Further Education and Training
Grade 12 (FET)
Bright Ideas
Physical Science
Revision Booklet
Physical Science
Grade 12
REVISION BOOKLET
1
2
1
Foreword
5
2
How to use this Revision Booklet
6
2.1
For every topic you need to:
6
2.2
General strategy for problem-solving in Physics
6
3
Key Physics Concepts
7
3.1
Newton’s Laws
7
3.2
Momentum and Impulse
8
3.3
Vertical Projectile Motion
9
3.4
Work Energy and Power. Work-Energy Theorem. Law of Conservation of Mechanical Energy.
10
3.5
Doppler Effect
11
3.6
Electrostatics
12
3.7
Electric Circuits
15
3.8
Electrodynamics
16
3.9
Photoelectric Effect
18
4
Revision Questions- Set 1
19
4.1
Newton’s Laws
19
4.2
Momentum and Impulse
22
4.3
Vertical Projectile Motion
27
4.4
Work Energy Power
29
4.5
Doppler Effect
34
4.6
Electrostatics
34
4.7
Electric Circuits
38
4.8
Electrodynamics
42
5
Revision Questions - Set 2 (master an additional 20%)
46
5.1
Newton’s Laws
46
5.2
Momentum and Impulse
47
5.3
Vertical Projectile Motion
48
5.4
Work Energy Power
50
5.5
Doppler Effect
51
5.6
Electrostatics
51
5.7
Electric Circuits
52
5.8
Electrodynamics
54
5.9
Photoelectric Effect
55
6
Check your answers - Set 1
6.1
Newton’s Laws
56
6.2
Solutions: Momentum and Impulse
58
6.3
Vertical Projectile Motion
58
6.4
Work Energy Power
62
6.5
Electrostatics
65
6.6
Electric Circuits
67
6.7
Electrodynamics
69
7
Check your answers - Set 2
7.1
Newton’s Laws
71
7.2
Momentum and Impulse
76
7.3
Vertical Projectile Motion
79
7.1
Work Energy Power
82
3
7.2
Doppler Effect
82
7.3
Electrostatics
83
7.4
Electric Circuits
85
7.5
Electrodynamics
87
7.6
Photoelectric Effect
88
Chemistry
4
1
How to use this book
89
2
Key subject Concepts
89
3
The Representation Model
89
4
Tips on specific topics
90
5
Study and examination tips
90
6
Message to Grade 12 learners from the writers
90
7
The Mole Concept
91
8
Stoichiometry
91
8.1
Excess and Limiting Reagents
92
9
Practice activities from Grade 10
93
10
Dynamic Chemical Equilibrium Graphs
93
11
Concentration-time graphs
94
12
fertilizers
96
13
ORGANIC CHEMISTRY
101
13.1
BASIC ORGANIC CHEMISTRY
101
13.2
NOMENCLATURE OF ORGANIC COMPOUNDS
102
14
ELECTROCHEMISTRY
124
1 Foreword
Message from the Minister of Basic Education
Message to Grade 12 Learners from the Minister of Basic Education
Matric (Grade12) is perhaps the most important examination you will
prepare for. It is the gateway to your future; it is the means to enter
tertiary institutions; it is your opportunity to pursue the career of your
dreams.
It is not easy to accomplish but it can be done with hard work and
dedication, prioritising your time and effort to ensure that you cover as
much of the curriculum content as possible in order to be well-prepared
for the examinations.
I cannot stress the importance and value of revision in preparing for the
examinations. Once you have covered all the content and topics, you
should start working through the past examination papers, thereafter
check your answers against the memoranda. If your answers are not
correct, go back to the Mind the Gap (MTG) series and work through
the content again. Then retest yourself, and continue with this process
until you get all the answers right.
The Bright Idea… getting exam ready booklet will allow you to do this in a systemic way. It has been
developed to assist you to achieve a minimum of 40% in the examinations – if you work hard and follow the
advice and guidance provided. I also urge you to continue with the next section in this booklet that deals with
an additional 20%, which will ensure you have covered the basics required to achieve at least a 60% pass.
Use this valuable resource which has been developed especially for YOU. Work hard, persevere, work every
day, read and write every day to ensure that you are successful.
I have faith that you can do this. Remember ‘SUCCESS’ depends on the second letter, ‘U’.
Best Wishes
MRS AM MOTSHEKGA, MP
MINISTER OF BASIC EDUCATION
DATE: 24/02/2017
5
2 How to use this Revision Booklet
2.1 For every topic you need to:
•
Learn the content.
•
Study the strategy you need to use to answer questions on that topic.
•
Study the worked examples. First try to answer the worked examples yourself by writing down your
own solution. Then compare your solution with the solution given in the book.
•
Do the exercises without looking at the solutions.
•
Compare your solutions with the given solutions and identify what your mistakes were (if any).
•
Go back to your answer and correct any errors.
•
Do another exercise. You can find more exercises in this book, previous NSC question papers, Mind
the Gap (MTG) study guides, the Grade 12 Siyavula textbook and many other sources.
2.2 General strategy for problem-solving in Physics
(a)
Read the whole question as many times as you need to, in order to understand what is given and what
you need to find out or calculate.
•
If a diagram is given, look at the diagram while reading the question and try to understand the
diagram.
•
(e)
If a diagram is not given, represent the problem in a rough sketch and put all the information given into
the diagram, e.g. if the velocity of a car is given, write it down, e.g. car = 2 m∙s-1 to the left
Write the given information in symbolic form (data) using the correct scientific symbols, e.g. i for initial
velocity, net for net force, etc.
Select /choose the appropriate equation /formula /principal /law /theorem /concept.
Substitute the given data in the selected equation and do the calculations where calculations are
required.
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•
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•
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•
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with
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𝑭𝑭
𝑨𝑨𝑨𝑨 object
⃗𝑭𝑭
⃗direction
velocity
unless
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of the
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and
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onofthe
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first
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on
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tional
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non-zero
directly
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velocity
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unless
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toproportional
the
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⃗directly
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directly
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the 𝑭𝑭
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the
𝑭𝑭𝑭𝑭
=
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⃗
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⃗
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𝒎𝒎
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⃗
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𝑭𝑭
⃗
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𝑭𝑭
=
𝟎𝟎
B
A
on
proportional
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𝟏𝟏
𝟐𝟐
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𝑨𝑨𝑨𝑨
𝑭𝑭
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force
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resultant/net
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𝑩𝑩𝑩𝑩
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𝑭𝑭
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⃗
⃗
A
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acts
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to
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non-zero
proportional
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the
proportional
to𝟏𝟏the𝟐𝟐 square of𝑮𝑮𝒎𝒎
the 𝟐𝟐
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= centres.
𝑭𝑭𝑩𝑩𝑩𝑩
𝑮𝑮𝒎𝒎between
𝒎𝒎𝟐𝟐 𝑭𝑭𝑭𝑭
acts on it. resultant/net
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of ⃗directly
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and
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their
𝑩𝑩𝑩𝑩
𝟏𝟏
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𝟐𝟐
on
it.
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to
of
⃗the mass
= 𝒎𝒎𝒂𝒂
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⃗⃗⃗𝑨𝑨𝑨𝑨
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the
object.
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𝑮𝑮𝒎𝒎𝟏𝟏𝟏𝟏𝒎𝒎
𝒎𝒎𝟐𝟐
𝒏𝒏𝒏𝒏𝒏𝒏
𝒓𝒓=between
𝑮𝑮𝒎𝒎
𝒎𝒎
𝑭𝑭
=⃗resultant/net
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𝑭𝑭
𝑭𝑭𝑭𝑭 =
acts
on
it.
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acts
onit.it.it.
proportional
tothe
of ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = 𝒎𝒎𝒂𝒂
⃗⃗the mass
𝟏𝟏
𝟐𝟐
=
=
⃗proportional
the
object.
force
force
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inversely
centres.
𝟐𝟐 𝑭𝑭𝑭𝑭distance
𝑮𝑮𝒎𝒎
𝒎𝒎
𝑮𝑮𝒎𝒎
⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨
⃗⃗𝑩𝑩𝑩𝑩
𝒓𝒓𝟐𝟐 their𝑭𝑭𝑭𝑭
⃗𝟎𝟎⃗ on
⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 =
the object.
acts
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mass
of
𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗𝟎𝟎proportional
B
acts
to
mass
A
𝟏𝟏𝟏𝟏𝒎𝒎
𝟐𝟐𝟐𝟐
𝑭𝑭𝑭𝑭
=
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=
−𝑭𝑭
=
⃗
⃗
B
⃗
⃗
the
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⃗
⃗
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=
𝟎𝟎
B
A
⃗
⃗
𝟐𝟐
𝑭𝑭
=
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⃗
⃗
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=
⃗
⃗
the
object.
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=
⃗
⃗
⃗
⃗
the
object.
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𝑮𝑮𝒎𝒎
𝒎𝒎
𝒓𝒓
𝑭𝑭
=
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A
𝑭𝑭
=
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the
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𝒓𝒓𝒓𝒓𝟐𝟐𝟐𝟐
𝑭𝑭
𝒎𝒎𝒂𝒂
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𝒏𝒏𝒏𝒏𝒏𝒏 FORcES
⃗⃗⃗⃗B
⃗⃗𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩
⃗⃗𝑩𝑩𝑩𝑩
=
𝒎𝒎𝒂𝒂
𝒏𝒏𝒏𝒏𝒏𝒏=
𝑭𝑭𝑭𝑭 =
⃗𝑭𝑭⃗force
⃗B
⃗𝑩𝑩𝑩𝑩
𝑭𝑭
= −𝑭𝑭
𝒏𝒏𝒏𝒏𝒏𝒏
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⃗A
⃗TyPES
⃗𝒏𝒏𝒏𝒏𝒏𝒏 the
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==
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⃗⃗
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NormaldIFFERENT
Force
weight
Frictional
⃗𝟎𝟎⃗Force
𝑨𝑨𝑨𝑨
𝑭𝑭
⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏OF
object.
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⃗
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Frictional
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𝑭𝑭
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DIFFERENT
TYPES
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o
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o The the
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the object.
across
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Newton's
Commented [U22]:
across the surface.
Impulsesecond
law in
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o the
If the
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(𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ),),aaof
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3.2
Momentum
and Impulse
momentum
Law
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lineal momentum
resultant
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o
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terms of
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Impulse
theNewton's
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object.
momentum
momentum
Impulsesecond law in
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momentum
Law
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Momentum is the
The impulse
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⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
3.2
Momentum
and
Impulse
⃗⃗of
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
= ∆𝒑𝒑
terms
change in
Impulse
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product of an object
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external
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⃗
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momentum
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 momentum
= 𝑚𝑚∆𝑣𝑣⃗
⃗⃗
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mass and its
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𝐹𝐹
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momentum
and
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𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛of
∆𝑡𝑡a The resultant/
Momentum is the
The impulse
An isolated (closed) system in physics is a system
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
7
8
∑ after is the total momentum after collision
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planet:
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pse insert bullet points with
planet:planet:
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o If a force ( F ) applied to a body parallel to the surface does not cause the object lists.
to
= 𝟐𝟐
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Commented
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o
If
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force
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F
)
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with all equivalent items, including in tables and other inserts.
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𝑠𝑠
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Commented [U22]: Pse check/complete.
acrossforce
the surface.
o Theo static
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athe
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Commented [U22]: Pse check
across
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Thefrictional
static frictional
is a maximum
) just before
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If the applied force exceeds (𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ),𝑠𝑠 a𝑚𝑚𝑚𝑚𝑚𝑚
resultant force accelerates the movement of
Commented [U22]: Pse check/complete.
acrossacross
theosurface.
3.2 Momentum and Impulse
If the applied
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Commented
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theosurface.
𝑠𝑠
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the
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o If the
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𝑠𝑠
the object.
the object.
Physical quantities
3.2
Momentum and Impulse
3.2
Momentum and Impulse
Physical
quantities
Newton’s
3.2
Momentum
and
Impulse
Newton'ssecond
Physical
quantities
3.2
Momentum
and Impulse
Impulse-momentum
Newton's
Impulselaw
in terms
of
Law of conservation of lineal momentum
Physical
quantities
Impulsesecond
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and
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theorem
second
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Law of conservation of lineal momentum
momentum
and
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momentum
momentum
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of
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in
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second
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change in
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and
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momentum theorem
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momentum
momentumImpulse
terms
of
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terms of
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theorem
momentum
momentum
Momentum is the
The impulse
of a The resultant/theorem 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
An isolated (closed) system in physics is a system
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = ∆𝒑𝒑
momentum
⃗⃗
momentum
is the
of aresultant/
The resultant/
AnAn
isolated
(closed)
system
in physics
is a system
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Momentum is theMomentum
The
impulseThe
of impulse
The
net
isolated
(closed)
system
in physics
is a system on
⃗⃗ which
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
= ∆𝒑𝒑
product
ofThe
an object
force
isThe
theresultant/
net force 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
acting = ∆𝒑𝒑
on
the resultant/net
external
force is zero.
Momentum
is
the
impulse
of
a
An
isolated
(closed)
in physics
is a system
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛
∆𝑡𝑡
=
𝑚𝑚∆𝑣𝑣
⃗ isolated
ofaThe
an
object
force
theresultant/
net force
acting⃗𝐹𝐹
onsystem
which
the
resultant/net
external
force
is zero.
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isproduct
the
impulse
of a isThe
An
(closed)
system
in physics
isexternal
a system
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
product
of an
object
force
is
the
force
acting
on
an
which
resultant/net
force
is
zero.
⃗
⃗
⃗
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
=
∆𝒑𝒑
⃗
⃗
∆𝒑𝒑
𝐹𝐹
∆𝑡𝑡
=
𝑚𝑚∆𝑣𝑣
⃗
and its
product
offorce
the acting
on an object
is
product
of anmass
object
force
is
net
on𝑛𝑛𝑛𝑛𝑛𝑛
which the resultant/net
zero.is zero.
⃗⃗ OR external
⃗⃗ ∆𝒑𝒑⃗⃗force isforce
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛
= resultant/net
0
𝟎𝟎
⃗𝑛𝑛𝑛𝑛𝑛𝑛object
⃗𝑛𝑛𝑛𝑛𝑛𝑛
andproduct
itsthe is the
product
offorce
the acting
on 𝐹𝐹
an
is
product
of
anmass
object
force
net
the
∆𝑡𝑡to
𝑚𝑚∆𝑣𝑣
⃗=∆𝑡𝑡𝑚𝑚∆𝑣𝑣
𝐹𝐹
⃗⃗ =
⃗𝑛𝑛𝑛𝑛𝑛𝑛
mass
andits
its
velocity.
of the
object
is
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =∆𝒕𝒕0
OR external
= ⃗𝟎𝟎⃗
𝐹𝐹=
∆𝑡𝑡
⃗ ∆𝑡𝑡 on which
⃗⃗
∆𝒑𝒑
velocity.
force
acting
on
equal
to equal
the
rate
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛
mass and
product
of the
on anacting
object
is
∆𝒕𝒕
OR
⃗
⃗
⃗⃗
⃗
⃗
∆𝒑𝒑
⃗
force
on
equal
to
the
rate
mass
and itsvelocity.
product
of
the
on
an
object
is
𝐹𝐹
=
0
OR
=
𝟎𝟎
⃗
= 𝑚𝑚𝑣𝑣⃗𝑓𝑓 − 𝑚𝑚𝑣𝑣
⃗𝑖𝑖 𝐹𝐹⃗ = ∆𝒕𝒕0
⃗⃗ OR =∑
⃗⃗ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏
𝑛𝑛𝑛𝑛𝑛𝑛
𝐹𝐹of
∆𝑡𝑡
𝟎𝟎
force
acting
onand
anto thethe
𝑛𝑛𝑛𝑛𝑛𝑛change
an object
of rate
change
of 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡
𝑛𝑛𝑛𝑛𝑛𝑛⃗𝑖𝑖
Momentum
is a vector
= 𝑚𝑚𝑣𝑣
⃗𝑓𝑓 − 𝑚𝑚𝑣𝑣
velocity.
force
acting
on
equal
rate
∆𝒕𝒕
Momentum
is
a vector
an
object
andto
of
change
of
∑ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏
Momentum
is a vector
velocity.
force
acting
on
equal
the rate
= 𝑚𝑚𝑣𝑣
⃗𝑓𝑓of
−of
𝑚𝑚𝑣𝑣
⃗𝑖𝑖 − 𝑚𝑚𝑣𝑣⃗
object
and
the
time
of
momentum
the
time
the
momentum
quantity.
=
𝑚𝑚𝑣𝑣
⃗
𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎
an object
and andthe
of change
of
∑ when
Momentum
is a vector
𝑖𝑖
⃗𝟎𝟎⃗
⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 =
time
𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏
an object
of the
change
of momentum 𝑓𝑓of
∑ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎
Momentum
is quantity.
a vector
∑𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏
quantity.
⃗
⃗
force
acts
on the the
theobject
objectin
in the
the
⃗⃗ = the
⃗⃗ time
when ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 = ⃗𝟎𝟎⃗
𝒑𝒑
𝒎𝒎𝒗𝒗
acts
on
the
momentum
quantity.
force
acts
onofthe
the object
in the
⃗⃗ =the
⃗⃗force
the
time the
momentum
of
𝒑𝒑
𝒎𝒎𝒗𝒗
quantity.
⃗𝟎𝟎
⃗
⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 =
Then
When
∑
when
−1
⃗⃗
⃗
⃗
∑
object.
direction
of
the
when
𝑭𝑭
𝟎𝟎
Then
SI
units⃗⃗[𝑘𝑘𝑘𝑘force
∙ 𝑚𝑚 ∙ 𝑠𝑠the
] object.
acts
on
the
object
in
the in direction
⃗⃗ = 𝒎𝒎𝒗𝒗
⃗𝒑𝒑
𝒑𝒑
direction
of the of the
−1theonobject.
𝒆𝒆𝒆𝒆𝒆𝒆 =
force
acts
the
the
object
the
⃗⃗⃗ =
𝒎𝒎𝒗𝒗
SI
units
[𝑘𝑘𝑘𝑘
∙
𝑚𝑚
∙
𝑠𝑠
]
⃗⃗
Then
∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0
−1 in momentum
resultant/net
Impulse
is
a
⃗⃗
Then
Change
object.
direction
of
the
SI unitsSI[𝑘𝑘𝑘𝑘
∙
𝑚𝑚
∙
𝑠𝑠
]
∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0
Then
force.
resultant/net
object. is a vector
Impulsedirection
is aresultant/net
of the
Change
momentum
units [𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚
∙ 𝑠𝑠 −1 ] in Impulse
⃗⃗
⃗⃗
∆𝑝𝑝
⃗
=
0
𝑝𝑝
⃗
−
𝑝𝑝
⃗
=
0
force.
vector quantity.
⃗⃗
𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑝𝑝⃗ 𝑇𝑇𝑇𝑇= 0
∆𝑝𝑝⃗ = 𝑝𝑝⃗𝑓𝑓 − 𝑝𝑝⃗Impulse
𝑇𝑇𝑇𝑇
resultant/net
is
a
⃗⃗
Change
in
momentum
𝑓𝑓
𝑝𝑝⃗𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 0
𝑛𝑛𝑛𝑛𝑛𝑛
resultant/net force.
is a vector quantity.
∆𝑝𝑝⃗ = 𝑝𝑝⃗𝑓𝑓 − 𝑝𝑝Impulse
⃗𝑓𝑓
Change in momentum
⃗⃗
⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
∆𝒑𝒑
⃗∆𝒕𝒕
𝑝𝑝⃗𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇 =−0
force.
quantity.
⃗=
⃗ −vector
𝑚𝑚𝑣𝑣⃗ quantity.
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
= ⃗𝑭𝑭force.
⃗⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇
∆𝑝𝑝⃗ = 𝑝𝑝⃗∆𝑝𝑝
𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 0
𝑓𝑓 −
⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
∆𝒑𝒑
⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇
⃗ =𝑝𝑝⃗𝑓𝑓𝑝𝑝⃗𝑓𝑓∆𝑝𝑝−
𝑝𝑝⃗𝑓𝑓𝑚𝑚𝑣𝑣
𝑇𝑇𝑇𝑇
∆𝑝𝑝𝑓𝑓⃗ = 𝑚𝑚𝑣𝑣⃗𝑓𝑓𝑓𝑓 −vector
𝑚𝑚𝑣𝑣⃗𝑓𝑓 quantity.𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
= ⃗𝑭𝑭⃗∆𝒕𝒕 ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗⃗
⃗⃗
𝑭𝑭∆𝒕𝒕
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
∆𝒑𝒑
𝑝𝑝⃗𝑖𝑖𝑖𝑖=+𝑝𝑝⃗𝑝𝑝⃗𝑖𝑖𝑖𝑖 = 𝑝𝑝⃗𝑓𝑓𝑓𝑓 + 𝑝𝑝⃗𝑓𝑓𝑓𝑓
𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇
⃗∆𝒕𝒕units⃗⃗ [𝑁𝑁 ∙ 𝑠𝑠]
SI
𝒏𝒏𝒏𝒏𝒏𝒏 =
∆𝑝𝑝
𝑚𝑚(𝑣𝑣
⃗𝑓𝑓 −𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
𝑣𝑣⃗𝑓𝑓 ) 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
∆𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣
⃗⃗𝑓𝑓 =
− 𝑚𝑚𝑣𝑣
𝑚𝑚𝑣𝑣
= ⃗𝑭𝑭
⃗
⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
∆𝒑𝒑
𝑝𝑝
⃗
+
𝑝𝑝
⃗𝑖𝑖𝑖𝑖 = 𝑝𝑝⃗𝑓𝑓𝑓𝑓 + 𝑝𝑝⃗𝑓𝑓𝑓𝑓
⃗
⃗
∆𝑝𝑝
⃗⃗𝑓𝑓⃗𝑓𝑓 =
− 𝑚𝑚𝑣𝑣
⃗
SI
units
[𝑁𝑁
∙
𝑠𝑠]
=
𝑭𝑭
∆𝒕𝒕
𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇
∆𝒕𝒕
∆𝑝𝑝𝑓𝑓⃗ = 𝑚𝑚(𝑣𝑣⃗𝑓𝑓 − 𝑣𝑣⃗𝑓𝑓 )
𝑖𝑖𝑖𝑖
𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗𝑭𝑭⃗ =
𝑚𝑚𝑝𝑝⃗𝑝𝑝𝐴𝐴⃗𝑓𝑓𝑓𝑓
𝑣𝑣⃗𝑖𝑖𝑖𝑖++𝑝𝑝⃗𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑓𝑓𝑓𝑓 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑓𝑓𝑓𝑓
𝑝𝑝⃗𝑖𝑖𝑖𝑖 + 𝑝𝑝⃗𝑝𝑝𝑖𝑖𝑖𝑖
=+𝑝𝑝⃗𝑝𝑝⃗𝑓𝑓𝑓𝑓 +
𝒏𝒏𝒏𝒏𝒏𝒏
units
∙ 𝑠𝑠] [𝑁𝑁 ∙ 𝑠𝑠]
∆𝒕𝒕
⃗⃗)= SI
⃗⃗ SI[𝑁𝑁units
𝒎𝒎∆𝒗𝒗
∆𝑝𝑝⃗ = 𝑚𝑚(𝑣𝑣
𝑣𝑣⃗𝑓𝑓⃗) −∆𝒑𝒑
⃗
=
∆𝒕𝒕
𝑚𝑚
⃗𝑖𝑖𝑖𝑖 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑓𝑓𝑓𝑓 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑓𝑓𝑓𝑓
∆𝑝𝑝⃗⃗𝑓𝑓=−𝑚𝑚(𝑣𝑣
𝑣𝑣
⃗
𝑖𝑖𝑖𝑖
𝑖𝑖𝑖𝑖
𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓
⃗⃗ = 𝒎𝒎∆𝒗𝒗
⃗⃗
∆𝒑𝒑
𝐴𝐴 𝑣𝑣
𝑓𝑓
𝑓𝑓
of
states: The
𝑚𝑚𝐴𝐴 𝑣𝑣The
⃗𝑖𝑖𝑖𝑖 +
𝑚𝑚⃗The
⃗𝑖𝑖𝑖𝑖conservation
= 𝑚𝑚
+ 𝑚𝑚
⃗momentum
⃗⃗
SI⃗⃗units of ∆𝒑𝒑
⃗⃗ = 𝒎𝒎∆𝒗𝒗
∆𝒑𝒑
𝐵𝐵 𝑣𝑣
𝑓𝑓𝑓𝑓
𝐵𝐵 𝑣𝑣
𝑓𝑓𝑓𝑓
𝑚𝑚law
𝑚𝑚
⃗𝐴𝐴𝑖𝑖𝑖𝑖𝑣𝑣⃗conservation
= 𝑚𝑚
⃗of
⃗momentum
of
states: The
⃗⃗
⃗⃗ = 𝒎𝒎∆𝒗𝒗
SI⃗⃗−1
units of ∆𝒑𝒑
∆𝒑𝒑
𝐴𝐴 𝑣𝑣
𝑖𝑖𝑖𝑖 +law
𝐵𝐵 𝑣𝑣
𝐴𝐴 𝑣𝑣
𝑓𝑓𝑓𝑓 + 𝑚𝑚of
𝐵𝐵 𝑣𝑣
𝑓𝑓𝑓𝑓
total linearofmomentum
of
an isolated
(closed)
∙ 𝑚𝑚
∙
𝑠𝑠
]
The
law
of
conservation
momentum
states:
The
⃗⃗[𝑘𝑘𝑘𝑘
∆𝒑𝒑
SI unitsSIofunits
−1
total linearofmomentum
an isolated
(closed)
The law of conservation
momentum of
states:
The
⃗
⃗
of
∆𝒑𝒑
[𝑘𝑘𝑘𝑘
∙
𝑚𝑚
∙
𝑠𝑠
]
system remains
constant(closed)
(is conserved).
total linear
momentum
of an
isolated
[𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚[𝑘𝑘𝑘𝑘
∙ 𝑠𝑠 −1
] ∙ 𝑠𝑠 −1 ]
system
remains
constant(closed)
(is conserved).
total linear
momentum
of an isolated
∙ 𝑚𝑚
∑
∑
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
⃗(is
⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
𝒑𝒑
= 𝒑𝒑
system
remains
constant
∑ 𝒑𝒑
∑ 𝒑𝒑
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
⃗(is
⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
system
remains
constant
conserved).
=conserved).
∑ 𝒑𝒑
∑∑
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
⃗⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
𝑝𝑝⃗=before
is the total momentum before collision
𝒑𝒑
∑ 𝒑𝒑
∑
⃗⃗=
⃗
⃗
𝒑𝒑
∑𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
𝑝𝑝
⃗ beforelaw
is the
total momentum
collisionstates: The total
𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
The
of
conservation
ofbefore
momentum
∑ 𝑝𝑝⃗ before
the total
before before
collision
∑ is
𝑝𝑝⃗ before
is themomentum
total
momentum
collision
linear
momentum
of
an isolated (closed)
system
7
7
remains constant (is conserved).
7
7
VERTICAL PROJECTILE MOTION
Free
Projectile
Projectile
fall
gravitational force.
acceleration due to
constant
0
free fall.
is
(constant
acceleration)
(speed decreases)
Vertically upwards
00
(speed decreases)
(speed decreases)
increases)
(speed
decreases)
(speed
decreases)
(speed decreases)
experiences
(constant
(constant
(constant
(constant object
(constant
acceleration)
acceleration)
acceleration)
acceleration)
negligible
air
acceleration)
(constant
acceleration)
(constant
(constant (constant (constant
acceleration)
acceleration)
acceleration)
acceleration)
resistance 0 0 and
𝑡𝑡
0 𝑡𝑡 00 0
𝑥𝑥⃗
0
𝑡𝑡
𝑡𝑡0
𝑡𝑡 0
00 𝑡𝑡
𝑡𝑡𝑡𝑡
0
𝑡𝑡
0 𝑡𝑡
𝑡𝑡 0 0
0
𝑡𝑡
𝑣𝑣⃗
0
𝑡𝑡
𝑡𝑡0
𝑡𝑡 0
00𝑡𝑡
𝑡𝑡 𝑡𝑡
0
𝑡𝑡 0
𝑡𝑡
0𝑡𝑡
𝑡𝑡
0
𝑔𝑔⃗ 9
𝑡𝑡
𝑡𝑡0
𝑡𝑡 0
9
𝑡𝑡
𝑡𝑡
𝑡𝑡
𝑡𝑡
99
9
𝑡𝑡
9
9
9
9
9
Vertical
Motion
Vertical
3.3 Projectile
Projectile
Vertical
3.3VERTIcaL
Motion
Projectile
VerticalPROjEcTILE
Motion
Projectile Motion
mOTION
VERTIcaL
VERTIcaL
PROjEcTILE
PROjEcTILE
mOTION
mOTION
VERTIcaL
PROjEcTILE
mOTION
Equations
of motion mOTION
VERTIcaL
PROjEcTILE
Projectile
Equations
of
motion
Projectile
Projectile
Equations
Equations
of
motion
of motion
Projectile
Equations
of
motion
Equations
of motion
VERTIcaL
PROjEcTILE
mOTION
VERTIcaL
VERTIcaL
PROjEcTILE
VERTIcaL
PROjEcTILE
mOTION
PROjEcTILE
mOTION
mOTION
Projectile
is an object upon
Position Projectile
Displacement
Velocity
Acceleration
Position
Velocity
acceleration
[U23]:
Formatting:
pseComment
ensure
all
Projectile
Position
Position
displacement
displacement
Velocity
Velocity
acceleration
acceleration acceleration Commented
Commented
[U23]
Position
displacement
Velocity displacement
acceleration
Commented
[U23]:
Formatting:
pse
ens
Projectile
Projectile
is anisProjectile
andisplacement
Projectileis isan
an
Position
Velocity
is
an
Projectile
Equations
of
motion
Projectile
Projectile
Projectile
Equations
Equations
of
motion
Equations
of
motion
of
motion
appearappear
equivalent,
including
capital
letters
or not.
appearcapital
equivalent,
appear
equi
incl
equivalent,
including
letters
o
1
1
1
1
which the only force acting is the 𝑦𝑦⃗𝑓𝑓 =𝑦𝑦⃗𝑦𝑦𝑓𝑓⃗𝑖𝑖 =+ 𝑦𝑦∆𝑦𝑦
Magnitude
constant:
⃗
𝑣𝑣
=
𝑣𝑣
+
𝑔𝑔∆𝑡𝑡
Magnitude
constant:
𝑦𝑦
⃗
=
𝑦𝑦
⃗
𝑦𝑦
⃗
+
=
∆𝑦𝑦
⃗
𝑦𝑦
⃗
+
∆𝑦𝑦
⃗
𝑣𝑣
=
𝑣𝑣
𝑣𝑣
+
=
𝑔𝑔∆𝑡𝑡
𝑣𝑣
+
𝑔𝑔∆𝑡𝑡
Magnitude
Magnitude
constant:
constant:
⃗
+
∆𝑦𝑦
⃗
𝑣𝑣
=
𝑣𝑣
+
𝑔𝑔∆𝑡𝑡
Magnitude
constant:
𝑓𝑓 ∆𝑦𝑦
2 𝑦𝑦
𝑓𝑓 ∆𝑦𝑦 Position
𝑓𝑓
𝑖𝑖 Position
𝑖𝑖 1
𝑓𝑓
𝑖𝑖
𝑖𝑖
𝑓𝑓𝑖𝑖
2 𝑓𝑓
2 𝑦𝑦
⃗𝑓𝑓 =
+
𝑣𝑣𝑓𝑓acceleration
= 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡acceleration
Magnitude constant:
object
uponupon
which
object
upon
which
which
object
which object
Position
displacement
Commented
Comm
Projectile
is
an
Position
displacement
Velocity
Velocity
acceleration
acceleration
Commented [U23]:
[U23]
=
𝑣𝑣which
Projectile
Projectile
is upon
an
Projectile
is an
is𝑖𝑖∆𝑦𝑦
∆𝑦𝑦
= ⃗𝑣𝑣𝑖𝑖 𝑖𝑖∆𝑦𝑦
∆𝑡𝑡
=
+⃗ 𝑣𝑣
∆𝑡𝑡 displacement
+𝑖𝑖2 𝑔𝑔∆𝑡𝑡
=an𝑣𝑣+
+displacement
𝑔𝑔∆𝑡𝑡
𝑖𝑖 ∆𝑡𝑡
𝑖𝑖𝑔𝑔∆𝑡𝑡
𝑖𝑖 ∆𝑡𝑡𝑔𝑔∆𝑡𝑡
object
upon
∆𝑦𝑦Velocity
= 𝑣𝑣𝑖𝑖 ∆𝑡𝑡 + Velocity
𝑔𝑔∆𝑡𝑡 2
2
2
2
2
appear
equivalent,
appear
incl
appear equivalent, incl
force of gravity.
2 𝑔𝑔∆𝑡𝑡
𝑦𝑦𝑦𝑦
⃗⃗𝑓𝑓𝑓𝑓 =
⃗⃗𝑖𝑖𝑖𝑖 +
Magnitude
constant:
= 𝑦𝑦𝑦𝑦
+ ∆𝑦𝑦
∆𝑦𝑦
𝑦𝑦⃗⃗𝑓𝑓⃗ = 𝑦𝑦⃗𝑖𝑖 + ∆𝑦𝑦
𝑦𝑦⃗𝑓𝑓⃗ = 𝑦𝑦⃗𝑖𝑖 + ∆𝑦𝑦⃗2 11 2 2 22 2 1 𝑣𝑣𝑣𝑣𝑓𝑓𝑓𝑓2=
= 𝑣𝑣𝑣𝑣𝑖𝑖1
+
𝑔𝑔∆𝑡𝑡
𝑣𝑣 2 = 𝑣𝑣2𝑖𝑖 + 𝑔𝑔∆𝑡𝑡
𝑣𝑣𝑓𝑓 =−2
Magnitude
𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude
constant:
Magnitude
constant: constant:
𝑖𝑖 +
the only
forceforce
acting
2 2𝑓𝑓
the ∆𝑦𝑦
only
force
acting
force
acting
the only
acting
object
upon
which
object
upon
object
which
upon
object
=
𝑔𝑔⃗2𝑔𝑔∆𝑦𝑦
=𝑣𝑣𝑖𝑖𝑔𝑔
9,8
∙ 𝑠𝑠𝑚𝑚2∙ 𝑖𝑖𝑠𝑠 −22𝑔𝑔⃗ = 9,8𝑔𝑔⃗𝑚𝑚=∙ 9,8
∆𝑦𝑦
= 𝑣𝑣𝑣𝑣𝑖𝑖𝑖𝑖∆𝑡𝑡
∆𝑡𝑡𝑣𝑣+
∆𝑦𝑦
+
=
𝑔𝑔∆𝑡𝑡
𝑣𝑣𝑖𝑖+
∆𝑡𝑡𝑣𝑣2𝑔𝑔∆𝑦𝑦
∆𝑦𝑦
+ =2𝑔𝑔∆𝑦𝑦
𝑔𝑔∆𝑡𝑡
𝑣𝑣𝑖𝑖 ∆𝑡𝑡
𝑠𝑠 −2𝑚𝑚 ∙ 𝑠𝑠 −2𝑔𝑔⃗ = 9,8 𝑚𝑚 ∙ 𝑠𝑠 −2
⃗+
=𝑚𝑚
9,8
⃗ the
−
Displacement,
the
only
force
acting 2 ∆𝑦𝑦⃗ −∆𝑦𝑦
∆𝑦𝑦⃗which
−upon
Displacement,
∆𝑦𝑦
⃗ − which
Displacement,
∆𝑦𝑦
⃗ only
− Displacement,
Displacement,
=
𝑣𝑣
𝑣𝑣𝑓𝑓2+=2𝑣𝑣𝑔𝑔∆𝑡𝑡
2𝑔𝑔∆𝑦𝑦
𝑣𝑣𝑔𝑔∆𝑡𝑡
which
𝑓𝑓 2
𝑖𝑖𝑣𝑣𝑓𝑓+=
𝑓𝑓 𝑖𝑖2=
𝑖𝑖 +
Displacement,
𝑣𝑣
=
𝑣𝑣
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centre of the Earth).
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3.3 Vertical Projectile Motion
3.3
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VERT
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9
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𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇)
SI
unit:
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unit: joule
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In
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𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇)
Commented
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SI unit:
[J]
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𝟐𝟐𝑬𝑬
∆𝒕𝒕𝒏𝒏𝒏𝒏𝒏𝒏
In
isolated
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gravitational
plus
the𝑲𝑲kinetic
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𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇)
∆𝒕𝒕
In
isolated
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Comment
Ifbody
body
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relative
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In
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Commente
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unit:
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Commented
[G30]:
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plus
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Commente
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the point
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In
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theofdisplacement
conservation
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Kinetic
energy
Power
force multiplied
by𝒎𝒎𝒎𝒎𝒎𝒎 to some
SI
unit:
[J]
𝑬𝑬
= joule
𝒎𝒎𝒗𝒗relative
If the body
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gravitational
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𝑾𝑾𝒏𝒏𝒏𝒏𝒏𝒏 conservative
rectilinear
𝑾𝑾
𝑲𝑲
If
the
body
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∆𝒕𝒕
In
anconservative
isolated
system, Commented
only
conservative
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
only
SI
unit:
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[J]𝑬𝑬relative
possesses
OR
reference
point.
𝒏𝒏𝒏𝒏𝒏𝒏
𝑬𝑬𝑬𝑬𝑴𝑴𝑲𝑲to
=
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+body
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only
𝒑𝒑 =relative
=
𝜟𝜟𝑬𝑬
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
𝒏𝒏𝒏𝒏𝒏𝒏 forces
energy
energy
force
multiplied
kinetic
energy
of
position
in
a
SI unit:
joule
potential
𝑬𝑬𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊)
= 𝑬𝑬𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇)
reference
point.
are
acting.
=some
+Ifsome
= body
𝜟𝜟𝑬𝑬
the
Comment
𝑲𝑲
𝒑𝒑 energy.
𝑲𝑲𝑾𝑾
If the
theapplication
displacement
𝟐𝟐energy.
forces
are acting.
relative
some
𝑾𝑾𝒏𝒏𝒏𝒏𝒏𝒏 𝑾𝑾
point of
𝑴𝑴to
𝒑𝒑 energy.
the
displacement
𝑲𝑲 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
Commente
only
conservative
application
of
the
product
ofreference
the by
the of
point
of displacement
energy
of theof
point
𝑷𝑷
= total
If −
the
body
possesses
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displacement
relative
topossesses
some
reference
point.
energy.
𝑬𝑬𝑴𝑴[J]
=plus
𝑬𝑬𝑴𝑴
+
𝑬𝑬
𝑾𝑾
=
𝜟𝜟𝑬𝑬
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
point.
uniform
motion
𝑬𝑬
=
𝑬𝑬
+
𝑬𝑬
=
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(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
possesses
only
conservative
forces
are
acting.
reference
point.
=
𝑬𝑬
+
𝑬𝑬
𝑾𝑾
=
𝑬𝑬
−
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=
𝜟𝜟𝑬𝑬
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
𝑲𝑲
𝒑𝒑
𝒏𝒏𝒏𝒏𝒏𝒏
forces
are
acting.
𝑲𝑲
SI
unit:
joule
[J]
If
the
body
posThe
mechanical
rectilinear
𝑴𝑴
𝑲𝑲
𝒑𝒑
𝑬𝑬
=
𝒎𝒎𝒎𝒎𝒎𝒎
forces
are
acting.
𝑲𝑲
SI
unit:
joule
[J]
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝑲𝑲
𝒑𝒑
𝑲𝑲
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲
𝑲𝑲𝑲𝑲
the
object.
rectilinear
gravitational
field
the
kinetic
SI
unit:
joule
In
an
isolated
sys𝑬𝑬
=
𝒎𝒎𝒎𝒎𝒎𝒎
OR
the
displacement
− 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝒑𝒑
of
point
of
∆𝒕𝒕
an acting.
isolated
system,
SI unit:
joule [J] 𝑬𝑬𝑴𝑴 = 𝑬𝑬Total
reference
point.
= 𝜟𝜟𝑬𝑬𝑲𝑲OR
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
ofapplication
𝒑𝒑Kinetic
of
themultiplied
point
forces
are
of the
force
by
the
𝑲𝑲 +=𝑬𝑬𝑬𝑬
𝒑𝒑 +possesses
force
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thethe point
by
ofthe
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Commente
energy
Power
is the
work
done
by In
Total
mechanical
Gravitational
Work
doneof𝑬𝑬by
a 𝒎𝒎𝒎𝒎𝒎𝒎
possesses
application
of theof
rectilinear
reference
point.
𝑬𝑬
𝑬𝑬+𝒑𝒑 𝑊𝑊rectilinear
𝑬𝑬
=
𝒎𝒎𝒎𝒎𝒎𝒎
=The
𝜟𝜟𝑬𝑬
the
point
of
rectilinear
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
(v
constant),
SI unit:
[J]
rectilinear
forces
are acting.
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
OR
𝑬𝑬[J]
𝑴𝑴 motion
𝑲𝑲motion
𝒑𝒑relative
joule
𝑲𝑲 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
of
an
object
uniform
𝒑𝒑 =
SIjoule
unit:
[J]
OR
𝑊𝑊
=
𝑊𝑊
𝒑𝒑= =
− 𝑬𝑬−
sesses
If𝒏𝒏𝒏𝒏𝒏𝒏
the𝑾𝑾
body
uniform
to𝒎𝒎𝒎𝒎𝒎𝒎
some
SIjoule
unit:SI
joule
[J]
energy.
The
total mechanical
𝑾𝑾
𝑾𝑾
=energy
𝑬𝑬𝑲𝑲𝑲𝑲=−𝑬𝑬OR
𝑬𝑬
of the
of
𝑛𝑛𝑛𝑛𝑛𝑛joule
𝑐𝑐rectilinear
𝑛𝑛𝑛𝑛
SI unit:
[J]unit:𝑬𝑬
tem,
only
conseronly
conservative
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲
𝑲𝑲𝑲𝑲𝒏𝒏𝒏𝒏𝒏𝒏𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
The
total
mechanical
=potential
𝒎𝒎𝒎𝒎𝒎𝒎
application
of point
the of
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
ofby
the
𝑲𝑲𝑲𝑲
SI
unit:
joule
[J]
OR
application
the
𝒑𝒑
SI
unit:
joule
[J]
cosine
ofapplication
the
cosine
of theforce
angle
the
displacement
mechanical
the
resultant
(net)
energy
in
an
isolated
force
by
the
is
the
energy
rate
at
which
rectilinear
uniform
motion
constant
force
is
𝑬𝑬
=
𝒎𝒎𝒎𝒎𝒎𝒎
the
motion
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝑾𝑾
=
𝑬𝑬
−
𝑬𝑬
then:
application of the
SI
unit:
joule
[J]
SI unit:
joule
[J]uniform
uniform
motion
The
total
mechanical
OR
𝑾𝑾
=
𝑬𝑬
−
𝑬𝑬
𝒑𝒑 joule
SI
unit:
joule
[J]
The
total
mechanical
𝑾𝑾
=
𝑬𝑬
−
𝑬𝑬
(v
=
constant),
remains
constant
as
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲
𝑲𝑲𝑲𝑲
SI
unit:
[J]
The
total
mechanical
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲
𝑲𝑲𝑲𝑲
Using
the
workpossesses
(v
=
constant),
reference
point.
an object
𝑬𝑬𝑴𝑴 = 𝑬𝑬𝑲𝑲uniform
+ 𝑬𝑬
𝒏𝒏𝒏𝒏𝒏𝒏𝑲𝑲energy
𝑲𝑲𝑲𝑲
𝜟𝜟𝑬𝑬
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
motion
(vmotion
=𝑊𝑊𝑛𝑛𝑛𝑛
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛
=
+
𝑊𝑊=
uniform
motion
forces
areforces
acting.are
energy
of
anof
object
application
of the
=force
𝑬𝑬
𝑬𝑬𝑲𝑲𝑲𝑲
𝑊𝑊
=
𝑊𝑊
+
force by
the
vative
SI unit: joule
[J]=(v
The
mechanical
𝑐𝑐𝑾𝑾
𝑛𝑛𝑛𝑛
force
the
force
the
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲 −
𝑲𝑲𝑲𝑲
𝑛𝑛𝑛𝑛𝑛𝑛
angle
formed
bycosine
energy
is𝑊𝑊equal
acting
on
antotal
system
remains
of
theby
point
is𝑭𝑭𝒗𝒗
the
an joule
object [J]
has
work
is𝑐𝑐object
done
or
formed
by by
the
linecosine
uniform
(v𝒑𝒑 ==constant),
= constant),
defined
asofthe SI unit:
𝑾𝑾𝑊𝑊
𝑬𝑬
𝑬𝑬𝑲𝑲𝑲𝑲
𝑷𝑷𝒂𝒂𝒂𝒂𝒂𝒂
ofofthe of the
SIenergy
joule
[J] SIenergy
(v
=Using
constant),
The
total
mechanical
energy
of
object
force by
the
energy
of
an
object
𝑊𝑊
=
+
𝑊𝑊
the
moves,
then:joule
𝒏𝒏𝒏𝒏𝒏𝒏
energy
of
an
object
𝑊𝑊
+the
𝑊𝑊
Commented
[G27]:
Space
corrected
theorem:
rectilinear
𝑊𝑊
==
𝑊𝑊
+−
𝑊𝑊
𝑬𝑬unit:
𝑛𝑛𝑛𝑛𝑛𝑛−remains
𝑐𝑐 remains
𝑛𝑛𝑛𝑛
constant
asan
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐 then:
𝑛𝑛𝑛𝑛
(v
constant),
the
workunit:
[J]=
OR
constant
as
𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐𝑲𝑲𝑲𝑲
𝑛𝑛𝑛𝑛
𝒑𝒑 = 𝒎𝒎𝒎𝒎𝒎𝒎𝒂𝒂𝒂𝒂𝒂𝒂then:
constant),
force
by
the
Using
workenergy
of
an
object
cosine
of
the
𝑊𝑊
=
𝑊𝑊
+
𝑊𝑊
acting.
cosine
of
the
cosine
of
the
object
is
equal
to
constant
(is
energy
an
object
due
its
motion.
to
the
sum
of
the
energy
is
the
line
of
action
𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐
𝑛𝑛𝑛𝑛
(v = provided
constant),
of the
then:
formed
by application
then:
action of theangle
forceangle
SI
unit:
watt
[W]𝑷𝑷 𝑊𝑊 =
energy
of
an
then:
remains
constant
asobject
formed
by cosine
remains
constant
as
𝑊𝑊
=𝑬𝑬work𝑊𝑊
+ 𝑬𝑬
𝑊𝑊
ofscalar
the quantity
Using
the
workthatUsing
the
net
𝑷𝑷
==𝑭𝑭𝒗𝒗
remains
constant
as[G27]:[G27]:
Using
the
workuniform
motion
+
𝑊𝑊
∆𝐸𝐸
𝑭𝑭𝒗𝒗
the
𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐moves,
𝑛𝑛𝑛𝑛
the
object
moves,
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
𝑾𝑾
=
−
Commented
Space correc
then:
energy
theorem:
SI
unit:
joule
[J]
The
total
mechanical
the
object
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
Commented
Space corrected
energy
theorem:
𝒏𝒏𝒏𝒏𝒏𝒏
𝑲𝑲𝑲𝑲
𝑲𝑲𝑲𝑲
remains
constant
as
cosine
of
the
angle formed
by
Using
the
work𝟏𝟏
angle
formed
byofline
the
change
inthe
theobject
conserved).
Comme
has due to𝑷𝑷𝒂𝒂𝒂𝒂𝒂𝒂
its = 𝑭𝑭𝒗𝒗gravitational
angle
formed
expended.
then:
𝑷𝑷
=
𝑭𝑭𝒗𝒗
ofand
the the
force
and
force
by
thetheby
the
of action
remains
constant
as corrected
moves,
𝑷𝑷
=
𝑭𝑭𝒗𝒗
𝟐𝟐
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
equal
to
direction
of
the
object
moves,
Commented
[G27
the line
action
Using
the
workenergy
theorem:
𝒂𝒂𝒂𝒂𝒂𝒂
SI
unit:
watt
[W]
work
done
by
the
object
moves,
Commented
[G27]:
Space
angle formed
by
energy
theorem:
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
(v
=
constant),
Commente
SI
unit:
watt
[W]
But:
𝑊𝑊
=
−∆𝐸𝐸
energy
theorem:
provided
that
the
net
𝑬𝑬
=
𝒎𝒎𝒗𝒗
𝑷𝑷
=
𝑭𝑭𝒗𝒗
energy
of
an
object
+
𝑊𝑊
=
∆𝐸𝐸
provided
that
the
net
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑝𝑝+
𝑊𝑊
=
𝑊𝑊
+
𝑊𝑊
𝑲𝑲
𝑊𝑊
𝑊𝑊
=
∆𝐸𝐸
Using
the
workenerthe
object
moves,
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
Commented
[G27
angle
formed
by
energy
theorem:
𝑾𝑾
𝑛𝑛𝑛𝑛𝑛𝑛
𝑐𝑐
𝑛𝑛𝑛𝑛
the
line
of
action
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
OR
the
line
of
action
kinetic
energy
of
position
in
a
potential
energy
𝑬𝑬
=
𝑬𝑬
SI
unit:
watt
[W]
the
line
of
action
𝑷𝑷
=
𝑭𝑭𝒗𝒗
SI
unit:
watt
[W]
𝟐𝟐
the
direction
of
cosine
of
the
𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊)
𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇)
SI
unit:
watt
[W]
the
object
moves,
provided
that
the
net
Comme
of
the
force
and
𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
Comment
provided
that
the
net
product
of
the
energy
theorem:
external
non𝑊𝑊
+
𝑊𝑊
=
∆𝐸𝐸
of
the
force
and
provided
that
the
net
the displacement.
𝑊𝑊
+SI
𝑊𝑊
=then:
∆𝐸𝐸
the line of action
𝑊𝑊
+
𝑊𝑊
=
∆𝐸𝐸
work
done
by
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
𝑷𝑷
=
−∆𝐸𝐸
+
𝑊𝑊
=
∆𝐸𝐸
unit:
watt
[W]
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
remains
constant
as
But:
𝑊𝑊
=
−∆𝐸𝐸
work
done
by
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
Using
the
workBut:
𝑊𝑊
=
−∆𝐸𝐸
provided
that
the
net
𝑝𝑝
𝑛𝑛𝑛𝑛
𝐾𝐾
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑊𝑊
𝑝𝑝the
gy
theorem:
+
𝑊𝑊
=
∆𝐸𝐸
the
line
of
action
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑝𝑝
SI
unit:
joule
[J]
of
the
force
and
object.
gravitational
field
plus
the
kinetic
of
the
force
and
SI
unit:
watt
[W]
𝑐𝑐
𝑛𝑛𝑛𝑛
𝐾𝐾
of
the
force
and
∆𝒕𝒕
In
an
isolated
system,
the displacement.
provided
work work
done
by that
formed
by by
work
done
the direction
𝑊𝑊𝑐𝑐 +
𝑊𝑊
conservative
forces
But:
𝑊𝑊
=by
−∆𝐸𝐸
done
by the net
force
multiplied
But: 𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =𝑷𝑷−∆𝐸𝐸
−∆𝐸𝐸
the direction
of of of
𝑭𝑭𝒗𝒗
Comme
external
𝑊𝑊
=𝑝𝑝∆𝐸𝐸
−∆𝐸𝐸
the angle
force
and
𝑛𝑛𝑛𝑛 =
𝐾𝐾non𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑝𝑝 +
the
object
moves,
external
non𝒂𝒂𝒂𝒂𝒂𝒂
𝒂𝒂𝒂𝒂𝒂𝒂
𝑝𝑝
Commente
energy
theorem:
work
done
by
−∆𝐸𝐸
𝑊𝑊=
= 𝑊𝑊
∆𝐸𝐸
But:
𝑊𝑊But:
=𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
−∆𝐸𝐸
watt
[W]
𝑝𝑝body
𝑛𝑛𝑛𝑛
𝐾𝐾
of the
force and
the
relative to some Hence:
energy.SI unit:
𝑾𝑾
the direction
of
The
total
mechan𝑝𝑝If+
𝑛𝑛𝑛𝑛
𝐾𝐾= ∆𝐸𝐸
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑝𝑝= −∆𝐸𝐸
the 𝒄𝒄𝒄𝒄𝒄𝒄
direction
of
only
conservative
direction
of
𝒏𝒏𝒏𝒏𝒏𝒏
work
done
by
external
non𝑾𝑾 = 𝑭𝑭∆𝒙𝒙
𝜽𝜽 displacement.
the
line
external
nonis
zero.
But:
𝑊𝑊
the displacement.
external
nonthe
displacement
SI
unit:
watt
[W]
−∆𝐸𝐸
+
𝑊𝑊
=
∆𝐸𝐸
the
conservative
forces
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑝𝑝
−∆𝐸𝐸
+
𝑊𝑊
=
∆𝐸𝐸
the direction
ofof action
provided
that
the
net
conservative
forces
−∆𝐸𝐸
+
𝑊𝑊
=
∆𝐸𝐸
𝑝𝑝
𝑛𝑛𝑛𝑛
𝐾𝐾
𝑾𝑾
=
∆𝑬𝑬
+
∆𝑬𝑬
𝑊𝑊
+𝑝𝑝𝜟𝜟𝑬𝑬
𝑊𝑊𝑛𝑛𝑛𝑛∆𝐸𝐸
=𝑛𝑛𝑛𝑛
∆𝐸𝐸𝐾𝐾 𝐾𝐾externalforces
𝑛𝑛𝑛𝑛 Hence:
𝐾𝐾
non𝒏𝒏𝒏𝒏 = 𝑬𝑬𝑲𝑲 +
𝑲𝑲𝑝𝑝 𝑬𝑬Hence:
reference point.
𝑬𝑬𝑴𝑴
−∆𝐸𝐸
+𝑐𝑐∆𝑬𝑬
𝑊𝑊
the
direction
of
=
(𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘
the
are
acting.
ical
energy
of an
displacement.
𝒑𝒑 𝒑𝒑 possesses
𝑛𝑛𝑛𝑛 =
external
nonforces
the
displacement.
SI unit: the
joule
[J]𝑾𝑾 = 𝑭𝑭∆𝒙𝒙
the
force
and
0 =conservative
∆𝑬𝑬𝑲𝑲𝑝𝑝+
conservative
𝑾𝑾 =𝒄𝒄𝒄𝒄𝒄𝒄
𝑭𝑭∆𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄displacement.
𝜽𝜽of
−∆𝐸𝐸
+𝑲𝑲forces
𝑊𝑊𝐾𝐾=
∆𝐸𝐸conservative
is
zero.
of
the
point
of
𝒑𝒑
𝜽𝜽the
Hence:
work
done
byforces
is
zero.
displacement.
𝑝𝑝𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑛𝑛𝑛𝑛 =
𝐾𝐾
Hence: can
But:
𝑊𝑊
−∆𝐸𝐸
Hence:
conservative
forces
This
equation
𝑾𝑾∆𝑬𝑬
=+∆𝑬𝑬
+
∆𝑬𝑬
𝑝𝑝
rectilinear
𝑾𝑾
=
∆𝑬𝑬
𝑬𝑬
=
𝒎𝒎𝒎𝒎𝒎𝒎
𝒏𝒏𝒏𝒏
𝑲𝑲
𝒑𝒑
−
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
the
displacement.
SI
unit:
joule
[J]
Hence:
OR
𝑾𝑾
=
𝑭𝑭∆𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄
𝜽𝜽
𝒏𝒏𝒏𝒏
𝑲𝑲
𝒑𝒑
𝒑𝒑
conservative
forces
is
zero.
𝑾𝑾
=
𝑭𝑭∆𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄
𝜽𝜽
object
remains
is
zero.
𝑾𝑾
=
𝑭𝑭∆𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄
𝜽𝜽
is
zero.
the
direction
of
SIjoule
unit: joule
0
=
∆𝑬𝑬
+
∆𝑬𝑬
unit:
[J] 𝑾𝑾 [J]
application
of
the
SI unit: jouleSI[J]
Hence:
𝑾𝑾
=
∆𝑬𝑬
+
∆𝑬𝑬
external
non0
=
+
∆𝑬𝑬
𝑾𝑾
=
∆𝑬𝑬
+
∆𝑬𝑬
= 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽
𝑲𝑲
𝒑𝒑
=+𝑊𝑊
∆𝑬𝑬
+∆𝐸𝐸
∆𝑬𝑬
𝒑𝒑
𝒑𝒑 motion
−∆𝐸𝐸
+
𝒏𝒏𝒏𝒏
𝒑𝒑 zero.
This equation
𝑝𝑝𝑲𝑲
𝑛𝑛𝑛𝑛𝑲𝑲
𝑾𝑾
𝑬𝑬=𝒑𝒑𝒑𝒑𝑲𝑲𝑲𝑲𝑲𝑲
−𝐾𝐾is𝑬𝑬
This𝑲𝑲uniform
equation
can
SI unit: joule [J] be used to𝒏𝒏𝒏𝒏solve
𝑾𝑾𝒏𝒏𝒏𝒏
=𝑾𝑾∆𝑬𝑬
∆𝑬𝑬
∆𝑬𝑬
SI unit:
joule
[J] 𝒄𝒄𝒄𝒄𝒄𝒄
𝑾𝑾
= 𝑭𝑭∆𝒙𝒙
𝜽𝜽
zero.
SI unit: joule [J]
𝒏𝒏𝒏𝒏𝒏𝒏
=isThe
∆𝑬𝑬
+∆𝑬𝑬
∆𝑬𝑬𝑲𝑲mechanical
SI
unit:
joule
𝒏𝒏𝒏𝒏
𝑲𝑲
0can
=
+=
∆𝑬𝑬
constant
as
the
0𝑲𝑲=total
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𝒑𝒑+
the
displacement.
𝒑𝒑
𝑾𝑾𝒏𝒏𝒏𝒏𝑲𝑲
=
∆𝑬𝑬
∆𝑬𝑬𝒑𝒑𝑲𝑲𝑲𝑲0
𝒑𝒑
forces
force
by
SI unit:
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[J]the [J]
equation
can
𝑲𝑲 +can
This equation
0 conservative
= energy
∆𝑬𝑬
+ ∆𝑬𝑬
conservation
of used
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constant),
𝒑𝒑an
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used
toThis
solve
Hence:
be
to
solve
of
SI
unit:
joule
[J]
𝑊𝑊
=
𝑊𝑊
+
𝑊𝑊
This equation
can𝑐𝑐
0𝑲𝑲 = ∆𝑬𝑬
+ object
∆𝑬𝑬𝒑𝒑
𝑛𝑛𝑛𝑛𝑛𝑛
𝑛𝑛𝑛𝑛
𝑲𝑲
object
moves,
𝑾𝑾cosine
= 𝑭𝑭∆𝒙𝒙
𝒄𝒄𝒄𝒄𝒄𝒄
𝜽𝜽
is
zero.
of the
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used
to
solve
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problems
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then:
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of
𝑾𝑾
=
∆𝑬𝑬
+
∆𝑬𝑬
conservation of
remains constant as
𝒏𝒏𝒏𝒏 solve
the𝑲𝑲work- 𝒑𝒑
be usedUsing
to
SIangle
unit: joule
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∆𝑬𝑬𝒑𝒑the
provided
that
be used
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ofto solve
formed[J]by
conservation
and is called
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conservation
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𝑷𝑷of
𝑭𝑭𝒗𝒗
energy
equation
can
energy
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the object
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𝒂𝒂𝒂𝒂𝒂𝒂 = problems
𝒂𝒂𝒂𝒂𝒂𝒂 This
Comme
energy
theorem:
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of
conservation
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net
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the line of action
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SI
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[W]
and
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the
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is
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𝑊𝑊
This
equation
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𝑐𝑐 + 𝑊𝑊
𝑛𝑛𝑛𝑛 = ∆𝐸𝐸
𝐾𝐾 be provided that the net
energy
problems
and
is
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and is principle
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But:
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𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =conto
and
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of conservation
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of the 𝑝𝑝
the direction of
ofprinciple
servative
forces is
energy
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conservation
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−∆𝐸𝐸
+
𝑊𝑊
=
∆𝐸𝐸
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𝑛𝑛𝑛𝑛
𝐾𝐾
servation
of
energy
principle
conservation
of of the
conservation
of
conservation
of
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the displacement.
and
is called
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conservative
forces
zero.
conservation
of
Hence:
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energy.
problems
and
isof
energy.
energy.
𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽
principle
of
is zero.
energy.
𝑾𝑾
=
∆𝑬𝑬
+
∆𝑬𝑬
𝒏𝒏𝒏𝒏 principle
𝑲𝑲
energy.
called
the
SI unit: joule [J]
conservation
of 𝒑𝒑
0 = ∆𝑬𝑬𝑲𝑲 + ∆𝑬𝑬𝒑𝒑
10
This equation
can
of energy.
conservation
of
be used to solve
10 10
energy.
10
conservation 10
of
10
10
10
energy problems
and is called the
10
principle of
conservation of
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11
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3.5 3.5Doppler
Effect
Doppler
Effect
Sound
Light
dOPPLER EFFEcT
3.5
Doppler Effect
dOPPLER EFFEcT
Sound
Light
Theory
Equations
Applications
Red shift
Blue shift
dOPPLER
EFFEcT
dOPPLER
EFFEcT
Sound
Light
Equations
applications
Red
shift
Blue
shift
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3.5Theory
Doppler Effect
dOPPLER
EFFEcT
Sound Equations
Light
LightBlue shift
applications
Red shift
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3.5Theory
Doppler Effect Sound
(dEFINITION)
(DEFINITION)
Sound
Light
Theory
Equations
applications
Red
shift
Blue
shift
(dEFINITION)
Theory
Equations
applications
Red shift
Blue shift
pages in text.
dOPPLER
EFFEcT
Theory
applications
Red
shift
shift
± 𝑣𝑣𝐿𝐿
(dEFINITION)
The
Doppler
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ais
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)
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of
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𝑠𝑠
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is(or
aSound
If
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light
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source
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change
in frequency
The
Doppler
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is a Equations
𝐿𝐿
the
rate
change
in
frequency
the
observer
(positive
velocity),
then
the
rate
of
blood
server
(negative
velocity), the observed
away
from
the
observer
towards
the
observer
If
the
light
source
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moving
If
the
source
is
moving
Theory
applications
Red
shift
Blue
shift
𝐿𝐿

