a table 1.2 m high. If the
mass of the book is 2.5 kg, how much work have you done?
6. A bricklayer lifts 1000 bricks each of mass 2.5 kg over a height of 1.2 m. How
much work does he do on the bricks?
7. A varying force F acts on a mass. If the force is parallel to the displacement and
varies as indicated below, find the work done by the force:
a. over the first 4 m.
b. over the whole 10 m.
F (N)
25
20
15
10
5
0
66
1
2
3
4
5
6
7
8
9 10 11
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d(m)
Chapter 5: Work and Energy
8. A boy pushes a crate of mass 35 kg up an incline of 20° with a force of 200 N. He
pushes the crate over a distance of 2.0 m.
a. How much work does he do?
b. How much work is done by the normal force?
c. How much work is done by gravity?
9. A man pushes a crate with a horizontal force of 100 N across the floor over a
distance of 5.5 m.
d. How much work does he do?
e. If he now pushes with the same magnitude of force but at an angle of 30° to
the horizontal, how much work does he do (over the same distance)?
10. What is Hooke’s law for a spring? What does the minus sign indicate?
If experimental equipment is free, go to the back of the book and carry out
Experiment 3.
11. A horizontal spring with a spring constant of 30 N.m-1 has a mass at the end of it.
The spring is stretched by 0.15 m, then by a further 0.30 m.
a. Calculate the force required to do this.
b. Calculate the work done by the spring in each case.
12. Calculate the kinetic energy of:
a. a person of mass 70 kg running at 10 m.s-1
b. a car of mass 1250 kg travelling at 100 km.h-1
c. a Saturn V rocket of mass 2.9 x 105 kg travelling at 11.2 km.s-1
13. A branch of mass 5.5 kg falls from a height of 7.5 m to the ground.
a. Calculate the work done by the force of gravity on the branch.
b. What is the speed of the branch just before it hits the ground?
14. A motor is used to lift an 850 kg elevator a height of 110 m.
a. How much work is done by the motor on the elevator?
b. If the cable snaps and the elevator falls from that height, calculate the speed of
the elevator just before impact.
15. A 1.0 kg mass is at the end of a horizontal spring. The spring is extended 20 cm
from its relaxed state and then released. If the spring constant is 50 N.m-1.
a. Calculate the work done by the spring in moving the mass from the extended
to the relaxed state.
b. What is the kinetic energy of the mass when it is released from the extended
state?
c. What is the kinetic energy as the mass passes the relaxed state?
d. What is the speed of the mass as it passes the relaxed state?
16. What is potential energy? Give two examples of potential energy.
17. What is the law of conservation of mechanical energy? When does it apply and
when doesn’t it apply? Give examples to illustrate both cases.
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18. A 2 kg mass falls onto a vertical spring of spring constant 2500 N.m-1. The spring
compresses by 0.12 m before momentarily coming to rest. While being
compressed what work is being done by:
a. the force of gravity?
b. the spring force?
c. What is the total work done on the mass?
d. What is the initial speed of the mass?
e. What height does it start at?
19. You throw a ball vertically into the air. If there is no air resistance explain:
a. what happens to the initial kinetic energy of the ball;
b. what is the speed of the ball when it reaches your hand again?
If there is air resistance explain:
c. what happens to the initial kinetic energy of the ball;
d. what is the speed of the ball when it reaches your hand?
e. Compare the height the ball reaches with the first case.
20. A plumber drops a wrench of mass 1.5 kg, which falls a height of 25 m.
a. What is the kinetic energy of the wrench just before it hits the ground?
b. What is its speed?
21. A bobsled track begins at a height of 2500 m above sea level and ends at 2250 m
above sea level. Neglecting friction what is the speed of the bobsled at the finish
(assuming no push start)?
22. Consider a frictionless roller coaster travelling left to right along the path shown
below.
A
v
C
E
2h
B
h
D
At what points is the speed;
a. the highest;
b. the lowest.
c. Find the speed at point C, if the speed at A is v.
23. A 5.0 kg pot plant falls from a balcony 7.5 m above the ground. Take the
potential energy to be zero at ground level.
a. What is the initial potential energy of the pot plant?
b. What is its kinetic energy just before it reaches the bottom?
c. What is its speed?
24. A tennis ball is thrown at a speed of 25 m.s-1 vertically upwards. If it was released
at a height of 1.5 m above the ground what height does it reach?
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Chapter 5: Work and Energy
25. If a pendulum bob is raised by 50 cm and allowed to swing what is the speed of
the bob as it passes the lowest point of the swing? Draw a diagram.
26. A spring gun with spring constant 5.0 N.cm-1 is compressed 10.0 cm and a ball of
mass 25 g is placed in the barrel. If the gun is fired what is the speed of the ball as
it leaves the barrel?
27. Consider the pendulum below.
C
B
A
a. Identify the points at which the kinetic energy and potential energy are the
highest and lowest.
b. Where is the speed the highest and lowest?
28. A pendulum consists of a 2.0 kg stone swinging on a 4.0 m string. The stone has
a speed of 0.5 m.s-1 when it passes through the lowest point.
a. To what height will the stone rise?
b. What is the angle the string will make with the vertical?
c. If the stone is raised so the pendulum makes an angle of 60° with the vertical,
and then released, what will be the maximum speed of the stone?
29. A block of mass 5.0 kg is moving with a speed of 12 m.s-1 over a frictionless
surface. It collides with a spring of spring constant 750 N.m-1. How much is the
spring compressed before the block comes momentarily to rest?
30. An archer pulls back the string of his bow by 25 cm. If the spring constant of the
string is 1500 N.m-1 and the mass of the arrow is 150 g what is the speed of the
arrow as it leaves the string?
31. A 61.0 kg bungee-cord jumper is on a bridge 45.0 m above a river. In its relaxed
state the bungee-cord has a length of 25.0 m. We assume the cord obeys Hooke’s
law and has a spring constant of 160 N.m-1. The jumper steps off the bridge.
a. What are the various forms of energy the jumper has initially?
b. What happens to each of them during the fall?
c. Calculate the height above the water when the jumper comes to rest
momentarily.
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Student Learning
Which new term in this chapter is the most confusing to you? Which new term
do you feel most comfortable with?
Tell a classmate and find out what they think.
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CHAPTER 6
MOMENTUM
Linear momentum
Momentum, like energy, is a familiar term that is very useful in Physics. The linear
momentum of an object is defined as
p = mv
linear momentum
where m is the mass of the object and v is the velocity. We normally call this just the
momentum but sometimes we want to distinguish it from angular momentum if there
could be any confusion. Momentum is a vector quantity and its direction is the same
as that of the velocity.
Our everyday use of the term momentum normally relates to an object’s tendency to
keep moving and in fact Newton originally stated his second law of motion in terms
of momentum
Newton’s second law: The rate of change of momentum of an object
is proportional to the net force acting on that object and is in the
direction of that force.
The equation for this form becomes
åF
=
dp
dt
We can show that this is equivalent to our previous equation. Substituting our
definition for the momentum we obtain
åF
=
dp
d(mv)
dv
=
= ma
=m
dt
dt
dt
Linear momentum for a system of particles
Now we want to extend our definition of momentum to a system of n particles. Each
of the particles has its individual linear momentum and so we define the total linear
momentum, P, to be the vector sum of the individual momenta. We write this as
P = p1 + p2 + p3 + . . . + pn
= m1v1 + m2v2 + m3v3 + . . . + mnvn
We can rewrite this as
P = Mvcm
where M is the total mass of the system and vcm the velocity of the centre of mass of
the system (don’t confuse the subscript of vcm with centimetres).
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Newton’s second law can also be written for a system of particles
åF
ext
=
dP
= Ma cm
dt
Conservation of linear momentum
Suppose that the sum of all the external forces acting on our system of particles is
zero (we say the system is isolated) and that no particles enter or leave our system
(the system is closed). Putting  Fext = 0 in the above equation we obtain
dP/dt = 0 or
P = constant
(for a closed, isolated system)
This is known as the law of conservation of linear momentum. Sometimes it is
written as
Pi = Pf
where Pi is the total momentum at some initial time and Pf the total momentum at
some time later (final) i.e. initial momentum equals final momentum.
The law of conservation of momentum is like the law of conservation of energy, it
applies to all situations and to every event that happens.
Momentum is a vector quantity so that all our equations can be written in equivalent
form for each of the x, y and z components. This in turn means that if there was no
force in the x direction (say) then the x component of momentum will be constant. A
familiar example is projectile motion. For a projectile the only force acting is
gravity, which is in the y direction. This means that the momentum in the x direction
is constant. Since the mass of the projectile is constant this means the x component
of the velocity is constant, which we knew to be true.
Q.
Bullets each of mass m = 3.8 g are fired horizontally with a speed v = 1100
m.s-1 into a large wooden block of mass M = 12 kg that is initially at rest on a
table. If the block is free to slide without friction what is the speed of the
block after it has absorbed eight bullets?
rifle
v
M
A.
Before we can solve the problem we must decide what is our system. We
normally keep this to a minimum only including objects relevant to the
problem. In this case we choose our system to be the bullets and the block of
wood. The rifle is not relevant once the bullets are in flight.