To
find
the
rate
𝑣𝑣 ± 𝑣𝑣
of blood flow away from the observer
change
in effect
frequency
𝑓𝑓𝐿𝐿 =𝑓𝑓(𝐿𝐿 = 𝑣𝑣
the is observer
( ±)𝑣𝑣𝑓𝑓𝐿𝐿𝑠𝑠𝑠𝑠 ) 𝑓𝑓𝑠𝑠
The
Doppler
is a Equations
If theshift
light source is moving
Iftowards
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Theory
applications
Red
Blue
shiftsource
(Doppler

To
find
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rate
(dEFINITION)
𝑣𝑣
±
𝑣𝑣
𝑣𝑣
±
𝑣𝑣
of
blood
flow
of
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𝑓𝑓𝐿𝐿 = (
)𝑠𝑠 𝑓𝑓
change
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ed by
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3.5
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3.6 Electrostatics
13
Electric field pattern
Positive like charges
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14
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
It
R
𝜀𝜀 = I R  I
transferre
𝑷𝑷 = 𝑾𝑾Parallel: ∆𝒕𝒕𝑷𝑷 = 𝑷𝑷 =
𝜀𝜀 =electrode
positive
ina the
differencecurrent
across
Δ𝑡𝑡
𝑞𝑞IT IR
directly
𝑊𝑊
proportional
current
current
in
the
inproportional
the
2 I213 II321  I 2
current
in the
∆𝒕𝒕 The
∆𝒕𝒕 rate at
𝑞𝑞21IIR
𝑞𝑞31
W

It
∆𝒕𝒕General:
𝑞𝑞 total R
R1to the
The
Work
done
3 RI 3 1R
1
1
1
1
whole
circuit:
𝑷𝑷
=
= Potential
𝑷𝑷
=
𝑽𝑽𝑽𝑽
d.
in the𝜀𝜀battery.
conductor
is directly
conductor
at theatcurrent
1
1in the
R
+2R21V
+
R
11 3 1 2  11 R3 1 
𝑷𝑷 = 𝑽𝑽𝑽𝑽
𝑷𝑷 =which
𝑽𝑽𝑽𝑽
to
conductor
at in the 𝑷𝑷 = ∆𝒕𝒕
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toconstant
the constant
𝟐𝟐
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T1 1 V1
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defference
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21
32
3
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Vper
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the
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act
𝟐𝟐
1
2
3
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work
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per
unit
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3 unit
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temperature.
defference
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temperature.
work
work
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per
per
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as
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1 temperature.
2
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act
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per
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t
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circuit:
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=
V
+
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charge between
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current
current
dividers.)
dividers.)
I
current
dividers.)
V
Whole
circuit:
circuit:
T
1
2
3
Whole
circuit:
𝑊𝑊
𝑊𝑊
Whole
𝑊𝑊
circuit.
V
R circuit: transferre
=
V
=
V
positive
electrode
I

I

I

I
difference
across
a
𝑉𝑉circuit.
= 𝑉𝑉 = 𝑉𝑉 = (Resistors act as
Δ𝑡𝑡
points
in
a
1
2
3
T
1
2
3
A
real
battery
has
R

potential
𝑉𝑉 = 𝑊𝑊in a
two points
current
dividers.)
𝑄𝑄
Whole
circuit:has
A realAR
battery
real
battery
has hasis directly
=conductor
A real battery
whole circuit:
d.
in
𝑄𝑄 the𝑄𝑄battery.
(Resistors act as
I resistance.
internal
resistance.
𝑉𝑉 = 𝑄𝑄
circuit.
dividers.)
A
real
battery
has
internal
internal
resistance.
resistance.
internal
I
R

R

R

R
𝑾𝑾
𝑊𝑊
𝑄𝑄𝑊𝑊
proportional to the
T
1
2
3
W

It
current
dividers.)
Whole
circuit:
𝑷𝑷 =
𝜀𝜀 =

internal
resistance.
I
=
I
+
I
+
I



current
in
the
𝑉𝑉 =
T
1
2
3
I

I

I
∆𝒕𝒕
𝑞𝑞
Whole circuit:
I 1 has
real battery
1
2
3
𝑄𝑄
IA
1 
1 current
𝑷𝑷 = 𝑽𝑽𝑽𝑽
at
Potential
RI1r RI  rRconductor
act
resistance.
as

VT  V1  V2(Resistors
 V3 Iinternal

r
R

r

𝑽𝑽𝟐𝟐
defference is the
Rr1 R2 R3 A realconstant
dividers.) RER 
battery has internal
(Resistors act as
𝑷𝑷 =
temperature.
work done per unit
𝑹𝑹
IV1 V2  V3
potential dividers.)
charge between
RI r I  I resistance.V
I

T
1
2
3
R
two points in a
(Resistors act as
I
circuit.
current dividers.)
Whole circuit:
𝑊𝑊
𝑉𝑉 =
A real battery has
𝑄𝑄
internal resistance.
3.7 Electric Circuits
•
Strategy to Solve Problems on Electric Circuits
•
Read the problem carefully as many times as needed.
•
If not given, draw a circuit diagram.
•
Write down the data in symbolic form.
•
Indicate the conventional direction of the current from high-potential to low-potential (+ to -).
•
Re-draw the circuit diagram to simplify it if necessary.
•
Identify the type of connection (series/parallel).
Analysing circuits:
1 – The algebraic sum of the changes in potential in a complete transversal of any loop of a circuit must be
zero. (ε= Ir+ Ir)
2. The sum of the currents entering any junction must be equal to the sum of the currents leaving that
junction.
•
i = i1+i2+…+ in
•
Write down the formula/ equation that solves the question.
•
Find the unknowns if needed (multi-concept problems).
•
Do the calculations and write down the final answer.
-
Check your answer. Check if it makes sense, e.g.:
-
Is the unit correct?
Is the numerical value reasonable?
15
16
Elements of the AC generator:
Split-ring (commutator).
Two poles of magnet (permanent magnets)
3.
The thumb shows the direction of the force, the first
finger shows the direction of current and the second
finger shows the direction of the magnetic field. The
thumb, the first finger and the second finger must be
at a right angle to (900) each other.
Fleming’s left hand (motor) rule
Carbon brushes.
2.
The DC generator or dynamo is similar
to the AC generator, but the slip rings are
replaced by split ring (commutator) to get
direct current (current that flows in one
direction).
The DC generator (Dynamo)
The thumb represents the direction of motion of the conductor.
The first finger represents the direction of the magnetic field (north to south).
The second finger represents the direction of the induced or generated current (the direction of
the induced current will be the direction of conventional current - from positive to negative).
Fleming’s right hand (dynamo) rule (for generators)
4. Two poles of magnet (permanent magnets).
3. Carbon brushes.
2. Slip-rings.
1. Armature.
The generator is based on the principle of
“electromagnetic induction”
Elements of the DC-motor:
Armature.
Generators convert mechanical energy into electrical
energy.
Motors convert electrical energy into (rotational)
mechanical energy.
1.
AC and DC Generators
Direct current motor (DC–motor)
Electrical machines
3.8 Electrodynamics
17
18
18
18
18 18
Alternating
current
(AC)
is an
oscillating
electric
current
thatthat
varies
sinusoidally
with
time,
reversing
its its
direction
of flow
Alternating
current
(AC)
is an
oscillating
electric
current
varies
sinusoidally
with
time,
reversing
direction
of flow
aLTERNaTING
cuRRENT
aLTERNaTING
cuRRENT
periodically.
periodically.
Alternating
current
(AC)
is
an
oscillating
electric
current
that
varies
sinusoidally
with
time,
reversing
its
direction
of
flow
Alternating currentGraphs
(AC) is an oscillating electric current that varies sinusoidally
with time, reversing its direction Average
of flow power
rms values
aLTERNaTING
cuRRENT a
periodically.
periodically.
Graphs
a
rms
values
Average
Graphs aelectric current that varies sinusoidallyrms
power
Alternating current (AC) is an oscillating
withvalues
time, reversing its direction ofAverage
flow power
Advantages
of AC
over
Advantages
of AC
overGraphs a
Advantages of AC over DC periodically.
rms
values
Average
power
Graphs
values
Average power
DCDC
Currenta versus
time
Voltage
versus time rms
Current
Voltage
Voltage
Voltage
Advantages
Current versusCurrent
time versus time
Voltage versus
timeversus time
CurrentCurrent
Voltage
AdvantagesofofAC
ACover
over
Graphs
aversus time
rms values
Average power
DC
Current
Voltage
versus
time
Voltage
Current
versus
time cycle
Voltage
versus
time Current
Current
Voltage
The
root-meanThe
rootForFor
one
complete
For
oneone
complete
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎
= 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
 DC
Easy
to be
transformed
For one complete
cycle
•
Easy to be transformed
up
ortransformed
For
one
complete
cycle
The
root-meanThe
root-meanThe
root-meanThe
rootone
complete
cycle
For
complete
= 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟
 (step
Easy
to
Advantages
ofbe
AC
over
square
current
mean-square
cycle
(step
up
or
step
down
square
current
mean-square
cycle
step down usinga DC
transformer).
square
current
square
voltage
(step
up
or
step
down
Current
versus
time
Voltage
versus
time
Current
Voltage
The
root-meanThe
rootFor
one
complete
cycle
For
one
complete
𝑃𝑃
=
𝐼𝐼
𝑉𝑉
totobe
transformed
𝑎𝑎𝑎𝑎𝑎𝑎
𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟𝑟𝑟
The
root-meanThe
rootFor one complete cycle
For one complete
𝑃𝑃 = 𝑉𝑉=𝑟𝑟𝑟𝑟𝑟𝑟
𝐼𝐼𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
 Easy
Easy
be
is
the
value
voltage
is the
𝑟𝑟𝑟𝑟𝑟𝑟𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
using
a
transformer).
isvalue
the
value
voltage
is the
using
atransformed
transformer).
issquare
the
ofof of
is the
value
of 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑎𝑎𝑎𝑎𝑎𝑎
square
current
mean-square
cycle
(step
up
oror
step
down
𝑟𝑟𝑟𝑟𝑟𝑟 =
𝑅𝑅 𝑅𝑅
current
mean-square
cycle
(step
up
step
down
the
current
in
value
of
the
using
Easier
convert
from
ACFor
the
current
in
value
of the
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
 AC
Easier
to convert
from
AC one complete cycle
•
Easier to convert from
tototransformed
DC
The
root-meanThethe
rootFor one complete
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎
𝐼𝐼 𝑉𝑉
is
value
of
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Easy
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the
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voltage
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𝑉𝑉=
OR
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isthe
the
value
of anvoltage
voltage
using
transformer).
an
AC
circuit
voltage
inthe
an
𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟==
𝑅𝑅
to DC
than
from
DCDC
to to
an
AC
circuit
voltage
in an𝑟𝑟𝑟𝑟𝑟𝑟
toto
DC
than
from
square
current
mean-square
cycle
the
current
in
value
of
the
(step
up
or
step
down
than from DC to AC.
𝑅𝑅
𝑉𝑉
=
𝐼𝐼
𝑅𝑅 𝑅𝑅
Easier
convert
from
AC
AC
circuit
that
an
AC
circuit
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟
= 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟𝑟𝑟
the
current
in
value
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the thatthat
OR
 Easier
to convert from AC
that
willwill
have
ACAC
circuit
AC.a
that
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is
the
value
ofhavevoltage
is
the
an
AC
circuit
voltage
in
an
using
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totoDC
than
from
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to
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will
have
the
that
will
have
anthe
AC
circuit
voltage
in
an
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than
from
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same
will
have
thethe𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
𝑅𝑅𝐼𝐼 𝑅𝑅𝑅𝑅
𝑟𝑟𝑟𝑟𝑟𝑟
the
same
will
have
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to generate.
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DCEasier
to generate.
the
current
in
value
of
the
that
will
have
AC
circuit
that
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to convert
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series
𝑟𝑟𝑟𝑟𝑟𝑟
•
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For
series
that
will
have
AC
circuit
that
same
heating
the
same
heatOR𝑟𝑟𝑟𝑟𝑟𝑟
AC.
heating
effect
same
heating
heating
effect
same
heating
OR

It
can
be
transmitted
at
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AC
circuit
voltage
inthe
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the
same
will
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from
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connection:
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Easier
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generate.
connection:
the
same
will
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the
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=
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𝑅𝑅
For
series
effect
as
a
DC
ing
effect
as
a
as
a
DC
circuit.
effect
as
a
DC
Easier
to
generate.
𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟𝑟𝑟
as have
a DC circuit.
effect
as a DC
high
voltage
andand
low
For series
that
will
AC
circuit
that
high
voltage
heating
same
heating
•
It can be transmitted
at
high
voltage
  AC.
ItItcan
be
atatlow
heatingeffect
effect
same
heating connection:
circuit.
circuit.
DC
circuit.
cancurrent
betransmitted
transmitted
circuit.
current
over
long
connection:
the
samecircuit. will
have a
the
For
series
overlow
long
as
effect
)2 𝑅𝑅 2
 Easier
todistances
generate.
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎𝑃𝑃
= (𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟
(𝐼𝐼connection:
high
voltage
and
and low current over
long
asaaDC
DC circuit.
effectas
as aDC
DC For series
𝑎𝑎𝑎𝑎𝑎𝑎 =
𝑟𝑟𝑟𝑟𝑟𝑟 ) 𝑅𝑅
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
high
voltage
and
lowless
distances
with
less
energy
heating
effect
same
heating
distances
with
energy
circuit.
𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓
= =
Itcurrent
can beover
transmitted at
2
circuit.
with less energylost.
connection:
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎
𝑃𝑃𝑃𝑃 ==(𝐼𝐼(𝐼𝐼
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎
2
current
overlong
long
𝑟𝑟𝑟𝑟𝑟𝑟 ) )𝑅𝑅
lost.
OROR
√𝟐𝟐 √𝟐𝟐effect
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
as a DC circuit.
as
a=DC
𝑅𝑅
lost.
𝑎𝑎𝑎𝑎𝑎𝑎
𝑟𝑟𝑟𝑟𝑟𝑟
𝑽𝑽𝒓𝒓𝒓𝒓𝒓𝒓
For𝑎𝑎𝑎𝑎𝑎𝑎
parallel
connection:
high
voltage
and
lowenergy
distances
with
less
𝑽𝑽
=
𝑰𝑰
For
parallel
connection:
𝒓𝒓𝒓𝒓𝒓𝒓
𝒎𝒎𝒎𝒎𝒎𝒎
𝑰𝑰
=
distances
with
less
energy
OR
where
𝒓𝒓𝒓𝒓𝒓𝒓
2

High
frequency
used
in
√𝟐𝟐
OR
where
𝑽𝑽
circuit.
𝑰𝑰
=
2
 High
frequency
used in
𝒎𝒎𝒎𝒎𝒎𝒎 √𝟐𝟐
𝒓𝒓𝒓𝒓𝒓𝒓
𝑉𝑉
2
current
over
long
lost.
𝑽𝑽
𝑉𝑉
OR
√𝟐𝟐
𝑟𝑟𝑟𝑟𝑟𝑟
(𝐼𝐼connection:
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 =
•
High frequency usedlost.
in AC
makes
it
= 𝒎𝒎𝒎𝒎𝒎𝒎 For
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎
OR
√𝟐𝟐
𝒓𝒓𝒓𝒓𝒓𝒓Where
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 𝑽𝑽Where
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎
= )=𝑅𝑅 𝑟𝑟𝑟𝑟𝑟𝑟
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
AC
makes
suitable
for for
𝑽𝑽
Forparallel
parallel
connection:
𝑃𝑃𝑟𝑟𝑟𝑟𝑟𝑟
AC
makes
it suitable
𝒓𝒓𝒓𝒓𝒓𝒓 = √𝟐𝟐
𝑎𝑎𝑎𝑎𝑎𝑎
OR
where
distances
withitless
energy
2
𝑰𝑰
=

High
frequency
used
in
𝑅𝑅
𝑰𝑰
=
For
parallel
connection:
𝒎𝒎𝒎𝒎𝒎𝒎
𝑰𝑰
=
OR
where
𝑅𝑅
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹
2
suitable for motors.
𝒓𝒓𝒓𝒓𝒓𝒓
 High
frequency used in
√𝟐𝟐
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎
=𝒎𝒎𝒎𝒎𝒎𝒎
𝑰𝑰=
𝑹𝑹 𝑹𝑹
motors.
𝑽𝑽 𝑽𝑽
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
𝒎𝒎𝒎𝒎𝒎𝒎
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
𝑹𝑹 Where
motors.
𝑽𝑽√𝟐𝟐
lost.
𝒎𝒎𝒎𝒎𝒎𝒎
OR
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎
AC
𝑽𝑽
𝑽𝑽𝒓𝒓𝒓𝒓𝒓𝒓 =𝒎𝒎𝒎𝒎𝒎𝒎
For parallel
Where
𝒎𝒎𝒎𝒎𝒎𝒎
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎=connection:
= 𝑅𝑅
ACmakes
makesititsuitable
suitablefor
for OR
𝑰𝑰
=
𝒎𝒎𝒎𝒎𝒎𝒎
where
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑹𝑹
𝑽𝑽𝑽𝑽
𝑹𝑹
2 𝑅𝑅
 High
frequency used in
OR
motors.
𝒎𝒎𝒎𝒎𝒎𝒎 ==𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎
OR
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
𝑰𝑰√𝟐𝟐
𝑹𝑹
motors.
𝒎𝒎𝒎𝒎𝒎𝒎
𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎
Where
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 =
AC makes it suitable for
𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 =
𝑅𝑅
𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹
𝑹𝑹
motors.
cuRRENT
Alternating current (AC) isaLTERNaTING
anaLTERNaTING
oscillatingcuRRENT
electric
current that varies sinusoidally with time, reversing its direction of flow periodically.
ALTERNATING CURRENT
18
Photoelectric Effect
19
19
Photoelectric EffectOPTIcaL PHENOmENa
3.9 3.9Photoelectric Effect
OPTIcaL PHENOmENa
ENERGy RadIaTION
EFFEcT
SPEcTRa
3.9
Photoelectric EffectPHOTOELEcTRIc
OPTICAL PHENOMENA
OPTIcaL
PHENOmENa
ENERGyENERGY
RadIaTION
PHOTOELEcTRIc
EFFEcT
SPEcTRa
Theory
Einstein’s equation
Emission
RADIATION
PHOTOELECTRIC
EFFECT absorption
SPECTRA
3.9
Photoelectric Effect
ENERGy
RadIaTION
PHOTOELEcTRIc
EFFEcT
SPEcTRa
Theory
Einstein’s
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E = WoOPTIcaL
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Ek(max) PHENOmENa
The energy radiated by hot Photoelectric
An
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Einstein’s
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Theory
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absorption
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o + Ek(max)
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RadIaTION
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19
illustrates the particle nature of light.
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19
3.9
4 Revision Questions- Set 1
4.1 Newton’s Laws
Strategies for problem-solving questions on Newton’s Laws
Step 1: Read the problem as many times as needed.
Step 2: Sketch the problem, if necessary.
Step 3: Draw a force diagram for the situation.
Step 4: Draw a free-body diagram and resolve the forces into component on the Cartesian plane.
Step 5: List all the given information and convert these to SI units, if necessary.
Step 6: Determine which physical principle can be used to solve the problem.
Step 7: Use the principle to solve the question, often by substituting numerical values into an appropriate
equation.
Step 8: Check that the question has been answered and that the answer makes sense.
19
Worked Example 1
A block of mass M is held stationary by a rope of negligible mass. The block rests on a frictionless plane that
is inclined at 30o to the horizontal.
Worked Example 1
A block of mass M is held stationary by a rope of negligible mass. The block rests on a frictionless
plane that is inclined at 30o to the horizontal.
1.1
State Newton’s first law of motion in words.
1.2
Draw a free-body diagram that shows all the forces acting on the block.
1.3
Calculate the magnitude of the gravitational force of the block when the force in the rope is 8 N.
1.4
1.2 Draw
a free-body
shows
all theon
forces
on the
block. reaching the horizontal rough
Now the rope
is cut
and thediagram
block that
slides
down
the acting
inclined
plane,
1.3 Calculate the magnitude of the gravitational force of the block when the force in the rope-2is 8 N.
surface and then continuing to move with constant acceleration of magnitude 1,5 m·s .
1.1
State Newton’s first law of motion in words.
1.4 Now the rope is cut and the block slides down on the inclined plane, reaching the horizontal rough
1.4.1
-2
and then
continuing
to move
withinconstant
Statesurface
Newton’s
second
law of
motion
words.acceleration of magnitude 1,5 m·s .
Commented [U53]: Pse ch
as the editor is an English langu
There are many problems with
specialist must check the revis
ensure it is also technically cor
Comment applies to all sugges
1.4.1 State Newton’s second law of motion in words.
1.4.2 Calculate
the acceleration
of block
the block
whileititis
is sliding
sliding down
thethe
inclined
plane.plane.
1.4.2 Calculate
the acceleration
of the
while
down
inclined
1.4.4. Calculate the coefficient of kinetic friction between the block and the surface.
1.4.4 Calculate the coefficient of kinetic friction between the block and the surface.
Solutions for Worked
Example
1 Example 1
Solutions
for Worked
1.1
An object continues
a state inofa rest
uniform
motion
(motion
constant
velocity),
it is
1.1 An objectincontinues
state or
of rest
or uniform
motion
(motion with
with aaconstant
velocity),
unless unless
it
acted on by an(net
unbalanced
(net or resultant)
acted on by anisunbalanced
or resultant)
force. force.
To calculate
the gravitational
we can use
1.3 1.3
To calculate
the gravitational
force, we canforce,
use the
the parallel component:
parallel component:
𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃
We do not have the component, but we can calculate this
We do not have the component, but we can
calculate this by applying Newton’s second law in
are going to calculate only the magnitude, because the
the x direction. We are going to calculate only the
direction is known.
magnitude, because the direction is known.
by applying Newton's second law in the x direction. We
21
On
the x-axis:
On the x-axis:
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
𝑇𝑇 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0
𝑇𝑇 − 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 = 0
8 − 𝐹𝐹⃗ 𝑔𝑔 sin 30𝑜𝑜 = 0
𝐹𝐹𝐹𝐹 =
8
sin 30𝑜𝑜
𝐹𝐹𝐹𝐹 = 16 𝑁𝑁
1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the
resultant force. The acceleration of the body will be directly proportional to the resultant force and
inversely proportional to the mass of the body.
201.4.2 Free body diagram, coordinate system and components.
Applying Newton’s second law of motion
Commented [G54]: Better like this
𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝑜𝑜
8 −𝑅𝑅𝑅𝑅𝐹𝐹⃗ 𝑔𝑔 sin 30
=0
𝑇𝑇 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0
8
𝐹𝐹𝐹𝐹
= 𝐹𝐹𝐹𝐹 sin𝑜𝑜 𝜃𝜃 = 0
𝑇𝑇 −
sin 30
𝐹𝐹𝐹𝐹
=𝐹𝐹⃗16
𝑁𝑁 30𝑜𝑜 = 0
8−
𝑔𝑔 sin
8
𝐹𝐹𝐹𝐹 = If a𝑜𝑜 resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant
1.4.1
sin 30
1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the
The acceleration of the body will be directly proportional to the resultant force and inversely
𝐹𝐹𝐹𝐹 = 16force.
𝑁𝑁
resultant force. The acceleration of the body will be directly proportional to the resultant force and
proportional to the mass of the body.
inversely proportional to the mass of the body.
1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the
1.4.2 Free body diagram, coordinate system and components.
1.4.2
diagram, coordinate
system
and components.
resultantFree
force.body
The acceleration
of the body will
be directly
proportional to the resultant force and
inversely proportional to the mass of the body.
Applying Newton’s second law of motion
1.4.2 Free body diagram, coordinate system and components.
1.4.3
1.4.3
Applying Newton’s second law of motion
Free
body diagram
1.4.3
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗
(1)
There is only one force acting on the block,
Applying
Newton’s
which
is the
frictionalsecond
force: law of motion
𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗
(2)
𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗
(1)
𝜇𝜇𝐾𝐾𝑅𝑅𝑁𝑁 = 𝑚𝑚𝑚𝑚
(3)
There is only one force acting on the block,
The normal force is unknown and we need to
which is the frictional force:
calculate it.
𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗
(2)
(3)
𝜇𝜇𝐾𝐾 𝑁𝑁 = 𝑚𝑚𝑚𝑚
The normal force is unknown and we need to
calculate it.
⃗⃗
𝑁𝑁
Free body diagram
⃗⃗
𝑁𝑁
𝑓𝑓⃗𝑓𝑓
𝐹𝐹⃗𝑔𝑔
Commented [G54]: Better like this
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗
Applying
in the x direction
ApplyingitNewton’s
second law of motion
𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗
𝑜𝑜
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚30
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ = 𝑚𝑚𝑎𝑎⃗
𝑜𝑜
𝑎𝑎⃗Applying
= 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔30
it in the x direction
𝑜𝑜
𝑎𝑎⃗𝐹𝐹⃗= =
9,8𝑚𝑚𝑎𝑎
𝑠𝑠𝑠𝑠𝑠𝑠30
⃗
𝑔𝑔𝑔𝑔
𝑎𝑎⃗𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚30
= 4.9 m𝑜𝑜∙ 𝑠𝑠=−2𝑚𝑚𝑎𝑎⃗
𝑎𝑎⃗ = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔30𝑜𝑜
𝑎𝑎⃗ = 9,8 𝑠𝑠𝑠𝑠𝑠𝑠30𝑜𝑜
𝑎𝑎⃗ = 4.9 m ∙ 𝑠𝑠 −2
Free body diagram
𝐹𝐹⃗𝑔𝑔
Commented [G54]: Better like this
𝑓𝑓⃗𝑓𝑓
working
working in
in the
the y-direction:
y-direction:
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check––perhaps
perhaps“Work
“Workininthe
thedire
dire
of
ofy.”
y.”
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Psecheck
checkall
allsuch
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instances.
working
in
thevertical
y-direction:
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“Work
in the“Work
direction
working
in thedirection,
y-direction:
Commented
Pse check
– perhaps
in the dire
In
the
the
of
In
the
vertical
direction,
the acceleration
acceleration is
is zero,
zero, therefore
therefore we
we can
can apply
apply Newton’s
Newton’s first
first law
law22
ofCommented
of y.”
of y.”
Pse
check
all
such
instances.
motion.
Pse check all such instances.
motion.
Working in the y-direction:
In
thevertical
vertical
direction,
the is
acceleration
iswe
zero,
therefore
canlaw
apply
Newton’s
first law of motion.
In the vertical
direction,
the
acceleration
zero, is
therefore
can we
apply
first
of law
In the
direction,
the acceleration
zero, therefore
canNewton’s
applywe
Newton’s
first
of
motion.motion.
⃗𝟎𝟎⃗⃗⃗
⃗𝑭𝑭⃗𝑭𝑭
⃗⃗𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) =
𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) = 𝟎𝟎
22
⃗⃗⃗⃗
⃗⃗⃗
⃗⃗𝒈𝒈 =
𝑵𝑵
𝟎𝟎
⃗⃗⃗
𝑵𝑵 +
+ ⃗𝑭𝑭⃗𝑭𝑭
𝒈𝒈 = 𝟎𝟎
⃗
⃗+
⃗y-positive
⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) Taking
⃗𝟎𝟎⃗ ⃗⃗⃗
⃗⃗ direction
⃗⃗⃗
⃗𝑭𝑭⃗𝒈𝒈 = 𝟎𝟎
=⃗𝑭𝑭⃗⃗𝟎𝟎
𝑵𝑵 + ⃗𝑭𝑭⃗𝒈𝒈 the
=
= positive
𝑵𝑵 𝟎𝟎
𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚)as
Taking
as
the
y-positive
direction
Taking
aspositive
positive
the
y-positive
direction
Commented
Commented [U56]:
[U56]: Pse
Psecheck.
check.
Commented
Commented [G57]:
[G57]: Beter
Beterlioke
liokethis
this
Taking 𝑵𝑵
as
the𝑵𝑵
y-positive
Taking
as𝟎𝟎𝟎𝟎positive
the =
y-positive
direction
𝑭𝑭𝑭𝑭𝒈𝒈 =
−
𝑵𝑵 −
−positive
=
𝑵𝑵
− 𝒎𝒎𝒎𝒎
𝒎𝒎𝒎𝒎
= 𝟎𝟎𝟎𝟎direction
𝒈𝒈
𝑵𝑵 − 𝑭𝑭𝒈𝒈 =
𝑵𝑵
−
=−
𝟎𝟎 𝒎𝒎𝒎𝒎 = 𝟎𝟎
𝑵𝑵
=
𝟎𝟎 𝒎𝒎𝒎𝒎 𝑵𝑵
𝒈𝒈 𝒎𝒎𝒎𝒎
∴∴𝟎𝟎−
𝑵𝑵
𝑵𝑵𝑭𝑭=
=
𝒎𝒎𝒎𝒎
(4)
(4)
Substituting
3: 4 in 3:Hence
Substituting
𝜇𝜇𝜇𝜇
𝐾𝐾𝐾𝐾𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚4=
=in𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚
Hence
𝜇𝜇𝜇𝜇𝐾𝐾𝐾𝐾𝑔𝑔𝑔𝑔 =
= 𝑎𝑎𝑎𝑎
4 3:
in 3:(4)
∴ 𝑵𝑵 = Substituting
𝒎𝒎𝒎𝒎
∴ 𝑵𝑵 = 𝒎𝒎𝒎𝒎 44 in
Substituting
Substituting
in 3:
Commented
[U56]: Pse[U56]:
check. Pse check.
Commented
Commented
[G57]: Beter
lioke this
Commented
[G57]:
Beter lioke this
(4)
𝜇𝜇𝐾𝐾 𝑚𝑚𝑚𝑚 =Substituting
𝜇𝜇𝐾𝐾 𝑔𝑔 = 𝑎𝑎𝜇𝜇𝐾𝐾 𝑔𝑔 = 𝑎𝑎
𝜇𝜇𝑚𝑚𝑚𝑚
HenceHence
𝐾𝐾 𝑚𝑚𝑚𝑚 = 𝑚𝑚𝑚𝑚Hence
Substituting
Substituting
𝜇𝜇Substituting
(9,8) =
= 1,5
1,5
𝜇𝜇𝐾𝐾𝐾𝐾(9,8)
Substituting
𝜇𝜇𝜇𝜇𝐾𝐾𝐾𝐾 =
= 0,153
0,153
=
𝜇𝜇𝐾𝐾 (9,8) Exercise
(9,8) =11𝜇𝜇
1,5
𝜇𝜇𝐾𝐾 = 0,153
𝜇𝜇𝐾𝐾1,5
𝐾𝐾 = 0,153
Exercise
Exercise
AExercise
A 1block
block of
of1mass
mass 10
10 kg
kg is
is pulled
pulled to
to the
the right
right at
at constant
constant velocity
velocity with
with aa force
force of
of magnitude
magnitude 30
30 N
N that
that
oo
acts
at
an
angle
of
37
with
the
horizontal.
acts at an angle of 37 with the horizontal.
A block A
of block
mass of
10mass
kg is 10
pulled
the right
at constant
velocity velocity
with a force
magnitude
30 N that
kg istopulled
to the
right at constant
with of
a force
of magnitude
30 N that
acts at an
angle
of angle
37o with
theo with
horizontal.
acts
at an
of 37
the horizontal.
⃗⃗𝑭𝑭
⃗⃗ =
𝑭𝑭
= 𝟑𝟑𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑𝟑𝟑
10
10 kg
kg
10 kg
1.1
1.1
10 kg
37o
o
37
⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑𝑭𝑭
⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑
37o𝑭𝑭
37o
Draw
Draw aa free-body
free-body diagram
diagram with
with all
all the
the force
force action
action on
on the
the block.
block. Use
Use aa Cartesian
Cartesian plane
plane and
and
21
uting
) = 1,5
se 1
𝜇𝜇𝐾𝐾 = 0,153
Exercise 1
k of mass 10 kg is pulled to the right at constant velocity with a force of magnitude 30 N that
block
of mass 10 kg is pulled to the right at constant velocity with a force of magnitude 30 N that acts at an
an angle of 37o A
with
the horizontal.
angle of 37o with the horizontal.
10 kg
⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑
𝑭𝑭
37o
Draw a free-body
withaall
the force action
on the
block.
Cartesian
plane
1.1diagram
Draw
free-body
diagram
with
all Use
the aforce
action
onand
the block. Use a Cartesian plane and represent
represent the component
of component
the force.
the
of the force.
Calculate the magnitude and direction of the normal force.
Calculate the coefficient
of friction. the magnitude and direction of the normal force.
1.2
Calculate
The force is removed and the block comes to rest after a time. Calculate the acceleration of
1.3forceCalculate
the coefficient
of friction.
the block after the
has been removed
if the frictional
force remains constant.
1.4
The force is removed and the block comes to rest after a time. Calculate the acceleration of the block
after the force has been removed if the frictional force remains
constant.
23
Exercise 2 Exercise 2
o
o
to the30
horizontal.
A block,
massis20pulled
kg, is pulled
upwards
force 𝐹𝐹⃗ along
with an
incline
alonga plane
a plane
with
an 30
incline
to the horizontal. It
A block, of mass
20of kg,
upwards
bybyforce
-22 m·s-2. The surface of the incline exerts a friction force on the
It
accelerates
along
the
incline
plane
at
accelerates along the incline plane at 2 m·s . The surface of the incline exerts a friction force on the block.
block.
2.1
Draw a diagram showing all the forces acting on the block (force diagram) as it moves up
the incline.
2.1 aDraw
a diagram
showingfor
all the
acting
on the block
diagram)
as Cartesian
it moves up the
Also draw
free-body
diagram
theforces
block.
Represent
the(force
forces
on the
plane and resolve
incline. Also draw a free-body diagram for the block. Represent the forces on the Cartesian
the gravitational
force into components.
plane and resolve the gravitational force into components.
2.2
Calculate
component
of the gravitational
force
exerted
theblock
block
parallel
toincline.
the incline.
2.2 the
Calculate
the component
of the gravitational
force
exertedon
on the
parallel
to the
2.3
2.3 Calculate the normal force.
Calculate
the normal force.
2.4
Calculate
the magnitude of the friction force if the magnitude of is equal to 200 N.
2.5 Calculate the coefficient of kinetic friction for the block.
2.5
Calculate the coefficient of kinetic friction for the block.
Calculate the magnitude of the friction force if the magnitude of 𝐹𝐹⃗ is equal to 200 N.
2.4
4.2 Momentum
and Impulse
4.2
Momentum and Impulse
Momentum and impulse problem solving strategy
Momentum and impulse problem solving strategy
1
Read 1the Read
problem
as many
you
to; to;
then
draw
ordiagram
diagram
situation described
the problem
as time
many as
time
as need
you need
then
drawaasketch
sketch or
of of
thethe
situation
Commented [U58]: Formatting: Pse
all equivalent items, including line spaci
described if not provided.
if not provided.
etc.
2
2
From the reading, collect the data and write it in symbol form (not only numbers are part of the
data) e.g. the object starts moving from rest means the initial speed is zero.
Commented [U59]: Not understood
3 Draw
a freecollect
body diagram
for each
define
the (not
system
younumbers
are going toare
work
with.
Commented
From the
reading,
the data
andobject
writeand
it inclearly
symbol
form
only
part
of the data)
e.g.[G60]: It could be inclu
 If possible, choose a system that is isolated ( F net  0 ) and closed (m = constant). If the
the object
starts moving from rest means the initial speed is zero.

interactions are sufficiently short and intense you can ignore external forces.
If it is not possible to choose an isolated system, try to divide the problem into parts (scenarios).
Commented [U61]: Pse introduce b
3
to ensure legibility.
Draw a free body diagram for each object and clearly define the system you are going to work statement
with. applies
Comment
to all such instances.
•
Commented [U62]: Formatting: pse
numbers, including font type, font size,
alignment, etc. Repetitive problem seen
 If choose
the mathematical
is based
law closed
of conservation
of momentum
(m = constant).
If the interactions
If possible,
a systemrepresentation
that is isolated
( F net in
= 0the) and
,
write
it
in
component
form.
p

p
 initial
and
are sufficiently
short
finalintense you can ignore external forces.
•
the text, as the editor will not know all o
If it is not possible to choose an isolated system, try to divide the problem into parts (scenarios).
4 Select the law (principle), theorem, equation or formula that will answer your question.


5 Substitute the values into the equation or formula (the system of units must be homogeneous).
Commented [U63]: Pse check all sc
6 Check your answer


Is the value of the answer reasonable?
Is the unit correct?
Commented [G64]: I did the correc
worked example 1
22
24
4 Select the law (principle), theorem, equation or formula that will answer your question.
If the mathematical representation is based in the law of conservation of momentum
•
(∑ p
initial
= ∑ p final
), write it in component form.
5 Substitute the values into the equation or formula (the system of units must be homogeneous).
6 Check your answer
•
Is the value of the answer reasonable?
•
Is the unit correct?
Worked example 1
Trolley A, with a mass of 120 kg, is moving at 20 m·s-1 ; it collides head-on with trolley B, with a mass of 150
kg, moving at 30 m·s-1 in the opposite direction. If A rebounds at 25 m·s-1:
State the law of conservation of momentum in words.
1.1
Trolley A, with a mass of 120 kg, is moving at 20 m·s-1 ; it collides head-on with trolley B, with a mass
-1
in the opposite
direction.
If A rebounds at 25 m·s -1:
of
kg, moving at 30
m·s
1.2150
the
rebound
velocity
of-1 B.
Trolley
A, withCalculate
a mass of 120
kg,
is moving
at 20 m·s
; it collides head-on with trolley B, with a mass
of 150 kg, moving at 30 m·s-1 in the opposite direction. If A rebounds at 25 m·s -1:
1.1 State the law of conservation of momentum in words.
1.3
Calculate
the velocity
impulse
1.2 Calculate
the rebound
of of
B. the force exerted by trolley B on trolley A.
1.1 1.3
State
the law the
of conservation
of force
momentum
in by
words.
Calculate
impulse of the
exerted
trolley B on trolley A.
1.2 Calculate the rebound velocity of B.
1.3Solution:
Calculate the impulse of the force exerted by trolley B on trolley A.
Solution:
Solution:
data
Data
data
ma = 120 kg
ma =𝒗𝒗
kg m · 𝑠𝑠 −1 to the right
⃗⃗ 120
= 20
𝒊𝒊𝒊𝒊
−1
⃗⃗𝒊𝒊𝒊𝒊 m
=B20
m · 𝑠𝑠kg
to the right
𝒗𝒗
= 150
150 kg = 30 m · 𝑠𝑠 −1 to the left
mB =𝒗𝒗
⃗⃗(𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃)𝑩𝑩
⃗⃗(𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃)𝑩𝑩 = 30 m · 𝑠𝑠 −1 −1
to the left
𝒗𝒗
⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 to the left
𝒗𝒗
⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 −1 to the left
𝒗𝒗
⃗⃗(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂)𝑩𝑩 =?
𝒗𝒗
⃗⃗(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂)𝑩𝑩 =?
𝒗𝒗
1.1 The total linear momentum of an isolated system is constant.
1.11.1
TheThe
total linear
momentum
of an isolated
system
is constant.
total
linear
momentum
an
isolated
system
is constant.
1.2 When
we work
in Physics
it is veryof
important
to represent
a problem
using a diagram; here we are
working with vector quantities, so it is important to indicate direction with all elements (positive or
1.2 When
we work in Physics it is very important to represent a problem using a diagram; here we are
negative)
1.2
When
work
in Physics
it is
very important
represent
problem(positive
using aordiagram; here we are working
working
withwe
vector
quantities,
so it is
important
to indicate to
direction
with alla elements
negative)
with vector quantities,Positive
so it direction
is important to indicate direction with all elements (positive or negative)
y
y
direction
-1
VA = 20 m.sPositive
VB = 30 m.s-1
VA = 20 m.s-1
VB = 30 m.s-1
Before
B

VA = 25 m.s-1
B
A
A
VA = 25 m.s-1

B
A
A
After
B
x
After
x
Before
In order to find the velocity of trolley B after the collision, the law of conservation of momentum must
be applied. Learners' attention must be addressed to the general characteristics of this kind of problem,
In order
to find
velocity
of
trolleyof
B trolley
after
theBcollision,
thecollision,
law of conservation
momentum
mustofwe
i.e.order
when
anthe
interaction
between
two
bodies
takes
Firstly,
wethe
have
to identify
which bodies
In
to
find
the velocity
afterplace.
the
law
ofof conservation
momentum must be applied.
be applied.
Learners'
attention
be addressed
to theimportant
general characteristics
of this kind
of problem,
are going
to consider
as ourmust
“system”.
This is a very
step, since momentum
is only
conserved
attention
musttwo
beIn
addressed
to the
characteristics
of this
kind of problem, i.e. when an
i.e.Learners’
when
an interaction
between
bodies
takes place.
Firstly,
weBhave
to identify
which Once
bodies
in
closed
and
isolated
systems.
this problem,
trolleys
Ageneral
and
integrate
the system.
wewe
have
are identified
going to consider
as of
ourthe
“system”.
This
is a very
important
step, since between
momentum
is
only conserved
the
bodies
system,
we
have
to
identify
the
interaction
the
bodies
of
the
system
interaction
between
two bodies
takes trolleys
place. AFirstly,
we have
to identify
which
bodies we are going to consider
in closed
and
isolated with
systems.
In this
problem,
and
integrate
the system.
Once
we have a
and the
interaction
external
bodies.
Learners know
thatBwhen
we have
any kind
of interaction,
identified
the
bodies
of
the
system,
we
have
to
identify
the
interaction
between
the
bodies
of
the
system
force will appear (as a measurement of the intensity of any particular interaction). During the collision,
andthe
thetwo
interaction
with external
bodies.
Learners
knowmove
that when
wethe
have
anyand,
kindofofcourse,
interaction,
a
trolleys interact
with the
surface
where they
on, with
Earth
with each
Commented [G65]: corrected
force
will appear
(as aacting
measurement
of the intensity
of any particular
interaction).
other.
The forces
on each trolley
are represented
in the next
diagram: During the collision,
Commented [U66]: Pse clarify (surface of what?)
the two trolleys interact with the surface where they move on, with the Earth and, of course, with each
Commented [G65]: corrected
23
other. The forces acting on each trolley are represented in the next diagram:
Commented [U66]: Pse clarify (surface of what?)
25
as our “system”. This is a very important step, since momentum is only conserved in closed and isolated
systems. In this problem, trolleys A and B integrate the system. Once we have identified the bodies of the
system, we have to identify the interaction between the bodies of the system and the interaction with external
bodies. Learners know that when we have any kind of interaction, a force will appear (as a measurement of
the intensity of any particular interaction). During the collision, the twoytrolleys interact with theFsurface where
FBA
AB
they move on, with the Earth and, of course, with each other. The forces acting on eachNtrolley
are
represented
A
y
FBA
y
in the next diagram:
FAB
FBAFAB
yy
y NNA
FAB
FAB
y
A
yy y
NA
AA
y
FBA
FBA
FAB
A
BB
FFABABFABNA
FFBABAFBA
NA
NB
NNAANA
NNB BNB
F
F
F
BA
BAAB
NNB B
A
B
A
NA
NB
BB BNB
AA A
A
B
FAB FgA
NB
A
NB
B
B
FgA
B
FgB
FBA
x
FgB
x
FgA
F
NgB
xx x
FFN
A
BFF
FgA
FgB
gAgA
gBgB
xx
FgA
FgB
FgA
FgB
wA
wB
x
FgA
FgB
x
FgA
FgB
B
A
w
wB
A
wA
wB
wwAw
wwBw
A A
B B


wwA A
wwB B

wA
wB The forces FAB and FBA are the forces exerted by the trolleys on each o
wA
wB

x
F
F
and
gA FBA are the forces exerted by the
gB trolleys on each other during the interaction, so
The
 are


 forces
 FABthe
FBA
The forces FAB
and
forces
exerted by the trolleys
on
each
other during the interaction, so
trolleys
on
FFABAB
FFBABA
The
forces
are
the
forces
the
each
other
during
the
interaction,
so


exerted
by
Fand
FBA
The
The
forces
forces
and
and
are
are
the
the
forces
forces
exerted
exerted
byby
the
the
trolleys
trolleys
ontrolleys
on
each
each
other
other
during
during
the
the
interaction,
interaction,
so
so forces
N A and
Nthey
they
are
internal
forces.
The
so-called “norma
AB
The
forces
and
are
the
forces
exerted
by
the
on
each
other
during
the
interaction,
so
are
B are
N
N
and
they
are
internal
forces.
The
forces
are
so-called
“normal
forces”,
that
is,
the
reaction



and
 during
  the
internal
forces.
A interaction,
B “normal
the
 are
FF
exerted
by
the
on
each
so
Nother
Nduring
they
The forces
so-called
andFF
are
theforces
forces
exerted
by
thetrolleys
trolleys
on
each
interaction,
so forces”, that is, the reaction

AB
BA
Aother
B arethe
ABand
BA are
N
N
and
they
are
internal
forces.
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forces
are
so-called
“normal
forces”,
that
is,
the
reaction
Fare
FBAforces.
forces
exerted
by
the
surface
where
the trolleys
move[G67]:
on. These
NAN
NBeach
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and
and
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internal
internal
forces.
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forces
forces
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so-called
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“normal
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that
that
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the
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reaction
The
and
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theThe
forces
exerted
by
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on
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during
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Fthey
The forces FAB
andforces
are
forces
exerted
by
trolleys
on
other
during
the
interaction,
so
forces
exerted
by
the
surface
where
the
trolleys
move
on.
These
two
forces
are
insoequilibrium
with
Commented
Better like two
this
AN
AB
B
ABthe
BA forces.
w
weach
internal
The
forces
and
so-called
forces”,
that
the
reaction
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exerted
N
exerted
the
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in
with
Commented
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this
B“normal

inequilibrium
is,
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are
surface
Awhere
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einternal
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 are
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the
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Commented
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Better
like
this
forces
forces
the
the
surface
surface
where
where
the
the
trolleys
trolleys
move
move
on.
on.
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These
two
two
forces
forces
are
are
inis,
inequilibrium
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withforces
Commented
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[G67]:
[G67]:
Better
Better
likelike
thisthis Pse check.
Commented
[U68]:
N
N
and
they
are
internal
forces.
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forces
are
so-called
“normal
forces”,
that
the
reaction
F
F
and
.
Finally,
the
gravitational
force
applied
by
the
Earth
on
trolleys
A
and
B,
w


N
N
and
they
are
internal
forces.
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forces
are
so-called
“normal
forces”,
that
is,
the
reaction
F
the
gravitational
force
applied
by
the
Earth
on
trolleys
A
and
B,
A




Commentedforce
[U68]: Pse
check.
A forces
Bare
gA
gB Better
Aon.
gA
Fwith
F
and
.
Finally,
forces
gravitational
force
applied
byBmove
thetwo
Earth
on
trolleys
A and
B, with
w
A
xerted by
where
the
trolleys
move
Commented
[G67]:
like
this
exerted
bythe
thesurface
surface
where
the
trolleys
move
on.These
These
two
forces
areininequilibrium
equilibrium
the the
surface
where
the
trolleys
on.
These
two
forces
are
in
equilibrium
with
the
gravitational
applied
Commented
[G67]:
Better
like
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gA
gB
Commented
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[U68]:
PsePse
check.
check.
FFgAequilibrium
Fare
Finally,
forces
the
gravitational
applied
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trolleys
AAThese
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wwAw
the
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onon
Fand
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F.gBin
and
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Finally,
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the
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gravitational
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applied
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trolleys
trolleys
Aand
and
B,
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A A [G67]:
exerted
the
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two
forces
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Commented
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 force
surfacebywhere
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by
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the
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Commented
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gA
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

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

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[U68]:
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check.
and
are
the
forces
applied
by
the
trolleys
on
the
horizontal
surface
(sometimes
called
weights).
w
Ftrolleys
FFgBthe
B
forces
tational force
by
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AA
and
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wwother
FgAgAand
and
.Finally,
Finally,
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vitational
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applied
byEarth
the
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on
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and
B,exerted
AA
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FEarth
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andon
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by
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during
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so
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the
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w B .are
by
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and
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the
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gB
and
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the
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wthe
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w
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wcalled
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Pse check. [U68]:
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the
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the
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on
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B,surface
w A
trolleys
Fhorizontal
FgB
and
.F
Finally,
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the gravitational
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theapplied
A and
w
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and
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the
(sometimes
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wwBw
[U68]:
the
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and
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trolleys
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the
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horizontal
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(sometimes
(sometimes
called
called
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weights).
gA and
gB Finally,
gA
Bapplied
B
F
F
N
N
All
the
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(
,
,
and
)
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so
there
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









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
gA
gB
A
B
F
F
N
N
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the
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(
,
,
and
)
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so
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NB are so-called
they
aretrolleys
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“normal
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that is,forces
the
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(sometimes
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arethe
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forcesapplied
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the
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on
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horizontal
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(sometimes
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weights).
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All
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the
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(force
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gB
gA N A and
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A , NB , FgA and FgB )) are in
F,gAWe
Fcalled
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, ,AN
,the
and
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w
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 All
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All
All
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forces
( AN
,BBN
,the
and
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there
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and
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and w B are
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gA
gAand
gB
A
BF
on
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can
then
say
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exerted
by
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where
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trolleys
move
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These
twocalled
forcesweights).
are in called
equilibrium
with
Commented [G67]: Better like this


on
the
trolleys.
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can
then
say
that:





F
F
Nforces
xternal forces
,
and
)
are
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equilibrium,
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F
NAA, ,acting
N
external
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,
and
)
are
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equilibrium,
so
there
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gB
gAgAon so
BB
acting onWe
thecan
trolleys.
then say that:
gB
there
isWe
no
net
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force
acting
on theso
trolleys.
then We
saycan
that:
acting
the
trolleys.
can
then
say
that:
acting
acting
on
the
the
trolleys.
trolleys.
We
We
can
then
say
say
that:
that:
FgB
N
Ncan
All
the
external
forces
, then
insoequilibrium,
is noforce
net external force
All the external
forces
(onN
, FgA( applied
and
)Fare
in
equilibrium,
there
netthere
external
gA and
A, F
Bby
Commented [U68]: Pse check.
gB
Bforce
A, N
the
gravitational
the
Earth
on) are
trolleys
A andisB,noF
n the
then
say
gA and FgB . Finally, forces w A
on
thetrolleys.
trolleys.We
Wecan
can
then
saythat:
that:
⃗
⃗
onWe
thecan
trolleys.
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acting on the acting
trolleys.
then say
that:then say that:
0
=
𝑭𝑭
∑
𝒆𝒆𝒆𝒆𝒆𝒆
⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0
∑on
and w B are the forces applied by the trolleys
the
horizontal
surface (sometimes called weights).
⃗𝑭𝑭⃗⃗𝑭𝑭
⃗𝒆𝒆𝒆𝒆𝒆𝒆
⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆
00 0
===
∑
∑
𝒆𝒆𝒆𝒆𝒆𝒆
∑ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0
∑

 
⃗
⃗
⃗
⃗
0
=
𝑭𝑭
∑
0
𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆
𝒆𝒆𝒆𝒆𝒆𝒆 (=N A∆𝑝𝑝
All the external∑
forces
, ⃗NB ,⃗⃗FgA and FgB ⃗⃗) are in0equilibrium, so there is no net external force
∆𝑝𝑝⃗
∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 = 0 ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 =
Since:
∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒
Since:
=
⃗ ⃗∆𝑝𝑝⃗
∆𝑝𝑝
∆𝑡𝑡 then say that:
∆𝑝𝑝
acting
on
We can
⃗∑
∆𝑡𝑡trolleys.
⃗𝑒𝑒𝑒𝑒𝑒𝑒
⃗the
Since:
∑
Since:
𝐹𝐹
=
∆𝑝𝑝⃗
∑
Since:
Since:
𝐹𝐹
𝐹𝐹
=
=
𝑒𝑒𝑒𝑒𝑒𝑒
𝑒𝑒𝑒𝑒𝑒𝑒 ∆𝑡𝑡∆𝑡𝑡∆𝑡𝑡
⃗⃗
∆𝑝𝑝
∆𝑝𝑝
Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 =
=
=
𝑒𝑒
∆𝑡𝑡
𝑒𝑒𝑒𝑒𝑒𝑒 ∆𝑡𝑡∆𝑡𝑡
∆𝑝𝑝⃗
∆𝑝𝑝⃗
Then:
Commented [U69]: Pse check all for
⃗
Since: ∑ 𝐹𝐹𝑒𝑒𝑒𝑒𝑒𝑒 =
Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒Then:
=
Commented [U69]: Pse check all formulas and equations i
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text, Pse
including
brackets
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and
2
Then:
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allall
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equation
∑ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0
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∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0
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⃗𝑛𝑛𝑛𝑛𝑛𝑛
00 0
===
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knowwhat
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and
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not.
⃗𝑛𝑛𝑛𝑛𝑛𝑛
⃗𝑛𝑛𝑛𝑛𝑛𝑛
∆𝑝𝑝
Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = ∆𝑝𝑝
𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
not know
what not.
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∆𝑡𝑡
𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
− 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
= 0− 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 0
⃗𝑛𝑛𝑛𝑛𝑛𝑛
∆𝑝𝑝
⃗𝑛𝑛𝑛𝑛𝑛𝑛==00
∆𝑝𝑝
⃗
0
∆𝑝𝑝
=
⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
00 0 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
−−
==
Content editing is required.
𝑛𝑛𝑛𝑛𝑛𝑛 Content editing is required.
⃗0𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
=
𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑝𝑝𝑛𝑛𝑛𝑛𝑛𝑛
∆𝑝𝑝
=𝑝𝑝−
∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
= 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
==00
𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇−−𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
==𝑝𝑝=
⃗
⃗
⃗
⃗
𝑝𝑝
𝑝𝑝
Then:
[U69]: Pse check all formulas and equations
⃗
⃗
⃗
⃗
+
𝑝𝑝
=
𝑝𝑝
+
𝑝𝑝
𝑝𝑝
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗
⃗
0Commented
−
𝑝𝑝
=
𝑝𝑝
⃗⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
= 0 = 𝑝𝑝⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝𝑝𝑝
𝑝𝑝⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
0 𝑝𝑝⃗+𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
− 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
=−
𝑝𝑝⃗=𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇=
⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
text, including brackets (1 bracket and 2 brackets), as the edit
⃗𝑝𝑝𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗⃗𝑝𝑝𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑝𝑝𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝+
==
𝑝𝑝
++
𝑝𝑝⃗𝑝𝑝𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑚𝑚
⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝑝𝑝⃗𝑝𝑝⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝑝𝑝⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝⃗𝐴𝐴𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑣𝑣+
+=𝑚𝑚
𝑣𝑣=
=𝑝𝑝+
𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚𝐴𝐴+
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗
𝑝𝑝
=
⃗
⃗
⃗
⃗
⃗
⃗
𝑚𝑚
𝑣𝑣
+
𝑚𝑚
𝑣𝑣
𝑚𝑚
𝑣𝑣
+
𝑚𝑚
𝑣𝑣
𝑝𝑝
=
𝑝𝑝
⃗
⃗
𝑝𝑝
=
𝑝𝑝
not
know
what
is
correct and what not.
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗𝑚𝑚
⃗+
+
𝑝𝑝⃗𝑝𝑝
⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)=
⃗𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗+𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)++𝑝𝑝⃗𝑝𝑝𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
=𝑝𝑝
𝑝𝑝
+𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑝𝑝𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑣𝑣𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑣𝑣𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚
𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
==𝑚𝑚
++𝑚𝑚
𝑚𝑚
⃗𝑝𝑝
⃗𝐵𝐵𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐵𝐵𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+
𝑚𝑚
𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
=𝑚𝑚𝐴𝐴𝑚𝑚
𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+
𝑚𝑚𝐵𝐵𝑚𝑚
𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚
𝐵𝐵𝑚𝑚
Content editing is required.
𝐴𝐴𝑣𝑣
𝐴𝐴⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵
𝐴𝐴𝑣𝑣
𝐵𝐵𝑣𝑣
0𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛
=
⃗
⃗
⃗
+
𝑝𝑝
=
𝑝𝑝
+
𝑝𝑝
⃗
⃗
⃗
⃗
+
𝑝𝑝
=
𝑝𝑝
+
𝑝𝑝
𝑝𝑝
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗
⃗
⃗
+
𝑝𝑝
=
𝑝𝑝
𝑝𝑝
𝑣𝑣⃗𝑣𝑣
𝑣𝑣⃗𝑣𝑣
++𝑚𝑚𝑚𝑚𝐵𝐵𝐵𝐵𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)++𝑚𝑚𝑚𝑚𝐵𝐵𝐵𝐵𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)==𝑚𝑚𝑚𝑚
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴𝐴𝐴
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐴𝐴
When
the
data
inwe
the
thesubstituting
in the
problem,
have:
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)we will have: 𝑚𝑚 𝑣𝑣⃗
+data
𝑚𝑚𝑝𝑝⃗𝐵𝐵𝑚𝑚
𝑣𝑣⃗given
=
𝑣𝑣⃗𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+will
𝑚𝑚problem,
+𝑚𝑚
𝑚𝑚𝐴𝐴𝐵𝐵𝑣𝑣⃗𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
=
𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+𝑚𝑚
𝑚𝑚𝐴𝐴𝐵𝐵given
𝑚𝑚𝐴𝐴 𝑣𝑣⃗When
𝐵𝐵 𝑣𝑣
⃗
0
−
=
𝑝𝑝
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)substituting
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
⃗
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+
𝑚𝑚
𝑣𝑣
=
𝑚𝑚
When
substituting
the
data
given
ininthe
problem,
we
will
have:
𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴 𝑣𝑣
When
When
substituting
substituting
the
the
data
data
given
given
inthe
the
problem,
problem,
we
we
will
will
have:
have:
⃗
⃗
𝑝𝑝
=
𝑝𝑝
tuting the
we
will
have:
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
tituting
thedata
datagiven
givenininthe
theproblem,
problem,
we
will
have:
(120)(20) =
(120)(−25)
+ (150)(−30)
= 150𝑣𝑣
+ 150𝑣𝑣⃗
(120)(20)
(120)(−25)
⃗𝐵𝐵(𝑎𝑎+
+ (150)(−30)
+
When
substituting
the
given
in we
the
problem,
we𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
will have:𝐵𝐵(𝑎𝑎
When substituting
the
data
in data
the
have:
𝑝𝑝⃗(120)(−25)
=will
𝑝𝑝⃗in
𝑝𝑝⃗(150)(−30)
𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
(120)(20)
(150)(−30)
When
substituting
the+problem,
data
the
we willWhen
have:substituting the data given in the problem, we will have:
⃗+
++given
=
+
150𝑣𝑣
(120)(20)
(120)(20)
(150)(−30)
(120)(−25)
⃗problem,
⃗150𝑣𝑣
+
=
=(120)(−25)
+
+
150𝑣𝑣
150𝑣𝑣
𝐵𝐵(𝑎𝑎
2400
−given
4500
=
−300
𝐵𝐵(𝑎𝑎
𝐵𝐵(𝑎𝑎 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗
2400
−
4500
=
−300
+
150𝑣𝑣
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+
𝑚𝑚𝐵𝐵=
𝑣𝑣⃗=𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
=
𝑚𝑚
𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+ 𝑚𝑚𝐵𝐵−1𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚+𝐴𝐴+𝑣𝑣
⃗𝐵𝐵(𝑎𝑎
150𝑣𝑣
(150)(−30)==(120)(−25)
(120)(−25)
⃗−
+(150)(−30)
150𝑣𝑣
𝐴𝐴
𝐵𝐵(𝑎𝑎
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
2400
4500
−300
++
150𝑣𝑣
⃗
⃗
2400
2400
−
−
4500
4500
=
−300
−300
+
150𝑣𝑣
150𝑣𝑣
⃗
𝑣𝑣
=
+6
𝑚𝑚
∙
𝑠𝑠
−1
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
(120)(20) +
(150)(−30)
⃗
=+(120)(−25)
(120)(20)
(150)(−30)
(120)(−25)
⃗ 𝑚𝑚 ∙ 𝑠𝑠+ 150𝑣𝑣
150𝑣𝑣
𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
= +6
−1 −1𝐵𝐵(𝑎𝑎
⃗=
2400
−−4500
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
2400
4500=+
=−300
−300++150𝑣𝑣
150𝑣𝑣
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
==𝐵𝐵(𝑎𝑎
+6
𝑚𝑚𝑚𝑚∙𝑚𝑚
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑣𝑣
=+6
+6
∙𝑠𝑠𝑠𝑠∙−1
𝑠𝑠
(120)(20) + (150)(−30) = (120)(−25) + 150𝑣𝑣⃗𝐵𝐵(𝑎𝑎
2400
−the
4500
=
−300
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
2400
−+6
4500
−300
+ 150𝑣𝑣
−1
−1
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) we will have:
When
substituting
data
given +
in 150𝑣𝑣
the ⃗problem,
𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
==+6
𝑚𝑚𝑚𝑚∙ ∙𝑠𝑠𝑠𝑠=
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
Since
the
is
positive,
wetrolley
can say
thatmove
trolley
will
move direction
in the original
direction
of trolley ⃗a
Since the
answer
positive,
can
B will
in B
the
original
trolley
𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
= +6
𝑚𝑚say
∙ 𝑠𝑠 −1that
𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
=is+6
𝑚𝑚 answer
∙ 𝑠𝑠 −1we
2400
−of4500
=a
−300
+ 150𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
Since
the
answer
isispositive,
we
can
say
that
trolley
BBwill
move
ininthe
original
direction
ofofof
trolley
aaa
before
the collision.
Since
Since
the
the
answer
answer
ispositive,
positive,
we
we
can
can
say
say
that
that
trolley
trolley
Bwill
will
move
move
inthe
the
original
original
direction
direction
trolley
trolley
before
the
collision.
(120)(20)
(150)(−30)
(120)(−25)
⃗
+
=
+
150𝑣𝑣
⃗
𝑣𝑣
=
+6
𝑚𝑚 ∙ 𝑠𝑠 −1
𝐵𝐵(𝑎𝑎
nswer
is
positive,
we
can
say
that
trolley
B
will
move
in
the
original
direction
of
trolley
a
answer is positive, we before
can
say
that
trolley B will move in the original direction of trolley a
the
collision.
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
before
before
the
the
collision.
collision.
Since
the
answer
is
positive,
we
can
say
that
trolley
B
will
move
in
the
original
direction
of
trolley
a
Since
the
answer
is
positive,
we
can
say
that
trolley
B
will
move
in
the
original
direction
of
trolley
a
2400
−
−300
+ original
150𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
he
hecollision.
collision.
=4500
6original
𝑚𝑚 ∙ =
𝑠𝑠 −1
in the
direction
of trolley a
𝑣𝑣⃗in
𝑓𝑓𝑓𝑓the
= 6 𝑚𝑚 ∙the
𝑠𝑠 −1−1
direction
of trolley
a
before
collision.
−1
before the𝑣𝑣⃗𝑓𝑓𝑓𝑓
collision.
𝑣𝑣⃗original
=direction
+6
𝑚𝑚 ∙ of
𝑠𝑠of
==6=6𝑚𝑚6
ininthe
direction
trolley
aaa
𝑣𝑣⃗𝑣𝑣𝑓𝑓𝑓𝑓
𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝑓𝑓𝑓𝑓
𝑚𝑚∙𝑚𝑚
∙𝑠𝑠𝑠𝑠∙−1
𝑠𝑠 −1
inthe
the
original
original
direction
oftrolley
trolley
𝑣𝑣⃗𝑓𝑓𝑓𝑓
Since the answer is positive, we can say that trolley B will move in the
−1
−1in the originalSince
the
direction
ofofanswer
trolley
aais positive, we can say that trolley B will move in the original direction of trolley A before the
𝑠𝑠 in the original
direction
trolley
1.3
−1
−1
before the collision.
6 𝑚𝑚
∙ 𝑠𝑠 indirection
the original
direction
𝑣𝑣⃗𝑓𝑓𝑓𝑓
in =
the
original
of trolley
a of trolley a
𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 1.3
Since
1.3
1.3
1.3the answer is positive, we can say that trolley B will move in the original direction of trolley a
collision.
before the collision.
𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 −1 in the original direction of trolley a
1.3
1.3
26
26
−1
in the
theoriginal
original
direction
of trolley
A
direction
of trolley
a
𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 in
26
2626
26
26
1.3
26
26
1.3
Then:
1.3
ma = 120 kg
mA = 120 kg
⃗⃗𝒊𝒊𝒊𝒊 = 20 m · 𝑠𝑠
𝒗𝒗
−1
26
to the right
⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 −1 to the left
𝒗𝒗
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 =?
Let’s take positive to the right:
24𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = ∆𝑝𝑝
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝑚𝑚(𝑣𝑣𝑓𝑓 − 𝑣𝑣𝑖𝑖 )
Commented [U70]: Pse check.
⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠
𝒗𝒗
to the left
⃗⃗𝒊𝒊𝒊𝒊 =
𝒗𝒗
20 m · 𝑠𝑠 −1 to the right
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
=?
⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂)
𝒗𝒗
= 25tomthe
· 𝑠𝑠 −1
to the left
Let’s take
positive
right:
Commented [U70]: Pse check.
𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
= take
∆𝑝𝑝 =?positive to the right:
Let’s
Let’s
− 𝑣𝑣𝑖𝑖 ) to the right:
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
= take
𝑚𝑚(𝑣𝑣𝑓𝑓positive
Commented [U70]: Pse check.
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
∆𝑝𝑝 − 20)
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
= 120=(−25
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
− ∙𝑣𝑣𝑠𝑠𝑖𝑖 )
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼= =−5400
𝑚𝑚(𝑣𝑣𝑓𝑓 𝑁𝑁
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
∙ 𝑠𝑠 𝑡𝑡𝑡𝑡
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼= =5400
120𝑁𝑁(−25
− 𝑡𝑡ℎ𝑒𝑒
20)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = −5400 𝑁𝑁 ∙ 𝑠𝑠
Worked Example
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 =2 5400 𝑁𝑁 ∙ 𝑠𝑠 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
-1
Two Worked
cars (S andExample
T) are travelling
2 on a straight road. They approach a robot at velocities of 10 m·s
East and 18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction.
worked Example 2
Commented [U71]: Formatting: pse standardise appearance of
all equivalent items, including font colour.
Commented [G72]: Better like this
Two cars (S and T) are travelling on a straight road. They approach a robot at velocities of 10 m·s-1
East and 18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction.
Commented [U71]: Formatting: pse standardise appeara
all equivalent items, including font colour.
Commented [G72]: Better like this
Car T suddenly stops and car S collides with car T. After the collision, the two cars move off together
-1
cars
(S andmass
T) are
travelling
straight
road.
as a Two
unit. The
combined
of each
car withon
theadriver
is 1500
kg. They approach a robot at velocities of 10 m·s East and
18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction.
2.1 State the law of conservation of linear momentum in words.
(2)
2.2 Calculate
the speed
of the
immediately
the collision.
(4) off together
Car T suddenly
stops
andtwo
car cars
S collides
with carafter
T. After
the collision, the two cars move
Research
has
shown
that
a force
greater
than
000the
N
during
collision
fatal the
injuries.
as aT
unit.
The combined
mass
of
each
car85with
driver
1500
kg. may
Car
suddenly
stops
and
car
S collides
with
carisa T.
After
the cause
collision,
two cars move off together as a
The collision described above lasts for 0,08 s.
unit.
The the
combined
mass of each
car
with
theindriver
1500
kg.in a fatal injury.(4)
2.3 Determine,
by
means
of calculations,
the
collision
aboveis
could
result
2.1 State
law of conservation
of whether
linear
momentum
words.
(2)
2.2 Calculate the speed of the two cars immediately after the collision.
(4)
The cars
have crumple
zones,
belts,
air bags
interiors
that canmay
reduce
thefatal
chance
Research
has shown
thatseat
a force
greater
thanand
85 padded
000 N during
a collision
cause
injuries.
2.1
State
theduring
law
of
conservation
(2)
of death
serious
injury
accidents.
Commented [U73]: Part
of 2.3 or 2.4?
Theorcollision
described
above
lasts for 0,08 s.of linear momentum in words. 2.4 Use
of by
Physics
toof
explain
how airwhether
bags can
reduce
theabove
risk ofcould
injuryresult
or death.
2.3 principles
Determine,
means
calculations,
the
collision
in a fatal injury.(4)
2.2
Calculate the speed of the two cars immediately after the collision. (4)
Solution
The cars have crumple zones, seat belts, air bags and padded interiors that can reduce the chance
of death or serious injury during accidents.
Commented [U73]: Part of 2.3 or 2.4?
2.1 The
momentum
of an
and
closed
system
remains
magnitude
and cause fatal injuries. The
2.4 total
Uselinear
principles
of Physics
toaisolated
explain
air bags
can
theconstant
risk of injury
or death.may
Research
has
shown
that
force how
greater
than
85reduce
000 N
during
aincollision
direction. OR In an isolated system, the total linear momentum of a system before a
collision/interaction
is equal to
the total
linear
collision
above
lasts
formomentum
0,08 s. of the system after the collision.
Solution described
Commented [U74]: Formatting/re-typing: pse standardise line
2.2 What to look for:
breaks, indentation, etc. of all equivalent items.
2.1 The total linear momentum of an isolated and closed system remains constant in magnitude and
27 result in a fatal injury.(4)
2.3
Determine,
by means
of calculations,
whether
collision
could
direction.
OR In an isolated
system, the
total linear momentum
of athe
system
beforeabove
a
collision/interaction is equal to the total linear momentum of the system after the collision.
Commented [U74]: Formatting/re-typing: pse standardis
2.2 What to look for:
breaks, indentation, etc. of all equivalent items.
The cars have crumple zones, seat belts, air bags and padded interiors that can reduce the chance of death
27
or serious injury during accidents.
2.4
Use principles of Physics to explain how air bags can reduce the risk of injury or death.
Solution
2.1
The total linear momentum of an isolated and closed system remains constant in magnitude and
direction. OR In an isolated system, the total linear momentum of a system before a collision/interaction
is equal to the total linear momentum of the system after the collision.
2.2
What to look for:
(i)
In which direction are the cars travelling?
(ii)
What is the mass of each car?
(iii)
Is there a possibility of the cars moving together or moving separately after the collision?
(iv)
What will happen to the speed of the cars after collision?
(v)
What is the most appropriate equation for the scenario?
25
h direction are the cars travelling?
the mass of each car?
(i)
In
which direction
are
the cars travelling?
a possibility of the cars moving
together
or moving
separately
after the
(ii)
What is the mass of each car?
?
Is there
possibility of the cars moving together or moving separately after the
ill happen to the speed of(iii)
the cars
afteracollision?
(i)
In which
direction are the cars travelling?
collision?
the most appropriate equation
for
the
scenario?
(ii)
What will
is the
mass to
of the
each
car? of the cars after collision?
(iv)
What
happen
speed
(iii)
Is
there
possibility
of the cars
moving
together
or is:
moving separately after the
general
equation
for conservation
of momentum
(v) is:
What
is athe
most
appropriate
equation
for
the scenario?
or conservation The
of momentum
collision?
(i)
In which direction are the cars travelling?
(iv) equation
happen
speed
of the cars
(ii)
What will
isfor
the
mass to
of the
each
car?
The general
conservation
of
momentum
is: after collision?
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
⃗𝒑𝒑⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
(iii)
Is
there
possibility
of the cars
moving for
together
or moving separately after the
(v) = ∑
What
is athe
most appropriate
equation
the scenario?
∑ 𝒑𝒑
collision?
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
𝒑𝒑
= cars
∑momentum
∑
𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 collision?
Thetwo
general
equation
forhappen
conservation
of
is: ⃗𝒑𝒑⃗after
(iv) were
What
will
to the
speed
of the
blem statement the
cars
moving
separately
before
collision
but
However,
from
the
problem
statement
the two
cars
were moving separately before collision but moved
(v)
What
is
the
most
appropriate
equation
for
the
scenario?
nit after the collision so the most appropriate equation to use is:
Commented [U75]: Pse check/re-write – not understood.
However, from
problem
the
separately
before collision
butis:
⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
𝒑𝒑
∑ two
∑ ⃗𝒑𝒑⃗moving
together
as athe
unit
afterstatement
the collision
socars
the=were
most
appropriate
equation
use
𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
Commented
[G76]:to
I think
it is better
in this way
general
equation
for conservation
of momentum
is:appropriate equation
moved
together
as a unit
after the collision
so the most
to use is:
Commented [U75]: Pse check/re-write – not understood.
𝑚𝑚𝑠𝑠 + 𝑚𝑚 𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) The
Commented [G76]: I think it is better in this way
= 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) (1500+1500)
However, from the problem statement the two cars were moving separately before collision but
⃗𝒑𝒑⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂
−1 𝑚𝑚𝑠𝑠 𝑣𝑣𝑣𝑣𝑠𝑠 + 𝑚𝑚 𝑇𝑇 𝑣𝑣𝑖𝑖 𝑖𝑖 𝑇𝑇 = ( 𝑚𝑚𝑠𝑠 + 𝑚𝑚 𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) ∑ 𝒑𝒑
= ∑appropriate
𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃
Commented
[G77]:
moved together as a unit after the collision⃗⃗so
the most
equation
to use
is: Symbols corrected
Commented [U75]: Pse check/re-write – not understood.
(𝑆𝑆𝑆𝑆) = 5 𝑚𝑚 ∙𝑠𝑠
(1500)(10)+(1500)(0) = 𝑣𝑣
(1500+1500)
hat the accident was fatal, the net force has𝑓𝑓(𝑆𝑆𝑆𝑆)
to be calculated
using the
Commented [G76]: I think it is better in this way
−1
=statement
5 𝑚𝑚
∙𝑠𝑠
Commented [G77]: Symbols corrected
𝑓𝑓(𝑆𝑆𝑆𝑆)
However,
𝑚𝑚𝑠𝑠 𝑣𝑣𝑣𝑣𝑠𝑠 + 𝑚𝑚from
= problem
(𝑣𝑣𝑚𝑚
𝑇𝑇 𝑣𝑣𝑖𝑖 𝑖𝑖 𝑇𝑇the
𝑠𝑠 + 𝑚𝑚
𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) the two cars were moving separately before collision but
2.3
In
order
to
prove
that
the
accident
was
fatal,
the
net
force
has
to
be
calculated
using
the
moved
together
as
a
unit
after
the
collision
so
the
most
appropriate
equation
to
use
is:
Commented [U75]: Pse check/re-write – not understood.
(1500)(10)+(1500)(0) = 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) (1500+1500)
equation: −1 N, then the accident
em statement, if the answer obtained 𝑣𝑣exceeds
85
000
Commented [G76]:
think it is
better in this way
=
5
𝑚𝑚
∙𝑠𝑠
[G77]: ISymbols
corrected
𝑓𝑓(𝑆𝑆𝑆𝑆)
⃗𝑛𝑛𝑛𝑛𝑛𝑛
⃗𝑖𝑖 𝑖𝑖mass
∆𝑡𝑡
= 𝑚𝑚
𝑚𝑚∆𝑣𝑣
𝐹𝐹
𝑚𝑚
𝑚𝑚𝑠𝑠prove
+the
𝑚𝑚accident
In
to
that
the
was
fatal,
net
force has
tothe
be calculated using the equation:
𝑠𝑠 𝑣𝑣𝑣𝑣
𝑠𝑠 +
𝑇𝑇 𝑣𝑣
𝑇𝑇 = (and
𝑇𝑇 )𝑣𝑣
𝑓𝑓(𝑆𝑆𝑆𝑆)
he accident was 2.3
not
fatal.
The
velocity
of
either
offatal,
the cars
before
2.3
In
order
toorder
prove
that
wasaccident
the net
force
hasthe
to be
calculated
using
According
to the problem
statement,
if the answer obtained exceeds 85 000 N, then the accident
(1500+1500)
(1500)(10)+(1500)(0)
= 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆)
be used.
equation:
was fatal. If not, then𝑣𝑣the accident
was
not fatal. The mass and velocity of either of the cars before
−1
=
5
𝑚𝑚
∙𝑠𝑠
Commented [G77]: Symbols corrected
⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣⃗
𝑓𝑓(𝑆𝑆𝑆𝑆)
𝐹𝐹
and
after collision can be used.
2.3
In ordertotothe
prove
that the
accidentif was
fatal, theobtained
net force
has to be
using
the
According
problem
statement,
the answer
exceeds
85 calculated
000 N, then
the accident
equation:
)
was
fatal. If not, then the accident was not fatal. The mass and velocity of either of the cars before
𝐹𝐹⃗ ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣⃗
⃗to the
and
after
collision
canproblem
be used. statement, if the answer obtained exceeds 85 000 N, then the accident was fatal. If
∆𝑡𝑡
= 𝑚𝑚∆𝑣𝑣
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛
According
𝑛𝑛𝑛𝑛𝑛𝑛
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) =
(5-0) statement, if the answer obtained exceeds 85 000 N, then the accident
According
to 1500
the problem
⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08)
not,
then
accident
was not
TheThe
mass
either
of cars
the before
cars before and after collision can
𝐹𝐹
was
fatal.
If=the
not,
then the accident
wasfatal.
not fatal.
massand
and velocity
velocity ofof
either
of the
⃗7500
𝑚𝑚∆𝑣𝑣
𝐹𝐹
𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 =
7500
⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08)
and
after
collision
can
be
used.
𝐹𝐹
=
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛used.
be
0,08= 1500 (5-0)
greater than 85 000
N, so the accident will be fatal.
7500
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08)
= 93 =
750
𝑁𝑁
𝑚𝑚∆𝑣𝑣⃗ ⃗
𝐹𝐹⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 =
7500
𝐹𝐹
=
than 85 000 N, so the accident will be fatal.
In
both
cases,
𝐹𝐹depends
𝑛𝑛𝑛𝑛𝑛𝑛the collision
𝑛𝑛𝑛𝑛𝑛𝑛 is greater
of the impact during
Commented [U78]: Pse check.
0,08
⃗
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) = 1500
(5-0) on the relationship between the Fnet
ollision when the𝐹𝐹⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛
momentum
is
constant.
When
the air bag inflates during the
=
93
750
𝑁𝑁
[G79]: I think it is better in this way
(0,08)
= 7500 of the impact during the collision depends on theCommented
𝐹𝐹
𝑛𝑛𝑛𝑛𝑛𝑛 The magnitude
2.4
relationship between the Fnet
Commented [U78]: Pse check.
tact time of the passenger
or driver
with
the airthan
bag 85
is longer
without
air will be fatal.
7500
iscollision
greater
000
N,than
so the
accident
In
both
cases,
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛of
⃗𝑛𝑛𝑛𝑛𝑛𝑛
𝐹𝐹
=
and
the
time
when
the
momentum
is
constant.
When
the
air
bag
inflates during the
Commented [G79]: I think it is better in this way
force on the passenger0,08
or driver is reduced according to the equation (F net
collision, the contact time of the passenger or driver with the air Commented
bag is longer
than
air
[U80]:
Psewithout
check.
⃗𝑛𝑛𝑛𝑛𝑛𝑛 collision.
=The
93magnitude
750 𝑁𝑁
p is constant for 𝐹𝐹
the
2.4
of
the
impact
during
the
collision
depends
on
the
relationship
between
the
F
Commented [U78]: Pse check.
net
bag and thus the force on the passenger or driver is reduced according to the equation (F net
⃗𝑛𝑛𝑛𝑛𝑛𝑛ofiscollision
and
the
time
when
the
momentum
is
constant.
When
the
air
bag
inflates
during
the
greater
than
85
000
N,
so
the
accident
will
be
fatal.
In
both
cases,
𝐹𝐹
∆𝑝𝑝
Commented [G79]:
I think
it is better in this way
[U80]: Pse
check.
In both
greater
than
N, so the accident will be fatal.
= ∆𝑡𝑡 cases,
), becauseisΔp
is constant
for85
the000
collision.
collision,
the contact time of the passenger or driver with the air bag is longer than without air
and
thus below
theofforce
on
theinitially
passenger
or driver
is reduced
according
to thebetween
equationthe
(F net
2.4 bag
The
magnitude
the
during
the collision
depends
on the
relationship
Fnet
Commented [U78]: Pse check.
s time graph is shown
in the
graph
forimpact
a car
moving
horizontally
∆𝑝𝑝 the time of collision when the momentum is constant. When the air bag inflates during the
Commented
[U80]:
check.
shown below. 1.4 and
The
magnitude
of the impact
during the collision depends on the relationship between
the[G79]:
Fnet Pse
= ∆𝑡𝑡
), because
Δp is constant
for the collision.
Commented
Iand
think
it the
is better in this way
The momentum
versus
time
graph
is passenger
shown in the
forair
a bag
car initially
moving
horizontally
collision, the
contact
time
of the
or graph
driver below
with the
is longer
than without
air
towards
the
EAST.
Is
below.
time
of
collision
the momentum
constant.
When
theequation
air bag(Finflates
during the collision, the
bag
and
thus
theshown
forcewhen
on the passenger
or driver is
is reduced
according
to the
net
∆𝑝𝑝
Commented [U80]: Pse check.
=
),
because
Δp
is
constant
for
the
collision.
contactversus
time of
thegraph
passenger
driver
with
theforair
bag
is longer
without air bag
and thus the force
The momentum
time
is shown or
in the
graph
below
a car
initially
movingthan
horizontally
∆𝑡𝑡
towards the EAST. Is shown below.
on the passenger or driver is reduced according to the equation (Fnet = ), because Δp is constant for the
The momentum
versus time graph is shown in the graph below for a car initially moving horizontally
collision.
towards the EAST. Is shown below.
The momentum versus time graph is shown in the graph below for a car initially moving horizontally towards
the EAST. Is shown below.
28
28
28
28
26
1.1
Define momentum. (2)
1.2
What is the value of the initial momentum? (1)
1.3
In which time interval is the momentum of the car constant.
(1)
1.4
At what time (s) did the car stop?
(3)
1.5
Compare the direction in which the car is moving from: 0 to10 s; 10 to 15 s; and 15 to 25 s.
(2)
1.6
There are many reasons for the change in momentum. Give any two possible reasons that could have
caused the momentum of the car to change from 10 s to 20 s.
(2)
1.7
Define impulse of a force. (2)
1.8
Calculate the change in momentum of the car from the starting point to 15s. (3)
[16]
4.3 Vertical Projectile Motion
PROJECTILE MOTION PROBLEM SOLVING STRATEGY
Step 1: Read the problem as many times as you need to, then draw a sketch of the situation (translate the
words into a sketch).
Step 2: List all the given information and convert it to SI units, if necessary.
Step 3: Draw a free-body diagram.
•
If the only force acting on the object is the gravitational force, then it is a projectile and the equations of
uniformly accelerated motion can be used.
Step 4: Select the formula or equation that will be used to answer the question. In some cases, more than one
equation is needed to calculate the answer.
Step 5: Substitute numerical values into an appropriate equation.
Step 6: Check your answer once you have and answer, check if it makes sense:
•
Does it has the correct unit?
•
Is the numerical value reasonable?
27
Exercise 1
A boy is standing on top of a building. He throws a ball vertically upwards from a position 3.5 m above
the ground, with an initial velocity of 10 m∙s-1. Ignore the effects of air resistance and answer the following
questions:
1.1
What is the magnitude and direction of the acceleration of the ball?
1.2
Calculate the maximum height above the ground that the ball reaches.
1.3
What was the velocity of the ball at its maximum height?
1.4
Calculate the time taken by the ball to reach its maximum height.
1.5
How much time did the ball take to reach its maximum position and return to the position from which it
was thrown?
1.6
Calculate the total time taken by the ball to reach the ground.
1.7
Calculate the velocity with which the ball hits the ground.
1.8
Draw a rough sketch of the velocity-time graph. Show relevant points on the velocity and time axis.
1.9
Draw a position-time graph.
Exercise 2
A boy is standing on the top of a building and throws a tennis ball vertically upwards. The graph below shows
the velocity of the ball from the moment it is thrown until it hits the ground for the second time. Ignore air
resistance.
Graph of velocity versus time
2.1
Describe the motion of the tennis ball in words.
2.2
What is the initial velocity of the tennis ball?
2.3
Calculate the height of the building.
2.4
Is the collision of the tennis ball with the ground elastic or inelastic? Explain.
28
2.5
Draw the position-time graph.
4.4 Work Energy Power
PROBLEM-SOLVING STRATEGY FOR WORK ENERGY POWER
STEP 1: Read and model the situation
•
Read the problem as many time as you need to, in order to understand it.
•
Draw a sketch of the situation (if not given) and identify which objects are parts of the system. Some
problems may need to be sub-divided into two or more parts (scenarios).
STEP 2: Visualize
STRATEGY FOR WORK ENERGY POWER
Draw aPROBLEM-SOLVING
free-body diagram
that show all the forces acting on the object (objects). Identify the type of force
acting on
the object (conservative
or non-conservative)
PROBLEM-SOLVING
STRATEGY FOR WORK
ENERGY POWER
STEP 1: Read and model the situation
 Read
the problem STRATEGY
as many time
youENERGY
need to,POWER
in order to understand it.
PROBLEM-SOLVING
FOR as
WORK
STEP
Solve
STEP
Readofand
the(ifsituation
 3:
Draw
a1:
sketch
themodel
situation
not given) and identify which objects are parts of the system. Some
needastomany
be sub-divided
into
twoto,
or in
more
parts
(scenarios).it.
 problems
Read the may
problem
time as you
need
order
to understand
STEP
2:sketch
Visualize
 STEP
Draw a1:
of themodel
situation
(if
not given) and identify which objects are parts of the system. Some
Read
and
the
situation
•
Collect
the
data
andbewrite
it in symbol
form.
Draw
athe
free-body
diagram
that
allinto
thetwo
forces
acting
onunderstand
the
object (objects).
Identify the type of
problems
may
need
sub-divided
(scenarios).
PROBLEM-SOLVINGSTRATEGY
FOR
WORK
ENERGY
POWER
Read
problem
asto
many
timeshow
as you
need
to,orinmore
orderparts
to
it.
force
acting
on the object (conservative or non-conservative)
STEP
2:
Visualize