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Chapter 6: Momentum
We must be sure we can apply the conservation of momentum i.e. the system
must be closed and isolated. The system is certainly closed, the bullets and
wood remain but it is not isolated. The force of gravity acts on the bullets and
wood. In the case of the wood this is balanced by the normal force of the table
but not so for the bullets. However, since gravity acts in the vertical direction
and there are no external forces in the horizontal direction, then momentum in
the horizontal direction is conserved. The bullets and block do exert forces on
each other in the horizontal direction but these are internal forces.
The horizontal component of momentum is simply
Px = mvx
Now we need to decide what is the initial state and what is the final state. This
is usually obvious from the question. In our case the initial is when the bullets
are in flight (wood at rest) and final is when they are embedded in the wood
(bullets and wood move together).
Pi = Pf
(in x direction)
n(mv) = (M + nm)V
where V is the velocity of the wood and bullets. We can rearrange to find V
nmv
V =
M + nm
8 (3.8 ´10-3 kg) (1100 m.s-1 )
=
12 kg + 8 (3.8 ´10 -3 kg)
= 2.8 m.s-1
Q.
Two blocks of masses 1.0 kg and 3.0 kg are connected by a spring resting on a
horizontal table. The masses are pulled apart and released. When the spring
returns to the relaxed state the velocity of the 1.0 kg mass is 1.7 m.s-1. What is
the velocity of the other mass at this time?
3.0 kg
1.7 m.s-1
1.0 kg
A.
Our system consists of the 2 masses and the spring. The external forces acting
are gravity and the normal force of the table. These are in the vertical direction
but should be equal and opposite anyway. The initial state is when the masses
are at rest and then released. The force the spring exerts is an internal force so
we don’t need to consider it. The final state is when the 1.0 kg mass is moving
at 1.7 m.s-1. Conservation of momentum gives
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Pi = 0 = (1.0 kg)(1.7 m.s-1) + (3.0 kg)v
-(1.0 kg)(1.7 m.s-1 )
v =
3.0 kg
=  0.57 m.s-1
In this case the minus sign means the mass is moving opposite to the first mass.
Collisions
In the world around us we see many events that we classify as collisions. The
collision between two cars, two billiard balls, a ball and tennis racquet and so on.
There are two characteristics that are common to all these collisions. They involve
large forces and last for a short time. We use these to form a definition of a collision
A collision is an isolated event in which a relatively strong force acts
on each of two or more colliding bodies for a relatively short time.
Note that by saying isolated we assume there is no net external force on the system.
We also restrict the system to include only the colliding bodies. Again it is important
to identify what we mean by before and after the collision. By before we mean just
before impact and by after we mean just after.
Elastic collisions in 1-dimension
Consider a simple head-on collision between two bodies of different masses. For the
moment let us assume that one of the bodies is at rest before the collision. We
normally refer to this as the ‘target’ and the other as the ‘projectile’. We assume that
this system is closed and isolated so that linear momentum is conserved. We also
assume that the kinetic energy of the system is conserved. Now we know that the
total energy must be conserved but there is no law that says the kinetic energy must
be conserved, it is one of our assumptions. Collisions in which the kinetic energy
does not change are called elastic collisions.
From the conservation of linear momentum we can write
m1v1i = m1v1f + m2v2f
and the conservation of kinetic energy gives
1 m1 v 2 = 1 m1 v 2 + 1 m2 v 2
1f
2f
1i
2
2
2
Note the use of numbered subscripts (1 and 2) to refer to the two bodies and i and f to
distinguish between the initial (before) and final (after) values. This is standard
notation. If we know the masses of the bodies and the initial velocity v1i we are left
with just 2 unknowns, v1f and v2f. Since we have 2 equations we should be able to
solve for these.
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Chapter 6: Momentum
Q.
Two blocks are sliding along a frictionless surface. The masses and
velocities are shown in the figure below.
a) What is the final velocity of the 1.6 kg mass?
b) Is the collision elastic?
c) Suppose the initial velocity of the 2.4 kg mass was in the reverse
direction what would be the final velocity of the other mass?
5.5 m.s
2.5 m.s-1
-1
1.6 kg
2.4 kg
before collision
4.9 m.s-1
v
1.6 kg
2.4 kg
after collision
A.
a) We assume in collisions that external forces have no effect on the objects
of interest. Remember the time for a collision is short and the internal forces large.
At any rate the net external force on the above masses is zero. We can now use the
conservation of momentum to write
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the appropriate values we obtain
(1.6 kg)(5.5 m.s-1) + (2.4 kg)( 2.5 m.s-1)
= (1.6 kg)v + (2.4 kg)(4.9 m.s-1)
v = 1.9 m.s-1
b) For the collision to be elastic the kinetic energy before the collision must
be equal to the kinetic energy after
Ki = 1 m1 v1i2 + 1 m2 v 2i2
2
2
1
=
(1.6 kg)(5.5 m.s-1)2 +
2
1 (2.4 kg)( 2.5 m.s-1)2
2
= 31.7 J
Kf = 1 m1 v12f + 1 m2 v 22 f
2
2
1
=
(1.6 kg)(1.9 m.s-1)2 +
2
1 (2.4 kg)( 4.9 m.s-1)2
2
= 31.7 J
The kinetic energies before and after the collision are equal so the collision
is elastic.
c) Now let’s assume that the initial velocity of m2 was in the opposite
direction. The conservation of momentum gives
(1.6 kg)(5.5 m.s-1) + (2.4 kg)( 2.5 m.s-1)
= (1.6 kg)v + (2.4 kg)(4.9 m.s-1)
v = -5.6 m.s-1
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Inelastic collisions in 1-dimension
An inelastic collision is one in which the kinetic energy of the system of colliding
bodies is not conserved. Most collisions are inelastic to some degree. But this varies
greatly…
Consider bouncing different types of balls. When you drop a ball to the ground this is
a collision between the ball and Earth. If you drop a superball it will bounce back to
almost the same height. This means the kinetic energy of the ball just after the
collision is almost the same as just before. A tennis ball on the other hand will only
bounce to about 55% of its original height (International Tennis Federation rules:
“Each ball shall have a bound of more than 134.62 cm and less than 147.32 cm when
dropped 254.00 cm upon a flat, rigid surface e.g. concrete”). A considerable fraction
of the kinetic energy is lost after the collision. If you drop some putty it will stick to
the ground - it does not bounce. The two colliding bodies coalesce after the collision
and this is called a totally inelastic collision. It represents the maximum kinetic
energy loss from the system.
If we assume both bodies are in motion before a totally inelastic collision then the
conservation of momentum can be written as
m1v1 + m2v2 = (m1 + m2)V
Note that there is only one final velocity since the bodies stick together.
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Chapter 6: Momentum
Q.
A block of wood with mass M = 5.4 kg is hanging from two cords as shown
below. A bullet of mass m = 9.5 g is fired into the block and the block and
wood swing upward to a height h = 2.3 cm above the initial position. What is
the speed of the bullet before the collision?
v
h
m
A.
M
Just after the collision the bullet and block coalesce and their speed is V.
From our equation for a totally inelastic collision we have
mv = (M + m)V
Now even though the kinetic energy is not conserved before and after the
collision the total mechanical energy is conserved while the wood and bullet
swing together. This is because gravity is the only force doing work. So we
have
1 (M + m)V2 + 0 = 0 + (M + m)gh
2
V = 2gh
Substituting this into the equation for conservation of momentum we obtain
M+m
2gh
m
(5.4 kg + 0.0095 kg)
=
2 (9.8 m.s-2 ) (0.063 m)
0.0095 kg
= 630 m.s-1
v =
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Questions and Problems
If experimental equipment is free, go to the back of the book and carry out
Experiment 4.
1. What is the law of conservation of momentum?
2. What is the linear momentum of a 1200 kg car travelling at 55 km.h-1?
3. A vessel of mass 32x103 kg at rest explodes, breaking into 2 pieces. One piece
(m = 10x103 kg) travels at a speed of 120 km.h-1. What is the speed and
direction of the other piece?
4. You are running along at 10 km.h-1, heading north. You then turn and run at
10 km.h-1 heading south. Feel free to approximate your own mass.
a. What is the change in your kinetic energy?
b. What is the change in your momentum? Is momentum conserved? Explain.
5. A man finds himself in the middle of a frictionless ice surface. He wants to
reach the nearby edge of the surface. All he has with him is a backpack.
a. Should he throw the backpack in the direction he wants to go or in the
opposite direction?
b. If he throws his 10.0 kg backpack with a speed of 5.5 m.s-1 and the mass of the
man is 70 kg what is his speed?
6. A 750 kg cannon fires a 12 kg shell with a speed of 65 m.s-1. What is the
recoil speed of the cannon?
7. A 50 g ball is thrown from ground level into the air with an initial speed of
16 m.s-1 at an angle of 30 above the horizontal.
a. Draw a diagram illustrating this motion.
b. What are the values of kinetic energy of the ball initially and just before it hits
the ground?
c. Find the corresponding values of the linear momentum (magnitude and
direction)?