Draw
a
sketch
of
the
situation
(if
not
given)
and
identify
which
objects
are
parts
of
the
system.
Some
•
Select
equation,
law
(principle)
orforces
theorem
that
you
can(objects).
apply to
solve
the
problem.
STEP
Solve
Draw a3:the
free-body
diagram
that
show all
thetwo
acting
on the
object
Identify
the
type
of
problems
may need
to be sub-divided
into
or more
parts
(scenarios).
 model
Collect
the
data
and
write
it
in
symbol
form.
force
acting
on
the
object
(conservative
or
non-conservative)
STEP 1: Read and
the
situation
STEP 2: Visualize
• asIfSelect
the asystem
isdiagram
isolated
non-dissipative,
theobject
total
mechanical
energy
STEP
3:
Solve
the
equation,
law to,
(principle)
or
theorem
thatit.
youthen
can
to solve
the problem.
 Read the problem
many
time
as you
need
inand
order
toall
understand
Draw
free-body
that
show
the forces
acting
onapply
the
(objects).
Identify
the typeisofconserved:
 situation
If
the system
is given)
isolated
and
non-dissipative,
thenare
theparts
totalof
mechanical
energy
Collect
the
data
and
write
it in
symbol
form.
 Draw a sketch of the
(if not
and
identify
which
objects
the system.
Someis conserved:
force
acting
on
the
object
(conservative
or
non-conservative)
𝐸𝐸3:
=equation,
𝐸𝐸𝐾𝐾𝐾𝐾 +into
𝐸𝐸𝑃𝑃𝑃𝑃law
= 𝐸𝐸(principle)
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
 to𝐸𝐸
Select
the
or theorem
that you can apply to solve the problem.
problems may need
be+sub-divided
two
or
parts
(scenarios).
𝐾𝐾𝐾𝐾
𝑃𝑃𝑃𝑃
𝑀𝑀 more
STEP
Solve
STEP 2: Visualize
the system
is isolated
then theortotal
mechanicalthen
energy
conserved:
IfIf there
are data
non-conservative
forces
acting
dissipative),
use is
equation:
 Collect
the
and writeand
it innon-dissipative,
symbol
form.(external
Draw a free-body
show
all
the
forces
acting
ontheorem
the object
typethe
of problem.
𝐸𝐸𝐾𝐾𝐾𝐾
+that
𝐸𝐸the
=equation,
𝐸𝐸𝐾𝐾𝐾𝐾
+=𝐸𝐸
=+
𝐸𝐸𝑀𝑀
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑊𝑊
∆𝐸𝐸
∆𝐸𝐸=
𝑃𝑃𝑃𝑃are
𝑃𝑃𝑃𝑃
𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐾𝐾
𝑝𝑝
• diagram
IfSelect
there
non-conservative
forces
acting
(external
dissipative),
then use equation:
law
(principle)
or
that(objects).
you
can Identify
applyor
tothe
solve
force acting on the object
(conservative
or non-conservative)
If the
there
are
non-conservative
forces acting (external
or
dissipative),
then
For
both
conservative
and
forces
acting,
you can use
theuse
work
energy theorem:
If
system
is isolated
andnon-conservative
non-dissipative,
then the
total
mechanical
energy
isequation:
conserved:
STEP 3: Solve
𝑊𝑊
∆𝐸𝐸=
+𝑀𝑀∆𝐸𝐸
∆𝐸𝐸=𝐾𝐾 𝐸𝐸𝐾𝐾𝐾𝐾 + =
𝑊𝑊
𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐾𝐾 𝐸𝐸
𝑛𝑛𝑛𝑛𝑛𝑛
𝐸𝐸
+=𝐸𝐸𝑃𝑃𝑃𝑃
𝐸𝐸𝑃𝑃𝑃𝑃
=𝑝𝑝𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐾𝐾𝐾𝐾
 Collect the data and
it inare
symbol
form. and
Kinematics
equations
other
laws(external
must
be acting,
used
with
problems.
For
both
conservative
andsome
non-conservative
forces
yousome
can use
theuse
work
energy theorem:
 write
If
there
non-conservative
forces
acting
or dissipative),
then
equation:
 Select the equation,
(principle)
or
theorem
that
you canequation.
apply to solve the problem.
 law
Substitute
the
values
in
the
selected
=
∆𝐸𝐸
𝑊𝑊
𝑛𝑛𝑛𝑛𝑛𝑛
𝐾𝐾
𝑊𝑊
=
∆𝐸𝐸
+
∆𝐸𝐸
𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐾𝐾
𝑝𝑝 non-conservative forces acting, you can use the work energy theorem:
For
both
conservative
and
 If the system•is isolated
and
non-dissipative,
thensome
the total
energy
conserved:problems.
Kinematics
equations and
and
othermechanical
laws must
beacting,
usediswith
 For
both conservative
non-conservative
forces
you some
can use the work energy theorem:
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 Substitute
= 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾the values in the selected equation.
 If there are non-conservative
forces
acting (external
dissipative),
use equation:
 Kinematics
equations
and someorother
laws mustthen
be used
with some problems.
wORKEd
ExamPLE
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =
𝐾𝐾 + ∆𝐸𝐸𝑝𝑝 the
 ∆𝐸𝐸
Substitute
values in the selected equation.
•
Kinematics
equations
and
some
other
laws
must
be used
with some problems.
 For both conservative and non-conservative forces acting, you can use the work energy
theorem:
1. The
sketch ExamPLE
below shows a 2,0 kg block sliding from A to B along a frictionless surface. When the block
wORKEd
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾
reaches
B, other
it the
continues
to slide
along
a horizontal
surface (BC), where a kinetic frictional force acts on
•
Substitute
values
inbethe
selected
equation.
 Kinematics equations
and some
laws
must
used
with
some
problems.
the
block.
As
a result,
theablock
slows
at C.aThe
kinetic energy
the block
at A
1. wORKEd
The
sketch
below
shows
2,0 kg
blockdown,
slidingcoming
from Ato
torest
B along
frictionless
surface.ofWhen
the block
 Substitute the values
in
the
selected
equation.
ExamPLE
is
40,0 J;B,
theit height
of Atoand
B is
14,0am
and 8,0 m
above(BC),
the ground,
reaches
continues
slide
along
horizontal
surface
where arespectively.
kinetic frictional force acts on
WORKED
EXAMPLE
the block.
As
a result,
thea block
to B
rest
at C.
The kinetic surface.
energy of
the block
at A
1. The
sketch
below
shows
2,0 kgslows
block down,
slidingcoming
from A to
along
a frictionless
When
the block
is 40,0 J;B,the
height of A
is 14,0
m and 8,0 surface
m above(BC),
the ground,
reaches
it continues
toand
slideBalong
a horizontal
where arespectively.
kinetic frictional force acts on
WORKED EXAMPLEthe block. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A
1.
The sketch below shows a 2,0 kg block sliding from A to B along a frictionless surface. When the block
is 40,0 J; the height of A and BAis 14,0 m and 8,0 m above the ground, respectively.
1. The sketch below shows
a 2,0 kg
sliding fromtoA slide
to B along
a frictionless
surface.
When the
blockwhere a kinetic frictional force acts on
reaches
B,block
it continues
along
a horizontal
surface
(BC),
reaches B, it continues to slide along a horizontal
force acts on
A m(BC), where a kinetic frictional
hAsurface
= 14,0
C kinetic energy of the block at A is
B
As adown,
result,
the block
slows
down,
rest
at atC.A The
the block. As a result,the
theblock.
block slows
coming
to rest at
C. The
kineticcoming
energy oftothe
block
hB= hc= 8,0 m
is 40,0 J; the height of A and B is 14,0 m and 8,0h m
above
ground, respectively.
14,0
40,0 J; the height of A and
B ismthe
14,0
m and 8,0 m above
A= A
C respectively.
B the ground,
hB= hc= 8,0 m
1.1. Is the total mechanical
hA=energy
14,0 mof the block conserved as Bthe block goes from
C A to B? Why or
why not?
h
B= hc= 8,0 m
A the block reaches point B, has its kinetic energy INCREASED, DECREASED or
1.2.
1.1. When
Is the total
mechanical energy of the block conserved as the block goes from A to B? Why or
REMAINED
why not? THE SAME? Provide a reason for your answer.
= 14,0 the
mtheblock
A
1.3.
Calculate
speed
of the block
when
it reaches
B. Cenergy INCREASED, DECREASED or
1.2. hIs
When
reaches
point
B,Bblock
has
its
kinetic
1.1.
the total mechanical
energy
of the
conserved
as the block goes from A to B? Why or
hB=conserved
hfor
manswer.
1.4. Is
the total mechanical
energy
of the
block
as the block goes from B to C? Justify
c= 8,0
REMAINED
THE
SAME?
Provide
a
reason
your
why not?
your
answer.
1.3. When
Calculate
speed
of the block
it reaches
B. energy INCREASED, DECREASED or
1.2.
thethe
block
reaches
point when
B, has
its kinetic
1.5.
How
does the
kinetic
do during
the block
BC segment
of the
1.4. REMAINED
Is themuch
total work
mechanical
energy
of frictional
the
blockforce
conserved
as the
goes from
B totrip?
C? Justify
THE SAME?
Provide
a reason
for your answer.
your
answer.
1.1. Is the total mechanical
energy
of
block
as the
block goes
1.3. Calculate
thethe
speed
of conserved
the block when
it reaches
B. from A to B? Why or
SOLuTION:
1.5. Is
How
work does the
kinetic
frictional
force
do during
theblock
BC segment
of B
the
why not?
1.4.
themuch
total
mechanical
energy
of
conserved
as the
from
totrip?
C?
Justify
1.1. Is
the
total
mechanical
energy
ofthe
theblock
block
conserved
as thegoes
block
goes
from
A to B? Why or why not?
STEP
1:
Read
and
1.2. When the block
reaches
point
B,model
has its kinetic energy INCREASED, DECREASED or
your answer.
There
is a Provide
sketch ofa the
situation,
so answer.
we do not need to draw the sketch.
SOLuTION:
REMAINED THE
SAME?
reason
for
your
1.5. How much work does the kinetic frictional force do during the BC segment of the trip?
The
block
isblock
interacting
the surface
and the Earth.
STEP
and
model
1.3. Calculate
of1:
theRead
when
itwith
reaches
B. B, has
1.2.the speed
When
the
block
reaches
point
its kinetic energy INCREASED, DECREASED or REMAINED THE
There
is
a
sketch
of
the
situation,
so
we
do
need
to draw
1.4. Is the total mechanical
energy
of
the
block
conserved
as
thenot
block
goes
from the
B tosketch.
C? Justify
SOLuTION:
31
The
block
is
interacting
with
the
surface
and
the Earth.
SAME?
Provide
a
reason
for
your
answer.
your answer. STEP 1: Read and model
1.5. How much workThere
does is
thea kinetic
force dosoduring
segment
ofthe
thesketch.
trip?
sketch frictional
of the situation,
we dothe
notBC
need
to draw
31
The block is interacting with the surface and the Earth.
SOLUTION:
STEP 1: Read and Model
31
There is a sketch of the situation, so we do not need to draw the sketch.
29
The block is interacting with the surface and the Earth.
31
1.3.
Calculate the speed of the block when it reaches B.
1.4.
Is the total mechanical energy of the block conserved as the block goes from B to C? Justify your
answer.
1.5.
How much work does the kinetic frictional force do during the BC segment of the trip?
SOLUTION:
STEP 1: Read and Model
There is a sketch of the situation, so we do not need to draw the sketch.
The block is interacting with the surface and the Earth.
System: block-surface-Earth
STEP 2: Visualize
System: block-surface-Earth
STEP 2: Visualize
Free-body
Free-body
diagram diagram
Part 1 (from A to B) the only two forces acting are gravitational force and normal force.
Commented [U88]: Pse check/clarify.
System: block-surface-Earth
Sliding down
Sliding up
Commented [GI89]: Better like this
Part
1 (from A to B) the only two forces acting
are gravitational force and normal force.
STEP
2: Visualize
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
Free-body diagram
𝑁𝑁
𝑁𝑁
Part 1 (from A to B) the only two forces acting are gravitational force and normal force.
Commented [U88]: Pse check/clarify.
Sliding down
Sliding
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑖𝑖𝑖𝑖 up
𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Sliding(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
down
Sliding up
𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Commented [GI89]: Better like this
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐹𝐹⃗𝑔𝑔 - Conservative
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁
𝐹𝐹⃗𝑔𝑔 - Conservative
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐹𝐹⃗𝑔𝑔 - Conservative
From A to B there are \only conservative forces acting therefore
the system is isolated.
Part 2 Sliding𝐹𝐹⃗on
a horizontal surface (from B to C)
- Conservative
𝑔𝑔
From A to B there are \only conservative forces acting therefore the system is isolated.
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁
Part 2 Sliding on a horizontal surface
(from B to C)
⃗⃗⃗⃗
𝑓𝑓𝑓𝑓
Non- conservative
⃗⃗⃗⃗
𝑓𝑓𝑓𝑓
Non- conservative
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐹𝐹⃗𝑔𝑔 - Conservative
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obscured on a page or disappears off a page.
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Commented
[GI92]:
like thispse ensure that no text is
Commented
[U90]:Better
Formatting:
obscured on a page or disappears off a page.
Commented [U91]: Pse check/clarify.
Commented [GI92]: Better like this
From BFrom
to C the
frictional
force
is acting
the block which
is not acting
conservative
therefore
system is isolated.
A to
B there
are
\only on
conservative
forces
therefore
thethe
system
is not isolated.
Commented [U93]: Pse check/clarify.
𝐹𝐹⃗𝑔𝑔 - Conservative
STEP 3: Solve
Commented [GI94]: Better like this
Data Part 2 Sliding on a horizontal surface (from B to C)
From B to C the frictional force is acting on the block which is not conservative therefore the system
m= 2,0 kg
is not isolated.
Commented [U93]: Pse check/clarify.
EKA= 40 J
STEP 3: Solve
Commented
[GI94]: Better
this
m B to C the frictional force is acting on the block which is not conservative therefore
hA= 14,0
From
the system
is like
not
Data
hB= hc= 8,0 m
m= 2,0 kg
vB-? isolated.
EKA= 40 J
Wff-?
h = 14,0 m
1.1.A The total mechanical energy is conserved if the net work done by the non- conservative forces
= 8,0 m3: Solve
hB= hSTEP
isc zero, or only conservative forces act on the object, or W nc = 0J. Only two forces act on the
vB-?
block during its trip from A to B: the gravitational force, which is conservative; and the normal
Wff-?
force, which is conservative in this case. Thus, we conclude that W nc = 0 J, with the result being
1.1. Data
The total mechanical energy is conserved if the net work done by the non- conservative forces
that the total mechanical energy is conserved during the AB part of the trip.
is zero, or only conservative forces act on the object, or W = 0J. Only two forces act on the
1.2. As we have seen, the total mechanical energy is the sum of thenckinetic and gravitational potential
block during its trip from A to B: the gravitational force, which is conservative; and the normal
energy,
and
remains constant from A to B. Therefore, as one type of energy decreases, the
m=
2,0which
kgit is
force,
conservative in this case. Thus, we conclude that W nc = 0 J, with the result being
other must increase, in order for the sum to remain constant. Since
B is lower than A, the
that the total mechanical energy is conserved during the AB part of the trip.
gravitational potential energy at B is less than that at A. As a result, the kinetic energy at B must
1.2. EAs =
we40
have
the total mechanical energy is the sum of the kinetic and gravitational potential
J seen,
be KA
greater
than
that at A.
energy, and it remains constant from A to B. Therefore, as one type of energy decreases, the
1.3. From A to B, total mechanical energy is conserved.
other must increase, in order for the sum to remain constant. Since B is lower than A, the
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
that at A. As a result, the kinetic energy at B must
= 14,0 m potential energy at B is less than
hgravitational
1
A
be
greater than that at A. 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵
2
1.3. From A to B, total mechanical energy is conserved.
1
𝐸𝐸
=
𝐸𝐸
+
𝐸𝐸𝑃𝑃𝑃𝑃
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐸𝐸𝐾𝐾𝐾𝐾𝐸𝐸+
+
𝑚𝑚𝑚𝑚ℎ
=
𝑚𝑚𝑣𝑣=2 𝐸𝐸
+𝑀𝑀𝑚𝑚𝑚𝑚ℎ
𝑃𝑃𝑃𝑃
𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾
𝐴𝐴
𝐵𝐵
2 1 𝐵𝐵
2
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵
2
32
30
1
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵
2
A to B there are \only conservative forces acting therefore the system is isolated.
Sliding on a horizontal surface (from B to C)
Commented [U90]: Formatting: pse ensure that no text is
obscured on a page or disappears off a page.
Commented [U91]: Pse check/clarify.
Commented [GI92]: Better like this
⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
hB= hc= 8,0 m
⃗⃗⃗⃗
𝑓𝑓𝑓𝑓
Non- conservative
vB-?
Wff-?
𝐹𝐹⃗𝑔𝑔 - Conservative
B to C the frictional force is acting on the block which is not conservative therefore the system
isolated.
Commented [U93]: Pse check/clarify.
1.1. The total mechanical energy is conserved if the net work done by the non- conservative forces is zero,
3: Solve
Commented [GI94]: Better like this
0 kg
40 J
4,0 m
= 8,0 m
or only conservative forces act on the object, or Wnc = 0J. Only two forces act on the block during its trip
from A to B: the gravitational force, which is conservative; and the normal force, which is conservative
in this case. Thus, we conclude that Wnc = 0 J, with the result being that the total mechanical energy is
conserved during the AB part of the trip.
The total mechanical energy is conserved if the net work done by the non- conservative forces
1.2. As we
have
seen,
the total
mechanical
energy
is the
sum of the kinetic and gravitational potential
is zero, or only conservative
forces
act on
the object,
or W nc
= 0J. Only two
forces act
on the
block during its trip from A to B: the gravitational force, which is conservative; and the normal
energy, and it remains constant from A to B. Therefore, as one type of energy decreases, the other must
force, which is conservative in this case. Thus, we conclude that W nc = 0 J, with the result being
that the total mechanicalincrease,
energy is conserved
the sum
AB part
the trip. constant. Since B is lower than A, the gravitational potential
in orderduring
for the
to ofremain
As we have seen, the total mechanical energy is the sum of the kinetic and gravitational potential
energy from
at BAistoless
than thatasatone
A.type
As aof result,
the kineticthe
energy at B must be greater than that at A.
energy, and it remains constant
B. Therefore,
energy decreases,
other must increase, in order for the sum to remain constant. Since B is lower than A, the
gravitational potential energy at B is less than that at A. As a result, the kinetic energy at B must
1.3. From A to B, total mechanical energy is conserved.
be greater than that at A.
From A to B, total mechanical energy is conserved.
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
1
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵
2
1
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵
2
32
1
× 2 × 𝑣𝑣𝐵𝐵2 + (2 × 9,8 × 8)
2
40 + 274,4 = 𝑣𝑣𝐵𝐵2 + 156,8
314,4 − 156,8 = 𝑣𝑣𝐵𝐵2
1
𝑣𝑣𝐵𝐵2 =
157,6
40 + (2 × 9,8
× 14)
= × 2 × 𝑣𝑣𝐵𝐵2 + (2 × 9,8 × 8)
2
𝑣𝑣𝐵𝐵40
=+
√157,6
274,4 = 𝑣𝑣 2 + 156,8
40 + (2 × 9,8 × 14) =
𝑣𝑣𝐵𝐵 = 12,55 m·s-1
𝐵𝐵
314,4 − 156,8 = 𝑣𝑣𝐵𝐵2
𝑣𝑣𝐵𝐵2 = 157,6
1.4.
During the trip from B to C, the frictional force acts
on the block. This force is non-conservative
1.4. During
the
trip
from
B
to
C,
the
frictional
force
acts
on the block. This force is non-conservative and does
𝑣𝑣
𝐵𝐵 =
and does work on
the block, consequently. The
net√157,6
work done by the non–conservative force
-1
𝑣𝑣
=
12,55
m·s
𝐵𝐵
is
not
zero
(W
≠
0),
so
the
total
mechanical
energy
is
notby
conserved
during the BC part force
of the is not zero (W ), so
work on the block,ncconsequently. The net work done
the non–conservative
nc
trip.
the
mechanical
isBmechanical
not
during
the
part
of
the
trip.
1.4.
During
theenergy
trip
to C,conserved
the frictional
on BC
the block.
This
is non-conservative
1.5. total
During
the
BC part,
thefrom
total
energyforce
is notacts
conserved,
because
aforce
kinetic
frictional
and does work on the block, consequently. The net work done by the non–conservative force
force is present.
is not zero (W nc≠ 0), so the total
energy is not conserved during the BC part of the
𝑊𝑊𝑛𝑛𝑛𝑛mechanical
= ∆𝐸𝐸𝑀𝑀
1.5. During the BC
energy is not conserved, because a kinetic frictional force is
trip.part, the total mechanical
𝑊𝑊𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝
1.5.
During the BC part,
the=total
energy
is not conserved, because a kinetic frictional
present.
(𝐸𝐸𝐾𝐾𝐾𝐾mechanical
− 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ
𝑊𝑊𝑛𝑛𝑛𝑛
𝐶𝐶 − ℎ𝐵𝐵 )
force is present.
1
1
𝑊𝑊𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2𝑊𝑊+
= ∆𝐸𝐸𝐶𝐶𝑀𝑀− ℎ𝐵𝐵 )
𝑛𝑛𝑛𝑛𝑚𝑚𝑚𝑚(ℎ
2
2
𝑊𝑊𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝
1
1
2 × (12,55)2 + 2 × 9,8 × (8 − 8)
𝑊𝑊𝑛𝑛𝑛𝑛 = ( 2 × 02 −𝑊𝑊𝑛𝑛𝑛𝑛
2
2 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 )
12 + 02
𝑊𝑊
= 01− (12,55)
2
𝑊𝑊𝑛𝑛𝑛𝑛
𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶 − 𝑚𝑚𝑣𝑣𝐵𝐵 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 )
2 𝐽𝐽
𝑊𝑊𝑛𝑛𝑛𝑛 =2 −157,50
1
1
2 points in the opposite direction of
The work done by frictional force𝑊𝑊
is –157,50
J, 2because
this force
𝑛𝑛𝑛𝑛 = ( 2 × 0 − 2 × (12,55) + 2 × 9,8 × (8 − 8)
2
2
the displacement.
𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 + 0
Alternative solution:
𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽
Applying the work energy theorem:
The work done by frictional force is 𝑊𝑊
–157,50
J, because this force points in the opposite direction of
𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾
the displacement.
+ 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊J,
(𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾this
)
The work done
by frictional force 𝑊𝑊
is𝑓𝑓𝑓𝑓 –157,50
force points in the opposite direction of the
𝑁𝑁 =because
Alternative solution:
1
1
2
0
0
displacement.Applying the work 𝑊𝑊
90 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁90 = ( 𝑚𝑚𝑣𝑣𝐶𝐶 − 𝑚𝑚𝑣𝑣𝐵𝐵2 )
energy
theorem:
𝑓𝑓𝑓𝑓 + 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
2
2
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛1 = ∆𝐸𝐸𝐾𝐾
1
2
(12,55)
(
−
𝑊𝑊𝑓𝑓𝑓𝑓 + 0 + 0 =
2
×
0
2
×
𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊𝑁𝑁 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 2𝐸𝐸𝐾𝐾𝐾𝐾 )
2
2
2
1
1
−0(12,55)
𝑛𝑛𝑛𝑛 = 090
𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
+ 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁900 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 )
2
2
𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽
1
1
2 opposite direction of
The work done by frictional force is –157,50
in the
0 = ( 2this
− 2points
𝑊𝑊𝑓𝑓𝑓𝑓 + 0J,+because
× 02force
× (12,55)
2
2
the displacement.
𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2
STEP 4: Assess
−157,50
𝐽𝐽 they are reasonable and answer
𝑛𝑛𝑛𝑛 =for
work (J),
The results have the correct units for speed (m·s-1)𝑊𝑊and
The work done by frictional force is –157,50 J, because this force points in the opposite direction of
the questions.
the displacement.
STEP 4: Assess
The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer
the questions.
31
𝑊𝑊𝑛𝑛𝑛𝑛 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 )
1
1
𝑊𝑊𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 )
2
2
1
1
𝑊𝑊𝑛𝑛𝑛𝑛 = ( 2 × 02 − 2 × (12,55)2 + 2 × 9,8 × (8 − 8)
2
2
𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 + 0
𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽
Alternative solution:
The work done by frictional force is –157,50 J, because this force points in the opposite direction of
the displacement.
Alternative
solution:
Applying the work
energy
theorem:
Applying the work energy theorem:
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾
𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊𝑁𝑁 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 )
1
1
𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 900 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁900 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 )
2
2
1
1
𝑊𝑊𝑓𝑓𝑓𝑓 + 0 + 0 = ( 2 × 02 − 2 × (12,55)2
2
2
𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2
𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽
The work done by frictional force is –157,50 J, because this force points in the opposite direction of
the displacement.
STEP
Assess force is –157,50 J, because this force points in the opposite direction of the
The work done
by4:frictional
The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer
displacement.the questions.
STEP 4: Assess
The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer the
questions.
33
32
Exercise 1
A donkey pulls a cart of mass 600 kg from rest along a horizontal road. The donkey applies a constant force
of magnitude 191,7 N at an angle of 30° to the horizontal. The cart accelerates and reaches a speed of 3
m.s-1 in 5 minutes. The average frictional force that acts on the cart is 160,02 N. Ignore the effect of the
rotation of the wheels of the cart.
Exercise 1
A donkey pulls a cart of mass 600 kg from rest along a horizontal road. The donkey applies a
constant force of magnitude 191,7 N at an angle of 30° to the horizontal. The cart accelerates and
reaches a speed of 3 m.s-1 in 5 minutes. The average frictional force that acts on the cart is 160,02
N. Ignore the effect of the rotation of the wheels of the cart.
1.1
State the work-energy theorem in words.
(2)
1.2
Use the WORK-ENERGY THEOREM to calculate the distance covered by the cart in 5
minutes.
(5)
1.3
Calculate the power developed by the donkey in 5 minutes.
(3)
The donkey now applies a force of the same magnitude on the cart, during the same
time period and over the same distance, but at a SMALLER ANGLE to the horizontal.
1.1
1.2
State the work-energy theorem in words.
(2)
Commented [U95]: Plse check.
1.4 Use the WORK-ENERGY
How does the THEOREM
power developed
donkey
nowbycompare
developed
to calculateby
thethe
distance
covered
the cart into the power
Commented [U96]: Formatting/re-typing: pse standardise
5 minutes.
(5) SMALLER
appearance of terminology in text. Repetitive problems seen.
by the donkey in QUESTION 1.3? Write down only GREATER THAN,
1.3
Calculate the power developed by the donkey in 5 minutes.
(3)
THAN or EQUAL TO.
(2)
The donkey now applies a force of the same magnitude on the cart, during the same
time period and over the same distance, but at a SMALLER ANGLE to the
Give a reason for the answer. (No calculations are required.)
horizontal.
1.4
How does the power developed by the donkey now compare to the power developed
by the donkey in QUESTION 1.3? Write down only GREATER THAN, SMALLER
THAN or
Exercise
2 EQUAL TO.
Give a reason for the answer. (No calculations are required.)
[12]
(2)
[12]
A 2 kg block, initially at 4 m height, is released and slides downhill from rest on
a frictionless ramp, and then
Exercise 2
moves along a horizontal surface. It then moves on a 10 m length rough horizontal surface with coefficient of
A 2 kg block, initially at 4 m height, is released and slides downhill from rest on a frictionless ramp,
=0,2, until
reaches
roughonramp
therough
same
coefficient
of friction and slides on it for 5 m
kinetic
friction
and then
moves
along aµkhorizontal
surface.
It thenamoves
a 10 mwith
length
horizontal
surface
with coefficient
of kinetic friction µk =0,2, until reaches a rough ramp with the same coefficient of friction
until it stops.
and slides on it for 5 m until it stops.
5m
θ
A
2.1.
2.2.
2.3.
2.4.
10 m
h
B
State the law (principle) of conservation of mechanical energy in words.
State
the lawCONSIDERATIONS
(principle) of conservation
ofspeed
mechanical
energy
in words.
Use ENERGY
to calculate the
of the block
at the bottom
of the ramp.
State the work energy theorem in words.
Use the WORK ENERGY THEOREM to calculate the speed of the block when passing position
2.1.
B.
Use ENERGY CONSIDERATIONS to calculate the speed of the block at the bottom of the ramp.
2.3.
Use the WORK ENERGY THEOREM to calculate the speed of the block when passing position B.
2.4.
34
Use the LAW OF CONSERVATION OF ENERGY to calculate the height
reached by the block until it
comes to rest.
2.5.
Calculate the angle at θ.
2.5. Use the LAW OF CONSERVATION OF ENERGY to calculate the height reached by the block
2.2.
Statetothe
until it comes
rest.work energy theorem in words.
2.6. Calculate the angle at θ.
33
4.5 Doppler Effect
4.5
Doppler Effect
Worked Example
worked Example
Doppler Effect
AAman
is standing on a pavement when he hears an ambulance approaching. The siren of the ambulance is
man is standing on a pavement when he hears an ambulance approaching. The siren of the
ked Example emitting
ambulance
is emitting
a wave
with a frequency
of 500
Hz.ambulance
The ambulance
is moving
a wave
sound
withsound
a frequency
of 500 Hz.
The
is moving
at ata aconstant speed of 30 m·s-1.
constant speed of 30 m·s-1. Calculate the frequency of the sound the man hears. Take the speed of
the frequency
of the sound the man hears. Take the speed of the sound in the air as 340 m·s-1.
an is standing Calculate
on
pavement
theasound
in thewhen
air ashe
340hears
m·s-1.an ambulance approaching. The siren of the
ulance is emitting a wave sound with a frequency of 500 Hz. The ambulance is moving at a
stant speed of 30Solution
m·s-1. Calculate the frequency of the sound the man hears. Take the speed of
Solution
sound in the air as 340 m·s-1.
𝑣𝑣 ± 𝑣𝑣𝐿𝐿
𝑓𝑓𝐿𝐿 = (
) 𝑓𝑓
ution
𝑣𝑣 ± 𝑣𝑣𝑠𝑠 𝑠𝑠
(
𝑣𝑣 ± 𝑣𝑣𝐿𝐿
) 𝑓𝑓
𝑣𝑣 ± 𝑣𝑣𝑠𝑠 𝑠𝑠
(
fL  (
340  0
)  500
340  30
340  0
f L  1,10  500
)  500
340  30
f L  550Hz
 1,10  500
 550Hz
4.6
Electrostatics
4.6
Electrostatics
worked example
Worked example
Electrostatics QuESTION 1 (Question 3 - Spring School material, Northern cape)
ked example
Two identically1 charged
spheres
(a and B)
are 30 material,
cm apart, with
a chargeCape)
of +3 x 10-4 C and
QUESTION
(Question
3 - Spring
School
Northern
ESTION 1 (Question
3
Spring
School
material,
Northern
cape)
-4
-8
–2 x 10 C, respectively. A small positive charge (c) of +10 C is placed 10 cm from sphere a, as
Two
identically
charged
(A with
andaB)
are 30
cmx apart,
with a charge of +3 x 10
shown
in the diagram
below.
identically charged
spheres
(a and
B)
are spheres
30 cm apart,
charge
of +3
10-4 C and
C and –2 x 10-4 C,
Bis placed
respectively.
A small
of +10
10ascm from sphere A, as shown in the diagram
10-4 C, respectively.
A small positive
charge
(c) ofCcharge
+10-8 C is(C)
placed
10 cmCfrom
sphere a,
A positive
wn in the diagram
below.
below.
-4
-8
A
1.1
1.2
C
10 cm
B
30 cm
10 cm
State
Coulomb’s law in words.
30 cm
Calculate the magnitude of the force exerted on the small positive charge c by charge a.
1.3law in
Calculate
State Coulomb’s
words. the magnitude of the force exerted on the small positive charge c by charge B.
1.1
State Coulomb’s law in words.
1.4
Calculate
the magnitude
ofthe
thesmall
resultant
forcecharge
exerted
small
Calculate the magnitude
of the force
exerted on
positive
c on
by the
charge
a.positive charge c.
1.5
Define
electric
at a on
point
words.
Calculate the magnitude
of the
forcefield
exerted
theinsmall
positive charge c by charge B.
1.2
Calculate the magnitude of the force exerted on the small positive charge C by charge A.
1.6
Calculate
the magnitude
of the electric
at positive
point c,charge
10 cm from
Calculate the magnitude
of the resultant
force exerted
on thefield
small
c. a, as a result of charge a.
thein
small
charge c is removed. Calculate the resultant electric field at a point situated at
Define electric1.7
field atNow
a point
words.
1.3
Calculate the magnitude of the force exerted on the small positive charge C by charge B.
the of
centre
point between
a andc,B.10 cm from a, as a result of charge a.
Calculate the magnitude
the electric
field at point
Now the small charge c is removed. Calculate the resultant electric field at a point situated at
1.4
Calculate the magnitude of the resultant force exerted on the small positive35charge C.
1.5
Define electric field at a point in words.
1.6
Calculate the magnitude of the electric field at point C, 10 cm from A, as a result of charge A.
1.7
Now the small charge C is removed. Calculate the resultant electric field at a point situated at the centre
point between A and B.
1.8
Charge C makes contact with charge B until the total charge between B and C is distributed. Calculate
the new charge on C.
the centre point between a and B.
34
35
SOLUTION
1.8
Charge c makes contact with charge B until the total charge between B and c is distributed.
Charge
c makesof
contact
with charge B force
until the
total charge
between
B and
c is distributed.
1.11.8 Calculate
The
magnitude
theonelectrostatic
exerted
by one
point
charge
(Q1) on another point charge (Q2)
the new charge
c.
Calculate
the
new
charge
on
c.
is directly proportional to the product of the magnitudes of the charges and inversely proportional to the
SOLuTION
square of the distance (r) between them.
SOLuTION
1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point
1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point
charge (Q2) is directly proportional to the product of the magnitudes of the charges and
1.2 Data
charge (Q2) is directly proportional to the product of the magnitudes of the charges and
inversely proportional
to the square of the distance (r) between them.
inversely proportional to the square of the distance (r) between them.
1.2 Data
1.2
QAData
= +3x10-4 C
-8
A = +3x10
qc =Q+10
C -4 C
q
=
+10
C x 10-2 m
rAC =c 10 cm-8=10
9 cm =10
2
r
x2 10-2 m
AC = 10
k = 9 x10 N9 m  C
2
2
k ==9?x10 N  m  C
FAonC
FAonC = ?
KQAqC
FAonC  KQ
2 q
FAonC (rAC ) A 2C
(rAC )
9 x10 9  3  10 4  10 8
FAonC  9 x10 9  3 21024  10 8
FAonC  (10  10 )2 2
(10  10 )
27  10 3 3
FAonC 
27  10
FAonC100
  10 4
100  10 4
FAonC = 0, 27x101N = 2,70 N
FAonC = 0, 27x101N = 2,70 N
1.3
Data
Data
1.31.3Data
QB = – 2 x 10-4 C
QB = – 2 x 10-4 C
qc = +10-8 C
= +10
rBCq=c 20
cm-8=C20 x 10-2 m
9 cm = 220 x210-2 m
r
=
20
BC
m C
k = 9  10 N
9
2
2
k = =9 
FAonC
? 10 N  m  C
FAonC = ?
KQB qC
FBonC  KQ
(rBC ) 2 B q2C
FBonC 
(rBC )
9 x10 9  2  10 4  10 8
FBonC  9 x10 9  2 21024  10 8
FBonC  (20  10 )2 2
(20  10 )
18  10 3 3
FBonC 
18  10
4
FBonC 400
  10 4
400  10
FBonC = 0,045x101 N = 0,45 N
FBonC = 0,045x101 N = 0,45 N
1.4 There are two electrostatic forces acting on point charge C; the gravitational force is
1.4 There are two electrostatic
forcesit acting
on point
charge to
C;the
theelectrostatics
gravitational
force
is ignored because it is
ignored because
is too small
in comparison
forces,
36 therefore we
too small in comparison
toapply
the electrostatics
forces, therefore
we must apply the principle
of superposition of
36
must
the principle of superposition
of forces.
forces.
F  F AC  F BC (it is a vector equation)
F R = F AC + F BC (it is Ra vector equation)
We must draw a free body diagram showing the electrostatic forces acting on the point
chargediagram
and choose
the positive
We must draw a free body
showing
the direction.
electrostatic forces acting on the point charge and choose
the positive direction.
𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴
𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
Xx
Working with the
magnitudes
Working with the magnitudes
of the
forces:of the forces:
FR = FAC + FBC
Substitution:
FR = 2,70 N + 0,45 N
𝐹𝐹 R = 3,15 N
35
𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴
𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴
𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
Xx
Xx
Working with the magnitudes of the forces:
Working with the magnitudes of the forces:
FR = FAC + FBC
FR = FAC + FBC
FR = FAC + FBC
Substitution:
Substitution:
Substitution:
= 2,70
0,45 N
N
NN++0,45
F =FR2,70
FR = 2,70 RN + 0,45 N
𝐹𝐹 R = 3,15 N
= 3,15 N
𝐹𝐹 R = 3,15
RN
2.3
2.4
2.3 The electric field at a point is the electrostatic force experienced per unit of positive
The
field
at a point
is athe
electrostatic force experienced per unit of positive
2.3electricThe
electric
fieldatat
charge
placed
thatpoint
point.is the electrostatic force experienced per unit of positive charge placed at that point.
charge placed at that point.
2,7
F
-1
-1
= 2,7
108 N∙C
x 10x8 N∙C
2.3F2.4 E = 2,7 E = 8 E = E82,7
E = q8
E 10
= 2,7 x 10 N∙C-1
E=
10
q
Each charge has an electrostatic field around it, therefore there are two electric fields at
1.71.7
Each
charge has an electrostatic field around it, therefore there are two electric fields at the point, and we
1.7 Each charge has an electrostatic field around it, therefore there are two electric fields at
the point, and we must apply the principle of superposition of fields.
must
apply
principle
of superposition
the point,
and we
mustthe
apply
the principle
of superpositionofoffields.
fields.
 EA A+ EBB (it is a vector equation)
E R E=R E
is a vector equation)
equation)
E R  E A  E B (it is a vector (it
We must draw the vector electric field of each charge at a point.
must
draw electric
the vector
electric
field at
ofaeach
We mustWe
draw
the vector
field of
each charge
point.charge at a point.
𝐸𝐸⃗⃗𝐴𝐴
𝐸𝐸⃗⃗𝐴𝐴
𝐸𝐸⃗⃗𝐵𝐵
X x
𝐸𝐸⃗⃗𝐵𝐵
X x
Both
electric
point
in the
same
direction,
therefore
we add
them together: Commented [U97]: Pse check – not understood.
Both
electricfiled
filed vectors
vectors point
in the
same
direction,
therefore
we add them
together:
Both electric filed
vectors
point
in
the
same
therefore
we
add
them
together:
Commented [U97]:
Pse check[G98]:
– not understood.
9
2direction,
2
4
5
2
Commented
Corrected
KQ
EA E
=R = E2AA+ EB EA =
ER = EA + EB
KQ
r
9 x10 Nm C x3x10 C
(15 x10 2 m) 2
9 x10 9 Nm 2 C 2 x3x10 4 C
EA =
27 x10 Nm C
225 x10 4 m 2
Commented [G98]: Corrected
27 x10 5 Nm 2 C
A
EA = KQ
KQAEA = 9 x1099xNm
10 9 2Nm
x3x410
275x4Nm
10 52 2Nm
C 22xC3x210
C 4ECA = 27 x10
C 2C
EA =E225
EA =ErA2 =2 A 2 E9 A =EA = (15 x810 2 m2-1) 2 2 2
A =x10
m 24 2
4
EA = 0,12
(10
15 x10
) right
r rx 10 = 1,2 x 10
(15 xN∙C
mto
) mthe
225 x225
10 x10
m m
right
EA = 0,12 x 109 = 1,29 x 108 N∙C9-18 to the
2
-1 to2-1
= 0,12
= x1,2
10
N∙C
to
the
x 10x9 10
= 1,2
N∙CNm
the
right
EA =E0,12
AKQ
910xx810
C
x2 xright
10 4 C
B
18 x105 Nm2C
37
37
EB =
EB =
EB =
2
4
2
2 (15
2 x10  2m
4 )2
225
m2
KQBr
9 x10 9 Nm
C
C
18 x105 Nm
C x10
92 x
22 x10
2
4
9

2

4
EB = KQ
KQBEB = 9 x109 xNm
B=
10 Nm
x2 x10
185x4Nm
10522Nm
C
xC22 x10
C EC
18 x10
C 2C
B
2
2
EB =ErB =2 2 EB =EB = (15 x10 m2 ) 2 2
EB =E
225
B =x10 m  4
15tox10
)
r rx 1099 = 0,8 x (10
225 x225
10 x410
m2 m2
158x8(10
m
) mright
EEBB== 0,08
the
0,08 x 10 = 0,8 x 10 to the right
EB = 0,08 x 109 = 0,89 x 108 to the
right
= x0,8
to right
the right
EB =E0,08
x 10x98810
= 0,8
10x8 10
to888the
B = 0,08
10
EERR== 1,2 x 10
0,8xx10
10
++ 0,8
ER = 1,2 x 108 + 0,8
x 108
1,2
+ x0,8
ER =E1,2
10x8 10
+80,8
10x8 108
R= x
N∙C-1-1 to
tothe
theright
right
EER== 2 x 108 N∙C
R 2 x 108 N∙C-1 to the right
ER =
8 N∙C
-1 to-1
2 x8 10
to right
the right
ER =E2R x= 10
N∙C
the
1.8 Since
Since
system
issystem
isolated,
then
weapply
must
apply
the
law
of conservation
of charge.
Commented
[U99]:
Alignment
1.8
Since
isolated,
then
we
apply
theof law
of conservation
of charge.
1.8
thethe
system
isthe
isolated,
theniswe
must
the
lawmust
of
conservation
charge.
Commented
[U99]:
Alignment problem
/ missing
item problem / missing
problem?
Since
the system
is isolated,
we must
apply
the law
of conservation
of charge. problem?
Commented
[U99]:
Alignment
problem
/ missing
1.8 1.8 Since
the system
is isolated,
thenthen
we must
apply
the law
of conservation
of charge.
Commented
[U99]:
Alignment
problem
/ missing
item item
problem?
problem?
Commented [U100]:
Pse check/complete.
Commented
[U100]: Pse check/complete.
Commented
[U100]:
Pse check/complete.
Commented
[U100]:
Pse check/complete.
Commented [G101]:
Corrected [G101]: Corrected
Commented
Commented
[G101]:
Corrected
Commented
[G101]:
Corrected
Commented [U102]: Pse check.
Commented
[U102]: Pse check.
Commented
[U102]:
Pse check.
Commented
[U102]:
Pse check.
= 𝑄𝑄
+𝑐𝑐𝑄𝑄+
𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛
𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛
=𝑐𝑐 𝑄𝑄
𝑄𝑄𝐵𝐵𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐵𝐵 =
+=
𝑄𝑄𝐵𝐵𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛𝑄𝑄=
𝑛𝑛𝑛𝑛𝑛𝑛𝑄𝑄𝑐𝑐=+𝑄𝑄𝑄𝑄
𝑐𝑐 𝐵𝐵
After
being
in contact
(𝑄𝑄𝑐𝑐 =(𝑄𝑄𝑄𝑄𝐵𝐵 =
= 𝑄𝑄
𝑄𝑄 = 𝑄𝑄
After
being
contact
being
in
contact
(𝑄𝑄𝑐𝑐𝑐𝑐𝑄𝑄𝐵𝐵= =
𝑄𝑄𝐵𝐵
AfterAfter
being
inin
(𝑄𝑄𝑐𝑐 (=
𝐵𝐵𝑄𝑄= 𝑄𝑄 )
After
being
incontact
contact
𝑄𝑄𝑐𝑐 + 𝑄𝑄𝐵𝐵 = 2𝑄𝑄
+
𝑄𝑄
= 2𝑄𝑄
𝑄𝑄
+𝑄𝑄𝑄𝑄
=𝐵𝐵2𝑄𝑄
2𝑄𝑄
𝑄𝑄𝑐𝑐𝑐𝑐 +
𝑐𝑐𝑄𝑄𝐵𝐵
𝐵𝐵=
𝑄𝑄𝑐𝑐 + 𝑄𝑄𝐵𝐵
𝑄𝑄 = 𝑄𝑄
+ 𝑄𝑄𝐵𝐵
𝑄𝑄𝑄𝑄
𝑐𝑐𝑄𝑄
𝐵𝐵
𝑄𝑄2𝑐𝑐𝑐𝑐++
𝐵𝐵
𝑄𝑄
=
𝑄𝑄 =𝑄𝑄 =2 2
2
(110 8 )8 (
2 10 4 )4 04,000110 4 4 2 410 4 4 1,9999
10 4 4 4
4
- 1,00 x10-4 C -4
Qc = (110

(110
(10
2 10
0,0001
2
10
1
,9999 =10

)8 8()
2
)=4 )0,0001
10
1
,
9999
1010
 2
42 10
= -41,00
=
- 10
1,00
x10-4x10
C C
Qc =Q(c1=10 2)  (2 10 =) = 0,0001
2 10 4  2 10
1
,
9999
2 2
2 2
=
= - 1,00 x10-4 C
Qc =
2 2
2
2
2
Exercise 1
Exercise
Exercise
1 1
Two
small,
metal spheres (a and B), carrying charges of +30nC and -14nC,
Exercise
1identical
small,
identical
metal
spheres
(a and
carrying
charges
of +30nC
-14nC,
TwoTwo
small,
identical
metal
spheres
(a and
B), B),
carrying
charges
of +30nC
and and
-14nC,
respectively, are mounted on insulated stands; these are placed at a certain distance r from
respectively,
mounted
on
insulated
stands;
these
are
placed
a certain
rand
from-14nC,
Two
small, identical
metal
spheres
(a
and
B),
of distance
+30nC
respectively,
are are
mounted
on insulated
stands;
these
arecarrying
placed
at charges
aat
certain
distance
r from
each other, as shown below.
each
other,
as
shown
below.
each other, asare
shown
below. on insulated stands; these are placed at a certain distance r from
respectively,
mounted
each other, as shown below.r r
30 nC
30 nC30AnC
A
A
36
30 nC
A
r
r
B
B -B14 nC
- 14 nC
- 14 nC
B
- 14 nC
Qc =
Exercise 1
2
(110 8 )  (2 10 4 ) 0,000110 4  2 10 4 1,9999 10 4
=
= - 1,00 x10-4 C

2
2
2
Exercise 1
Two small, identical metal spheres (a and B), carrying charges of +30nC and -14nC,
respectively,
insulated
stands; these
are placed
at a certain
distance r from
Two small, identical metal
spheresare
(Amounted
and B),on
carrying
charges
of +30nC
and -14nC,
respectively,
are mounted
each
other,
as
shown
below.
on insulated stands; these are placed at a certain distance r from each other, as shown below.
r
30 nC
A
B
- 14 nC
1.1. Define electric field in words.
1.1.
Define electric field in words.
1.2.
The contact
magnitude
of the
electrostatic
force
between
them
is now
6,4original
x 10 N.position. The
Sphere B is moved position.
and makes
with
sphere
A. It is
then
moved
back
to its
magnitude of the1.2.1
electrostatic
force
between
them is
6,4 x 10-4 N.
Sketch the field
pattern
surrounding
thenow
charges.
1.2. Sphere B is moved and makes contact with sphere a. It is then moved back to its original
-4
1.2.2 Calculate the magnitude of the charge on each sphere after contact.
1.2.1 Sketch the field
pattern surrounding the charges.
1.3. Determine the original distance between the spheres.
38
1.2.2 Calculate the magnitude of the charge on each sphere after contact.
Exercise 2
1.3.
Determine the original
spheres.
Two smalldistance
objects (abetween
and B) arethe
equally
positively charged; these are placed in a vacuum, as
Exercise 2
shown in the figure below.
B
A
200 cm
Two small objects (A and B) are equally positively charged; these are placed in a vacuum, as shown in the
Charge B repels charge a with an electrostatic force of 3,0 x 10-6 N.
figure below.
2.1 State Coulomb’s law in words.
2.2 What is the magnitude of the electrostatic force
that A exerts on B?
Charge B repels charge
A with an electrostatic force of 3,0 x 10-6 N.
2.3 Draw the resultant electric field lines of the electric field around charges a and B.
2.1
2.4law
Calculate
the magnitude of charge a and B.
State Coulomb’s
in words.
2.2
What is the magnitude of the electrostatic force that A exerts on B?
2.3
STRATEGY
SOLVE
PROBLEMS
ON ELECTRIC
CURRENT
Draw the resultant electric
fieldTO
lines
of the
electric field
around charges
A and B.
2.4
Calculate the magnitude of charge A and B.
2.5
Calculate the electric
field
strength
atinasymbolic
point inform.
the centre of charge A and B.
 Write
down
the data
2.5 Calculate the electric field strength at a point in the centre of charge a and B.
4.7
Electric Circuits
 Read the problem carefully, as many times as you need to.
 If not given, draw a circuit diagram.
Commented [U103]: Pse c
 Indicate the conventional direction of the current, from high-potential to low-potential
Commented [GI104]: corr
(+ to -).
 Re-draw the circuit diagram to simplify it, if necessary.
 Identify the type of connection (series/parallel).
 Analysing circuits:
1 – The algebraic sum of the changes in potential in a complete transversal of any loop
of a circuit must be zero (ε= Ir+ Ir).
2. The sum of the currents entering any junction must be equal to the sum of the currents
leaving that junction:
39
37
2.6
4.7 Electric Circuits
STRATEGY TO SOLVE PROBLEMS ON ELECTRIC CURRENT

Read the problem carefully, as many times as you need to.

If not given, draw a circuit diagram.

Write down the data in symbolic form.

Indicate the conventional direction of the current, from high-potential to low-potential (+ to -).

Re-draw the circuit diagram to simplify it, if necessary.

Identify the type of connection (series/parallel).

Analysing circuits:
1 – The algebraic sum of the changes in potential in a complete transversal of any loop of a circuit must be
zero (ε= Ir+ Ir).
2. The sum of the currents entering any junction must be equal to the sum of the currents leaving that
junction:
i = i1+i2+…+ in

Write down the formula/ equation that solves the question.

Find the unknowns, if needed (multi-concept problems).

Do the calculations and write down the final answer.

Check your answer. Check if it makes sense, i.e.:
-
Is the unit correct?
-
Is the numerical value reasonable?
38
 Do the calculations and write down the final answer.
 Check your answer. Check if it makes sense, i.e.:
WORKED EXAMPLE
-
Is the unit correct?
-
Is the numerical value reasonable?
wORKEd ExamPLE
In the circuit diagram
below, the voltmeter V reads 12 V when switch S is open. When switch S is closed, the
In the circuit diagram below, the1 voltmeter V1 reads 12 V when switch S is open. When switch
S V.
is closed,
the readingofdrops
to 10 V. The
Ammeter and the wires is
reading drops to 10
The resistance
the Ammeter
andresistance
the wiresofisthe
negligible.
negligible.
𝑉𝑉1
1
𝑉𝑉2
S
R
A
3A
6Ω
𝑉𝑉3
1.1
4Ω
What is the reading of the ammeter when the switch is open?
1.2
What is the emf of the battery?
Commented [GI105]: corrected
1.1
What is the reading of the ammeter when the switch is open?
1.2
1.3 ofWhat
is the reading of voltmeter 2?
What is the emf
the battery?
1.3
What is the reading of voltmeter 2?
1.4
Calculateresistance
the internal of
resistance
of the battery?
Calculate the1.6
equivalent
the resistors
connected in parallel.
1.5
Determine the
reading of voltmeter 3?
1.8 What is the resistance of the resistor R?
1.6
How would the reading of voltmeter V2 change if the 6 Ω resistor is removed from the
Calculate the1.9
internal
resistance of the battery?
1.7
answer.
Determine the total
resistance of the circuit?
1.8
What is the resistance of the resistor R?
1.9
1.4
Calculate the equivalent resistance of the resistors connected in parallel.
1.5
Determine the reading of voltmeter 3?
1.7
Commented [U106]: Formatting/
points from blank lines in text – repet
Pse use either bullets or numbers – no
items/sub-items. Repetitive problems
Determine the total resistance of the circuit?
circuit? Write down INCREASE, DECREASE or REMAIN THE SAME. Explain your
40
How would the reading of voltmeter V2 change if the 6 Ω resistor is removed from the circuit? Write down
INCREASE, DECREASE or REMAIN THE SAME. Explain your answer.
SOLUTION
Commented [U107]: Formatting: pse use formatting to ma
1.1. Zero (when the circuit is open there is no flow of charges therefore no current in the
SOLUTION
items legible – repetitive problems seen, including use of bullet
point/not using bullet points, line spacing, alignment, indentatio
standardisation of formatting, etc. (It is difficult to understand w
belongs with what in the text because of the lack of attention to
formatting.)
12 V (when
(when there
is no electric
current
flowing
the circuit,
the reading
of the voltmeter
Commented [U108]: Pse check ‘bullet’ points at items
1.1.1.2.Zero
the circuit
is open
there
is noin flow
of charges
therefore
no current in the
throughout – pse standardise appearance of all with equivalent
1.2.
connected across the terminals of the battery is equal to the emf of the battery, because
items and remove bullet points from items that are not bullet
points. There appear to be repetitive problems in the text.
12 V (when there is no electric current flowing in the circuit, the reading of the voltmeter
connected across
Commented [GI109]: corrected
there is no energy dissipated.
the terminals of the battery is equal to the emf of the battery, because there is noCommented
energy dissipated.
[U110]: Pse check/ complete.
Pse re-check the text in its entirety: the editor is an English
specialist, not a Physics specialist, but there appear to be multip
problems with the drafting of the text.
Zero (there
in the
but thebut
resistance
is negligible:
V=IR=(I)(0)=0
V)
1.3.1.3.Zero
(thereisiscurrent
current
in circuit,
the circuit,
the resistance
is negligible:
V=IR=(I)(0)=0
V)
Commented [U111]: Pse check.
of of
6 Ω6and
4 Ω are
in parallelin parallel
1.4.1.4.The
The resistors
resistors
Ω and
4 Ωconnected
are connected
Commented [U112]: Formatting required: text should not b
obscured – repetitive problem seen in text.
1 solve
1
1 the following way
Where there are two resistors we can
in
=
+
𝑅𝑅𝑃𝑃 𝑅𝑅1 𝑅𝑅2
𝑅𝑅1 𝑅𝑅2
(6)(4)
𝑅𝑅𝑃𝑃 =
=
= 2,4 Ω
𝑅𝑅1 + 𝑅𝑅2
6+4
1 1
5
1
= + =
𝑅𝑅𝑃𝑃 6 4 12
𝑅𝑅𝑃𝑃 = 2,4 Ω
1.5 𝑉𝑉𝑝𝑝 = 𝐼𝐼𝐼𝐼
𝑉𝑉𝑝𝑝 = 3 × 2,4
𝑉𝑉𝑝𝑝 = 7,2 𝑉𝑉
39
1.5
𝑃𝑃 𝑅𝑅 + 𝑅𝑅
++4 4
1𝑅𝑅1 + 2𝑅𝑅2 6 6
1 1
5
1
= + =
𝑅𝑅𝑃𝑃 6 4 12
11 11 11 55
== ++ ==
4 4 1212
𝑅𝑅
6Ω
𝑅𝑅𝑃𝑃𝑅𝑅𝑃𝑃= 6
2,4
𝑃𝑃
= 𝐼𝐼𝐼𝐼
2,4
2,4ΩΩ
1.5 𝑅𝑅
𝑉𝑉𝑝𝑝𝑃𝑃𝑅𝑅=
𝑃𝑃 =
1.5
= 𝐼𝐼𝐼𝐼
1.5𝑉𝑉
𝑉𝑉𝑝𝑝𝑝𝑝𝑉𝑉𝑝𝑝==3 𝐼𝐼𝐼𝐼
× 2,4
𝑉𝑉𝑝𝑝𝑉𝑉𝑝𝑝==3 3××2,4
𝑉𝑉𝑝𝑝 = 7,2 𝑉𝑉 2,4
𝑉𝑉𝑝𝑝𝑉𝑉𝑝𝑝==7,2
7,2𝑉𝑉 𝑉𝑉 of the voltmeter 3 is 7,2 V.
The reading
The reading of the voltmeter 3 is 7,2 V.
ofofthe
The
reading
thevoltmeter
voltmeter3 3isis7,2
7,2V.V.
1.6 The
𝜀𝜀 =reading
𝑉𝑉
𝑒𝑒𝑒𝑒𝑒𝑒 + 𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖
1.6 1.6
++𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖
𝜀𝜀 𝜀𝜀===𝑉𝑉𝜀𝜀𝑒𝑒𝑒𝑒𝑒𝑒
𝑉𝑉−
𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖
1.6𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖
𝑒𝑒𝑒𝑒𝑒𝑒𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
𝑉𝑉
=−
𝜀𝜀 𝜀𝜀−𝑉𝑉−𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
𝑉𝑉=
𝐼𝐼𝐼𝐼𝑖𝑖𝑖𝑖𝑖𝑖
𝑖𝑖𝑖𝑖𝑖𝑖𝜀𝜀=
𝑒𝑒𝑒𝑒𝑒𝑒𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
𝐼𝐼𝐼𝐼
𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
𝜀𝜀−−
𝑉𝑉
3𝑟𝑟𝐼𝐼𝐼𝐼=
==𝜀𝜀12
−
10
𝑒𝑒𝑒𝑒𝑒𝑒
3𝑟𝑟
=1212−
−1010
𝑟𝑟 3𝑟𝑟
==0,67
Ω
𝑟𝑟 𝑟𝑟==0,67
0,67ΩΩ
1.7 𝜀𝜀 = 𝐼𝐼𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
1.7 1.7
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
1.7𝜀𝜀 𝜀𝜀==𝐼𝐼𝑅𝑅𝐼𝐼𝑅𝑅
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
12 = 3𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
1212==3𝑅𝑅
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
3𝑅𝑅
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
41
Rtotal = 4 Ω
Rtotal = 4 Ω
4141
1.8
1.8
Option 1
Option 2
𝑅𝑅𝑇𝑇 = 𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒 + 𝑟𝑟
𝑉𝑉𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑉𝑉2 = 𝑉𝑉1
𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑅𝑅𝑃𝑃
𝑅𝑅𝑇𝑇 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑅𝑅𝑃𝑃 + 𝑟𝑟
4 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 2,4 + 0,67
𝐼𝐼𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 7,2 = 10
3𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 7,2 = 10
𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = 0,93 Ω
𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = 0,93 Ω
INCREASE
1.8 1.8 INCREASE
If the bulb of 6 Ω is removed, the total resistance of the circuit increases, then the
If the bulb of 6 Ω is removed, the total resistance of the circuit increases, then the reading of the ammeter
reading of the ammeter decreases, because the current decreases due to the
decreases, because the current decreases due to the increasing external resistance; but the reading of the
increasing external resistance; but the reading of the voltmeter increases, because the
voltmeter
increases, because the drop of potential in the internal resistance also decreases; but the emf of the
drop of potential in the internal resistance also decreases; but the emf of the battery is
battery isconstant,
constant,
according
ε -load
Vint==εε -- V
Ir.int = ε - Ir.
according
to Vloadto= V
Exercise 11
Exercise
Three resistors ( 2 Ω, 4 Ω and Rx ) are connected to a battery, as shown in the circuit
diagram below. With switch S1 open and S2 closed, the reading on the voltmeter is 10 V.
With both switches closed, the reading on the voltmeter is 8 V and the ammeter is 1 A.
V
A
S1
4Ω
S2
40
Rx
2Ω
1.1
Write down the value of the emf of the battery.
1.2
Calculate the:
(1)
1.2.1 resistance of the unknown resistor Rx.
(7)
1.2.2 internal resistance of the battery.
(3)
drop of potential in the internal resistance also decreases; but the emf of the battery is
constant, according to Vload = ε - Vint = ε - Ir.
Exercise 1
2 xΩ,) are
4 Ω connected
and Rx ) are to
connected
to aas
battery,
as in
shown
in the circuit
Three resistors (Three
2 Ω, resistors
4 Ω and( R
a battery,
shown
the circuit
diagram below. With
diagram
below. With
switch Son
S2 closed,
theV.reading
on the
voltmeterclosed,
is 10 V.the reading on the
1 open
switch S1 open and
S2 closed,
the reading
the and
voltmeter
is 10
With both
switches
With
both
switchesisclosed,
voltmeter is 8 V and
the
ammeter
1 A. the reading on the voltmeter is 8 V and the ammeter is 1 A.
V
A
S1
4Ω
S2
Rx
2Ω
1.1
Write down the value of the emf of the battery.
1.2
Calculate the:
1.1
Write down the value of the emf of the battery.
1.2
Calculate the:
(1)
1.2.1 resistance of the unknown resistor Rx.
(7)
1.2.1 resistance
of the
unknown
resistor Rx.
1.2.2 internal
resistance
of the battery.
(7)
(3)
1.2.2 internal resistance of the battery.
1.3
(1)
(3)
How will the reading on the voltmeter be affected if the switch
42
S1 is closed while S2 is opened.
1.3
will the
reading
on the voltmeter
be affected if the
switch
S 1 is closed
while
Write How
down
only
INCREASES,
DECREASES
or
REMAINS
THE
SAME.
(4)
Briefly
explain
answer.
Write
down onlythe
INCREASES,
DECREASES or REMAINS THE SAME.
[15]
S2 is opened.
Exercise 2
Briefly explain the answer.
(4)
[15]
Exercise 2
In the circuit diagram below, the emf of the battery is 6 V and its internal resistance is 0,10 Ω. The resistance
In the circuit diagram below, the emf of the battery is 6 V and its internal resistance is 0,10 Ω.
(R) is UNKNOWN.
The resistance (R) is UNKNOWN.
V
6V
0,10 Ω
A
4Ω
R
2.1
Explain the term internal resistance.
2.2
Write down an equation for the terminal potential difference using the values given.
2.3
2.2 a sketch
Write down
an equation
for the
terminal potential
difference
the values
given.the follow(2)
Draw
graph
of terminal
potential
difference
versususing
current.
Indicate
(3)
ing2.3in theDraw
graph:
a sketch graph of terminal potential difference versus current. Indicate the
(3) Commented [U113]: Pse check.
2.1
Explain the term internal resistance.
(2)
(2)
(2)
following in the graph:
2.4
•
 Theofvalue
of the emf.
The value
the emf.
•
The current at which the terminal potential difference is zero.

2.4
The current at which the terminal potential difference is zero.
The energy dissipated in 4 Ω resistance is 40 J; the energy dissipated in resistance R
The energy dissipated in 4 Ω resistance is 40 J; the energy dissipated in resistance R is
is 60 J.
60 J.
Calculate:
Calculate:
2.5
2.4.1
Resistance R
(4)
2.4.2
Total current in the circuit
(3)
2.4.3
Reading of the voltmeter
(3)
A 7 Ω resistor is now connected in parallel to the 4 Ω resistor. How will this action
affect the reading of the voltmeter? Write down only INCREASES, DECREASES or
41
2.5
2.4.1
Resistance R
(4)
2.4.2
Total current in the circuit
(3)
2.4.3
Reading of the voltmeter
(3)
A 7 Ω resistor is now connected in parallel to the 4 Ω resistor. How will this action affect
the reading of the voltmeter? Write down only INCREASES, DECREASES or REMAINS
THE SAME.
Briefly explain your answer.
(4)
[21]
(4)
[21]
4.8 Electrodynamics
4.8
Electrodynamics
Problem
Solving
Strategy
for Electrodynamics
Problem
Solving
Strategy
for Electrodynamics



With
machines,
look at the
energy
to identifytothe
type ofthe
electrical
• electrical
With electrical
machines,
look
at theconversions
energy conversions
identify
type of electrical machine.
machine.
Re-draw
the circuit
diagram
to simplify
if necessary.
•
Re-draw
the
circuit diagram
toit,simplify
it, if necessary.
Identify the type of connection (series/parallel).
•
Identify the type of connection (series/parallel).
Worked example
Worked example
The simplified diagrams below represent an electric motor and a generator.
The simplified diagrams below represent an electric motor and a generator.
Y
X
1.1
1.1
Which
ONE
the
diagramsabove
aboverepresents
representsa asimplified
simplifieddiagram
diagramofofan
anelectric
electricmotor? Give
Which
ONE
of of
the
diagrams
motor? for
Give
a reason
for your answer.
a reason
your
answer.
1.2
1.2
What
type
of of
generator
(AC
oror
DC)
is is
represented
What
type
generator
(AC
DC)
representedininthe
thesimplified
simplifieddiagrams
diagramsabove? Give a
reason
for Give
your aanswer.
above?
reason for your answer.
1.3
1.3
1.4
1.4
1.5
1.5
State
ONE
method
of of
increasing
the
induced
State
ONE
method
increasing
the
inducedemf
emfofofthis
thisgenerator.
generator.
Write down ONE use of electric motors.
Write down ONE use of electric motors.
The maximum potential difference produced by this generator is 12 V
The maximum potential difference produced by this generator is 12 V and the
and the frequency is 50 Hz.
frequency is 50 Hz.
1.1. Sketch a graph of the induced potential difference versus time.
1.1. Calculate
Sketch a graph
of the
induced rms
potential
differencedifference.
versus time.
1.5.2.
the
induced
potential
1.5.2.
Calculate thethe
induced
rms potential
difference.
1.5.3.
Calculate
average
power
dissipated if a 5 Ω resistor is
1.5.3.connected
Calculate the average
dissipated if a 5 Ω resistor is connected to this
to thispower
generator.
SOLUTION
generator.
SOLUTION
42
44
1.1
X. It converts electrical energy into mechanical energy.
1.2
AC generator. It has slip rings.
W electrical
X.ItItconverts
converts
electricalenergy
energyinto
intomechanical
mechanicalenergy.
energy.
X.
1.1
1.1
1.31.2
ANY
ONE of the
following:
ACgenerator.
generator.
has
sliprings.
rings.
AC
ItIthas
slip
W
W
Increasing
the
speed/frequency of rotation
ANYONE
ONEofofthe
thefollowing:
following:
ANY
1.2
1.3
1.3
Increasingthe
the
speed/frequency
rotation
Increasing
the
number of coils
Increasing
speed/frequency
ofofrotation
Increasingthe
thenumber
numberofofcoils
coils
Increasing
Increasing
the
strength
of the magnetic field
Increasingthe
thestrength
strengthofofthe
themagnetic
magneticfield
field
Increasing
Insert a soft iron core
Insertaasoft
softiron
ironcore
core
Insert
1.4
ANY ONE of the following:
1.4
1.4
ANYONE
ONEofofthe
thefollowing:
following:
ANY
Pumps
Pumps
Pumps
Fans
Fans
Fans
Compressors
Compressors
Compressors
Hairdryers
dryers
Hair
dryers
Hair
1.5.1
1.5.1
1.5.1
1.5.2
1.5.2
1.5.2
VVmax/
max/maks
maks
VVrms
rms/ wgk
/ wgk 

22
12
Vrms / wgk  12 
Vrms / wgk  2 
2
Vrms / wgk  8,49 V
Vrms / wgk  8,49 V
1.5.3
1.5.3
1.5.3
45
45
V 2 / wgk
2
Pave / gem  V rms