8. An astronaut’s safety line breaks and he is left floating unattached to his
spaceship. He is at rest relative to the spaceship. In order to return to the ship
the astronaut throws a 500 g wrench that he is holding with a speed of 10.0
m.s-1.
a. If the astronaut has a mass of 120 kg what is his speed?
b. If he is 30 m away from the ship how long does it take him to get back?
9. A bullet of mass 25 g is fired into a piece of wood resting on a frictionless
surface. If the mass of the wood is 2.5 kg and the block and bullet move off
together at 0.5 m.s-1, find the initial speed of the bullet.
10. Two blocks of mass 1.5 kg and 3.5 kg are placed on a frictionless table and
connected to each other with a spring. The masses are pulled apart and
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Chapter 6: Momentum
released. At a given instant the velocity of the 1.5 kg mass is 4.5 m.s -1, what is
the velocity of the other mass?
11. A girl of mass 60 kg is riding on a skateboard of mass 5.5 kg at a speed of
6.0 m.s-1. If the girl jumps off so that she has no velocity (she jumps
backwards), what is the speed of the skateboard?
12. What is meant by elastic collision? What is an inelastic collision? What is a
completely inelastic collision?
13. Consider the following collision between two blocks.
5.5 m.s-1
2.5 m.s-1
2.4kg
3.6kg
before collision
4.9 m.s-1
v
2.4kg
3.6kg
after collision
a. What is the velocity v?
b. Is the collision elastic?
14. A frictionless cart of mass 340 g moving on a frictionless track at an initial
speed of 1.2 m.s-1 strikes a second cart of unknown mass initially at rest. The
collision is elastic. If the first cart continues after the collision with a speed of
0.66 m.s-1 in the original direction find the mass and speed of the second cart.
15. A 60.0 kg sled is coasting on ice at 9.0 m.s-1 when a 12 kg package is dropped
onto it from above. What is the new speed of the sled, neglecting any vertical
effects?
16. Two balls undergo an elastic head-on collision. The mass of ball A is 2.0 kg
and has an initial speed of 8.0 m.s-1, the mass of ball B is 4.0 kg and has an
initial speed of 5.0 m.s-1.
a. Find the total momentum of the balls.
b. Find the total kinetic energy of the balls.
c. Find the velocities of the balls after the collision.
17. A 25 000 kg truck moving at 80 km.h-1 has a head-on collision with a car of
mass 950 kg travelling at 70 km.h-1.
a. If the collision is totally inelastic, find the final velocity of the truck-car mass.
b. How much kinetic energy is lost in the collision?
18. A bullet of mass 10 g strikes a ballistic pendulum of mass 2.0 kg. The bullet
remains in the pendulum and the pendulum rises a vertical distance of 12 cm.
What is the bullet’s initial speed?
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Student Learning
Write down what you think are the key concepts in this topic.
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CHAPTER 7
OSCILLATIONS AND WAVES
Vibrations are all around us; the swing of a pendulum, the vibrations of a guitar string
or the diaphragms of speaker systems. And in less obvious contexts, too, oscillations
are the mechanisms for air molecules to transmit sounds, define the heat of solids, and
describe the electromagnetic waves of light, microwaves and radio waves. We are
now going to study these periodic oscillations, often called harmonic motions.
The frequency
A simple oscillating system consists of a particle moving repeatedly back and forth
about a point. An important property of this kind of system is its frequency, which is
the number of completed oscillations (or cycles) per second. We usually use the
symbol f and the unit of the frequency is called the Hertz (abbreviated Hz). So
1 hertz = 1 Hz = 1 oscillation per second = 1 s-1.
The period
Another important property of an oscillatory motion is its period T, which is the time
for one complete cycle (or oscillation). The period is related to the frequency via the
formula
T=
1
.
f
Obviously, the unit for the period is the second s.
Simple harmonic motion
Any motion that repeats itself at regular intervals of time is called periodic motion or
harmonic motion. We will start by examining a simple case of motion, where the
displacement of an object (x) follows a sinusoidal pattern in time. In this case, known
as Simple harmonic motion (SHM), the position can be described by either a sine or
cosine relationship. Sine and cosine waves have the same shape and differ only in
their starting point or phase. We choose cosine here to make later maths easier:
x(t) = xm cos (t)
In this formula, the amplitude x varies as the time t varies, as shown in Figure 7.1.
The other values are constant: xm, the maximum displacement; , the angular
frequency, and  , the phase. These are explained below.
.
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The amplitude
The amplitude of a sinusoidal motion, like the one described in the previous formula,
is given by xm. The subscript m stands for maximum because the amplitude is the
magnitude of the maximum displacement of the particle in either direction about the
origin. Since the cosine function oscillates between -1 and +1, then the displacement
x(t) varies between -xm and +xm, as shown in Figure 7.1.
Figure 7.1: Simple harmonic motion.
The phase
The equation of the displacement contains the term ( t) which is called the phase
of the motion, and the constant  is called the phase constant (or phase angle). The
value of the phase constant depends on the displacement and the velocity of the
particle at the time t = 0 s, and its SI unit is the radian (not degrees - in a circle there
are 2 p radians – equivalent to 360 degrees). The plot of x(t) in Fig. 7.1 has a phase
constant f = 0.
The angular frequency
The angular frequency is the constant w . The displacement x(t) must return to its
initial value after one period T (or any integer number of the period) of the motion. In
the case of , we can write
(t + T) t,
so then the angular frequency can be written as
w=
2p
= 2pf .
T
The SI unit of the angular frequency is the radian per second.
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The velocity
If we differentiate the equation of the displacement with respect to time, we can find
the velocity of the particle moving with the simple harmonic motion. That is,
v(t) =
dx
= -wx m sin(wt + f ).
dt
Similarly to the displacement, the positive quantity xm is called the velocity
amplitude vm. So then the velocity of the particle can vary between the values of -vm
and +vm.
The acceleration
We know that the acceleration of a particle is simply the derivative, with respect to
time, of the velocity of that particle. Knowing the velocity v(t) for simple harmonic
motion, we can calculate the acceleration as
a(t) =
dv
= -w 2 x m cos(wt + f ).
dt
Again the positive quantity 2xm is called the acceleration amplitude am. That is, the
acceleration of the particle can vary between the values -am and +am.
We can see that if we combine the displacement and acceleration we obtain
a(t) = -w2 x(t)
This means that the acceleration is proportional to the displacement but opposite in
sign, and the two quantities are related by the square of the angular frequency.
The force law
If we want to know what force must act on the particle, we need to know its
acceleration and how it varies with time. Using Newton’s second law and the formula
for the acceleration of a particle in a simple harmonic motion, we get
F = ma = - (m2)x,
which is very similar to Hooke’s law for a spring which gives
F = -kx
We can say then that the spring constant is
k = m2
Now we can have an alternative definition of the simple harmonic motion, taking into
account the equation relating the force and the displacement. This states
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Simple harmonic motion is the motion executed by a particle of mass m
subject to a force that is proportional to the displacement of the particle
but opposite in sign.
Linear simple harmonic oscillator
Let us consider a block-spring system, like the one shown in Fig. 7.2.
k
m
x
-xm
x=0
xm
Figure 7.2: A simple harmonic oscillator.
This system forms a linear simple harmonic oscillator, where linear indicates that the
force F is proportional to the displacement x rather than to some other power of x (x2,
or x3 etc). As we have seen before, the angular frequency  of the simple harmonic
motion of the block is related to the spring constant k and the mass m of the block via
the formula k = mw2, which can be written as
w=
k
m
The period of the linear oscillator can be obtained then as
T = 2p
m
k
Energy considerations
The potential energy of a linear oscillator is associated entirely with the spring. Its
value depends on how much the spring is stretched or compressed. It is
1
1
U(t) = kx 2 = kx m2 cos2 (wt + f ).
2
2
The kinetic energy of the same system is associated entirely with the block, because
its value depends on how fast the block is moving. It is given by
1
1
K(t) = mv 2 = m(-wx m ) 2 sin 2 (wt + f )
2
2
but we know that k=m2, so then the kinetic energy becomes
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1
1
K(t) = mv 2 = kx m2 sin2 (wt + f ).
2
2
The mechanical energy E(t) is given by E = U + K, which is
1
1
E(t) = kx m2 cos2 (wt + f ) + kx m2 sin 2 (wt + f ).
2
2
We know that cos2 + sin2=1, so the mechanical energy can be simplified to
1
E = U + K = kx m2 ,
2
which is a constant, independent of time. In other words, mechanical energy is
conserved, as we would expect. At full stretch of the spring, the block is stationary,
and the energy is entirely stored in the spring as potential energy. But as the spring
contracts and the block begins to move, the potential energy is converted into kinetic
energy, until, at the point of no stretch, the block is at its maximum speed and the
spring is relaxed, containing no potential energy. As it moves past the centre point the
spring begins to compress and the block slows down, the kinetic energy is being
converted back into potential energy. Thus it will continue, with the total energy
remaining constant but switching from kinetic to potential and back again.
Of course in reality the energy will dissipate due to friction and the spring will
eventually stop.
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Simple pendulum
A pendulum will swing back and forth on a string in a hypnotic fashion. What makes
it move? The answer is gravity. But what governs how fast or slow it swings (its
period)? Length? Mass? How high it swings?