Pave / gem  rmsR/ wgk 
R
(8,49) 2
Pave / gem  (8,49) 2 
5 
Pave / gem 
5
Pave / gem  14,42 W
Pave / gem  14,42 W
Exercise 1
Exercise 1
A team of grade 12 learners conducted a practical investigation to investigate the
A team of grade 12 learners conducted a practical investigation to investigate the
relationship between the root mean square value of voltage generated and the frequency
relationship between the root mean square value of voltage generated and the frequency
of rotation of the armature of the generator using the device shown in the figure below.
of rotation of the armature of the generator using the device shown in the figure below.
43
ave / gem
5
Pave / gem  14,42 W
Exercise 1
Exercise 1
A team of grade 12 learners conducted a practical investigation to investigate the
A team of grade 12 learners conducted a practical investigation to investigate the relationship between the root
relationship between the root mean square value of voltage generated and the frequency
mean square value of voltage generated and the frequency of rotation of the armature of the generator using
of rotation of the armature of the generator using the device shown in the figure below.
the device shown in the figure below.
The following table
shows
thetable
results
obtained.
The
following
shows
the results obtained.
Frequency Hz)
Frequency Hz) Vrms(V)
Vrms(V)
1.1.
0
00
0
10
1 10
1
20
2
30
20
3
2
40
4
50
305
3
40
50
4
5
1.1.
What was
type used
of generator
used during
by thethe
learners
during
the practical Give a reason for
What type of
generator
by the was
learners
practical
investigation?
investigation? Give a reason for your answer.
your answer.
1.2.
Write an investigative question for this investigation.
1.2.
Write an investigative
this investigation.
1.3.
Whatquestion
could the for
learner’s
hypothesis have been for this experiment?
1.3.
What could1.5.
the learner’s
thispaper
experiment?
Use the hypothesis
data to draw ahave
graphbeen
on thefor
graph
provided.
1.4.
What is the dependent variable in this experiment?
1.5.
Use the data to draw a graph on the graph paper provided.
1.6.
Write the conclusion for the results obtained.
1.7.
Calculate the root mean square value of the voltage when the amplitude of the voltage is 5 V.
1.8.
If a resistor of 5 Ω is connected to the generator, determine the root mean square value of the current
through the resistor when the frequency of rotation of the armature is 50 Hz.
1.9.
Calculate the average power dissipated through the resistor when the frequency of rotation of the
armature is 50 Hz.
1.4.
1.6.
What is the dependent variable in this experiment?
Commented [GI114]: corrected
Commented [U115]: Pse check.
Write the conclusion for the results obtained.
46
1.10. Draw the graph of instantaneous voltage vs time when the armature rotates with constant frequency of
50 Hz.
1.11. List the advantages of alternating current (AC) over direct current (DC).
44
Exercise 2
The diagram below shows one of the electrical machines studied in class.
2.1
Is this electrical machine a DC motor or a DC generator? Give a reason for your answer.
2.2
What is the direction of rotation of the loop: clockwise or anti-clockwise?
2.3
Label the parts of this electrical machine indicated by A, B and C.
2.4
Write three applications of this electrical machine.
45
5 Revision Questions - Set 2 (master an
additional 20%)
5.
5.
Revision
Questions
- Set
2 (master
an additional 20%)
Revision
Questions
- SetLaws
2 (master an
additional
20%)
5.1
Newton’s
5.1
Newton’s5.1
Laws Newton’s Laws
Exercise 1
Exercise 1
Exercise 1
Two
blocks
of
the are
same
materials
are
connected
by a
light,
inelastic
rope.
BlockofA
A has
of of 5 kg; block
Two blocks
blocks ofofthe
same
materials
connected
by
a light,
inelastic
rope.
Block
Arope.
has
a mass
Two
the
same
materials
are
connected
by
a light,
inelastic
Block
hasaamass
mass
5ofkg;
block
mass
of
3fixed
kg.
rope
isB fixed
toforce
block
a100N
forceis (is
𝐹𝐹⃗ applied
) of 100Nhorizontally.
is
The
B5 has
a mass
kg.
Another
to additional
block
B and
a aforce
(( B𝐹𝐹⃗ and
kg; block
a3mass
ofB3has
kg. aAnother
rope
is Another
fixed
to block
and
))ofof100N
5.B has
Revision
Questions
-rope
Set 2is
(master
an
20%)
5.1
Newton’s
Laws
applied
horizontally.
The blocks
are
moving
along
a horizontal
frictionless surface.
appliedare
horizontally.
The
blocks
are moving
along
a horizontal
frictionless
surface.
blocks
moving
along
a horizontal
frictionless
surface.
Exercise 1
𝐹𝐹⃗
𝐹𝐹⃗
aB
a
Two blocks of the same materials are
connected byBa light, inelastic rope. Block A has a mass of
1.1
1.1
1.2
1.3
1.2
5 kg; block B has a mass of 3 kg. Another rope is fixed to block B and a force ( 𝐹𝐹⃗ ) of 100N is
1.1
Draw
separate
labelled
free-body
diagrams
all the
action surface.
on the blocks.
Draw separate
free-body
diagrams
all the
force’saof
action
onforce’s
the
blocks.
appliedlabelled
horizontally.
The
blocks
areofmoving
along
horizontal
frictionless
Draw
separate
labelledthe
of all the force’s action on the blocks.
1.2
Calculate
acceleration
of the blocks.
Calculate
the acceleration
offree-body
the
blocks.diagrams
Calculateofthe
the
tension in the rope.
Calculate1.3
the magnitude
themagnitude
tension inof
the
rope.
a
Calculate the acceleration of the blocks.
𝐹𝐹⃗
B
Thenow
twopulled
blocksover
are now
pulled
over another
surfaceexperience
and the blocks
experience
friction. The blocks
The two blocks are
another
surface
and the blocks
friction.
The blocks
1.3
-2. The in
Calculate
the
magnitude
them·s
tension
the
rope.
exerted
by the
onN.
block A is 62,5 N.
accelerate
at of
8,972
force
exerted
byforce
the
rope
on block
A rope
is 62,5
accelerate at
8,972
m·s-2. The
1.1
Draw separate labelled free-body diagrams of all the force’s action on the blocks.
1.2
Calculate the acceleration of the blocks.
The
blocks1.4
arekinetic
now
pulled
another
blocks
experience friction. The blocks accelerate
Calculate
theover
kinetic
coefficient
of friction
forthe
block
A.
1.4 twoCalculate
the
coefficient
of friction
for surface
block
A. and
1.3
Calculate the magnitude of the tension in the rope.
-2
.1.5
Theblocks
force
exerted
by
the
rope
onsame
block surface
A is block
62,5
N.
at1.5
8,972If m·s
If thehave
two the
blocks
have
the
area,
will block
B have a different
the two
same
surface
area,
will
B have
a different
1.4
1.6
1.7
coefficient
friction?
Explain
your
answer.
coefficient
of blocks
friction?
your over
answer.
The two
areExplain
nowofpulled
another
surface
and the blocks experience friction. The blocks
Calculate
thefrictional
kinetic
coefficient
offorce
friction
foron
block
Calculate
theexacted
frictional
exacted
byblock
the A.
surface
on block B.
Calculate1.6
the
force
by
the
surface
B.
-2
accelerate at 8,972 m·s . The force exerted by the rope on block A is 62,5 N.
1.7action-reaction
Name two pairs
action-reaction
pairs in the system.
Name two
in the system.
If the two blocks have the same surface area, will block B have a different coefficient
of
1.4
Calculate the kinetic coefficient of friction for block A.
Explain
your answer.
Exercise
2
Exercise
2
1.5
If the two blocks have the same surface area, will block B have a different
1.5
friction?
coefficient of friction? Explain your answer.
kg frictional
block on
a force
horizontal,
surface
is
tobyan
block by a light inextensible
A 4 kgCalculate
block on Aa 4horizontal,
rough
surface
isrough
connected
anconnected
8 kg block
a8
light
1.6
the
exacted
by
thetosurface
on block
B.kg inextensible
1.6
Calculate the frictional force exacted by the surface on block B.
string
passes over
a frictionless
shown
below.of
The
coefficient
of kinetic (dynamic)
string that passes
overthat
a frictionless
pulley,
as shownpulley,
below.asThe
coefficient
kinetic
(dynamic)
1.7
Name two action-reaction pairs in the system.
1.7
Name
two
action-reaction
pairs
in
the
system.
between
of 4 kg and
the surface is 0,6.
friction between friction
the block
of 4 kgthe
andblock
the surface
is 0,6.
Exercise 2
Exercise 2
4 kg
4 kg


A 4 kg block on a horizontal, rough surface is connected to an 8 kg block by a light inextensible
A 4 kg block on a horizontal, rough surface is connected to an 8 kg block by a light inextensible string that
string that passes over a frictionless pulley, as shown below. The coefficient of kinetic (dynamic)
passes over a frictionless pulley, as shown below. The coefficient of kinetic (dynamic) friction between the
friction between the block of 4 kg and the surface is 0,6.
block of 4 kg and the surface is 0,6.
8 kg
8 kg
4 kg

49
49
8 kg
2.1
Draw a free-body diagram of all the forces acting on both blocks.
2.2
Write down Newton’s second law of motion in words.
2.3
Calculate the acceleration of the system.
46
49
2.4
Calculate the magnitude of the tension in the string.
2.5
Calculate the magnitude of the frictional force that acts on the 4 kg block.
2.6
Calculate the apparent weight of the 8 kg block.
2.7
How does the apparent weight of the 8 kg block compare with its true weight? Write down only,
GREATER THAN, EQUAL TO or LESS THAN.
2.8 How does the apparent weight of the 4 kg block compare with its true weight? Write down only,
GREATER THAN, EQUAL TO or LESS THAN.
Exercise 3
ZACUBE-1 is the first South African satellite. It was designed and built by
postgraduate students, following the CubeSat Programme at the French
South African Institute of Technology (F’SATI), in collaboration with the
South African National Space Agency (SANSA). ZACUBE-1 was lunched on
21 November 2013 and is orbiting at a height of 600 km from the surface of
Earth. Weighing 1.2 kg, this CubeSat is about 100 times smaller than Sputnik
- the first satellite launched into space in 1957.
3.1 State Newton’s Law of Universal Gravitation in words.
3.2
(2)
How would the magnitude of the gravitational force exerted by the Earth on ZACUBE-1
compare with the magnitude of the gravitational force exerted by the Earth on Sputnik, if
both satellites were orbiting at the same height?
3.3
3.4
Write down only LARGER THAN, SMALLER THAN or THE SAME. Explain the answer.
(4)
Calculate the gravitational force exerted by the earth on the
satellite ZACUBE-1.
(5)
Calculate the height the satellite must be orbiting at for the value
of the acceleration due to gravity to be 25% of the gravitational
acceleration on the surface of the Earth.
(6)
[17]
5.2 Momentum and Impulse
Exercise 1 (Integration of: Law of Conservation of Momentum, Newton’s Laws and Equations of Motion)
The sketch below shows a missile launcher of mass 4800 kg at rest on a frictionless horizontal
surface
to thebelow
base of
a smooth
inclined launcher
plane. It fires
rocket4800
of mass
horizontally,
Thenext
sketch
shows
a missile
of amass
kg100
at kg
rest
on a frictionless horizontal surface next
and recoils up the inclined plane, rising to a height of 6 m.
to the base of a smooth inclined plane. It fires a rocket of mass 100 kg horizontally, and recoils up the inclined
plane, rising to a height of 6 m.
6m
𝜃𝜃
1.1.
State the law of conservation of linear momentum in words.
1.1.
State the law of conservation of linear momentum in words.
the system
formed
by the
missile
the
rocket
is said to be isolated?
1.2.1.2. Why Why
is the is
system
formed by
the missile
launcher
andlauncher
the rocket and
is said
to be
isolated?
1.3.
Calculate the initial speed of the rocket.
Answer: 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1
Exercise 2
47
1.1.
1.2.
1.3.
1.3.
State the law of conservation of linear momentum in words.
Why is the system formed by the missile launcher and the rocket is said to be isolated?
Calculate
thethe
initial
speed
of theof
rocket.
Calculate
initial
speed
the rocket.
Answer: 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1
Answer:
Exercise 2
Exercise 2
P= kg·m·s-1
A Aconstant
tothe
theright
rightis is
applied
to Trolley
a mass
10 kg3during
3 s and starts moving from rest
constant force
force to
applied
to Trolley
A; it A;
hasitahas
mass
10 kg during
s and starts
onmoving
a horizontal
surface.
Aftersurface.
the 3 s,After
thethe
trolley
moves
constant
velocity and collides with Trolley
from restfrictional
on a horizontal
frictional
3 s, the
trolleywith
moves
with constant
B,velocity
whichand
hascollides
a mass
8 kg B,
and
is athas
rest.
After
the After
two the
trolleys
stick together and move as a
withofTrolley
which
a mass
of 8the
kg collision,
and is at rest.
collision,
system.
The
graph
below
shows
how
the
momentum
of
Trolley
A
changes
with
time just before and after the
the two trolleys stick together and move as a system. The graph below shows how the momentum
collision.
of Trolley a changes with time just before and after the collision.
Momentum versus time graph for trolley A
P= kg·m·s-1
80
70
4,2
3,0 4,0
80
70
0,0
Momentum versus time graph for trolley A
7
51
t (s)
2.1 the term
Define
the of
term
impulse
of a force in words.
2.1 Define
impulse
a force
in words.
2.2 Use the information in the graph to calculate:
2.2
Use the information
in the graph
calculate:
0,0
3,0 4,0to4,2
7
t (s)
2.2.1. The impulse of the force exerted on the trolley during the three first seconds.
Commented [U119]: Is this one item? Is this complet?
Pse check all – repetitive problems are suspected, with items that
belong together being split into different items.
2.2.2 Velocity
of Trolley
A before
the collision.
2.2.1
The
impulse
of the force exerted on the trolley during the three first seconds.
Commented [GI120]: corrected
2.2.3 Velocity of the2.1
system
formed
by Trolley
and
B after
collision.
Define
the term
impulseAof
a force
in the
words.
2.2.4 The average
force
exerted
by
trolley
B
on
trolley
A
during
the
2.2.2 2.2
Velocity
of
Trolley
A
before
the
collision.
Use the information in the graph to calculate:collision.
Commented [U121]: Pse check/correct.
Commented [GI122]: corrected
[U119]: Is this one ite
2.2.1. The impulse of the force exerted on the trolley during the three first seconds.
Commented [U123]: Are theseCommented
sub-items of something?
2.2.3 Velocity of the system formed by Trolley A and B after the collision.
Pse check all – repetitive problems are
Vertical Projectile2.2.2
Motion
belong together being split into differe
Velocity of Trolley A before the collision.
5.3
Exercise 1
2.2.4
Commented [GI120]: corrected
2.2.3
Velocity of
the system
formed
by Trolley
A and
B after
the collision.
The
average
force
exerted
by trolley
B on
trolley
A during
the collision.
2.2.4 The average force exerted by trolley B on trolley A during the collision.
A falling stone takes 0,28 s to travel past a window that is 2,2 m in height. Ignore air friction.
Commented [U121]: Pse check/co
5.3 Vertical Projectile Motion
Exercise 1
5.3
Commented [GI122]: corrected
Commented [U123]: Are these su
Vertical Projectile Motion
Exercise 1
A falling stoneAtakes
to travel
a window
that isthat
2,2ism2,2inmheight.
Ignore
falling0,28
stonestakes
0,28 spast
to travel
past a window
in height.
Ignoreair
air friction.
friction.
1.1. From what height above the top of the window was the stone dropped?
52
1.1. From what height above the top of the window was the stone dropped?
1.1. From what height above the top of the window was the stone dropped?
52
48
1.2.
Draw the position vs time graph for the motion of the stone for the period of time from when it was
dropped until it reached the base of the window.
1.3.
Draw the velocity vs time graph for the motion of the stone for the period of time from when it was
dropped until it reached the base of the window.
Exercise 2
A very small rocket (A) is launched vertically upwards with an initial velocity of 100 m.s-1. At the same time,
a stone (B), which is initially at a height of 150 m, is dropped from the top of the very high building. Ignore air
resistance.
2.1
Calculate the velocity of stone B when it hits the ground.
2.2
Calculate the time taken for A and B to pass each other.
2.3
Calculate the fly time of the small rocket (A).
2.4
Draw the velocity versus time graph for the motion of the small rocket (A) from the moment it is launched
until it strikes the ground. Indicate the respective values of the intercepts on your velocity-time graph.
49
5.4 Work Energy Power
QUESTION 1
The diagram
below illustrates
illustrates a roller-coaster
track. Sections
A, B and
C are frictionless.
The diagram
below
a roller-coaster
track.
Sections
A, B and C are
There is friction on sections D and E. The sled has a total mass of 24,7 kg and is at rest at
frictionless.
There is friction on sections D and E. The sled has a total mass of
the top left position.
24,7 kg and is at rest at the top left position.
1.1
1.1
How does the total mechanical energy of the sled at the top left position compare to
the total mechanical energy of the sled at the end of section D. Write down
GREATER THAN, LESS THAN or EQUAL TO. Then explain your answer.
(3)
1.2 How
State
thethe
work-energy
theoremenergy
in words.
does
total mechanical
of the sled at the top left position compare (3) (2)
the speed
total mechanical
of the sled
at theBend
of section
1.3 to The
of the sled atenergy
the beginning
of section
is 24,25
m·s-1 ; D.
theWrite
speeddown
of the
-1
sled
at
the
end
of
section
D
is
18
m·s
.
GREATER THAN, LESS THAN or EQUAL TO. Then explain your answer.
1.3.1 Use the work-energy theorem to calculate the frictional force that acted on
the sled in section D.
1.3.2 Use the LAW OF CONSERVATION OF ENERGY (ENERGY PRINCIPLE) to
calculate the minimum length of section E required for the sled to come to
rest, if it has the same frictional force as calculated in 1.3.1.
(4)
(5)
(6)
[16]
5.5
Doppler Effect
Exercise 1
1.2
1.3
State the work-energy theorem in words.
(2)
The siren of a stationary police car emits sound waves at a frequency of 620 Hz. A
stationary
listener
watches
car approaching
at constant
onof
a straight
the speed
the
(4)
The speed
of the
sled atthe
thepolice
beginning
of section B him
is 24,25
m·s-1 ; velocity
-1
-1
sled
at the end
18 m·s
. is 340 m·s .
road.
Assume
that of
thesection
speedDofissound
in air
1.1
1.2
50
How does the wavelength of the sound waves heard by the listener compare to
the wavelength of the sound produced by the siren of the police car when it
approaches the listener? Write down LONGER THAN, SHORTER THAN or
EQUAL TO. Then explain your answer.
1.3.1
Use the work-energy theorem to calculate the frictional force that
Name the phenomenon observed in QUESTION 1.1.
acted on the sled in section D.
1.3.2
Use the LAW OF CONSERVATION OF ENERGY (ENERGY
PRINCIPLE) to calculate the minimum length of section E
required for the sled to come to rest, if it has the same frictional
force as calculated in 1.3.1.
(4)
(5)
(6) 54
[16]
(1)
5.5 Doppler Effect
Exercise 1
The siren of a stationary police car emits sound waves at a frequency of 620 Hz. A stationary listener watches the police car approaching him at constant velocity on a straight road. Assume that
the speed of sound in air is 340 m·s-1.
1.1
How does the wavelength of the sound waves heard by the listener compare to the
(4)
wavelength of the sound produced by the siren of the police car when it approaches the
listener? Write down LONGER THAN, SHORTER THAN or EQUAL TO. Then explain
your answer.
1.2
Name the phenomenon observed in QUESTION 1.1.
(1)
1.3
Calculate the wavelength of the sound waves detected by the stationary listener if the police car moves toward him at a speed of 110 km·h-1.
(6)
1.4
How will the answer to QUESTION 7.3 change if the police
car moves away from the listener at 120 km·h-1? Write down
INCREASES, DECREASES or REMAINS THE SAME.
(1)
1.5
Write down ONE application of the Doppler effect in medicine.
(1)
[13]
5.6 Electrostatics
Exercise 1
A group of grade 12 learners obtained the following set of values when investigating Coulomb’s
law.
Distance between charges (m)
1
2
4
8
Electrostatic force (N)
1600
400
100
25
1.1
Write down an investigative question for this investigation.
(2)
1.2
Write down the dependent variable.
(1)
1.3
Write down one controlled variable in this investigation.
(1)
1.4
Using the results in the table above, plot a graph of electrostatic
force versus distance between the charges.
(2)
51
1.5
1.6
Write down a conclusion for this investigation.
(2)
Two small identical positively charged spheres are used in this
investigation.
1.6.1 Draw the electric field pattern for the system of the two
(2)
charged spheres.
1.6.2 Calculate the magnitude of the charges used in this
(4)
investigation.
1.6.3 One of the charges is removed and the other one is mounted
on an insulated stand, as shown in the sketch below.
Calculate the electric field at a point 25 cm from the
charge.
+
(3)
25 cm
[19]
(3)
5.7
Electric Circuits
Exercise 1
+
(3)
25 cm
In the circuit diagram, below the emf of the battery is 12 V and the internal resistance is 0,4 Ω.
The ammeter reading is 3 A. The resistance of the three resistors is the same. Ignore the
[19]
5.7 Electric
Circuits
resistance of the wires.
Exercise 1
5.7
Electric Circuits
Exercise 1
[19]
V
V
In the circuit
diagram,
emf of
the battery
is 12
and the
the internal
resistance
is 0,4 Ω.
In the circuit diagram,
below
thebelow
emf the
of 12
the
battery
is 12
V Vand
internal
resistance
is 0,4 Ω. The
Theisammeter
reading
is 3 A. of
Thethe
resistance
of the three
is the
same. the
Ignore
the
ammeter reading
3 A. The
resistance
three resistors
isresistors
the same.
Ignore
resistance
of the
S
2
resistance of the wires.
wires.
A
V
R1
R2
R3
12 V
S2
S1
1.1
A
Write down TWO differences between electromotive force (emf) and terminal
potential difference.
1.1
R1
R2
R3
1.2
State ohm's law in words.
(2)
1.3
Calculate the current passing through resistor R 2.
(6)
1.4
S1
Determine the reading of the voltmeter.
(3)
1.5
How will the reading on the voltmeter be affected if switches S 1 and S2 are
1.1 Write down TWO differences between electromotive force (emf) and terminal
(2)
Write closed?
down TWO differences between electromotive force (emf) and
potential difference.
Write down INCREASES, DECREASES or REMAINS THE SAME.
1.2 State ohm's law in words.
terminal
potential difference.
Then explain the answer.
1.3 Calculate the current passing through resistor R 2.
1.2
State Ohm’s law in words.
1.3
Calculate the current passing through resistor R2.
1.4
(2)
1.4
Exercise 1.5
2
Determine the reading of the voltmeter.
(2)
(4)
[17]
(6)
(2)
(3)
(6)
How will the reading on the voltmeter be affected if switches S 1 and S2 are
Determine the reading of the voltmeter.
(3)
closed?
Write down INCREASES, DECREASES or REMAINS THE SAME.
Then explain the answer.
56
(4)
[17]
Exercise 2
52
(2)
56
1.5
How will the reading on the voltmeter be affected if switches S1 and
S2 are closed?
Write down INCREASES, DECREASES or REMAINS THE SAME.
Then explain the answer.
(4)
[17]
Exercise 2
Thabo is a grade 12 learner. He is investigating the relationship between potential difference and
Thabo is a grade
12 learner.
is investigating
relationship
between
potential
and current using
current
using a He
resistor
with unknownthe
resistance.
He sets
up the circuit
showndifference
in the diagram.
a resistor with unknown
resistance.
setscircuit
up the
circuit
shown He
in the
diagram.
Thabo
adjusts
Thabo adjusts
the currentHe
in the
using
the rheostat.
takes
the ammeter
reading
while the current in
the circuit usingthe
the
rheostat.
He takes the
reading
while the
voltmeter
is the
disconnected.
He then
voltmeter
is disconnected.
He ammeter
then measures
the potential
difference
across
unknown
measures the potential
difference
across
the
unknown
resistor
for
each
current
value
I.
From
the
readings
resistor for each current value I. From the readings taken from the ammeter and voltmeter, he
taken from the ammeter
and the
voltmeter,
could complete the following table.
could complete
following he
table.
Experiment
Current
Potential
difference
number
(A)
1
0,02
6
2
0,04
12
3
0,06
18
4
0,08
24
(V)
2.1. Writeinvestigative
a possible investigative
this practical
investigation.
2.1. Write a possible
questionquestion
for thisforpractical
investigation.
2.2. What could Thabo’s hypothesis have been for this investigation?
Commented [GI125]: Better lik
2.2. What could Thabo’s hypothesis have been for this investigation?
2.3. What is:
2.3. What is:
2.3.1. the dependent variable
Commented [GI126]: corrected
2.3.1. the dependent variable
Commented [U127]: Pse check.
2.3.2. the2.4.
independent
Plot a graph variable?
of current vs. potential difference with the experimental data collected.
Commented [GI128]: Better in
2.3.2. the independent variable?
Commented [U129]: Pse check
2.5. Calculate
thevs.
gradient
of thedifference
graph.
2.4. Plot a graph
of current
potential
with the experimental data collected.
2.6. What quantity does the gradient of the graph represent?
2.5. Calculate the gradient of the graph.
2.7. Write a conclusion for this investigation.
Commented [GI130]: Better fo
2.6. What quantity does the gradient of the graph represent?
5.8
Electrodynamics
Exercise 1 for this investigation.
2.7. Write a conclusion
57
53
5.8 Electrodynamics
Exercise 1
The diagram below represents an electrical machine; P is a split ring (commutator).
The diagram below represents an electrical machine; P is a split ring (commutator).
P
1.1
1.1
Identify
theoftype
of electrical
write conversion
down thethat
energy
Identify
the type
electrical
machine andmachine
write downand
the energy
takes conversion that
takes
place
in this
electrical machine.
place
in this
electrical
machine.
(2)
1.2
1.2
Explain
the function
of the component
P.
Explain
the function
of the component
P.
1.3
1.3
The
ring (commutator)
by slip
rings.
of the following voltThe
splitsplit
ring (commutator)
is replacedis
byreplaced
slip rings. Which
ONE
of theWhich
followingONE
voltageage-time
graphs
(Graph
A or
Graph B)with
corresponds
with the above change?
time
graphs (Graph
A or
Graph B)
corresponds
the above change?
(2)
(2)
(2)
Voltage
time
Voltage
12 V
12 V
time
Graph B
Graph A
Explain your answer.
Explain your answer.
1.4
1.4
(3)
The light bulb shown in the circuit dissipates energy of 6 J per second. An identical light (5)
The
light bulb
showntoin
the circuit
6 J the
pernew
second. An identical
bulb
is connected
in parallel
it. Calculate
the dissipates
rms current inenergy
the circuitofunder
light bulb
is connected
in parallel
to it. Calculate the rms current in [12]
the circuit under
conditions.
Assume
the emf remains
unchanged.
the new conditions. Assume the emf remains unchanged.
58
54
(3)
(5)
[12]
5.9 Photoelectric Effect
Exercise 1
The work function of potassium is 3,52 x 10-19 J. An electromagnetic radiation strike on the surface of this metal
produces the photoelectric effect. The maximum kinetic energy of the electrons ejected is 4,20 x 10-19 J.
1.1
Define work function.
1.2
Define photoelectric effect.
1.3
Calculate:
1.4
1.3.1
The frequency of the incident radiation.
1.3.2
The wavelength of the incident radiation.
1.3.3
The threshold/cut-off frequency.
The intensity of the incident radiation on the metal plate is increased, whilst
maintaining
constant frequency. State and explain what effect this change has
on the following:
1.4.1
kinetic energy of the emitted photo-electrons
1.4.2
number of photo-electrons emitted.
a
55
mber of photo-electrons emitted.
Commented [GI131]: Better in this way
6 Check your answers - Set 1
r answers - Set 6.1
1
Laws
Newton’s Laws
Exercise 1
Commented [U132]: Pse check right-hand item below.
1.1
1.2 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗
Applying Newton's second law on the y-direction.
y-axis. (upwards as positive
y-axis
⃗⃗
𝒇𝒇𝒌𝒌
⃗⃗⃗
𝑵𝑵
⃗⃗𝑨𝑨𝑨𝑨
𝑭𝑭
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦
⃗⃗
⃗
⃗⃗ +𝑭𝑭𝐹𝐹𝑨𝑨⃗y-axis
𝑁𝑁
𝐴𝐴𝐴𝐴 + 𝐹𝐹 𝑔𝑔 = 0
59
y-axis
𝟑𝟑𝟑𝟑𝒐𝒐
𝑁𝑁⃗⃗+ 𝐹𝐹 sin
37𝑜𝑜 − 𝑚𝑚𝑚𝑚 = 0
⃗⃗⃗x-axis
𝑭𝑭𝑨𝑨𝑨𝑨 𝑵𝑵
⃗𝑭𝑭⃗𝑨𝑨 𝑜𝑜
⃗𝑁𝑁
⃗𝑨𝑨𝑨𝑨= 𝑚𝑚𝑚𝑚⃗⃗⃗
𝑵𝑵− 𝐹𝐹 sin
𝑭𝑭
⃗𝑭𝑭⃗37
𝑨𝑨
⃗⃗𝑨𝑨𝑨𝑨
𝑭𝑭
⃗⃗
𝒇𝒇𝒌𝒌
𝒐𝒐
𝟑𝟑𝟑𝟑
⃗⃗
𝒇𝒇𝒌𝒌
𝟑𝟑𝟑𝟑𝒐𝒐 y-axis
⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨
x-axis
⃗⃗𝒈𝒈
𝑭𝑭
⃗⃗ ⃗⃗⃗
𝑵𝑵 x-axis
𝑭𝑭
⃗⃗
𝒇𝒇𝒌𝒌
1.3 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 =
𝑓𝑓𝑘𝑘
𝑁𝑁
𝑨𝑨𝑨𝑨
x-axis
⃗⃗𝒈𝒈
𝑭𝑭
⃗⃗𝒈𝒈
𝑭𝑭
⃗⃗𝒈𝒈
𝑭𝑭
To calculate the frictional force, we apply Newton's second law in the x-direction.
𝑓𝑓
1.3 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘
𝑁𝑁
= 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑘𝑘
1.3 as
𝑓𝑓𝑘𝑘 positive)
1.3
x-axis. (Right
𝑁𝑁
Commented [U133]: Pse check (including punctuation). (all)
To calculate
the frictional
force, we apply Newton's second law in the x-direction.
Exercise
2
⃗𝑅𝑅𝑅𝑅
To
thefrictional
frictional
force,
we apply
Newton’s
second
law in the x-direction.
To𝐹𝐹calculate
calculate
force,
we apply
Newton's
second law
in the x-direction.
= 𝑚𝑚𝑎𝑎⃗𝑥𝑥 the
𝑓𝑓⃗𝑘𝑘 x-axis.
+ 𝐹𝐹⃗𝐴𝐴𝐴𝐴 =(Right
0 2.1 as positive)
x-axis.
(Right
positive)
x-axis.
(Rightasas
positive)
𝑜𝑜
−𝑓𝑓𝑘𝑘 + 𝐹𝐹 cos 37 = 0
⃗ 𝑜𝑜= 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗⃗⃗
𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos𝐹𝐹𝑅𝑅𝑅𝑅
37
𝑵𝑵
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
⃗
⃗
+ cos
𝐹𝐹𝐴𝐴𝐴𝐴37
= 𝑜𝑜0
𝑓𝑓𝑘𝑘 =𝑓𝑓𝑘𝑘30
𝑓𝑓⃗𝑘𝑘 + 𝐹𝐹⃗𝑜𝑜𝐴𝐴𝐴𝐴 = 0
−𝑓𝑓𝑘𝑘 +
37 𝑁𝑁= 0
23,96
𝑓𝑓𝑘𝑘𝐹𝐹=cos
−𝑓𝑓𝑘𝑘 + 𝐹𝐹 cos 37𝑜𝑜 = 0 ⃗⃗
𝒇𝒇𝒔𝒔
𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos
37𝑜𝑜
𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos 37𝑜𝑜𝑜𝑜
𝑓𝑓𝑘𝑘
23,96
= 0,31
𝜇𝜇𝑘𝑘 = =
𝑓𝑓𝑘𝑘 = 30 cos 37
𝑁𝑁
79,95
𝑓𝑓𝑘𝑘 𝒐𝒐= 30 cos 37𝑜𝑜
𝑓𝑓𝟑𝟑𝟑𝟑
𝑘𝑘 = 23,96 𝑁𝑁
𝑓𝑓𝑘𝑘 = 23,96 𝑁𝑁 ⃗𝑭𝑭⃗𝒈𝒈
𝑓𝑓
1.4.1
Commented [U133]: Pse check (including punctuation). (a
Commented [U133]: Pse check (including punctuation). (
⃗⃗
𝑭𝑭
23,96
𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 = 23,96= 0,31
𝑁𝑁
79,95
𝜇𝜇𝑘𝑘 = 𝑘𝑘 =2.2 = 0,31
𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃
𝑁𝑁
79,95
We do not know the magnitude
of the gravitational
force,
butx we
can calculate it.
1.4.1
We apply Newton's
second law
in the
direction.
1.4.1
1.4.1
𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (20)(9,8) = 196 𝑁𝑁
(Right as positive)
Now we can calculatex-axis
the component:
𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30𝑜𝑜 = 196 sin 30𝑜𝑜 =
98apply
𝑁𝑁 Newton's second law in the x direction.
60
We
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 We apply Newton's second law in the x direction.
2.3
56
𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 x-axis (Right as positive)
x-axis (Right as positive)
−23,96 = 10𝑎𝑎⃗𝑥𝑥
2.4 We apply Newton's second 60
law in the
We apply Newton's second
60
⃗𝑥𝑥 y𝐹𝐹⃗𝑅𝑅𝑅𝑅law
= in
𝑚𝑚𝑎𝑎the
⃗
x-direction.
𝐹𝐹𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
direction.
𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦
−23,96 = 10𝑎𝑎⃗𝑥𝑥
𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
−23,96 = 10𝑎𝑎⃗𝑥𝑥
⃗⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 0
𝑁𝑁
𝐹𝐹⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
61
𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0
200 + (−98) + 𝐹𝐹⃗𝑓𝑓 = (20)(2)
𝑁𝑁 − 𝐹𝐹𝐹𝐹 cos 30𝑜𝑜 = 0
Commented [U134]: Below and all instance
Commented [GI135]: Corrected (in in both
Commented [U136]: Formatting: pse align
correctly (with the number). Repetitive problem
correct all.
y-axis
y-axis
y-axis
y-axis
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
⃗⃗⃗
𝑵𝑵 𝑵𝑵
⃗⃗⃗
𝑵𝑵y-axis
⃗⃗
𝒇𝒇y-axis
⃗⃗⃗
𝒌𝒌
𝑵𝑵
⃗⃗
𝒇𝒇𝒌𝒌
𝒇𝒇 ⃗⃗
⃗⃗
𝒇𝒇𝒌𝒌𝒌𝒌
⃗⃗
𝒇𝒇𝒌𝒌
⃗⃗⃗
𝑵𝑵
⃗⃗
𝒇𝒇𝒌𝒌
x-axis
x-axis
x-axis
x-axis
x-axis
x-axis 𝑭𝑭
⃗⃗𝒈𝒈
⃗𝑭𝑭⃗𝒈𝒈 ⃗𝑭𝑭⃗𝒈𝒈
⃗𝑭𝑭
⃗𝒈𝒈
⃗𝑭𝑭⃗𝒈𝒈
⃗⃗𝒈𝒈
𝑭𝑭
Exercise 2
Exercise
Exercise
2 2
Exercise 2 2
Exercise
Exercise 22.1
2.1 2.1 2
Exercise
2.1
2.1
2.1
2.1
⃗⃗
𝒇𝒇
⃗⃗
𝒇𝒇𝒔𝒔𝒔𝒔
⃗⃗
𝒇𝒇𝒔𝒔
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
𝑵𝑵
⃗⃗⃗
𝑵𝑵
⃗⃗
𝒇𝒇𝒔𝒔
⃗⃗
𝒇𝒇𝒔𝒔
⃗⃗
⃗𝑭𝑭⃗
𝑭𝑭
⃗⃗
𝑭𝑭
⃗⃗
𝑭𝑭
⃗⃗
𝑭𝑭
⃗𝑭𝑭⃗
𝒐𝒐
⃗⃗ 𝟑𝟑𝟑𝟑
𝟑𝟑𝟑𝟑𝒐𝒐𝒐𝒐 𝟑𝟑𝟑𝟑𝒇𝒇𝒐𝒐𝒔𝒔
⃗𝑭𝑭⃗𝒈𝒈
𝟑𝟑𝟑𝟑
⃗⃗
⃗𝑭𝑭⃗𝒈𝒈 𝑭𝑭
⃗𝑭𝑭
⃗𝒈𝒈 𝒈𝒈
𝟑𝟑𝟑𝟑𝒐𝒐
⃗⃗𝒈𝒈
𝑭𝑭
𝟑𝟑𝟑𝟑𝒐𝒐 2.2
𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃
⃗
⃗sin
2.2
𝐹𝐹
=
2.2
𝐹𝐹
=
𝐹𝐹𝐹𝐹
𝒈𝒈𝐹𝐹𝐹𝐹𝜃𝜃𝜃𝜃sin 𝜃𝜃
2.2
𝑔𝑔𝑔𝑔 = 𝑔𝑔𝑔𝑔
2.2
𝐹𝐹𝑔𝑔𝑔𝑔
𝐹𝐹𝐹𝐹𝑭𝑭sin
2.2
𝐹𝐹We
𝐹𝐹𝐹𝐹not
sinknow
𝜃𝜃
𝑔𝑔𝑔𝑔 =do
the magnitude of the gravitational force, but we can calculate it.
We
do
not
know
the
magnitude
ofof
the
gravitational
force,
butcan
we calculate
can we
calculate
We
do
not
know
the
magnitude
of the
gravitational
force,
but we
it. it.
We
do
not
know
the
the
gravitational
can
2.2
= 𝐹𝐹𝐹𝐹
sin
𝜃𝜃 magnitude
𝑔𝑔𝑔𝑔
We do not𝐹𝐹𝐹𝐹𝐹𝐹
know
the
magnitude
force, but force,
we can but
calculate
it. calculate it.
= 𝑚𝑚𝑚𝑚 = (20)(9,8)of
= the
196gravitational
𝑁𝑁
(20)(9,8)
𝐹𝐹𝐹𝐹
=
𝑚𝑚𝑚𝑚
=
=
196
𝑁𝑁
(20)(9,8)
𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚
=know
= 196 𝑁𝑁 of the gravitational force, but we can calculate it.
We
not=
the magnitude
(20)(9,8)
𝐹𝐹𝐹𝐹 =do𝑚𝑚𝑚𝑚
196 𝑁𝑁 the component:
Now
we can=calculate
Now
we
can
calculate
the
We
do
not
know
the
magnitude
of the gravitational force, but we can calculate it.
Now
we
can
calculate
the
component:
(20)(9,8)
𝐹𝐹𝐹𝐹 =we
𝑚𝑚𝑚𝑚can
= calculate=the
196
𝑁𝑁 component:
𝑜𝑜
Now
𝐹𝐹𝐹𝐹𝐹𝐹 =𝑜𝑜 𝐹𝐹𝐹𝐹 𝑜𝑜sin 30component:
= 𝑜𝑜196 𝑜𝑜sin 30𝑜𝑜 = 98 𝑁𝑁
Now
we
calculate
the
𝐹𝐹𝐹𝐹
sin
30196
196
sincomponent:
3098=𝑁𝑁98 𝑁𝑁
(20)(9,8)
𝐹𝐹𝐹𝐹
𝑚𝑚𝑚𝑚
==
196
𝑁𝑁
𝐹𝐹𝐹𝐹𝐹𝐹=𝐹𝐹𝐹𝐹𝐹𝐹
=
𝐹𝐹𝐹𝐹=can
sin
30
=
sin
30
=
Now
we
the
𝐹𝐹𝐹𝐹𝐹𝐹 =
𝐹𝐹𝐹𝐹can
sincalculate
30𝑜𝑜 = 196
sincomponent:
30𝑜𝑜 = 98 𝑁𝑁
𝑜𝑜
𝑜𝑜
Now
we
can
calculate
the
component:
𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30 = 196 sin 30 = 98 𝑁𝑁
𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30𝑜𝑜 = 196 sin 30𝑜𝑜 = 98 𝑁𝑁
Commented [U134]: Below and all instance
2.3
2.4 We apply Newton's second law in the
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all instances
: in or
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and all and
instances
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2.3 2.3
We apply Newton's second law in the
2.4 apply
We apply
Newton's
second
lawthe
in the
2.3
2.4y-We
Newton's
second
law in
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or on?
2.3 We apply
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in both
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Newton's
second
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in
the
yNewton's second law in the y2.4 We apply
Newton's second law in the
x-direction.
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We
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Newton's
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in
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ydirection.
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inFormatting:
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x-direction.
2.4
We
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direction.
x-direction.
direction.
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items
ve
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inalign
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items
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:
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or
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2.3 direction. 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦
correctly
(with
the
number).
Repetitive
problems
seen
inP
correctly
(with
the
number).
Repetitive
problems
seen
in
text.
correct all.
2.4
We
apply
Newton's
second
law
in
the
x-direction.
correctly
(with all.
the number). Repetitive problems seen in text. P
We
second law in the y𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
𝑚𝑚𝑎𝑎⃗
𝐹𝐹⃗ 𝑚𝑚𝑎𝑎=
direction.
⃗ Newton's
𝐹𝐹⃗𝑅𝑅𝑅𝑅 apply
=
correctcorrect
all.
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align
items
vertically
Commented
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Corrected
(in
in
both
cases)
⃗
𝐹𝐹⃗𝑅𝑅𝑅𝑅
= 𝑅𝑅𝑅𝑅
𝑚𝑚𝑎𝑎⃗𝑦𝑦𝑦𝑦 ⃗⃗ 𝑦𝑦 ⃗
⃗
correct
all.
𝐹𝐹 𝑚𝑚𝑎𝑎=
⃗ 𝑚𝑚𝑎𝑎⃗
𝐹𝐹𝑅𝑅𝑅𝑅 =
x-direction.
correctly (with the number). Repetitive problems seen in text. P
𝑁𝑁 + 𝐹𝐹𝑔𝑔𝑔𝑔 = 0
𝐹𝐹⃗𝑅𝑅𝑅𝑅
= 𝑅𝑅𝑅𝑅
𝑚𝑚𝑎𝑎⃗𝑥𝑥𝑥𝑥 ⃗ 𝑥𝑥 ⃗
direction.
⃗⃗ +=
Commented
[U136]: Formatting: pse align items vertically
⃗⃗𝑔𝑔𝑔𝑔
⃗𝑚𝑚𝑎𝑎
𝐹𝐹 + 𝐹𝐹𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
⃗𝐹𝐹⃗𝑦𝑦𝑔𝑔𝑔𝑔
𝐹𝐹
correct all.
+=
𝑁𝑁
𝑁𝑁
𝐹𝐹
0 =0
61 (with the number). Repetitive problems seen in text. P
⃗
⃗
⃗
⃗𝑁𝑁⃗𝑅𝑅𝑅𝑅+
⃗
⃗
⃗
⃗
++⃗𝐹𝐹𝑥𝑥𝑔𝑔𝑔𝑔
𝐹𝐹 = 0
𝐹𝐹⃗𝑅𝑅𝑅𝑅+ =
𝑚𝑚𝑎𝑎
⃗𝑥𝑥 𝑚𝑚𝑎𝑎⃗𝑥𝑥
𝐹𝐹𝐹𝐹
𝐹𝐹⃗𝑓𝑓 +
=𝐹𝐹𝑚𝑚𝑎𝑎
𝑓𝑓 =
𝑔𝑔𝑔𝑔
⃗𝑔𝑔𝑔𝑔
61 correctly
𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0
61
⃗
𝐹𝐹
+
𝐹𝐹
+
𝐹𝐹
=
𝑚𝑚𝑎𝑎
⃗⃗⃗𝑅𝑅𝑅𝑅 =⃗𝑔𝑔𝑔𝑔
𝑓𝑓
𝑥𝑥
⃗
𝐹𝐹
𝑚𝑚𝑎𝑎
correct all.
61
𝑁𝑁 −=𝐹𝐹𝑦𝑦𝑔𝑔𝑔𝑔
+ 𝐹𝐹⃗𝑓𝑓 = (20)(2)
𝑁𝑁 +
0 =0
− 𝐹𝐹
⃗⃗𝑅𝑅𝑅𝑅 =⃗ 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗200 + (−98)
𝐹𝐹
⃗(20)(2)
𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔
𝐹𝐹
+ 𝐹𝐹
++𝐹𝐹𝑓𝑓(−98)
=
200
𝑔𝑔𝑔𝑔 = 0
200
+
+ 𝑚𝑚𝑎𝑎
𝐹𝐹⃗⃗𝑓𝑓⃗+
=
𝑔𝑔𝑔𝑔(−98)
𝑥𝑥 𝐹𝐹
𝑓𝑓 = (20)(2)
𝑁𝑁 −𝑜𝑜 𝐹𝐹𝐹𝐹 cos
30𝑜𝑜 = 0
61
(−98)
(20)(2)
200
+
+
𝐹𝐹
=
⃗𝑁𝑁
⃗
⃗
𝑜𝑜
=
0030
𝑔𝑔𝑔𝑔
𝑁𝑁
cos =
300 = 0
𝑁𝑁 +
− 𝐹𝐹
𝐹𝐹
=𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹
cos
+𝑓𝑓𝐹𝐹
⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗102
𝑔𝑔𝑔𝑔−
⃗⃗𝑥𝑥𝑓𝑓 = 40
𝐹𝐹
= 𝑚𝑚𝑎𝑎
𝑁𝑁 − 𝐹𝐹𝐹𝐹
cos𝑁𝑁30=𝑜𝑜 𝐹𝐹𝐹𝐹
= 0cos 30𝑜𝑜
(−98)
200
61
102
=𝐹𝐹⃗40
102 +
𝐹𝐹⃗⃗𝑓𝑓 +
=𝑓𝑓𝐹𝐹⃗40
𝑓𝑓 = (20)(2)
𝑓𝑓+
𝑜𝑜
𝑜𝑜
102
+
𝐹𝐹
=
40
𝑜𝑜 30
𝑁𝑁
−
030
𝐹𝐹𝐹𝐹
cos
𝑓𝑓
=𝐹𝐹𝑁𝑁
𝐹𝐹𝐹𝐹
cos
𝑔𝑔𝑔𝑔==
𝑁𝑁
⃗𝑓𝑓 +
𝐹𝐹
=
40
−
102
⃗
𝑁𝑁 −
= 𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹 cos
cos 30
30𝑜𝑜 = 0
(−98)
(20)(2)
200
+
𝐹𝐹
=
𝑓𝑓
⃗𝑓𝑓−=
102
40
𝐹𝐹⃗40
40
− 102
𝐹𝐹⃗ =+
102
𝑜𝑜
𝑓𝑓𝐹𝐹=
𝑁𝑁 −
𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 40
− 102
= 𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹cos
cos30
30𝑜𝑜 = 0
⃗
102
𝐹𝐹𝑓𝑓−=102
40
𝐹𝐹⃗𝑓𝑓 =+40
𝑁𝑁 = 𝐹𝐹𝐹𝐹 cos 30𝑜𝑜
⃗
𝐹𝐹𝑓𝑓 = 40 − 102
𝑓𝑓
2.5 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑓𝑓𝑘𝑘𝑘𝑘 = 𝑁𝑁𝑘𝑘
2.5
𝑓𝑓
𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑁𝑁
2.5
62
𝜇𝜇𝑘𝑘62=
= 0,365
169,7
𝜇𝜇𝑘𝑘 =
= 0,365
169,7
6.2
Solutions: Momentum and Impulse
Solutions: Momentum and Impulse
Exercise 1 Solutions
Exercise 1 Solutions
1.1 Momentum is the product of an object’s mass and its velocity.
1.1 1.2
Momentum
is-1the
product
of an object’s
mass and its
velocity.
25 kgm.s
[Read
the graph;
initial momentum
is where
the graph starts].
-1
1.2 1.3
25 kgm.s
[Read
graph;
initial
is where
theisgraph
starts].or decrease; the slope
From 1s
to 10the
s the
graph
is amomentum
horizontal line
– there
no increase
1.3 Fromof1s
tograph
10 s the
graph is a horizontal line – there is no increase or decrease; the slope
the
is zero.
of the
1.4
Atgraph
15 s, is
25zero.
s and 30 s [These points correspond to the 0 value for both the x and y axis.]
1.4 1.5
At 15From
s, 250-15
s and
30 sobject
[These
points correspond
the 0 value
for both
x andaty15
axis.]
s the
is moving
in the sametodirection
(East);
thenthe
it stops
s; BUT it
1.5 Frommoves
0-15 sinthe
is moving
in the
same
direction
theobject
opposite
direction
from
15s to
25 s. (East); then it stops at 15 s; BUT it
moves
the opposite
direction
from 15sstopped
to 25 s. and changed direction. / The object could have
1.6
Theinobject
could have
decelerated,
1.6 The collided
object could
decelerated,
stopped
and changed
/ The
objectmoved
could in
have
with have
another
object at 10s,
and stopped
for 5 sdirection.
from 10-15
s, then
the
collided
with another
10s,s.and stopped for 5 s from 10-15 s, then moved in the
opposite
directionobject
from at
15-25
opposite
directionoffrom
15-25acting
s.
1.7
The product
the force
on an object and the time the force acts on the object.
1.7 1.8
The F
product
force acting
on anfrom
object
the time the force acts on the object.
= the
∆p [using
the values
theand
graph]
net ∆t of
1.8 FnetF∆t
=
∆p
[using
the
values
from
the
graph]
net ∆t = 0-25
6.2
Commented [U137]: Pse check item, includ
Commented [U137]: Pse check item, including bra
Commented [GI138]: corrected
Commented [GI138]: corrected
57
6.2 Solutions: Momentum and Impulse
Exercise 1 Solutions
1.1
Momentum is the product of an object’s mass and its velocity.
1.2
25 kgm.s-1 [Read the graph; initial momentum is where the graph starts].
1.3
From 1s to 10 s the graph is a horizontal line – there is no increase or decrease; the slope of the graph
is zero.
1.4
At 15 s, 25 s and 30 s [These points correspond to the 0 value for both the x and y axis.]
1.5
From 0-15 s the object is moving in the same direction (East); then it stops at 15 s; BUT it moves in
the opposite direction from 15s to 25 s.
1.6
The object could have decelerated, stopped and changed direction. / The object could have collided
with another object at 10s, and stopped for 5 s from 10-15 s, then moved in the opposite direction from
15-25 s.
1.7
The product of the force acting on an object and the time the force acts on the object.
1.8
Fnet ∆t = ∆p [using the values from the graph]
Fnet ∆t = 0-25
= -25 kgm.s-1
Impulse = 25 kgm.s-1 to the West
6.3 Vertical Projectile Motion
Exercise 1 Solutions
It is important to make a sketch of the situation and to select positive direction. In this case, let’s use positive
direction upwards. If a drawing is not provided, drawing the situation will help in solving the problem.
1.1
a =g= 9,8 (m.s-2 ) downwards.
1.2
H= hi + ∆y
v2= vi2+2g∆y
At the maximum height, the velocity of the projectile is zero:
58
02 = (10)2+2 (-9,8) ∆y
-100 = - (19,6) ∆y
∆y = 5,10 m
H = 3,5 + 5,10 = 8,60 m
1.3 1.1 Zero.
a turn, here so it stops momentarily.]
-2
) downwards.
a =g=[The
9,8 ball
(m.smakes
1.2 H= hi + ∆y
2
1.4
The timevrequired
is from point 0 of projection to the maximum height where the stone turns back = vi2+2g∆y
At the maximum
height,
thev velocity
of the projectile is zero:
= 0 m.s-1
towards
the ground
and the
f
2
2
0 = (10) +2 (-9,8) ∆y
-100 = - (19,6) ∆y
vf =(m.s
vi +-2 )gdownwards.
Δt
1.1 a =g= 9,8
∆y = 5,10 m
1.2 H= hi + ∆y
2
H = 3,5 + 5,10 = 8,60 m
= vi2+2g∆y
1.1 a =g= 9,8 (m.s-2 0
)vdownwards.
= 10+ (-9,8)
At the maximum
height,Δt
the velocity of the projectile is zero:
1.2
1.3
1.4
1.5
1.6
H= hi + ∆y
2
2
= (10)
+2 (-9,8)
0Zero.
v2=1.3
vi2+2g∆y
[The
ball ∆y
makes a turn, here so it stops momentarily.]
Commen
-100
=
(19,6)
∆y of the
square brac
At the maximum
the Δt
velocity
projectile
is zero:
-10height,
= (-9,8)
1.4
The
time
required
is
from
point
0
of
projection
to
the
maximum
height
where
the
stone
turns
back
∆y (-9,8)
= 5,10∆y
m
items.
02 = (10)2+2
-1
towards
the ground and the vf = 0 m.s
H = 3,5
-100 = - (19,6)
∆y + 5,10 = 8,60 m
Commen
Δtm= 1,02 s
∆y = 5,10
1.3 HZero.
ball= makes
Commented [U139]: Formatting/re-typing:
pse check u
vf = mvia+turn,
g Δthere so it stops momentarily.]
Commen
= 3,5[The
+ 5,10
8,60
square brackets – standardise use throughoutseen
text with
1.4 The time required0is=from
0 ofΔt
projection to the maximum height where the stone turns back
withequ
c
10+point
(-9,8)
items.
-1
1.5
Time
taken
from
the
point
of
projection
the
maximum
height
and
the
time
from
the
maximum
height
towards
the
ground
and
the
v
=
0
m.s
f stops momentarily.]
Zero. [The ball makes a -10
turn,=here
so
it
Commented [U139]: Formatting/re-typing: pse check use of
(-9,8) Δt
Commented
corrected
square brackets
– standardise[GI140]:
use throughout
text with equivalent
The time required
is from
point
0 of projection
the maximum height where the stone turns back
the
point
ofsprojection
aretoequal
items.
Δt
=
1,02
-1
v
=
v
+
g
Δt
Commented
[U141]:
Re-typing
required – repetitive pr
f
i
towards the ground and the vf = 0 m.s
with character
spacing – pse standardise throughout.
Commentedseen
[GI140]:
corrected
0 = 10+ (-9,8) Δt
-10
(-9,8)
Δt
+tdown
tΔt
==tup
vf = vi + g
Commented [U141]: Re-typing required – repetitive problems
= 1,02
s Δt taken from the point of projection the maximum height and the seen
with character spacing – pse standardise throughout.
0 =Δt
10+
(-9,8)
1.5
Time
time
from the maximum
Commen
-10 = (-9,8)t height
Δt= t
when the ball reaches the projection point
use of italic
up
downthe point of projection are equal
Δt = 1,02 s
t
=
t
+t
Commen
up
down
1.5 Time taken from the point of projection the maximum height and the time from the maximum
Commented [U142]: Formatting: Pse check
need for a
t = 2tup of projection are equal
use of italics (pattern not seen by editor).
height the point
tup= tdown when the ball reaches the projection point
Commen
= tuppoint
+tdownof
[U143]:Pse
Pse
check.
Time taken fromt t=
the
projection
the maximum height and the time from the maximum
CommentedCommented
[U142]: Formatting:
check
need for ad hoc
t ==
2t2,04
up
x 1,02
use of italics (pattern
not seen[GI144]:
by editor).corrected
=2tdown
when
the
ballsreaches the projection point
height the point oftupprojection
are
equal
Commented
t
=
2
x
1,02
=
2,04
s
t = tup+tdownt = 2tup
Commented [U143]: Pse check.
1.6 when
The
total
time
for the
motion
of the stone is the time taken by the stone from
the point of projection
t = 2total
x 1,02
=reaches
2,04
t
the
ball
themotion
projection
up= tdown The
1.6
time
for sthe
of point
the stone is the time taken by the stone from
the point
of corrected
projection to
Commented
[GI144]:
to
the
maximum
height,
plus
the
time
for
the
stone
from
the
maximum
height
to the
ground.
Commen
1.6 t The
total
time
for
the
motion
of
the
stone
is
the
time
taken
by
the
stone
from
the
point
of
projection
= 2tup
the
maximum
height,
plus
the
time
for
the
stone
from
the
maximum
height
to
the
ground.
to
the
maximum
height,
plus
the
time
for
the
stone
from
the
maximum
height
to
the
ground.
Commented
[U145]: Pse check.
t = 2 x 1,02 = 2,04 s
1
Commen
2
Δyof
viΔt
+ is g
1=the
The total time for the motion
stone
theΔt
time taken by the stone from the point of projection
Commented [GI146]: corrected
Δy = viΔt + g Δt2
2
to the maximum height, plus 2the time for the stone from the maximum height to the ground.
Commented [U145]: Pse check.
theheight
maximum
height to the ground:
1 the2 From
Commented [GI146]: corrected
From
maximum
to the ground:
Δy = viΔt + g Δt
1
1
2
2
2
-8,60 = (0-8,60
xheight
Δt) += (0
(-9,8)
Δt
xthe
Δt)
+ (-9,8) Δt
From
themaximum
maximum
to
ground:
From the
height
to the
2 ground: 2
2
1 Δt
-8,60 =- 4,9
-8,60 = (0 x Δt) + -8,60
(-9,8) Δt
=-2 4,9 Δt2
8,60 2
 1,76