To approach this question, let’s imagine a simple pendulum, consisting of a particle of
mass m suspended from an unstretchable, massless string of length L, such as the one
shown in Fig. 7.3. The mass is free to swing back and forth.
L

Tension
m
mg
Figure 7.3: A simple pendulum.
For small oscillations (  < 10o), we can work out a relation between m, g, L and k,
and this relation is
mg
k=
L
Substituting this equation in the period equation obtained previously, we obtain the
period of a simple pendulum as
T = 2p
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L
.
g
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Chapter 7: Oscillations and Waves
Waves
Until now we have only discussed the oscillation of isolated particles. However, if an
oscillating particle is connected to other particles – for example, an atom in a solid –
then the oscillation energy can be transferred from particle to particle. This is a wave,
travelling through a medium. The energy of the oscillating particles gets transferred
through the medium, leaving the medium unaffected.
There are two ways that a wave can be graphed.:
 A snapshot graph shows the whole wave at a given instant. It graphs
displacement against position.
 A history graph shows the motion of a single point in the medium, as the
wave travels past it. It graphs displacement against time.
The two representations often look identical, but not always. The graphs we have seen
so far in this chapter have all been history graphs, graphing an oscillating point
against time.
Two types of waves
A Transverse wave is when oscillation is at right angles to the wave’s direction of
travel. Waves on the surface of water, a vibrating guitar string and light are all
examples of transverse waves.
A Longitudinal wave is when the oscillation is in the same direction that the wave’s
energy is moving. Sound is a longitudinal wave.
Sinusoidal Waves
As with the oscillations we will start our discussion with the simplest of waves, the
sine wave. Waves can have many different shapes, but sine waves are the easiest for
us to deal with mathematically. (In fact Fourier theory shows us that any wave can be
represented as a combination of sine waves; however that is beyond the scope of this
course).
At the start of this chapter we defined the amplitude, the period, the frequency and the
phase for oscillations. Now we have a travelling wave, two other quantities become
important;
 The wavelength ( – greek letter lambda) is the distance between two points
at identical parts of two consecutive waves. Its units are metres.
 The speed (v) has units of metres per second. This is generally a property of
the medium and is not affected by the wave’s properties (frequency, amplitude
etc).
We already know that the period (T) tells us how long an oscillation takes. The
wavelength then tells us how far the wave travels in that time:
 = vT
We also know that T = 1/f, so
l=
v
or
f
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v = fl
Some examples of wave speeds are the speed of light in a vacuum – 2.98 x 108 ms-1
and the speed of sound in air – 330 ms-1.
Q
The radio waves for Triple J radio station in Sydney have a frequency of 105.7
MHz. Calculate their wavelength. The speed of light is 2.98 x 108 ms-1.





v 2.98 ´10 8
l= =

f 105.7 ´10 6
m
Two and three dimensional waves
The diagrams we have shown so far have only been in one dimension, similar to a
wave travelling along a string. However waves can travel across a two dimensional
plane, (e.g. the surface of water), or through three dimensional space, (e.g. sound or
light in a room). For example, light from a point source, such as a distant star, will
travel out in a spherical wave.
When drawing two dimensional waves, diagrams are sometimes shown as plan view,
seen from above. Lines denote the crests of the wave, called wavefronts. The
direction of travel is denoted by an arrow called a ray. Rays are always at 90 degrees
to the wavefronts.
Figure 7.4: Wavefronts and Rays for straight circular waves
Inverse square law
We know that distant sounds are faint, and distant stars are too. This is because the
waves from these sources are spherical: the energy from the oscillation is spreading
out in all directions. The power P of the original source does not change, so we define
the Intensity I as the power per unit area a.
I = P/a
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Because the waves are a regular, spherical shape, we can easily calculate the
relationship between the energy and the distance from the source. The energy has
been spread over the surface of a sphere, and we know the formula for the area of a
sphere is 4r2. Therefore for spherical sources:
I = P / 4r2
As the intensity drops off in relation to the square of the distance from the source r,
this is known as an inverse-square relationship. This is a very common relationship
and occurs in many areas of Physics, including electromagnetism and gravity as well
as light and sound.
Reflection
Light reflects from a mirror so you can see yourself; sound reflects off a cliff wall to
create an echo; When waves hit the edge of the medium they are travelling through
their energy is reflected and begins to travel in a different direction. The direction of
the reflected wave depends on the angle between the incident wave and the
discontinuity it is reflecting from, as shown in Figure 7.5
Figure 7.5: When waves reflect from a surface the angle of reflection r is
equal to the angle of incidence i.
It turns out that waves will not only reflect from the end of the medium, but from any
boundary within the medium. At such a boundary some of the wave energy will
continue on, and some will be reflected. For example, imagine light travelling through
air and then reaching a glass window. Most of the light passes into the glass, but a
small amount (usually about 5%) is reflected. This will happen at any boundary
within a medium, for example a light string attached to a heavy string, or where deep
water becomes shallower.
The reflected wave has the same wavelength as the incident wave, however the phase
is not always the same. If the speed of the wave decreases at the boundary, or if the
wave is completely reflected then the displacement changes sign, as shown in Figure
7.6. If the wave speed increases, then the reflected wave will have the same
displacement (as will the portion of the wave which continues forward).
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Figure 7.6: Reflection of a wave at a boundary. In this case, the wave speed is
faster in the thicker rope, on the right hand side. Note that the amplitudes of
both waves in the lower diagram are smaller than the original wave, to
conserve energy.
Superposition – the interaction of two waves
How do waves within a medium interact? Two waves overlapping in a medium will
pass through each other without affecting the direction of travel or the frequency or
phase of each other. However, at the moment that the waves are on top of one another
their amplitudes add up. In other words the displacement at a given point is the sum of
the displacements due to all the waves present at that point. This is known as the
principle of Superposition.
Remember that displacement is a vector, which means that a positive displacement
caused by one wave can be cancelled out by a negative displacement from another
wave. As waves all oscillate between positive and negative displacement, this makes
the phase very important. Two identical waves in phase will add up to a double-sized
amplitude, but if they are half a wavelength out of step they will exactly cancel each
other out.
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Chapter 7: Oscillations and Waves
Figure 7.7: Superposition of two identical waves: Top, in phase; Bottom, out
of phase.
Standing waves
The property of superposition – in particular the ability of waves to cancel each other
out – leads to all kinds of interesting and unexpected behaviour by waves. For
example, imagine a wave travelling along a guitar string. It is reflected at either end of
the string, and bounces back and forth, and as different parts of the wave overlap each
other while travelling back and forth their displacements add and subtract. The result
of these two waves travelling in opposite directions is that in some places the string
will vibrate a lot, while in other places the string will remain stationary. These
stationary points, known as nodes, remain in the same place, as do the points where
the vibration is maximum ( known as antinodes). This is known as a standing wave.
Standing waves can also occur in two and three dimensions, for example on the
surface of pool, or in the air inside a trumpet. There will be antinodes where the wave
has a large amplitude, and other points where the reflected waves completely cancel
out and leave no oscillation at all.
Interference
Similar patterns to standing waves can be generated by two separate sources of
waves. These “interference patterns” show areas where the two waves are in phase
and therefore add up to a large amplitude, while in other areas the waves are out of
phase, and so the result is no oscillation at all.
To determine whether a point will be a node or an antinode, one needs to calculate the
distance that each of the waves has traveled to that point, and compare if they are at
the same point of their cycle.
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Q.
Two students, Bill and Ben, are sitting in a lecture theatre, in a row 12 m from
the front bench, as their lecturer demonstrates interference of sound waves.
The experiment involves two speakers set 5 m apart on the bench, both
playing a note at a frequency of 500 Hz. Bill is sitting directly in front of the
left hand speaker, while Ben is halfway between the two speakers. Calculate
for each student whether they are in a node or an antinode.
A
To answer this questions we need to calculate the path length of the waves
from each speaker to the students, and work out whether the two waves will be
in phase or not. To do this, we need to know how many wavelengths each path
is. First we need to calculate the wavelength:
v = f
= v f
= (330 ms-1)/(500s-1) = 0.66 m
For Bill: We need to find how many wavelengths there
are between him and the speaker.
Left speaker : 12 m/0.66 m = 18.18 wavelengths
Right speaker: Using Pythagoras we can calculate the
path:
P = (5) 2 + (12) 2 =13m
13 /0.66 = 19.69 wavelengths
Therefore the difference between the two paths is
19.69 – 18.18 = 1.51 wavelengths
This means the two waves will be almost exactly out of phase and will cancel. Bill is
in a node and will not hear the speakers.
Ben is the same distance from both speakers. Therefore the waves from both speakers
will be in phase . Therefore we can deduce, without calculating the distance, that he is
in an antinode, and will hear the tone clearly.
Beats
The examples we have discussed so far have involved waves of the same frequency.
We will now make the situation more complex and consider two waves of differing
frequencies.
Imagine a loudspeaker is emitting two notes, with equal amplitudes and frequencies
only slightly different. Figure 7.8 shows the history graph of the two waves. The
waves begin in phase, so their amplitudes will add. However, as time passes the lower
frequency wave lags. After a time the wave is so far behind that its displacement is
the opposite sign to the higher frequency wave – completely out of phase – so the
resultant amplitude is zero. But then the wave lags even further, until it is back in
phase, and the whole process will cycle through again.
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Chapter 7: Oscillations and Waves
Figure 7.8. Beating frequencies. Top: Two waves of slightly different
frequencies. The solid line has a lower frequency and lags behind the dotted
line. Bottom: The resultant wave from the addition of the two waves. The
amplitude oscillates from double the original wave to zero and back.
The resultant wave (bottom, Figure 7.8) shows that the amplitude (which we perceive
as volume) varies from double to zero and back again. This oscillation in the volume
is known as beating, and is well known to musicians who listen for beating while
tuning their instruments
Refraction
Refraction refers to the change in direction that waves experience when they travel
from one medium to another with a different wave speed. This change of direction
only occurs when waves do not hit the boundary at right angles.
Viewed from a ray perspective this behaviour does not make sense, but if one views
the wavefronts, then an explanation becomes apparent. As shown in Figure 7.9, waves
travelling at an angle will not strike the boundary all at once. As one end enters the
slower medium, its drag effectively pulls the wave around towards the normal (90o).
Note that as the waves change medium, the frequency remains the same. Therefore,
as the speed changes, the wavelength must change to compensate. Waves have shorter
wavelengths in slower mediums, and longer wavelengths in faster mediums.
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Figure 7.9: Refraction of a wave passing from a fast medium into a slower
medium.
Diffraction
When a wave is partly blocked by an impenetrable medium an interesting situation
occurs. At the edge where the wave is blocked you might think that the wave would
just stop, but in turns out that the wave spreads into the region behind the block, in the
region of geometrical shadow. This is known as diffraction.
Figure 7.10. Diffraction of waves into an area of shadow.
One explanation for this behaviour was proposed by the Dutch physicist Christiaan
Huygens in the 1700s, who theorised that every point on the wave is acting as a point
source emitting wavelets. In free propagation, the superposition of all these wavelets
cancel out to leave only the advancing wavefront. However, if part of the wave is
blocked as discussed above, then the wavelets emitted from the end of the wave will
spread out into the area behind the blocker, as shown in Figure 7.10. Note that the
amplitude of the wave will drop off in the area behind the blocker – the greater the
angle the less energy is transmitted.
Diffraction will occur wherever waves encounter edges or apertures. The strength of
the effect depends on the relative sizes of the aperture and the wavelength.
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Questions and Problems
1. What is the period of a simple pendulum of length 1.0 m?
What should be its length for the period to be tripled in value?
2. What changes could you make in a harmonic oscillator that would double the
value of the maximum speed of the oscillating mass?
3. A force of 10.5 N is applied to pull a mass attached to a spring with a spring
constant of 70 N.m-1. What is the amplitude of the resulting oscillation?
4. A block of mass 920 g is fastened to a spring whose spring constant k is 85 N.m-1.
The spring is pulled a distance of 11 cm from its equilibrium position.
a. What force does the spring exert on the block just before the block is released?
b. What are the angular velocity , the frequency f, and the period T of the
resulting oscillation?
5. A 4 kg block hangs from a spring extending it 16 cm from its unstretched
position.
a. What is the spring constant?
b. How fast will it oscillate
6. What is the maximum acceleration of a platform that vibrates with an amplitude
of 2.20 cm at a frequency of 6.60 Hz?
7. A 20 N weight is hung from the bottom of a vertical spring, causing it to stretch
20 cm.
a. What is the spring constant?
b. The spring is now placed horizontally on a frictionless table. One end is held
fixed and the other end is attached to a 5 N weight. The weight is stretched
then released from rest. What is the period of oscillation?
8. A 0.10kg block oscillates back and forth along a straight line on a frictionless
horizontal surface. Its displacement from the origin is given by
p
é
ù
x(cm) = (10 cm)cos (10 rad.s-1) t + rad .
ë
2
û
a. What is the oscillation frequency?
b. What is the maximum speed acquired by the block? At what value of x does
this occur?
c. What is the maximum acceleration of the block? What value of x does this
occur?
d. What force, applied to the block, results in the given oscillation (as a function
of the displacement x)?
9. Find the mechanical energy of a block-spring system having a spring constant of
1.3 N.cm-1 and an amplitude of 2.4 cm.
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10. The end point of a spring vibrates with a period of 2 s when a mass m is attached
to it. When this mass is increased by 2 kg, the period is found to be 3 s.
What is the value of m?
11. Consider the simple harmonic oscillator shown in Fig 7.2. There is no friction or
heat loss in the system. Imagine it oscillating back and forth.
a) Describe when the potential energy is at a maximum.
b) Describe when the kinetic energy is at a maximum.
c) Which will reach the largest value, potential energy or kinetic energy?
d) Describe when the mechanical energy is at a maximum.
Figure 7.11
12. Figure 7.11 is a snapshot graph of a wave at t = 0. Draw the history graph for this
wave at x = 5 m for t = 0 s to 5 s.
Figure 7.12
13. Figure 7.12 shows two waves travelling in opposite directions at t = 0 s. Draw the
resultant waves as they pass through each other at t =2, 3 and 5 seconds.
14. The second string of a guitar is tuned to A an octave below middle C, which has a
frequency of 220 Hz. If the length of the string is 66 cm, what is the speed of the
wave in the string?
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Chapter 7: Oscillations and Waves
15. You are watching the New Year’s Eve fireworks over Sydney Harbour from the
Gladesville Bridge 40m above the water and you see the first rocket explode four
kilometres away, at an altitude of 660m. What is the time delay between when you
see the flash and hear the explosion?
16. The intensity of light hitting the earth’s atmosphere is 1.4 kW/m2
a) Calculate the power of the sun (the distance from the earth to the sun is 1.50
x 1011 m)
b) Calculate the intensity of radiation reaching Saturn, 1.43 x 1012 m from the
sun.
17. You can hear your friend talking whether they face you or turn away. What makes
this possible? But if your friend whispers, she has to face you. Why? (Hint how do the
frequencies of sound in a normal voice and a whisper compare?)
Student Learning
In what ways has this session increased your
understanding of this topic?
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CHAPTER 8
ATOMIC THEORY AND
ELECTROMAGNETISM
Structure of matter
The Greek philosopher Democritus, was the first known person to suggest that matter
is made up of smaller individual constituents: atoms. If you start with a lump of iron
and keep cutting it into smaller pieces, you will get to one atom of iron. If you cut it
even smaller, it will no longer be iron.
Atoms are grouped into 92 naturally occurring elements, conveniently described by
the periodic table, and all matter we see every day is made up of different
combinations of these elements. Initially people assumed atoms were solid spheres,
but various experiments around the 1900’s showed these assumptions had to be
incorrect.
J.J. Thomson’s experiments in the late 1800’s identified sub-atomic particles for the
first time. He discovered electrons are contained inside all atoms and showed that
they are about 2000 times lighter than the smallest known atom, hydrogen.
Furthermore electrons from every element are identical. Because electrons have a
negative electric charge and are so small, atoms had to also contain at least one other
positively charged thing to cancel out the negative charge and make up for the rest of
the mass and volume.
Ernest Rutherford in his famous experiment shot α-particles (positively charged
nuclear radiation) at a very thin gold foil. He found that 99.99% of the α-particles
went straight through, and only 0.01% (one in ten thousand) bounced back.
Figure 8.1 Rutherford’s α-particle scattering experiment
From this and other experiments people concluded that:
-
Atoms are 99.99% empty space (hence the particles going through undeflected).
It has all the positive charge and most of the mass concentrated in a small volume
in the middle, called the nucleus (what the one in ten thousand particles bounces
off).
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-
The nucleus has two types of particles in it, positively charged protons and
neutral neutrons. These have almost the same mass, and there are usually a few
more neutrons than protons.
The electrons exist outside the nucleus.
Although a lot of progress was made in terms of determining the structure of atoms, it
was instantly realised that this simple description could not be correct. Atoms were
known to emit specific types of radiation, and amongst other things this model could
not explain why only specific types are emitted.
Figure 8.2: A representation of an atom. Note that this is only a representation
showing key properties of atoms. Atoms look nothing like this in reality.
For example, the simplest atom hydrogen only emits light at several specific
wavelengths in the visible, rather than at all wavelengths.
λ
Figure 8.3: Line spectrum of hydrogen.
(http://en.wikipedia.org/wiki/File:Visible_spectrum_of_hydrogen.jpg)
Radiation
Radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and
gamma rays are all fundamentally the same thing: electromagnetic radiation (EMR).
They can be thought of as a wave but instead of particles in a medium oscillating,
they consist of an oscillating electric and magnetic field. Therefore they require no
medium through which to propagate and can propagate through a vacuum. They all
travel at the same speed through a vacuum: the speed of light, c.
c = 3 × 108 ms-1
c = fλ
Early in 1900’s it was realised by people like Max Planck and Albert Einstein, that
EMR could also be thought of as little packets of energy called photons. It is easiest to
think that EMR travels along as a wave, but when it interacts with something (e.g.
absorbed or emitted) it behaves like little particles – the photons. The energy of each
photon depends on the frequency of the radiation:
E = hf = hc/λ
h is the Planck constant, h = 6.63 × 10-34 Js.