8,60
4,9t 

t  Δt2
-8,60 =-4,9
t 
1.7
8,60Δt = 1,33 s
4,9
 1,76 
4,9 tt = t1 (upwards)
+
t
2 (downwards)
Δt = 1,33 s
1,76 
= 2,35 s
Δt = 1,33 stt = 1,02+ t1,33
t = t1 (upwards) + t2 (downwards)
tt = t1 (upwards) + t2 (downwards)
tt =s 1,02+ 1,33 = 2,35 s
1.7 tvt f==1,02+
vi + g 1,33
Δt = 2,35
vf = 0 + (-9,8)1,33
v f = vi +
g 1.7
Δt
vf = vi + g Δt
1.7
vf = 0 + (-9,8)1,33 vf = 0 + (-9,8)1,33
vf = 0 – 13,03
vf = - 13,03 m.s-1
vf = 13,03 (m.s-1 ) downwards
downwards
63
63
63
Thetime
timetaken
takenby
by the
the object
object to
to go
go up
up is
is less
less than
than the
the time
time taken
taken for
The
for itit to
to come
come down
down from
from the
the maximum
maximum height. Therefore, the area of the triangle representing motion of the stone up will be
height.
Therefore,
area
of the representing
triangle representing
of from
the stone
up will
be less
than the area of
less than
the areathe
of the
triangle
the objectmotion
moving
maximum
height
to the
ground.
the
triangle representing the object moving from maximum height to the ground.
1.8
59
ground. maximum height. Therefore, the area of the triangle representing motion of the stone up will be
less than the area of the triangle representing the object moving from maximum height to the
ground.
1.8
The position vs time graph is actually a mind-map of the actual path travelled by the object.
10
1.9 1.9 The
Theposition
position
graph
is actually
a mind-map
of path
the actual
travelled
vs vs
timetime
graph
is actually
a mind-map
of the actual
travelledpath
by the
object. by the object.
9
8
10
7
9
6
8
5
7
4
6
3
y (m)
1.9
1.8
y (m)
1.8
5
2
4
1
3
t (s)
2
0
0
0.5
1
Exercise 2
1
1.5
2
2.5
t (s)
0
0
0.5
1
1.5
2
2.5
Solution:Exercise 2
Exercise 2
2.1
Solution:
Solution:
2.1
64
64
2.1

From 0 to 1 s, the velocity of the ball decreases from 10 m∙s-1 to zero, and the acceleration of the ball
is constant.

From 1 to 2 s, the velocity of the ball increases from zero to (– 10 m∙s-1) with constant acceleration. At 2
s, the ball is at the point of projection.

From 2 to 3 s, the velocity of the ball increases from (– 10 m∙s-1) to (– 20 m∙s-1). At 3 s, the ball hits the
ground and bounces back.

From 3 to 4 s, the velocity of the ball decreases to zero at the maximum height.

From 4 to 5 s, the velocity of the ball increases with constant acceleration, until it hits the ground for the
second time.
60
ball
ballhits
hitsthe
theground
groundand
andbounces
bouncesback.
back.
-1
 From 0
to1From
s,From
the velocity
ballvelocity
decreases
10
m∙s
to zero, and to
the
acceleration
3 3toto4of4s,the
s,the
the
velocityoffrom
ofthe
the
ball
balldecreases
decreases
tozero
zeroatatthe
themaximum
maximumheight.
height.
of the ball
is
constant.
 From
From4 4toto5 5s,s,the
thevelocity
velocityofofthe
theball
ballincreases
increaseswith
withconstant
constantacceleration,
acceleration,until
untilit ithits
hitsthe
the
-1
 From 1 to 2 s, the velocity of the ball increases from zero to (– 10 m∙s ) with constant
ground
ground
for
for
the
the
second
second
time.
time.
acceleration. At 2 s, the ball is at the point of projection.
 From 2 to 3 s, the velocity of the ball increases from (– 10 m∙s-1) to (– 20 m∙s-1). At 3 s, the
-1 -1 -1
ball
hits the
ground
andupwards
bounces
back.
2.2
1010
m∙s
upwards
2.2
2.2
10
m∙s
m∙s
upwards
 From 3 to 4 s, the velocity of the ball decreases to zero at the maximum height.
 From 4 to 5 s, the velocity of the ball increases with constant acceleration, until it hits the
2.3
Option
11 1 time.
Option
2
ground
for
the second
2.3
2.3
Option
Option
Option
Option
22
|∆𝑦𝑦|
|∆𝑦𝑦|
𝐻𝐻𝐻𝐻=
-1 =
10 m∙s
upwards
2.2
𝑣𝑣𝑓𝑓𝑣𝑣𝑓𝑓++𝑣𝑣𝑖𝑖𝑣𝑣𝑖𝑖
2.3 Option 1
𝐻𝐻𝐻𝐻==( (
) ∆𝑡𝑡
) ∆𝑡𝑡
𝐻𝐻 = |∆𝑦𝑦|
22
𝐻𝐻 = (
Option 2
𝑣𝑣𝑓𝑓 + 𝑣𝑣𝑖𝑖
) ∆𝑡𝑡 −20
(−10)
−20++(−10)
2
(3(3
𝐻𝐻𝐻𝐻==( (
22
−20 + (−10)
𝐻𝐻 = (
) (3 − 2)
2
𝐻𝐻𝐻𝐻==1515𝑚𝑚𝑚𝑚
) ) −−2)2)
𝐻𝐻 = 15 𝑚𝑚
11
22
𝐻𝐻𝐻𝐻==𝐴𝐴 𝐴𝐴==𝐴𝐴1𝐴𝐴1++𝐴𝐴2𝐴𝐴2==𝑏𝑏ℎ𝑏𝑏ℎ++ 𝑏𝑏ℎ𝑏𝑏ℎ
1
𝐻𝐻 = 𝐴𝐴 = 𝐴𝐴1 + 𝐴𝐴2 = 𝑏𝑏ℎ + 𝑏𝑏ℎ
2
1
𝐻𝐻 = (1)(−10) + (1)(−10)
2
11
(1)(−10)++ (1)(−10)
(1)(−10)
𝐻𝐻𝐻𝐻==(1)(−10)
22
(−5)==1515𝑚𝑚𝑚𝑚
𝐻𝐻 = −10 + (−5) = 15 𝑚𝑚 𝐻𝐻𝐻𝐻==
−10
−10++(−5)
2.4
Inelastic, because after bouncing, the speed of the ball is less than the speed it hits the
2.4
2.4
Inelastic,
Inelastic,
because
because
after
after
bouncing,
bouncing,
thespeed
speed
ofof
the
the
ball
ball
isisthan
less
lessthe
than
than
the
thespeed
itPseithits
hits
the
thethen
2.4
Inelastic,
after
bouncing,
the the
speed
of the
ball
is
less
speed
itspeed
hits the
ground
Commented
[U147]:
check
– not understood.
ground then
kinetic
energy
isbecause
not
conserved.
2.5
ground
ground
then
thenkinetic
kineticenergy
energyisisnot
notconserved.
conserved.
kinetic energy is not conserved.
2.5
2.5
2.5
Position vs. time (taking ground as level
zero)
Position
Positionvs.
vs.time
time(taking
(takingground
groundasaslevel
level
zero)
zero)
25
20
y (m)
15
2525
10
2020
5
1515
y (m)
y (m)
0
-5 0
1
2
1010
3
4
5
55
Or
00
-5-50 0
Or
6
t (s)
11
Position vs. time (taking top of
building as level zero)
22
33
44
65 5 5
66
t (s)
t (s)
10
5
Y (m)
0
-5 0
1
2
3
4
5
6
6565
-10
-15
-20
t (s)
Work Energy Power
Exercise 1
1.1 The net (total) work (done on an object/particle) is equal to the change in kinetic energy
(of the object/particle).
1.2 Data
Free-body diagram
m = 600 kg
FA = 191,7 N
Θ = 300
ff = 160,02 N
61
Commented
Commente
-20
6.4
Work Energy Power
Exercise 1
6.4
t (s)
Work Energy Power
Exercise
1.1 The net (total) work (done
on1an object/particle) is equal to the change in kinetic energy
6.4 Work Energy 1.1
Power
The net (total) work (done on an object/particle) is equal to the change in kinetic energy
(of the object/particle).
Exercise 1
1.1
(of the object/particle).
The net (total) work (done on an object/particle) is equal to the change in kinetic energy (of the
object/particle).
1.2
Free-body diagram
1.2 Data
Data
Free-body diagram
1.2 Data
m = 600 kg
mkg
= 600 kg
m = 600
FA = 191,7 N
FA = 191,7 N
Θ = 300
FA = 191,7 N
ff = 160,02 N
Θ = 300
Δx = ?
Θ =N
30
ff = 160,02
0
Δx = ?
ff = 160,02 N
Δx = ?
The work-energy theorem can be applied for both conservative and non-conservative forces:
W net = ΔEK 
Wfr +be
W Aapplied
+W g+ Wfor
N = ΔE
K
The work-energy theorem can
both
conservative and non-conservative forces:
mv2f
mvi2
2
2
ΔE 
W
The=work-energy
theorem can be applied for both conservative and non-conservative
forces:
ff Δx cos 1800 + FΔx cos 300 + FgΔx cos 900 + NΔx cos 900 =
net

K
W net = ΔEK 
Wfr fr++WW
+W
= ΔE
A
K
+W
+g+
WW
=NΔE
W
A
g
N
K
ff Δx cos 1800 + FΔx cos 300 + FgΔx cos 900 + NΔx cos 900 =
mv2f
2

66
2
i
mv
2
600(0)
6003

160,02 Δx (-1) +(191,7) Δx (cos 30 ) + 0 + 0 =

2
2
2
2
0
(-160,02+ 166.017...) Δx= 2700
(5,997...) Δx =2700
Δx = 450,22
m
(5,997...)
Δx =2700
66
(-160,02+ 166.017...) Δx= 2700
6003
2
160,02
Δx (-1)
+(191,7)
Δx (cos
300is
) +done:
0+0P
=W  
1.3= 450,22
Power
is the
rate at which
work
Δx
m
2
t
600(0) 2
2

2
(-160,02+
166.017...)
Δx= 2700
600(0) 2
W0
6003
Fxcos


160,02
Δx
(-1)
+(191,7)
Δx
(cos
30
)
+
0
+
0
=
 answer
P
=
Alternative

1.3
Power
is
the
rate
at
which
work
is
done:
P
(5,997...)
Δx =2700
2
2
∆t

t
Δx = 450,22 m
(-160,02+ 166.017...) Δx= 2700
We must
convert
seconds:
P=F v 
θ minutes
F∆xcos
(5,997...) into
Δx =2700
W
= Power=is300
1.3
the srate
at
which
work
is
done:

P

5P minutes
Δx = 450,22 m
∆t
t

 cos
191
,
7
450
,
22
30 o is the rate at which work is done: P  W 
30
1.3
Power
F

xcos

P=191,7answer
cos 300 

P 
 )
t
Alternative
P
300
We must 
convert
minutesFinto
 2 
xcosseconds:

t
Alternative answer
P
P = must
249,03
W minutes
t seconds:
We
convert
into
P=F v 
Greater
than
P=F v  W
P = 249,03
5
minutes
== 300
ss We must convert minutes into seconds:
51.4
minutes
300
minutes
= 300 s
The work done will 5be
greater.
o
30 
191,7  450,22  cos 30,7  450,22  cos 30 o
0  0 3
As
increases;
therefore,
will
greater.
P=191,7
cosbe
30

P θ decreases, cosP θ 191
30   0)
P=191,7
cos
 the horizontal component
 )

300
300
 2 2 
P = 249,03 W
P
= 249,032W
Exercise
1.4 Greater than
P = 249,03 W
P
= Greater
249,03 W
1.4
than
P = 249,03
2.1 The total mechanical
energy
sum of gravitational potential
energyW
and kinetic energy) in
The work
done will(the
be greater.
The
work done
will be
greater.
As
θ decreases,
cos θ increases; therefore, the horizontal component will be greater.
an isolated
system
remains
constant.
1.4
thancos θ increases; therefore, the horizontal component will be greater.
As θGreater
decreases,
Exercise 2
2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in
Exercise
2
The
work done
will beangreater.
isolated system remains constant.
2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in
an θ
isolated
system
constant.
As
decreases,
cosremains
θ increases;
therefore, the horizontal component will be greater.
2.2 From top to bottom of the inclined plane:
Data
2.2 From top to bottom of the inclined plane:
m= 2 kg
Free body diagram
Data
hi=62
4m
m= 2 kg
Free body diagram
hi= 4 m
hf = 0 m
2.2
From top to bottom
of the inclined plane:
hf = 0 m
vi= 0 m·s-1
Data
vi= 0 m·s-1
300
5 minutes = 300 s
We must convert
minutes
seconds:
P = 249,03
W into
 cos
191,7 = 450
,Greater
22
30 o
5Pminutes
300
s
1.4
than


 2 
P=F v 
30
PP=191,7
= 249,03 W
cos 300 
 )
The
work done willo be greater.
300
 cos 30cos θ increases; therefore, the horizontal component will be greater. 0  32 0 
191,7  450
,θ22
As
decreases,
P=191,7 cos 30 

 )
PP =249,03 W
2 
300

1.4 Greater than
P = 249,03 W
2.1will
Thebe
totalgreater.
mechanical energy (the sum of gravitational potential energy and kinetic energy) in
The work done
1.4
than
an isolated system remains constant.
P = 249,03
W
As θGreater
decreases,
cos θ increases; therefore, the horizontal component
will be greater.
2.1
Thedone
total will
mechanical
energy (the sum of gravitational potential energy and kinetic energy) in an The work
be greater.
As
θ decreases,
cos θremains
increases;
therefore, the horizontal component will be greater.
isolated
system
constant.
Exercise
2
Exercise 2
Exercise
2 W
P = 249,03
2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in
Exercise
2 system
2.2
From
top to bottom
of the
inclined plane:
an isolated
remains
constant.
2.2 From top to bottom of the inclined plane:
2.1 The total
mechanical energy (the sum of gravitational potential energy and kinetic energy) in
Data
an
isolated
system
remains constant. Free body diagram
m=
2
kg
Data hi= 4 m
hf = 0 m
m= 2 kg vi= 0 m·s-1
g= 9,8 m·s-2
v
2.2 From topi=?
to bottom of the inclined plane:
hi= 4 m
Data
2.2
From top to bottom of the inclined
m= 2 kg
Freeplane:
body diagram
h= 0 m
Data
hi= 4f m
Only conservative energy acting
the diagram
object, the system is isolated therefore mechanical
m=
kg
Free on
body
hf= 02 m
-1
energy
is
conserved.
=
0
m·s
v
hi=
m -1
vi= 04i m·s
𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
h
=
0
m
f
g= 9,8 m·s-2
We can apply the principle of conservation of mechanical energy to solve the problem:
9,8-1m·s-2 𝐸𝐸𝑀𝑀𝑀𝑀 = 𝐸𝐸𝑀𝑀𝑀𝑀
vvii=
=?0g=m·s
𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝐾𝐾𝐾𝐾 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝐾𝐾𝐾𝐾
g= 9,8 m·s-2
𝑚𝑚𝑣𝑣𝑓𝑓2
𝑚𝑚𝑣𝑣𝑖𝑖2
vi=?vi=?
+ 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 =
+ 𝑚𝑚𝑚𝑚ℎ𝑓𝑓
2
Commented [U148]: Pse check –
used so that all readers can easily un
– repetitive problems seen.
2
Only conservative energy acting on the object, the system is isolated therefore mechanical energy is
Only
conservative energy acting on the object, the system is isolated therefore mechanical
conserved.
67
energy is conserved. Only
𝐸𝐸𝑀𝑀 =conservative
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 energy acting on the object, the system is isolated therefore mechanical
energy
conserved.
We canisapply
the principle
of conservation of mechanical energy to solve the problem:
𝐸𝐸𝑀𝑀
=
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= 𝐸𝐸𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀can
We
apply the principle of conservation of mechanical energy to solve the problem:
We
can
the
of conservation of mechanical energy to solve the problem:
𝐸𝐸𝐾𝐾𝐾𝐾apply
= 𝐸𝐸𝐾𝐾𝐾𝐾
+ principle
𝐸𝐸𝐾𝐾𝐾𝐾
𝐸𝐸𝐾𝐾𝐾𝐾 +
2
2
𝐸𝐸
𝑀𝑀𝑀𝑀 = 𝐸𝐸𝑀𝑀𝑀𝑀
𝑚𝑚𝑣𝑣𝑓𝑓
𝑚𝑚𝑣𝑣
𝑖𝑖
+𝐸𝐸𝑚𝑚𝑚𝑚ℎ
+ 𝑚𝑚𝑚𝑚ℎ𝑓𝑓
=
𝑖𝑖
+
=
𝐸𝐸
𝐸𝐸𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾 +2 𝐸𝐸𝐾𝐾𝐾𝐾
22
2
𝑚𝑚𝑣𝑣
𝑚𝑚𝑣𝑣𝑖𝑖
𝑓𝑓
+ 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 =
+ 𝑚𝑚𝑚𝑚ℎ𝑓𝑓
2
2
2𝑣𝑣𝑓𝑓2
2 × 02
+ 2 × 9,8 × 4 = 2
2𝑣𝑣𝑓𝑓 2 + 2 × 9,8 × 4
2
2
2×0
+ 2 × 9,8 × 4 =
+ 2 × 9,8 × 4
2
0 + 78,4 = 𝑣𝑣𝑓𝑓2 + 0 2
67
67
v2 = 78,4
v=
Commented
used so that al
– repetitive pro
Commente
used so that
– repetitive p
0 + 78,4 = 𝑣𝑣𝑓𝑓2 + 0
v = 78,4 v = 8,85 m∙s-1
v2 = 78,4
= The
8,85net/total
m∙s-1work done on an object is equal to the change in the object’s kinetic energy.
78,4 v2.3
2.4
From A to B.
Free
2.3 The
net/total
work done
an
object
is
equalistoequal
the change
the object’s
energy.
2.3 The
net/total
work on
done
onbody
andiagram
object
to the in
change
in thekinetic
object’s
kinetic energy.
Data:
2.4
= 0,2
From Aµkto
B.
2.4
Data:
Δx = 10 m
From
A m∙s
to B.
-1
vi = 8,85
Free body diagram
vi =
Data:
k = 0,2
0,2
µk = µ
Δx = 10 m
m∙s-1
vi = 8,85
Δx = 10
m
vi =
The work energy theorem can be applied to both conservative and non-conservative forces.
vi = 8,85 m∙s-1
vi =
Wnet  Ek
Wfr  WFg  WN  Ek f  Eki
f K x cos180 0  Fg cos 90 0  N cos 90 0  Ek f  Eki  K Nx(1) 
mv 2f
2

mvi2
2
2
mv 2f mvcan
The work energy theorem
be applied to both conservative and non-conservative forces.
i
  K mg.x 
2
Wnet  E k g.x  v  v
2
f
K

2
i
2
63
2.3 The net/total work done on an object is equal to the change in the object’s kinetic energy.
2.4
From A to B.
Free body diagram
The
work
energy
theorem
canapplied
be applied
both conservative
and non-conservative
forces.
Data:
The
work
energy
theorem
can be
to bothtoconservative
and non-conservative
forces.
µk = 0,2
Δx = 10 m
m∙s-1E k
viW
= 8,85
net 
vi =
Wfr  WFg  WN  Ek f  Eki
mv 2f
mvi2

f K x cos180  Fg cos 90  N cos 90  Ek f  Eki  K Nx(1) 
2
2
2
2
The work energy theorem
be applied to both conservative and non-conservative forces.
mv f canmv
  K mg.x 
 i
2
2
Wnet  Ek
2
2
Wfr  WFg  WN  vEkf f  Evki
  K g.x 
 i
mv 2f mvi2
0
0
2

f K x cos180  Fg cos 902
 N cos 90 0  Ek  Eki  K Nx(1) 
0
0
0
f
v 2f  (vi2  22g.x2 )
2
2
mv f
mv
 2K mg.x  2  i
v f  (8,85) 2  2 2 0,2  9,8 10)
v 2f vi2
2
vfK g.78
x ,32 39
,2
2
2
v 2v
vi2  39
2g,12
.x)
f  (
f
v2f  (8,85)2  2  0,21 9,8 10)
v f  6,25m.s
v 2f  78,32  39,2
2.5. From bottom to top.
2.5.
From bottom
to top.
Data:
1
v f  39,12
v f  6,25m.s
⃗⃗ (𝐼𝐼𝐼𝐼 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
𝑁𝑁
2.5. From bottom to top.
Data:
i
Data:
v  6,25m.s 1
⃗⃗ (𝐼𝐼𝐼𝐼 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
𝑁𝑁
vi  6,25m.s 1
𝑓𝑓⃗𝑓𝑓
𝑓𝑓⃗𝑓𝑓
(𝑛𝑛𝑛𝑛 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
(𝑛𝑛𝑛𝑛 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
m = 2 kg
𝐹𝐹⃗𝑔𝑔 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
68
68
𝐹𝐹⃗𝑔𝑔 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
hi = 0 m
vi  0 m.  s 1
==?2 kg
hm
f
𝜇𝜇𝑘𝑘 = 0,2
hi = 0 m
When
hf = ? there are non-conservative forces acting on a system, we can apply the principle of conservation of
When there
acting on a system, we can apply the principle of
energy
(laware
of non-conservative
conservation offorces
energy).
conservation of energy (law of conservation of energy).
𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝
𝑓𝑓𝑓𝑓 ∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 ) + (𝐸𝐸𝑃𝑃𝑃𝑃 − 𝐸𝐸𝑃𝑃𝑃𝑃 )
𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = (
𝑚𝑚𝑚𝑚𝑓𝑓2
2
𝜇𝜇𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = (
0,2 × 2 × 9,8 × 5(−1) = (
2.6
6.6
𝑚𝑚𝑚𝑚𝑓𝑓2
2
−
−
𝑚𝑚𝑚𝑚𝑖𝑖2
) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 )
2
𝑚𝑚𝑚𝑚𝑖𝑖2
) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 )
2
2×02
2×(6.25)2
−
) + (2 × 9,8 × ℎ𝑓𝑓 − 2 × 9,8 × 0)
2
2
−19.6 = 0 − 39,0625 + 19,6ℎ
h = 0,99 m
θElectrostatics
Exercise 1
Opposite
side
𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
ℎ𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
ℎ = ∆𝑥𝑥 sin 𝜃𝜃
0,99 = 5 sin 𝜃𝜃
sin 𝜃𝜃 = 0,198
θ = 11,450
1.1 electric field is a region of space in which an electric charge experiences a force. The
64
direction of the electric field at a point is the direction that a positive test charge would move
if placed at that point.
𝜇𝜇𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = (
0,2 × 2 × 9,8 × 5(−1) = (
𝑓𝑓
−
𝑚𝑚𝑚𝑚𝑖𝑖
) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 )
2
2×02
2×(6.25)2
−
) + (2 × 9,8 × ℎ𝑓𝑓 − 2 × 9,8 × 0)
2
2
−19.6 = 0 − 39,0625 + 19,6ℎ
h = 0,99 m
2.6
2.6
6.6
2
θElectrostatics
Exercise 1
Opposite
side
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
ℎ = ∆𝑥𝑥 sin 𝜃𝜃
0,99 = 5 sin 𝜃𝜃
sin 𝜃𝜃 = 0,198
θ = 11,450
𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
ℎ𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
6.5 1.1Electrostatics
Electric field is a region of space in which an electric charge experiences a force. The
direction of the electric field at a point is the direction that a positive test charge would move
Exercise 1
if placed at that point.
1.2.1Electric field is a region of space in which an electric charge experiences a force. The direction of the
1.1
electric field at a point is the direction that a positive test charge would move if placed at that point.
1.2.1
1.2.2 The system is isolated, so we must apply the law of conservation of charge:
69
1.2.2 The system is isolated, so we must apply the law of conservation of charge:
𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑄𝑄𝑄𝑄
𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
After being in contact (𝑄𝑄1 = 𝑄𝑄2 = 𝑄𝑄
Afterbeing
beinginincontact
contact(𝑄𝑄
(𝑄𝑄1==𝑄𝑄𝑄𝑄2==𝑄𝑄𝑄𝑄
After
1
2
𝑄𝑄1 + 𝑄𝑄2 = 2𝑄𝑄
1 + 𝑄𝑄2 = 2𝑄𝑄
𝑄𝑄𝑄𝑄
1 + 𝑄𝑄2 = 2𝑄𝑄
𝑄𝑄1 + 𝑄𝑄2
𝑄𝑄 =𝑄𝑄𝑄𝑄1++𝑄𝑄𝑄𝑄2
1 22
𝑄𝑄
=
𝑄𝑄 =
22
Commented [U149]: Pse check. (brackets) Pse check all ite
Commented
[U149]: Pse check. (brackets) Pse check all item
in text.
Commented
[U149]: Pse check. (brackets) Pse check all items
in text.
in text.
30nC  14nC
16nC
2
2
30nC 14
14nC
nC 16
16nC  8nC  8  109 C
(Charge on each) Q =30
 nC
(Chargeon
oneach)
each)QQ== nC  2
2  8nC  8  109 C

(Charge
2
2  8nC  8  10 C
9
1.3. To calculate the electrostatic force exerted by one charge on the other we must apply the
1.3.
To
calculate
theelectrostatic
electrostatic
forceexerted
exerted
byone
onecharge
charge
onthe
the
otherwe
wemust
must
applythe
the
1.3.
To
calculate
the
electrostatic
forceby
exerted
by one
charge
on
the apply
other
we must apply the mathematical
1.3.
To
calculate
the
force
on
other
mathematical
expression
of
Coulomb’s
law:
mathematical
expression
of
Coulomb’s
law:
mathematical expression of Coulomb’s law:
expression of Coulomb’s law:
kQ  QB
kQAA Q
Q
FE kQ
2 B
FFE E A r2r2 B
r
9 1099  8  1099  8 x1099
6,4  1044 9910
109 8810
1029 88x10
x109
10 4 
r2
66,4,410
2
r
r
r = 0,03 m
0,03mm
r r ==0,03
Because the charges are equal, we can also solve this in the following way:
Becausethe
thecharges
chargesare
are
equal,
wecan
can
alsosolve
solve
this
the
following
way:
Because
the
charges
are
equal,
wealso
can
alsothis
solve
this
in theway:
following
way:
Because
equal,
we
ininthe
following
kQ  QB
kQ2
kQAA Q
QB
FE kQ
QA=QB
FE kQ
kQ222
2
A=QB
E  r2
FFE E A r2r2 B
QQ
FF
A=QB
E 
r
r r2
9 2
9
9  109  (8  109 )2
6,4  1044 9910
109 (8(82 10
109 ) 2)
4 
6
,
4

10
r2
6,4  10 
2
r
r
r = 0,03 m
0,03mm
r r==0,03
Exercise 2
Exercise22
Exercise
2.1
The magnitude of the electrostatic force exerted by one point charge (Q1) on another
onanother
another
2.1 The
Themagnitude
magnitudeofofthe
theelectrostatic
electrostaticforce
forceexerted
exertedby
byone
onepoint
pointcharge
charge(Q
(Q)1)on
2.1
1
of the charges
point charge (Q2) is directly proportional to the product of the magnitudes
directlyproportional
proportionaltotothe
theproduct
productofofthe
themagnitudes
magnitudesofofthe
thecharges
charges
pointcharge
charge(Q
(Q)2)isisdirectly
point
2
and inversely proportional
to the square of the distance (r) between them.
andinversely
inverselyproportional
proportionaltotothe
thesquare
squareofofthe
thedistance
distance(r)(r)between
betweenthem.
them.
and
2.2
According to Newton’s third law, when one body exerts a force on a second body, the
2.2 According
AccordingtotoNewton’s
Newton’sthird
thirdlaw,
law,when
whenone
onebody
bodyexerts
exertsaaforce
forceon
onaasecond
secondbody,
body,the
the
2.2
second body exerts a force of equal magnitude in the opposite direction on the first
second
body
exerts
a
force
of
equal
magnitude
in
the
opposite
direction
on
the
first
second body exerts a force of equal magnitude in the opposite direction on the first
body, therefore:
body, therefore:
65
Exercise 2
2.1
The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge
(Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the
square of the distance (r) between them.
2.2
According to Newton’s third law, when one body exerts a force on a second body, the second body
exerts a force of equal magnitude in the opposite direction on the first body, therefore:
2.3
2.4
2.4
2.4
F=
kQ1Q2
2 kQ1Q 2
Fr =
r2
QA = QB = Q
QA = QB = Q
(9,0  109 )Q 2
3,0  10 6 
(29,0  109 )Q 2
3,0  10 6 2
22
(9,0  109 )Q 2
3,0  10 6 
(29,0  109 )Q 2
3,0  10 6 2
22
6
(3,0  10 ) 4
Q2 
(3,09  10 6 ) 4
Q92,0 10
9,0  10 9
QA = QB = 3,65 x 10-8 C
QA = QB = 3,65 x 10-8 C
2.5
2.52.5
We must draw the electric field vectors at the middle point:
We must draw the electric field vectors at the middle point:
We must draw the electric field vectors at the middle point:
𝐸𝐸⃗⃗𝐵𝐵
𝐸𝐸⃗⃗𝐵𝐵
𝐸𝐸⃗⃗𝐴𝐴
𝐸𝐸⃗⃗𝐴𝐴
X
x
X
x
E R  E A  E B (It is a vector equation.)
E R  E A  E B (It is a vector equation.)
ER= EA + EB
is a vector
equation.)
The vectors are pointing in(Itdifferent
directions
and positive is taken in the positive direction of
The vectors are pointing in different directions and positive is taken in the positive direction of
the X axis; therefore:
The
vectors
are pointing in different directions and positive is taken in the positive direction of the X axis;
the
X axis;
therefore:
E R  Etherefore:
A  BB
E R  EA  BB
R = E A −ofBthe
B charges are equal and the distance from the charge to the middle
TheE
magnitude
The magnitude of the charges are equal and the distance from the charge to the middle
point are the same; hence:
point are the same; hence:
71
66
71
The magnitude of the charges are equal and the distance from the charge to the middle point are the same;
EA  BB
hence:
EA  BB
KQ
E 2
r
KQ
E 2
r
9  109  3,65  108
E
12
9  109  3,65  108
E
12
E R 3 28,5  3 28,5
E R 3 28,5 -13 28,5
E R 0 N  C
E R  0 N  C-1
6.6 Electric Circuits
6.7
Electric Circuits
Exercise 1
Exercise
1 Circuits
6.7
Electric
Exercise
1.1
When1switch S1 is open, there is no flow of charges, and the current in the circuit is zero;
1.1
When switch S is open, there is no flow of charges, and the current in the circuit is zero; therefore,
1 reads the emf of the battery, which is 10 V.
therefore,
theSvoltmeter
1.1 When
switch
1 is open, there is no flow of charges, and the current in the circuit is zero;
the
voltmeter
reads
emf of theinbattery,
which
is 10
V.
1.2.1 Resistors
2 Ω and
4 Ωthe
are
parallel
and
both
therefore,
the voltmeter
reads
the connected
emf of the battery,
which
is
10 are
V. connected in series with
the resistor Rx.
1.2.11.1.1
Resistors
2 Ω and 42ΩΩ
areand
connected
in parallel
and both
connected
seriesare
withconnected in series with the
Resistors
4 Ω are
connected
inare
parallel
andinboth
resistor
Rext = Rx+ Rthe
(1) Rx. R .
parallel resistor
x
Rext = Rx+ Rparallel (1)
 R1 R2 (1)
=R+R
RRext

  (2)
ext  R xx   parallel
R R
 R2 R 1 
Rext  R x   1 2 R1 R
2(2) 
  (2)
R2 R1 
Rext = R x +

We have to calculate
the equivalent
external resistance and the equivalent resistance for the
 R2 + R1 
parallel
connection.
We
have
calculate
equivalent
external
resistance
and
the equivalent
resistance
for the parallel connection.
We
have
totocalculate
thethe
equivalent
external
resistance
and the
equivalent
resistance
for the
parallelVconnection. 8
Rext   Rext = 8 Ω
Rext  
I
1
V
8
Rext   Rext = 8 Ω
Rext  
1
1 I 1
1


R parallel R1 R2
1
1
1


R parallel R1 R2
1
1 1
  
Rp
2 4
1
1 1
  
Rp
2 4
Rparallel = 1,33 Ω
Rparallel = 1,33 Ω
We have only two resistors in
parallel, so we can use:
We have only two resistors in
parallel, so we can use:
 RR 
Rparallel =  1 2 
R R
 R21 R2 1 

 R2  R1 
 2 4 
Rparallel = 
 = 1,33 Ω
4 2
 2 4 
Rparallel = 
 = 1,33 Ω
4 2
Rparallel = 
72
Substituting the values of the external resistance and equivalent resistance in parallel in equation 1.
72
8 = Rx+ 1,33
Rx = 8 - 1,33 = 6,67 Ω
1.2.2 ε = Vext + I r
10 = 8 + 1r
r = 2 Ω
67
equation 1.
Commented [U150]: Pse check/complete.
8 = Rx+ 1,33
Rx = 8 - 1,33 = 6,67 Ω
1.3 INCREASES
1.2.2 ε = Vext + I r
Total10
resistance
increases.
= 8 + 1r
r = 2 Ω
Current in the circuit decreases. 
1.3 INCREASES
Total resistance increases.
The drop of potential within the battery (Vi= Ir) decreases, but the emf remains constant; therefore, according
Current in the circuit decreases. 
to the equation (Vext = ε- Ir), the voltage increases.
The drop of potential within the battery (Vi= Ir) decreases, but the emf remains constant;
therefore,
Exercise
2 according to the equation (Vext = ε- Ir), the voltage increases.
Exercise 2
2.1 When current flows through a voltage source (battery/generator), a resistance to current flow arises due
to the
resistance
of through
the materials
which tothe
source
2.1
When
current flows
a voltage(chemicals/conductors)
source (battery/generator), from
a resistance
current
flowis made.
arises due to the resistance of the materials (chemicals/conductors) from which the source
2.2 εis=made.
Vext +Ir
+I(0,10)
6 V=extV+Ir
2.2 ε =
ext
6 = Vext +I(0,10)
Vext = 6- (0,10)I
Vext = 6- (0,10)I
2.3
2.3
V (V)
To calculate the current at which
terminal potential difference is zero, we
need to substitute Vext with zero in the
equation.
6
Vext = 6 - (0,10)I
0 = 6- (0,10)I
I (A)
60 
I=
6
=60 A
0,1
2.4.1 Resistors 4 Ω and Rx are connected in series, therefore the current is the same for both.
2.4.1 Resistors 4 Ω and Rx are connected in series, therefore the current is the same for both.
W = I2RΔt 
W = I 2RΔt 
W = I RΔt 
(1)
(1)
For the resistor
(1) of 4 Ω
2
For the resistor of 4WΩ
4 = I RΔt = 40
For the resistor of 4 Ω
2
2
W
W44 == II2RΔt
RΔt==40
40
(4)Δt == 40
40
I22(4)Δt
I22 ==
I2(4)Δt = 40
I
2
10
=
∆𝑡𝑡
(2)
(3)
(2)(2)
(4)
(3)
(3)
W R = I2(4)
RΔt(4)
(5)
Substituting 4 in 5
The current
currentand
andthe
thetime
timeare
arethe
thesame,
same,
therefore
resistor
The
therefore
forfor
resistor
Rx :Rx :
10
(10)Δt = 60
∆𝑡𝑡
I RΔt
R
==
I2RΔt
WW
R
(5)(5)
2
same for both resistors, we can
solve
the problem
As the
current
and as
thefollows:
time are the
same
both resistors, we can
W for
= I2RΔt
solve the problem
as follows:
2
W4