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EMR of a particular frequency will be made up of a stream of identical photons.
Hence, the energy in EMR is said to be quantized, meaning it can only exist in
discrete amounts.
The energy of individual photons is very small, so instead of using the SI unit of
energy, Joules, it is more convenient to use electron-volts (eV) as the unit of the
energy.
1 eV = 1.6 × 10-19 J.
Even though EMR is all fundamentally the same, it is divided into categories
according to frequency or wavelength (and hence photon energy). Gamma rays have
the shortest wavelength and highest frequency and photon energy, whilst radio waves
have the longest wavelengths and lowest frequencies.
Figure 8.4: The electromagnetic spectrum. The visible part of the
spectrum ranges from red at the left hand end,(~700 nm) through the
spectrum to violet at the right hand end (~ 400 nm)
Energy levels in atoms
The next contribution to atomic theory was made by Niels Bohr who combined the
then current picture of the atom with the idea of quantization. He proposed that the
energy the electrons have inside atoms is quantized, i.e. can only take specific values.
Figure 8.5: Energy diagram for hydrogen
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An electron in an atom can occupy one of several allowed energy levels (also called
quantum states, allowed energy states, shells, orbitals). The more energy the electron
has (the higher the level it occupies) the further it is on average from the nucleus.
These energy levels can be described using a quantum number, n. The lowest possible
energy for the electron is called the ground state and corresponds to n = 1, this is the
energy level closest to the nucleus. The n = 2 is further away etc, and n = ∞
corresponds to a free electron outside the atom. The energy of these levels is different
for every element. For hydrogen, if we take n = ∞ to correspond to zero energy, then
the ground state n = 1 corresponds to -13.6 eV. (More complex models use four
quantum numbers.)
Absorbing and emitting radiation
The electron will normally sit in the ground state. If the incoming photon’s energy can
take it to another allowed energy level, then the photon will be absorbed and the
electron is said to be excited. After a while the electron will de-excite back to the
ground state and emit a photon. If the incoming photon cannot take the electron to an
allowed level then it cannot be absorbed. The energy of the photon must match the
energy change of the electron:
Ephoton = hf = ΔEelectron
For the simplest model of hydrogen – the Bohr model – the energy of the allowed
energy levels is given (in units of eV) by
E(n) = -13.64/n2 ,
and wavelengths emitted/absorbed in a transition from one energy level to another (n1
to n2) is given by the Rydberg Equation
æ1
æ1
1ö
1
1ö
where
= RH ç 2 - 2 ÷
= RH ç 2 - 2 ÷RH is the Rydberg
l
l
è n1 n 2 ø
è n1 n 2 ø
constant for hydrogen,
RH = 1.097 × 107 m-1.
1
This is derived from the energy of each allowed state and the relation E = hc/λ. This
simple model by Bohr was superseded by other more complex models which took
more interactions into account and used more quantum numbers, but which remained
qualitatively the same.
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Figure 8.6: Absorption (left) and emission (middle) of a photon. If the energy
of the photon cannot take the electron to an allowed state then it cannot be
absorbed (right).
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Questions and Problems
1. If the radius of a gold atom is around 1 × 10-10 m, what is the approximate radius
of a gold nucleus?
2. If atoms mostly contain empty space, why are solids not able to pass through each
other?
3. Drawings of atoms like Figures 8.2 and 8.5 can be misleading. What about them is
correct and what is misleading?
4. The energy of a photon depends on its frequency. Using Figure 8.4 calculate the
photon energy range (in eV) for the different types of radiation.
5. What is the photon energy for green light of wavelength 532 nm? If I had 1 W of
green light, how many photons are arriving per second? How would that compare
to 1 W of red light?
6. Why is it that that UV light causes skin cancer and we need sunscreen for
protection, but nobody worries about exposure to infra-red?
7. If a free electron, stationary in space, is taken to have zero energy, then an
electron in the ground state of hydrogen has an energy of -13.64 eV. What does it
mean for this energy to be negative?
8. Calculate the energy of the first six levels for hydrogen.
9. What is the minimum energy an electron in the ground state of hydrogen can
absorb? What is the maximum it can absorb and still remain in the atom? What
happens if it absorbs more than this?
10. What is the maximum energy an electron in hydrogen can emit? What type of
radiation is that?
11. The allowed energy levels are different for different elements. This is determined
by the nucleus. What property of the nucleus determines the energy levels? Will
the energy levels of other atoms have lower or higher energies than hydrogen?
What are the implications in terms of the types of radiation they can absorb or
emit?
12. Can hydrogen absorb or emit X-rays? Why or why not? What circumstances
would you need to produce X-rays?
13. How many different energy photons can be emitted by hydrogen if we only
consider the first six levels?
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14. The possible transitions in hydrogen that lead to emission of a photon are usually
presented in groups. All transitions that end at n = 1 are called the Lyman series,
ending in n = 2 is the Balmer series and in n = 3 is the Paschen series as in the
diagram below. For each series calculate the wavelength of each line shown in the
diagram. What is the shortest wavelength in each series and what type of radiation
is it? In the end, how many lines can we see and what colour will they be?
15. Diamond is made of carbon and is clear, but sometimes it contains very small
amounts of another element which makes pink, yellow or blue diamonds. What
can you say about the energy levels of carbon, and of the impurities? What can
you say in general about the energy levels and the transparency or colour of a
material?
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CHAPTER 9
ELECTRIC CHARGE
Electric charge
Electric charge is stored in all the familiar objects surrounding us and in our own
bodies. The discharge of lightning, small pieces of paper sticking to a plastic comb or
your clothing sticking to your body in dry weather are all effects due to the presence
of electric charge. Every object in the world contains an enormous amount of charge.
We can distinguish between two kind of electric charge: positive and negative
charges..
Usually, the objects contain equal amount of positive and negative charges, which
makes them electrically neutral, that is, they contain no net charge and therefore do
not interact with other charged objects. On the other hand, if there is an imbalance
between the two types of charges, then we say that an object is charged, meaning that
there is a net charge that will interact with other objects. (Note: any net charge is
always slight compared to the total amounts of positive and negative charges in the
object. Or else it would probably explode – we will see why…)
We can observe charge with a glass rod, by rubbing one end of the rod with a piece of
silk. At the points of contact between the silk and the rod, tiny amounts of charge are
transferred from one to the other, slightly upsetting the electrical neutrality of each of
them. In order to get the maximum transfer, we need to increase the number of
contact points, so we rub the silk over the rod.
a)
b)
F
Glass
Glass
F
Glass
-F
Plastic
-F
Figure 9.1: Forces between charged rods.
Let us consider the situation in Figure 9.1. In a) we suspend the charged glass rod
from a thread and bring a second, similarly charged, glass rod nearby. The two rods
repel each other. However, if we rub a plastic rod with fur and bring it near the
suspended glass rod (as shown in Figure 9.1 b), the two rods attract each other.
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Let us discuss this in terms of electric charge. When a glass rod is rubbed with silk,
the glass loses some of its negative charge onto the silk and therefore has a small
unbalanced positive charge (represented by the plus signs). However, when the plastic
rod is rubbed with fur, the plastic gains a small unbalanced negative charge from the
fur (represented by the minus signs). We can then deduce that
Like charges repel each other, and unlike charges attract each other.
Where “like charges” means they have the same sign, and “unlike charges” means
having opposite signs. The “positive” and “negative” labels and signs for electric
charge were chosen arbitrarily by Benjamin Franklin.
Conductors and insulators
Some materials are called conductors because some of the charged particles in them
can move freely, such as metals, tap water and the human body. Other materials,
where none of the charge can move freely, are called nonconductors or insulators.
Examples are glass, ceramics, chemically pure water and plastic.
Rubbing a copper rod with wool while holding the rod in your hand, will not charge
the rod, because both you and the rod are conductors. The rubbing will cause a charge
imbalance but the excess charge will immediately move through you and to the floor
(which is connected to the Earth’s surface), and the rod will quickly be neutralised.
So we say that we ground an object when we set up a pathway of conductors between
that object and the Earth’s surface, thus neutralizing the object and discharging it.
Now, if you hold the rod via an insulator, you eliminate the conducting path to the
Earth, and the rod can then be charged.
Conduction and insulation are due to how atoms are built. Atoms consist of positively
charged particles called protons, negatively charged particles called electrons and
electrically neutral particles called neutrons. The proton and the electron have the
same charge, but opposite in sign. The electrons of atoms of a solid which is a good
conductor do not remain attached to individual atoms but become free to wander
about within the solid. These electrons that are free to move are called conduction
electrons.
Semiconductors and superconductors
Semiconductors, such as silicon and germanium, are materials that have properties
which are intermediate between conductors and insulators. All the microelectronic
devices that have transformed our lives owe their existence to the semiconducting
properties of the materials they are made of.
Another type of material are superconductors. In a truly amazing twist of quantum
mechanics these have no resistance at all to the movement of electric charge through
them.
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Chapter 9: Electric charge
Coulomb’s law
The electrostatic force of attraction or repulsion between two particles (or point
charges) that have charge magnitudes q1 and q2 and are separated by a distance r has
the magnitude
F=k
q1q2
,
r2
where k is a constant, known as the electrostatic constant. This expression is called
Coulomb’s law after Charles Augustus Coulomb. It’s interesting to note that it’s the
same equation as that worked out by Newton for the magnitude of the gravitational
force between two particles with masses m1 and m2 that are separated by a distance r :
F =G
m1m2
r2
in which G is the gravitational constant.
F is the magnitude of the force acting on either particle owing to the charge on the
other. q1 and q2 are the magnitudes (or absolute values) of the charges of the two
particles. For practical reasons, the SI unit of charge is derived from the SI unit of the
electric current, the ampere (A). The SI unit of charge is called the coulomb (C), and
can be defined as:
One coulomb is the amount of charge that is transferred through the cross
section of a wire in 1 second when there is a current of 1 ampere in the wire.
For historical reasons, the electrostatic constant k is usually written as 1 4 pe . Then
0
Coulomb’s law becomes
F=
1 q1q2
.
4pe0 r 2
The constant in this equation has the value
1
= 8.99 ´10 9 N.m 2C-2 .
4pe0
The quantity 0 is called the permittivity constant and its value is
0 = 8.85 x 10-12 C2/N.m2 ..
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Charge is quantised
We now know that fluids, such as air or water, are not continuous but are made up of
atoms and molecules. Experiment shows that the electrical fluid is also not continuous
but is made up of multiples of a certain elementary charge. That is, any negative or
positive charge q that can be detected can be written as
q = ne,
n = ±1, ± 2, ± 3, ...
in which e, the elementary charge, has the value
e = 1.60 x 10-19 C e =1.60 ´10-19 C.
The elementary charge is one of the important constants of nature.
When a physical quantity such as charge can have only discrete values rather than any
value, we say that that quantity is quantised.
Charge is conserved
When we rubbed the glass rod with silk, a positive charge appeared on the rod.
Measurement shows that a negative charge of equal magnitude appears on the silk.
This means that rubbing does not create charge but only transfers it from one body to
another. This is the hypothesis of conservation of charge. Thus we can add electric
charge to the list of quantities that obey a conservation law.
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Questions and Problems
If experimental equipment is free, go to the back of the book and carry out
Experiment 5.
1. If matter is made of positive and negative charges why do we not see the effects
of these charges?
2. Explain what happens when we rub a glass rod with a piece of silk? How does the
rod become charged?
3. What is the purpose of the strips that hang from the rear of cars and touch the
ground?
4. What is a conductor? What is an insulator?
5. Write down Coulomb’s law and explain what it means.
6. A charge q1 of 1.2 C is 2.0 m away from a charge q2 of 2.5 C.
a. What is the electrostatic force of q1 and on q2?
b. Is it repulsive or attractive?
7. For the following charge configuration calculate the resultant electrostatic force
(magnitude and direction) on charge A.
C
+ 2.0 C
10 cm
+ 1.5 C
A
+ 1.0 C
10 cm
B
8. What separation must two equal charges of 1.0 C have for the force between them
to be 1.0 N?
9. Two fixed charges of 1.0 C and 3.0 C are 15 cm apart. Where would you place
a third charge so that no net electrostatic force acts on it?
15 cm
1.0 C
3.0 C
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10. Two equally charged particles, held 3.2 x 10-3 m apart, are released from rest. The
initial acceleration of the first particle (having mass 6.3 x 10-7 kg) is observed to
be 7 m.s-2 and that of the second to be 9 m.s-2.
a. What is the mass of the second particle?
b. What is the magnitude of the common charge?
11. Two charged spheres of equal mass, 50 g hang from a support. They are both
charged with the same charge and repel each other making an angle of 20° with
the vertical. The string is 1.0 m long. Find the charge of each sphere.
20º 20º
1.0 m
Q
50 g
1.0 m
Q
50 g
12. Two tiny water drops, with identical charges of -1.00 x 10-16 C are separated by
1.00 cm.
a. What is the magnitude of the electrostatic force acting between them?
b. How many excess electrons are on each drop, remembering that the charge of
an electron is -1.60 x 10-19 C?
Student Learning
Write down the three most important things you have
learnt in this topic.
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CHAPTER 10
CURRENT AND CIRCUITS
In the last session, we mainly dealt with charges that are slow-moving or at rest. Now
we will look at the Physics that explains what battery you need for your watch, how to
wire your house and what power you need for an electric chair: we will look at
electric current.
Electric potential
Let us consider a particle with charge q0. Similarly to the mechanics case, we must be
able to assign to this test charge an electric potential energy U. If the test charge
moves from an initial point i to a final point f, we can define the difference in its
electric potential energy as
U = Uf – Ui = - Wif
Here Wif is the work done by the electrical forces acting on the particle as it moves
from i to f. We define the electric potential V (or simply potential) at a point as the
potential energy per unit positive charge at that point. In other words V can be written
as
V=
U
.
q0
The SI unit is joule per coulomb (J/C) or volt (V). The potential difference between
any two points i and f can then be written as
Electric current
i
i
a)
Battery
b)
Figure 10.1: A loop of copper, with and without a battery.
Figure 10.1 a shows that an isolated conducting loop of copper is at the same potential
everywhere. Although conducting electrons are present, no net electric force acts on
them and thus there is no current.
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On the other hand, Figure 10.1 b shows that if we insert a battery in the loop, the loop
is no longer at the same potential. The conducting electrons are then forced to move
in one direction, reaching a steady-state flow and thus establishing a current.
If charge dq passes through a hypothetical cross section of the loop in time dt, then
the current through that section is defined as
I=
dq
.
dt
The SI unit for current is the coulomb per second or the ampere (A):
1 ampere = 1 A = 1 coulomb per second = 1 C/s.
For the direction of the current we follow the historical convention:
The current arrow is drawn in the direction in which positive carriers
would move, even if the actual carriers are not positive.
Resistance
If we apply the same potential difference between the ends of similar rods of copper
and of glass, we will observe very different currents. This is due to the characteristic
of the conductor that we call the resistance. We determine the resistance of a
conductor between any two points by applying a potential difference V between those
points and measuring the current I that results. The resistance R is then
R
V
I
The SI unit for resistance is volt per ampere. This combination occurs so often that we
give it the name ohm (symbol ), given as
1 ohm = 1  = 1 volt per ampere = 1 V/A
A conductor whose function in the circuit is to provide a specific resistance is called a
resistor.
Another way of calculating the resistance is given by the formula
R
L
A
Here  is called the resistivity of the material of the conductor and its SI unit is the
.m, and its values range from 1.62 x 10-8 for silver (good conductor) to ~101 6 for
fused quartz (insulator). L is the length (in m) of the conducting wire and A is its
cross-sectional area (in m2).
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Chapter 10: Current and Circuits
Ohm’s law
As we now know a resistor is a conductor with a specified resistance. This means that
it has the same resistance even if the magnitude and direction of an applied potential
difference change. We can say that
Ohm’s law is an assertion that the current flowing through a device is
directly proportional to the potential difference applied to the device.
But some devices do not obey that rule (like a pn junction diode), whereas others do
(like a resistor). We can state then that
A conducting device obeys Ohm’s law when its resistance is
independent of the magnitude and polarity of the applied potential
difference.
Power in electric circuits
Consider a simple circuit consisting of a battery (symbol
) that is connected
by wires to an unspecified conducting device. The wires have negligible resistance.
The battery maintains a potential difference of magnitude V across its terminals, and
thus across the device. Since there is a conducting path between the two terminals of
the battery, a steady current I flows around the circuit. The amount of charge dq that
moves between those terminals in time interval dt is equal to Idt. Thus the variation in
the electric potential energy is
dU = dqV = IdtV .
The power associated with this energy variation is given by dU/dt, which can be
written as
P = IV
The unit of the power is the volt-ampere, which can be written as
æ J öæ C ö
J
1 V.A = ç1 ÷ç1 ÷ = 1 = 1W.
è C øè s ø
s
For a resistor R that obeys Ohm’s Law, we can write the power as
P= I R
2
or
V2
P=
.
R
The last two equations apply only in regards to a resistor. The original equation,
P = IV , could be applied to all kinds of electric energy transfers.
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Kirchhoff’s loop rule
Let us consider a closed circuit. We start at any point in the circuit and mentally
proceed around the circuit in either direction, adding algebraically the potential
differences that we encounter. When we arrive at our starting point we must have
returned to our starting potential. So we can state Kirchhoff’s loop rule as
The sum of the changes in potential encountered in a complete traversal
of any circuit must be zero.
A rule that helps finding the potential differences in more complex circuits is the
resistance rule and it states that
If you imagine passing through a resistance in the direction of the
current, the change in potential is -IR; in the opposite direction it is
+IR.
Resistances in series
Let us consider n of resistances connected in series to an ideal battery (Figure 10.2).
R1
R2
R3
V
Figure 10.2: Resistances in series.
We state then that
Connected resistances are said to be in series when a potential
difference that is applied across the combination is the sum of the
resulting potential differences across the individual resistances.
This is equivalent to saying that the current through the each resistor is equal. The
equivalent resistance Req is then calculated as the sum of all the resistances in the
circuit. That is
n
Req = R1 + R2 + × × × + Rn = å R j .
j=1
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Chapter 10: Current and Circuits
Kirchhoff’s junction rule
A point in the circuit is considered as a junction if it is the connection point of more
than two branches in the circuit.
b
a
f
c
e
d
Figure 10.3: Junctions in a circuit.
Figure 10.3 shows that the points b and e are junctions, whereas the other points are
not. If charge travels from a towards b, at b the incoming charge will be divided
between the outgoing paths towards c and e, and because of the conservation of
charge, no charge is lost. Now if we consider the currents instead of the charge we
can state Kirchhoff’s junction rule as
The sum of the currents approaching any junction must be equal to the
sum of the currents leaving that junction.
Resistances in parallel
Let us consider n resistances connected in parallel to an ideal battery (Figure 10.4).
V
R1
R2
R3
Figure 10.4: Resistances in parallel.
We can say that
Connected resistances are said to be in parallel when a potential
difference that is applied across the combination is the same as the
resulting potential difference across the individual resistances.
The equivalent resistance Req is then given by
n
1
1
1
1
1
= +
+××× +
=å .
Req R1 R2
Rn j=1 R j
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Measuring instruments
The ammeter
The ammeter is an instrument used to measure the current in a circuit. To measure the
current in a device, the ammeter should be inserted in series with the device in the
circuit. It is essential that the resistance of the ammeter is much smaller than any other
resistance in the circuit. Otherwise, the very presence of the meter will change the
current to be measured.
The voltmeter
The voltmeter is an instrument used to measure the potential differences (or voltage
drops). To measure the potential difference across a device in a circuit, the voltmeter
should be inserted in parallel to that device. In the case of a voltmeter, it is essential
that its resistance is much larger than any other resistance in the circuit, in order not to
change the potential difference being measured.
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Chapter 10: Current and Circuits
Questions and Problems
1. What is electric current?
2. Explain Ohm’s law, and Kirchhoff’s two laws.
3. Draw a simple circuit diagram with a battery and two resistors in:
a. series,
b. parallel.
4. On the following circuit indicate where you would place:
a. an ammeter to measure the current through resistor R1,
b. a voltmeter to measure the potential drop across R2
R1
R2
5. When a potential difference of 200 V is placed across a motor, a current of 30 mA
flows. What is the resistance of the motor?
6. If the potential difference across R1 is 12 V and that across R2 is 5 V, what is the
potential difference across R3? Show all working.
R1
20V
R2
R3
7. Calculate the total resistance between A and B for the following combination of
resistors.
20 
A
17 
35 
B
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8. Calculate the total resistance between the points A and B.
20 
A
B
40 
50 
9. We now connect the ends A and B of the previous circuit to a 9 V battery.
a. What is the current through the circuit?
b. Calculate the current through the 20  resistor.
c. What is the power dissipated by the 50  resistor?
10. Find the current through each of the following resistors.
100 
12 V
50 
20 
11. What is the current, and in which direction does it travel through the circuit?
50 
20 V
12 V
100 
12. A home heater is rated at 2400 W.
a. If it is used with a standard 240 V power connection what current does it
draw?
b. If the maximum current drawn from that same power point is 10 A can you
run a 360 W electric oven from it?
Student Learning
What things about household electricity do you better understand now?
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Experiments
Experiment 1: The return of Galileo: Dropping Balls
Learning outcomes: Use of instruments to measure physical quantities. Performing
calculations, comparing with theory.
Aim: To measure the time a ball takes to drop from two different heights. If the height
is doubled what will happen to the time?
Equipment: Stopwatch, tape measure and ball.
Procedure: One person drops the ball from as high as possible, while another person
uses the stopwatch to measure the time for the ball to drop. Measure the height and
repeat 5 times from the same height. Write the measured time into the table below.
Now repeat the above experiment dropping the ball from half as high.
Now swap to other members of the group and repeat.
How precise are your measurements, do you think? Why? Are there any differences
between different people’s measurements? Why?
Data
Measuring team _____________________________________
Height
1
2
3
4
5
Av
Unc’ty
Measuring team _____________________________________
Height
1
2
3
4
5
Av
Unc’ty
Analysis
Enter the data into Excel and calculate the average. You do this by entering your
series of measurements into a row or column. Then select a new cell, in which you
want your average to appear, and type: “=AVERAGE(cell1:cellN)” where cell1 and
CellN are the first and last cells in your data series (e.g. A1:A6 or B2:G2). When you
hit <Return> the average will appear. Enter these values into the table above.
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Estimate your uncertainty, by looking at the scatter in the measurements.
When the height is halved, what happens to the time? Is that what you expected?
Rearrange the equation s=uv+½at2 to be in terms of t (for this experiment). Calculate
the expected values for t. How close are you?
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Experiments
Experiment 2: The Physicists toy, a spherical car
Learning outcomes: Measuring physical quantities, time distance, angle
Aim: Measure the acceleration of a ball down an inclined plane.
Equipment: Ball bearing, 1m quarter Al rail (V-shape cross section), stopwatch, retort
stand, tape measure
Procedure: Set up rail at a low angle using a retort stand. Measure rise over run for
the rail and calculate the angle between the table and the rail using trigonometry.
Mark off 5 roughly evenly spaced starting spots along the track and measure the
length of track from the start. Measure the time the ball takes to roll from start to each
point. Repeat each measurement 5 times.
Data
Rail Info: Height _________
Length _________ Angle________________
Data Info________________
Length
T1
T2
T3
T4
T5
TAv
Unc
Analysis
Enter the data into Excel and calculate the average, as in Experiment 1. Now plot the
average values on a graph, by:
1. Choose Insert  Chart.
2. Select XY(Scatter), sub-type Scatter (no lines)<Next>
3. Select the correct cells for the Data <Next> Ignore titles <Finish>. The graph
will appear.
What is the equation of motion for this situation?
What is the relationship between displacement and time?
Fit a line to your data with the following procedure:
4. Select a data point, right click and select “Add Trendline”.
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5. Choose Polynomial, Order 2. Under Options select, “Display Equation” and
“Display R2 value” <OK>.
The R2 value tells you how closely the line fits the data. 1 is perfect, 0 is no
correlation at all. How good is your fit?
From the graph’s equation write down your deduced value for the acceleration. Hint:
You’ll need to differentiate the equation twice.
Draw a force diagram for the ball and use this to calculate a value for the acceleration
due to gravity from your measured acceleration.
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Experiments
Experiment 3: Weights and Springs
Learning outcomes: Measuring distances and weights. Comparison of graphing with
averaging.
Aim: To determine the spring constant for a spring.
Equipment: 1 spring, suspended from a retort stand, 5 weights of different (pref
unlabelled) mass with hooks, electric balance, tape measure.
Procedure:
Hang the spring from the retort stand, and measure the position of its end (relative to
an unchanging point, such as the desk’s surface). Hang five different masses from the
spring and measure the displacement.
Data:
Measurement
Mass (kg):
s (m)
k (kg/s2)
1
2
3
4
5
6
na
Analysis
Enter the data into Excel and calculate the average value for k as per Experiment 1.
kav = __________________
Now use a Graph in Excel to determine k as per Experiment 2.
k = Slope of line (from Excel):____________
R2 factor_________________________
How do the 2 values for k compare? Which method do you think is better? Why?
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Experiment 4: The Hypnotic Pendulum
Learning outcomes
Time measurement, multiple periods to increase precision. Data manipulation in
Excel
Aim: To find the relationship between a pendulum’s length and its period.
Equipment: Pendulum with adjustable length, stopwatch, tape measure
Procedure: Measure the length of the pendulum. Set it swinging and time it for 10
oscillations. Work out the time for one oscillation. Write the data in the table below.
Adjust the length and repeat for four more lengths – varied as widely as possible.
Include uncertainties in the table
Data
1
2
3
4
5
Length (m)
Period (s)
Analysis
Enter data into Excel and display it on a Chart, as per Experiment 2.
Look up the equation for the period of a pendulum. Write it here:
What type of trendline should be used here? Add it to the graph.
How good is your fit? Why?
Rearranging the equation above, calculate a value for the acceleration due to gravity.
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Experiments
Experiment 5: Make Light Work
Learning outcomes: Measurements with different apparatus (electrical). Observing
variations from a rule (Ohms)
Aim: To measure the resistance of light bulb
Equipment: Light bulb, battery/LV power supply, ammeter, voltmeter, lots of leads,
variable resistor.
Procedure:
Build circuit, with powersupply, variable resistor, ammeter and globe in series. Add
in voltmeter across globe. Set power supply to 8V, and vary voltage across the globe
by adjusting the variable resistor. For five different voltages measure the current, and
calculate the resistance of the globe.
Data
Draw a schematic diagram of your circuit.
Data Info _________________________________
1
2
3
4
5
Voltage
Current
Resistance
Analysis
How good are your calculations of resistance? Why?
You may like to plot V vs I on a graph, to see a different representation of your data.
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Experiments
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