I R4 t
ε = I(Rext+r)  (1)
2
2
2
(4)t
W4 40 I 2 RI 42
t 
60
I
RR2t

2
WR I RR t
RR = 6 Ω
I 2 (4)t
40
 2

60
I RR t
R = 6 Ω
To calculate the emf, we need to know the value of theRexternal
resistance, as the
resistors are connected in series:

RR = 6 Ω
RR = 6 Ω
68

RR = 6 Ω
Substituting 4 in 5
Substituting 4 in 5 2.4.2
2.4.2
2.4.2
As the current
and the time are the
alternative
solution
I RR t
The current and the time are the same, therefore for resistor RW
x : =W
I2RRΔt
10
∆𝑡𝑡
10
(10)Δt = 60
∆𝑡𝑡
10)Δt
= 60 
73
alternative
solution
𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅
(3)
= 4 + 6 = 10 𝛺𝛺
ε = I(Rext+r) 𝑅𝑅𝑠𝑠 (1)
ε = I(Rext+r)  (1)
Substituting the value of the total resistance I series in (1)
To calculate the emf, we need to know the value of the external resistance, as the
6 = I(10+ 0,10)
resistors are connected
in series:
𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅
2.4.3
I = 0,594 A
(3)
We need to calculate the value of the terminal potential, as we have all the data.
Commented [U151]: Pse
RR = 6 Ω
2.4.2
ε = I(Rext+r)  (1)
Tocalculate
calculatethe
the
emf,
need
to know
value
of the
external
resistance,
asresistors
the
To
emf,
wewe
need
to know
thethe
value
of the
external
resistance,
as the
are connected in
resistors are connected in series:
series:
(3)
𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅
𝑅𝑅𝑠𝑠 = 4 + 6 = 10 𝛺𝛺
Substitutingthe
thevalue
value
total
resistance
I series
in (1)
Substituting
of of
thethe
total
resistance
I series
in (1)
Commented [U151]: Ps
I(10+0,10)
0,10)
66==I(10+
I = 0,594 A
I = 0,594 A
2.4.3
We need to calculate the value of the terminal potential, as we have all the data.
2.4.3
We need to calculate the value of the terminal potential, as we have all the data.
Vext = IRext
Vext = IRext
Vext = (0,594)(10) 
Vext = (0,594)(10) 
Vext = 5,94 V
= 5,94 V
V
ext
6.8
Electrodynamics
Exercise 1
6.7 Electrodynamics
1.2. What is the relationship between the root mean square voltage and the frequency of rotation
1.1. ACExercise
generator/alternator.
It has slip-rings.
1
of the armature /coil of the generator?
1.1.
ACroot
generator/alternator.
It has
slip-rings.
1.3. The
mean
voltage
generated
when
the frequency
of rotation
1.2.
What
is thesquare
relationship
between
the
rootincreases/decreases
mean square voltage and
the frequency
of rotation
1.2.
What
is the
relationship
between
thethe
root
mean
square
voltage
and
thethe
frequency
of of
rotation
1.2.
What
is the
relationship
between
root
mean
square
voltage
and
frequency
rotation
of the armature
/coilgenerator
of the generator?
of the armature
/coil
of/coil
the
increases/decreases.
of of
theis
armature
of of
thethe
generator?
the
armature
/coil
generator?
1.2. What
the
relationship
between
the root mean square voltage and the frequency of rotation of the
armature
/coil
the generator?
1.3.
The rootsquare
meanofsquare
voltage generated increases/decreases when the frequency of rotation
1.4. Root
mean
voltage.
1.3.
The
root
mean
square
voltage
generated
increases/decreases
when
thethe
frequency
of rotation
1.3.
The
root
mean
square
voltage
generated
increases/decreases
when
frequency
of rotation
1.3.
1.5.
74
of the armature /coil of the generator increases/decreases.
of of
thethe
armature
/coil
of of
thethe
generator
increases/decreases.
armature
/coil
generator
increases/decreases.
The root mean square voltage generated increases/decreases when the frequency of rotation of the
1.4.
Root mean
square
voltage.
armature
/coil
of the
generator increases/decreases.
1.4.
Root
mean
square
voltage.
1.4.
Root
mean
square
voltage.
6
1.5.
5
4
6
6 6
5
5 5
4
4 4
3
3 3
2
2 2
1
1 1
0 0
10
20
30
40
0 0
0
10
20
30 (Hz) 40
Frequency
0 0
10 10 20 20 30 30 40 40
Frequency (Hz)
Frequency
(Hz)(Hz)
Frequency
Root mean square voltage
Rootmean
meansquare
squarevoltage
voltage
Root
(V)
(V)
(V)
Root mean square voltage
(V)
1.5.
1.4. 1.5.
Root
1.5. mean square voltage.
3
2
1
0
50
60
50
50 50
60
60 60
1.6. The root mean square voltage generated is directly proportional to the frequency of rotation
1.6. The root mean square voltage generated is directly proportional to the frequency of rotation
The
root
mean
square
voltage
generated
is directly
to to
thethe
frequency
rotation
1.6.
The
root
mean
square
voltage
generated
is directly
proportional
frequency
rotation
1.6. 1.6.
The
root
mean
square
voltage
generated
is proportional
directly
proportional
to of
theof
frequency
of rotation of the
of the of
armature
/coil/coil
of the
generator.
the armature
of the
generator.
/coil
ofofthe
generator.
ofarmature
thethe
armature
/coil
thethe
generator.
of
armature
/coil
of
generator.
V
Vmax 5
5
Vmax  5 53,54
3,V
54V
=V 
1.7=Vrmsmax
1.7 V1.7
rms
   3,54
= =max
Vrms
3,V
54V
Vrms
rms1.7
2 2 22 2 2 22
V
5
V Vrms 5 5 1A
VIrmsrms
5
rms
I rms
 rms
1.8
1.8 I1.8
RR 5155A1A1A
1.8
rms  rms R
1.8 I
R
5
1.9. Pave = IrmsVrms = 1 x 5 = 5 W
1.9. 1.9.
Pave
= I=
= 1=x15x=5 5=W
1.9.
Pave
Vrms
5W
rmsIV
rms
rms
1.9. Pave = IrmsVrms = 1 x 5 = 5 W
1.10. Vmax = 2Vrms  2  5  7,07V
1.10.
Vmax
= = 2V2
 2 2557,07
VV
1.10.
Vmax
Vrms
7,07
rms
1.10. Vmax =
2Vrms  2  5  7,07V
69
1.8 I rms 
Vrms 5
  1A
R
5
1.9. Pave = IrmsVrms = 1 x 5 = 5 W
1.10. Vmax = 2Vrms  2  5  7,07V
1.10.
75
1.11ADVANTAGES
ADVANTAGES OF
OVER
DC DC
1.11
OFACAC
OVER
 Easy to be transformed (step up or step down using a transformer).
 Easier to convert from AC to DEC than from DC to AC.
Easy
to betotransformed
(step up or step down using a transformer).
 Easier
generate.
 It can be transmitted at high voltage and low current over long distances with less
Easier
to convert from AC to DEC than from DC to AC.
energy lost.
 High
used in AC makes it suitable for motors.
Easier
tofrequency
generate.
•
•
•
•
•
Exercise 2
It can be transmitted at high voltage and low current over long distances with less energy lost.
2.1
2.2
DC motor. Electrical energy is converted into mechanical energy.
High frequency used in AC makes it suitable for motors.
Clockwise.
Exercise 2
2.3
A - Armature
2.1
DC
motor.brushes
Electrical energy is converted into mechanical energy.
B - Carbon
2.2
Clockwise.
2.4
2.3
C - Split ring commutator (commutator)
Electric motors are used in pumps, fans, compressors, etc.
A - Armature
B - Carbon brushes
C - Split ring commutator (commutator)
2.4
Electric motors are used in pumps, fans, compressors, etc.
76
70
7.1
Newton’s Laws
SOLUTIONS
1.1
7 Check your answers - Set 2
1.1 Free-body diagram for block A:
7.
Free-body diagram for block B:
7.1 Newton’s
Laws 7. Check your answers - Set 2
Check your answers
7. -Check
Set 2 your answers - Set 2
7.1
x-axis
Newton’s
Laws7.1
SOLUTIONS
SOLUTIONS ⃗⃗⃗
𝑵𝑵𝑨𝑨
1.1
1.1
7.1
Newton’s Laws
SOLUTIONS
Newton’s Laws
SOLUTIONS
1.1
1.1
x-axis
⃗⃗⃗𝑩𝑩
𝑵𝑵
⃗⃗
𝑭𝑭𝑨𝑨
1.1 Free-body diagram for block A: Free-body diagram for block B:
1.1
Free-body
diagram
1.1
for
Free-body
block
A:
diagram
Free-body
for
diagram
block
A:
for
Free-body
block
B:
diagram
for
B: B:
1.1 Free-body diagram for block A:
Free-body diagram block
for block
x-axis
x-axis
x-axis
x-axis
⃗⃗
x-axis
x-axis
x-axis𝑻𝑻
x-axis
⃗⃗⃗𝑨𝑨
⃗⃗⃗𝑩𝑩
𝑵𝑵
𝑵𝑵
⃗⃗⃗
⃗⃗⃗
⃗⃗⃗
⃗⃗⃗
𝑵𝑵𝑨𝑨
𝑵𝑵𝑨𝑨
𝑵𝑵𝑩𝑩
𝑵𝑵
⃗𝑭𝑭⃗𝒈𝒈𝑨𝑨
⃗𝑭𝑭⃗𝒈𝒈𝑩𝑩 𝑩𝑩
⃗𝑭𝑭⃗𝑨𝑨
⃗⃗ 𝑭𝑭
𝑻𝑻
⃗⃗𝑨𝑨
⃗⃗𝑨𝑨
𝑭𝑭
⃗𝑻𝑻⃗
⃗𝑻𝑻⃗
x-axis
x-axis
⃗⃗
𝑻𝑻
x-axis
x-axis
⃗x-axis
⃗
⃗⃗ x-axis
𝑻𝑻
𝑻𝑻
1.2
⃗⃗
𝑻𝑻
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
⃗⃗𝒈𝒈𝑩𝑩
𝑭𝑭
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
We apply Newton's second law
to each
1.2
1.2
1.2
object:
1.2
⃗⃗𝒈𝒈𝑩𝑩
𝑭𝑭
⃗⃗𝒈𝒈𝑩𝑩
𝑭𝑭
Solving the system
We apply
Newton's second
We apply Newton’s second law to each
object:
T +law
FAto- each
T = m1a + m2a
We apply Newton's second
We apply
lawNewton's
to each second law to each
object: Solving the system Solving the system Solving the system
object:
object:
FA = (m1 + m
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗
T 2+ )a
FA - T = m1a + m2a
T + FA - T = m1a + m2aT + FA - T = m1a + m2a
FA = (m1 + m2 )a
100F ==(5
+ 3) a
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗
⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ (right positive)
FA = (m1 + m2 )a
(m1 + m2 )a
A
𝐹𝐹x-axis
𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗
100
= (5 + 3) a
x-axis (right positive)
100 = (5 + 3) a
(5 + 3) a· 𝑠𝑠 −2 to the right
𝑎𝑎⃗100
𝑥𝑥 == 12,5𝑚𝑚
x-axis
(right
positive)
𝑎𝑎⃗𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 −2 to the right
For block
−2
x-axis
(rightApositive) x-axis (right positive)
𝑎𝑎⃗𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 −2 to the𝑎𝑎⃗right
𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 to the right
For block A
T =block
mAa A
For
TT==
mAm
a Aa
For block B
For block B
For block B
a
FFAA --TT==mm
2a2
FA - T = m2a
1.3
1.3
For block A
T = mAa
For block B
FA - T = m2a
For block A
T = mAa
For block B
FA - T = m2a
1.3
To calculate the magnitude of the tension, we can use block A or block B.
To calculate the
1.3magnitude
To calculate
of the tension,
the magnitude
we canofuse
theblock
tension,
A orwe
block
canB.
use block A or block B.
To calculate the magnitude of the tension, we can use block A or77block B.
77
77
77
71
1.3
To calculate the magnitude of the tension, we can use block A or block B.
(UsingA)A)
(Using
(Using B)
(Using A)
(Using A)
(Using A)
x-axis
(right
positive)
x-axis
x-axis
x-axis
x-axis
⃗⃗⃗
𝑵𝑵𝑨𝑨
⃗⃗⃗
𝑵𝑵𝑨𝑨
⃗⃗⃗
⃗⃗⃗
𝑵𝑵𝑨𝑨
𝑵𝑵𝑨𝑨
⃗⃗
𝑻𝑻
⃗⃗
𝑻𝑻
⃗⃗
⃗⃗
𝑻𝑻
𝑻𝑻
x-axis
𝑇𝑇 = (5)(12,5)
𝐴𝐴 𝑥𝑥
⃗⃗⃗𝑨𝑨𝑻𝑻
⃗⃗
𝑵𝑵
x-axis
x-axis
⃗⃗
𝑻𝑻
⃗⃗𝒈𝒈𝑩𝑩
⃗⃗ 𝑭𝑭
𝑻𝑻
x-axis
⃗⃗⃗
𝑵𝑵𝑩𝑩
⃗⃗⃗
𝑵𝑵𝑩𝑩
⃗⃗𝑨𝑨
𝑭𝑭
⃗⃗𝑨𝑨
𝑭𝑭
⃗⃗
𝑻𝑻
⃗⃗
𝑻𝑻
x-axis
x-axis
⃗⃗𝒈𝒈𝑩𝑩
𝑭𝑭
⃗⃗𝒈𝒈𝑩𝑩
𝑭𝑭
⃗⃗𝒈𝒈𝑩𝑩x-axis
𝑭𝑭
(Using B)
x-axis
⃗𝑭𝑭⃗𝑨𝑨
⃗⃗𝑨𝑨
𝑭𝑭
𝑅𝑅𝑅𝑅𝑅𝑅
⃗⃗
𝑻𝑻
𝐵𝐵 𝑥𝑥
1.4
1.4
𝑇𝑇 = 62,5𝑁𝑁
−𝑇𝑇 = 37,5 − 100
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥
𝐹𝐹
𝑚𝑚𝐴𝐴𝐴𝐴𝑎𝑎𝑎𝑎⃗𝑥𝑥
𝑇𝑇 −⃗𝑅𝑅𝑅𝑅𝑅𝑅
𝑓𝑓𝑘𝑘 ==𝑚𝑚
(5)(8,972)
62,5
−
𝑓𝑓
=
𝑇𝑇 − 𝑓𝑓𝑘𝑘 =
𝑇𝑇 −
𝑘𝑘 𝑚𝑚𝐴𝐴 𝑎𝑎 62,5
𝐴𝐴 𝑎𝑎
(5)(8,972)
− 𝑓𝑓𝑓𝑓𝑘𝑘𝑘𝑘 =
= 𝑚𝑚
𝑅𝑅𝑅𝑅𝑅𝑅
Data:
𝑎𝑎 = 8,972𝑚𝑚 · 𝑠𝑠 −2
Data:
Data:
𝑎𝑎 = 8,972𝑚𝑚 · 𝑠𝑠 −2
−2
62,5𝑁𝑁 · 𝑠𝑠 −2 𝑎𝑎
𝑎𝑎 𝑇𝑇==8,972𝑚𝑚
𝑇𝑇 =
= 8,972𝑚𝑚
62,5𝑁𝑁 · 𝑠𝑠
1.4
Data:
x-axis
⃗𝑭𝑭⃗𝒈𝒈𝑩𝑩
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥
𝑇𝑇
𝐹𝐹𝐴𝐴 =
− 62,5𝑁𝑁
𝑇𝑇 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥
𝑇𝑇 = 62,5𝑁𝑁
100 − 𝑇𝑇 = (3)(12,5)
1.4
1.4
𝐹𝐹⃗
= 𝑚𝑚 𝑎𝑎⃗
1.4
Data:
Data:
⃗⃗𝑨𝑨
𝑭𝑭
x-axis (right positive)
𝐹𝐹⃗𝐴𝐴𝑅𝑅𝑅𝑅𝑅𝑅
− 𝑇𝑇==𝑚𝑚𝑚𝑚
⃗𝑥𝑥𝑎𝑎⃗𝑥𝑥 𝐹𝐹 𝐹𝐹⃗−
𝐵𝐵
𝐵𝐵 𝑎𝑎
𝑇𝑇 =
= 𝑚𝑚
𝑚𝑚𝐵𝐵𝐵𝐵 𝑎𝑎
𝑎𝑎⃗⃗𝑥𝑥𝑥𝑥
x-axis 𝐹𝐹(right
positive)
𝐴𝐴 𝑅𝑅𝑅𝑅𝑅𝑅
(3)(12,5)
100
−
𝑇𝑇
=
𝐹𝐹𝐴𝐴 − 𝑇𝑇 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 100
𝐹𝐹𝐴𝐴 −
− 𝑇𝑇𝑇𝑇 =
= (3)(12,5)
𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥
𝑇𝑇 = 62,5𝑁𝑁
−𝑇𝑇
=
37,5
−
100
𝑇𝑇 −
= 100
62,5𝑁𝑁
⃗
(3)(12,5)
(3)(12,5)
100
−
𝑇𝑇
=
100
−
𝑇𝑇
=
𝐹𝐹𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥
−𝑇𝑇 = 37,5
𝑇𝑇 = 62,5𝑁𝑁
𝑇𝑇 =−62,5𝑁𝑁
100
⃗𝑥𝑥= 37,5 − 100 −𝑇𝑇 = 37,5
𝑇𝑇 = 𝑚𝑚−𝑇𝑇
𝐴𝐴 𝑎𝑎
𝑇𝑇 = (5)(12,5)
⃗⃗⃗
𝑵𝑵𝑩𝑩
x-axis
x-axis
x-axis
positive)
⃗𝑭𝑭⃗𝒈𝒈(right
x-axis (right positive)
𝑨𝑨
x-axis (right positive)
x-axis (right positive)
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥
𝐹𝐹⃗
= 𝑚𝑚 𝑎𝑎⃗
x-axis (right positive)
x-axis (right positive)
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥
⃗𝑅𝑅𝑅𝑅𝑅𝑅
⃗
𝑎𝑎
⃗
𝑎𝑎
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅𝑇𝑇==𝑚𝑚𝑚𝑚
𝐹𝐹
= 𝑚𝑚
𝑚𝑚𝐴𝐴𝑎𝑎⃗𝑎𝑎⃗𝑥𝑥
𝐴𝐴
𝑥𝑥
𝐴𝐴 𝑥𝑥
𝑇𝑇 =
62,5𝑁𝑁
⃗𝑥𝑥
𝑇𝑇
𝐴𝐴 𝑎𝑎
(5)(12,5)
𝑇𝑇𝑇𝑇=
==𝑚𝑚
(5)(12,5)
𝑇𝑇𝑇𝑇==62,5𝑁𝑁
⃗⃗⃗
𝑵𝑵𝑩𝑩
x-axis
x-axis (right positive)
x-axis (right positive)
⃗𝑥𝑥
𝑇𝑇 𝑇𝑇==𝑚𝑚(5)(12,5)
𝐴𝐴 𝑎𝑎
x-axis
⃗⃗⃗
𝑵𝑵𝑩𝑩
(Using A)
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
(Using (right
B)
x-axis
positive)
x-axis
x-axis
(Using B)
x-axis
⃗⃗𝒈𝒈𝑨𝑨
𝑭𝑭
(Using
(Using B)
B)
𝐴𝐴 𝑥𝑥
𝑇𝑇𝐹𝐹⃗−
𝑓𝑓𝑘𝑘==𝑚𝑚𝑚𝑚𝐴𝐴𝐴𝐴𝑎𝑎⃗𝑎𝑎𝑥𝑥
𝑅𝑅𝑅𝑅𝑅𝑅
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥
44,8662,5
− 62,5
(5)(8,972)
62,5 − −𝑓𝑓
𝑓𝑓𝑘𝑘 𝑘𝑘==(5)(8,972)
−
𝑓𝑓𝑘𝑘𝑘𝑘 =
= 44,86
− 62,5
−𝑓𝑓
𝑇𝑇 − 𝑓𝑓𝑘𝑘 = 𝑚𝑚𝐴𝐴 𝑎𝑎
78
78
17,64𝑁𝑁
44,86
− 62,5−𝑓𝑓𝑓𝑓𝑘𝑘 =
− 62,5
−𝑓𝑓𝑘𝑘𝑓𝑓=
𝑘𝑘 =
= 44,86
17,64𝑁𝑁
𝑇𝑇 = 62,5𝑁𝑁
62,5 − 𝑓𝑓𝑘𝑘 = (5)(8,972)
𝑓𝑓𝑘𝑘 = 17,64𝑁𝑁
= 17,64𝑁𝑁
𝑎𝑎 = 8,972𝑚𝑚 ·𝑓𝑓𝑠𝑠𝑘𝑘−2
−𝑓𝑓𝑘𝑘 = 44,86 − 62,5
𝜇𝜇𝑘𝑘 =?
Calculating the normal
force:the normal force:
Calculating
𝑇𝑇 = 62,5𝑁𝑁
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
𝑓𝑓𝑘𝑘 = 17,64𝑁𝑁
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
Calculating the normal
force: the normal force:
Calculating
𝜇𝜇𝑘𝑘 =?
𝑓𝑓𝑘𝑘
𝑓𝑓
𝜇𝜇
𝑁𝑁
=
𝜇𝜇
𝑁𝑁
𝑓𝑓𝑘𝑘𝜇𝜇=
𝑓𝑓
y-axis
𝑘𝑘
=
𝑘𝑘 𝑘𝑘𝑁𝑁
y-axis
𝜇𝜇𝑘𝑘𝑘𝑘 = 𝑘𝑘
𝑁𝑁
𝑓𝑓
𝑓𝑓
Calculating the normal force:
y-axis
y-axis
𝜇𝜇𝑘𝑘 = 𝑘𝑘
𝜇𝜇𝑘𝑘 = 𝑘𝑘
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
𝑁𝑁
𝑁𝑁
𝐹𝐹⃗
= 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦
We must calculate
frictional
force
and
the
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦
Wethe
must
calculate
the
frictional
force
and𝑅𝑅𝑅𝑅𝑅𝑅
the force.
𝑓𝑓𝑘𝑘 normal
We must calculate
the frictional
force
and
y-axis
𝜇𝜇𝑘𝑘 =the
⃗
𝑁𝑁 𝑁𝑁
⃗⃗and
𝐹𝐹𝑅𝑅𝑅𝑅𝑅𝑅
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅
= 𝑚𝑚 𝑎𝑎⃗
= 𝑚𝑚 𝑎𝑎⃗
normal
+ 𝐹𝐹⃗the
We
must force.
calculateWe
themust
frictional
force the
andfrictional
the
calculate
force
𝑔𝑔 = 0𝐴𝐴 𝑦𝑦
⃗⃗ +
normal
force.
𝐹𝐹⃗𝑔𝑔 = 0 𝐴𝐴 𝑦𝑦
𝑁𝑁
⃗⃗𝑁𝑁+−𝐹𝐹⃗𝑔𝑔𝐹𝐹𝐹𝐹==0 0
⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0
normal force. the normal
force.
𝑁𝑁
Calculating
frictional
force:
𝑁𝑁𝑁𝑁−
𝐹𝐹𝐹𝐹 force
= 0 and the
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦
We must calculate
the frictional
Calculating the frictional
force:
Calculating
the frictional force:
(5)(9,8)
=
49𝑁𝑁
𝑁𝑁𝑁𝑁−=
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
==
0 𝑚𝑚𝑚𝑚 = 𝑁𝑁
−
𝐹𝐹𝐹𝐹
=
0
𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁
⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0
normal force.
𝑁𝑁
Calculating the frictional
force:the frictional force:
Calculating
x-axis
𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8)
𝑁𝑁 = 𝐹𝐹𝐹𝐹==49𝑁𝑁
𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁
x-axis
x-axis
𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0
𝑓𝑓𝑘𝑘
17,64
𝑓𝑓𝑘𝑘
17,64
𝜇𝜇𝑘𝑘 =the
= 49 =𝜇𝜇0,355
Calculating
force:
𝑁𝑁 frictional
𝑘𝑘 = 𝑁𝑁 = 49 = 0,355
x-axis
x-axis
𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁
𝑓𝑓
17,64
𝑓𝑓
17,64
Calculating the normal force:
𝜇𝜇𝑘𝑘 = 𝑁𝑁𝑘𝑘 = 49 = 0,355
𝜇𝜇𝑘𝑘 = 𝑁𝑁𝑘𝑘 = 49 = 0,355
x-axis
78 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 = 17,6478
= 0,355
y-axis
49
𝑁𝑁
𝑇𝑇𝜇𝜇𝑘𝑘==?
62,5𝑁𝑁
= 62,5𝑁𝑁
𝜇𝜇𝑇𝑇
𝑘𝑘 =?
𝜇𝜇𝑘𝑘 =?
𝑘𝑘
1.5
1.5
78
No, the boxes are made of the same material.
1.5
No, the boxes are made of the same material.
1.6
1.6
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
No, the boxes are made of the same material.
1.6
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
We must calculate the normal force:
We must calculate the normal force:
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑦𝑦𝐹𝐹⃗
1.7
1.7
⃗𝑦𝑦
𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎
⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0
𝑁𝑁
𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0
⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0
𝑁𝑁
(3)(9,8) = 29,4𝑁𝑁
𝑁𝑁
=𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹= =
𝑁𝑁 −
0 𝑚𝑚𝑚𝑚 =
𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁
Now we can calculate the frictional force:
Now we can calculate the frictional force:
(0,355)(29,4)
= 10,44 to the left.
𝑓𝑓𝑓𝑓𝑘𝑘 ==(0,355)(29,4)
= 10,44 to the left.
𝑘𝑘
The force from box A downward on the floor (weight); and the force upwards from the floor on box
72The force from box A downward on the floor (weight); and the force upwards from the floor on box
A (normal force).
A (normal force).
1.6
e material.
⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0
𝑁𝑁
𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁
We must calculate the normal force:
We
⃗⃗ +must
𝑁𝑁
𝐹𝐹⃗𝑔𝑔 = 0calculate the normal force:
𝑁𝑁
− 𝐹𝐹𝐹𝐹
= 𝐵𝐵0𝑎𝑎⃗𝑦𝑦
= 𝑚𝑚
𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅
1.7𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁
𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁
Now we can calculate the frictional force:
𝑓𝑓𝑘𝑘 = (0,355)(29,4) = 10,44 to the left.
Now we
the the
frictional
force: force:
Now
wecan
cancalculate
calculate
frictional
The force from box A downward on the floor (weight); and the force upwards from the floor on box
:
A (normal
force).
𝑓𝑓𝑘𝑘 = (0,355)(29,4)
= 10,44 totothe
theleft.
left.
The force from block A to the left by the rope on block B; and the force to the right by the rope on
1.7
block A.
the floor (weight); and the force upwards from the floor on box
The force from box A downward on the floor (weight); and the force upwards from the floor on box A (normal
force).
QuESTION 2
the rope on block B; and the force to the right by the rope on
2.1
The force from block A to the left by the rope on block B; and the force to the right by the rope on block A.
QUESTION 2
2.1
Let’s take positive in the direction of motion.
Commented [U152]: Pse check.
79
f motion.
Commented [U152]: Pse check.
79
Let’s take positive in the direction of motion.
2.2
2.3
If a resultant force acts on a body,
theforce
bodyacts
to accelerate
direction
ofto accelerate
t
h in the
e direction of
2.2it willIf cause
a resultant
on a body, it in
willthe
cause
the body
resultant force. The acceleration of the the
body
will beforce.
directly
to the
the body will
resultant
force
and
resultant
Theproportional
acceleration of
be directly
proportional
to the
inversely proportional to the mass of the
body. force and inversely proportional to the mass of the body.
resultant
Let’s apply Newton’s second law
to each
block.
2.3of motion
Let’s apply
Newton’s
second law of
∑ F = ma
Block of 4 kg (A)
In the x direction:
T − f f = mAa
T − µN = mAa
motion to each block.
 F  ma
Block of 8 kg (B)
In the y direction:
- T  Fg  mB a
Block of 4 kg (A)
 T  mB g  mB a (Equation 2)
In the x direction:
Solving the system of equations (adding 1
T  f f  mAa
plus 2)
T  N  mAa
In the Y direction
N  Fg  0
N  Fg  mg
T  mA g  T  mB g  mAa  mBa
 mA g  mB g  (mA  mB )a
 (0,6)(4)(9,8)  (8)(9,8)  (4  8)a
54.88  12a
73
a  4,57 m∙s-2
Block of 4 kg (A)
Block of 4 kg (A)
Solving
the system of equations (adding 1
In the
x direction:
In the x direction:
plus 2)In the x direction:
T  f m a
T  f f f  mAAa
T  f f  mAa
T  N 
m
a
T

T  N  mAA
amA g  T  mB g  mAa  mBa
T  N  mAa
 mA g  mB g  (mA  mB )a
 T  mBBg  mBBa (Equation 2)
 T  mB g  mB a (Equation 2)
Solving the system of equations (adding 1
Solving the system of equations (adding 1
plus 2) Solving the system of equations (adding 1
plus 2)
plus 2)
T  m g  T  m g  m a  m a
T  mAAg  T  mBBg  mAAa  mBBa
T  m g  T  m g  m a  mBa
 m g  mA g  (m Bm )a A
 mAAg  mBBg  (mAA  mBB)a
 mA g  mB g  (mA  mB )a
In the Y direction
In the Y direction
 (0,6In
)(4the
)(9,Y
8)direction
 (8)(9,8)  (4  8)a
 (0,6)(4)(9,8)  (8)(9,8)  (4  8)a
 (0,6)(4)(9,8)  (8)(9,8)  (4  8)a
 (0,6)(4)(9,8)  (8)(9,8)  (4  8)a
N F 0
N  Fgg  0
N  Fg  0 54.88  12a
N  Fg  mg
-2
N  Fg  mg
N  Fg  mga  4,57 m∙s
54.88  12a
54.88  12a
54m∙s
.88 -2 12a
a  4,57
a  4,57 m∙s-2
a  4,57 m∙s-2
T  m g  m a (Equation 1)
T  mAAg  mAAa (Equation 1)
T  mA g  mAa (Equation 1)
2.4
2.4
2.4
2.4
using equation
1
using equation 2
Usingequation
equation
2
using
2
T  mA g  mAa using equation 2
T  m g  m a
 T  m Bg  m Ba
using equation 1
Using
using equation
1 equation 1
using equation 1
T  m g  m a
T  m Ag  m Aa
A
 T) B mB g  mB a
T Am g  m a
T  (0,6)( 4)(9,8)  (4)(B-4T.57
 8  4,57  8  9,8
T  (0,6)( 4)(9,8)  (4)( 4.A57) A
T  (0,6)( 4)(9,8)  (4)( 4.57)
- T  8  4,57  8  9,8
 4),57  8  9,8
T  (0,6)( 4)(9,8)  (4)( 4.57)
T  (0,6)(T4)(
,8) -(TN
4)(48.57
= 941.84
T  (0,6)( 4)(9,8)  (4)( 4.57)
T = 41.84 N
T  (0,6)( 4)(9,8)  (4)( 4.57)
T
=
41.84
N
T  (0,6)( 4)(9,8)  (4)( 4.57)
T  41.8 0 N
T  41.8 0 N
T  41.8 0 N
T  41.8 0 N
2.5. f  N
2.5. f
2.5.
2.5.
,mg
ff  0
6  4  9,8
ff ff  23
0,6.52
 4N 9,8
f f  23.52 N
2.6
80
f  N
f f f  N
f f  N
f f  mg
f f  mg
f f  mg
80
80
80
2.6
2.6
 T  mB g  mB a
 T  mB g  mB a
T  8  4,57  8  9,8
T  8  4,57  8  9,8
Apparent weight = T = 41.84 N
Apparent weight = T = 41.84 N
Apparent weight = T = 41.84 N
2.7
Less than.
2.8
Equal to.
Exercise 3
3.1
Each particle in the universe attracts every other particle with a gravitational force that is directly proportional
to the product of their masses and inversely proportional to the square of the distance between their
centres.
(2)
74
81
2.8
Equal to.
3.1
Each particle in the universe attracts every other particle with a gravitational force that
is directly proportional to the product of their masses and inversely proportional to
Exercisethe
3 square of the distance between their centres.
3.1
Each particle in the universe attracts every other particle with a gravitational force that
is directly
proportional to the product of their masses and inversely proportional to
Smaller
than.
3.2
3.2
(2)
(2)
Smaller
than.
the square
of the distance between their centres.
The distance is the same as well as the mass of the Earth in both cases. The
Commented [U153]: Pse check
The distance is the same as well as the mass of the Earth in both cases. The magmagnitude
of the gravitational force exerted on the satellites is directly proportional to
Smaller
nitude
of than.
the gravitational force exerted on the satellites is directly proportional to the
the
mass
of
the satellites.
The
mass
of
ZACUBE-1is issmaller
smallerthan
thanthe
themass
mass of
of Sputmass
the satellites.
The
mass
of mass
ZACUBE-1
The of
distance
is the same
as well
as the
of the Earth in both cases.
The
Commented [U153]: Pse check.
Sputnik.
(4)
nik.
(4)
magnitude of the gravitational force exerted on the satellites is directly proportional to
3.2
the mass of the satellites. The mass of ZACUBE-1 is smaller than the mass of
GMm

FSputnik.

r2
3.3
3.3
3.3
F
GMm
(4)


24
11
6,67
r 2 10  6  10  1,2
F

(6,38  10116  600 2410 3 ) 2 
6,67  10
 6  10  1,2

6
3 2 Earth.
F =F9,85
Towards
the
(6,N
38 
 10
 600  10
) 
F = 9,85 N  Towards the Earth.
(5)
F = 9,85 N  Towards the Earth.
3.4
(5)
(5)
On the surface of the Earth.
3.4
On the surface of the Earth.
3.4
On the surface of the Earth.
GmM

F  GmM
2
F R 2 
R
GmM
2
(1) OR/OF R g  GM
mg  GmM
mg  R 2 2 (1) OR/OF R 2 g  GM
R
AA
height
heighth.h.
GmM
GmM

F1F1 
2
( R( Rhh))2
GmM
mg1  GmM 2 (2) OR/oF ( R  h) 2 g21  GM
mg1 
( R  h)2 (2) OR/oF ( R  h) g1  GM
( R  h)
From (1) and (2)
From (1) and (2)
g1
R2
(3)

g ( R  h) 2
82
82
g1
= 0,25
g
0,25 
R2
1
R2
 OR/OF

2
( R  h)
4 ( R  h) 2
From (3) we get
ℎ2 + 2𝑅𝑅ℎ + (𝑅𝑅 2 −
ℎ = −𝑅𝑅 ±
ℎ = −𝑅𝑅 +
𝑅𝑅 2
)=0
0,25
𝑅𝑅
√0,25
𝑅𝑅
0,5
ℎ = −3,38 × 106 +
(𝑅𝑅 + ℎ)2 = 4𝑅𝑅 2
𝑅𝑅 2 + 2𝑅𝑅ℎ + ℎ2 = 4𝑅𝑅 2
ℎ2 + 2𝑅𝑅ℎ − 3𝑅𝑅 2 = 0
(ℎ – 𝑅𝑅)(ℎ + 3𝑅𝑅) = 0
3,38 ×106
0,5
ℎ = 𝑅𝑅 = 3,38 x 106 m
7.2

(ℎ – 𝑅𝑅)(ℎ + 3𝑅𝑅) = 0
∴ ℎ = 𝑅𝑅
𝑜𝑜𝑜𝑜
− 3𝑅𝑅
(6)
(6)
[17]
[17]
Momentum and Impulse
Exercise 1
1.1. The total linear momentum of an isolated/a closed system remains constant (is conserved).
1.2. The resultant/net external force is zero.
1.3
F
75
ext
= 0
ℎ = 𝑅𝑅 = 3,38 x 106 m
∴ ℎ = 𝑅𝑅
𝑜𝑜𝑜𝑜
− 3𝑅𝑅
(6)
[17]
7.2
and
Impulse
7.2 Momentum
Momentum and
Impulse
Exercise
Exercise1 1
1.1.The
Thetotal
totallinear
linear
momentum
of an
isolated/a
closed
system
remains
constant
(is conserved).
1.1.
momentum
of an
isolated/a
closed
system
remains
constant
(is conserved).
1.2.The
Theresultant/net
resultant/net
external
force
is zero.
1.2.
external
force
is zero.
1.3
1.3
F
ext
= 0
Therefore: Δ p = 0
Hence:
p
p
after
before
p
= p
-
= 0
before
after
( p r + pb ) before = ( pr + pb )after
𝑚𝑚𝐿𝐿 𝑣𝑣⃗𝐿𝐿(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) + 𝑚𝑚𝑅𝑅 𝑣𝑣⃗𝑅𝑅(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 𝑚𝑚𝐿𝐿 𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝑅𝑅 𝑣𝑣⃗𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚𝐿𝐿 0 + 𝑚𝑚𝑅𝑅 0 = 4800𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚
0 +4800𝑣𝑣
𝑚𝑚𝑅𝑅 0 =⃗4800𝑣𝑣
+ 100𝑣𝑣𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+ 100
0 𝐿𝐿=
𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
0 = 4800𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
83
Wehave
have
two
unknowns,
soneed
we
need
to speed
calculate
the ofspeed
ofafter
thelauncher
missile
launcher
afterwas
thefired.
rocket
We
two
unknowns,
soneed
we
to calculate
the
speed
the
missile
the rocket
We
have
two
unknowns,
so we
to
calculate
the
of the
missile
launcher
the
rocket after
was
wasfired.
fired.
Using
equations
of motion
Using equations
of motion
Using equations of motion
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑥𝑥
𝑣𝑣𝑓𝑓2 do
= not
𝑣𝑣𝑖𝑖2 know
+ 2𝑎𝑎∆𝑥𝑥
We
the acceleration of the launcher on the inclined plane, therefore we must calculate it.
Commented [U154]: Is this a heading? (Pse re-check all
formatting and standardise all equivalent items to improve
th
Com
legibility of the text.
form
The
for the
missile launcher
on the inclined
planeinclined
is:
Wefree-body
do not diagram
know the
acceleration
ofmoving
the launcher
on the
plane, therefore we must calculate it.
The free-body diagram for the missile launcher moving on the inclined plane is:
𝐹𝐹⃗𝑔𝑔
⃗⃗
𝑁𝑁
𝐹𝐹⃗𝑔𝑔
𝐹𝐹⃗𝑔𝑔
𝐹𝐹⃗𝑔𝑔𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗
In the x direction
𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
⃗𝑔𝑔
−𝐹𝐹𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 𝐹𝐹
⃗⃗
𝑁𝑁
𝐹𝐹⃗𝑔𝑔
−𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗
𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃𝐹𝐹
In the x direction
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥
𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
ℎ
= ∆𝑥𝑥
sin 𝜃𝜃
−𝐹𝐹
𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ
−𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
02 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ
𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃
𝑣𝑣𝑖𝑖2 = 2𝑔𝑔ℎ
𝑣𝑣
𝑣𝑣𝑖𝑖2𝑓𝑓2==2(9,8)6
𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥
𝑣𝑣𝑖𝑖 = 10.84 𝑚𝑚 ∙ 𝑠𝑠 −1
ℎ = ∆𝑥𝑥 sin 𝜃𝜃
76Now we can calculate the initial speed of the rocket:
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ
84
legib
We have two unknowns, so we need to calculate the speed of the missile launcher after the rocket
was fired.
Using equations of motion
Commented [U154]: Is this a heading? (Pse re-check all
formatting and standardise all equivalent items to improve t
legibility of the text.
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑥𝑥
We
do not know the acceleration of the launcher on the inclined plane, therefore we must calculate it.
We do not know the acceleration of the launcher on the inclined plane, therefore we must calculate it.
Thefree-body
free-body diagram
for the
launcher
moving onmoving
the inclined
planeinclined
is:
The
diagram
formissile
the missile
launcher
on the
plane is:
⃗⃗
𝑁𝑁
𝐹𝐹⃗𝑔𝑔
𝐹𝐹⃗𝑔𝑔
𝐹𝐹⃗𝑔𝑔
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗
In the x direction
𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥
−𝐹𝐹𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
−𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚
𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥
ℎ = ∆𝑥𝑥 sin 𝜃𝜃
𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ
02 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ
𝑣𝑣𝑖𝑖2 = 2𝑔𝑔ℎ
𝑣𝑣𝑖𝑖2 = 2(9,8)6
𝑣𝑣𝑖𝑖 = 10.84 𝑚𝑚 ∙ 𝑠𝑠 −1
can calculate the initial speed of the rocket:
InNow
thewe
x direction
0 = 4800 × 10.84 + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
Now we can calculate the initial speed of the rocket:
100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = −52032
84
Exercise 2
Solution
𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1
Exercise 2
2.1
Impulse is the product of the resultant/net force (force) acting on an object and the time the resultant/net force (force) acts
Solution
on the object.
2.2
resultant/net force (force) acts on the object.
Data
2.2 Data
2.1 Impulse is the product of the resultant/net force (force) acting on an object and the time the
Commented [G155]: corrected
Commented [U156]: Pse check.
𝐹𝐹⃗ = 10 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
Δt= 3 s
Impulse -?
Let’s take positive to the right
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +10 × 3
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +30 𝑁𝑁 ∙ 𝑠𝑠
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 30 𝑁𝑁 ∙ 𝑠𝑠 to the right
2.3 Data
𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1
𝑚𝑚 = 10 𝑘𝑘𝑘𝑘
𝑣𝑣⃗ =?
Let’s take positive to the right:
𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣⃗
+ 80 = 10𝑣𝑣⃗
𝑣𝑣⃗ = + 8 𝑚𝑚 ∙ 𝑚𝑚−1
2.4 Data
𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘
𝑚𝑚𝐵𝐵 = 8 𝑘𝑘𝑘𝑘
= 8 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right
𝑣𝑣⃗
77
Commented [U157]: Formatting: pse ch
and standardise appearance of all eqivalent i
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡
𝐹𝐹⃗ = 10 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +10 × 3
Δt= 3 s
Impulse -?
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +30 𝑁𝑁 ∙ 𝑠𝑠
Let’s =
take
to the
right
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
30positive
𝑁𝑁 ∙ 𝑠𝑠 to the
right
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
= +10 × 3
2.3 2.3
Data
Data
𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
= 𝑘𝑘𝑘𝑘
+30 𝑁𝑁 ∙ 𝑠𝑠
𝑚𝑚 = 10
𝑣𝑣⃗ =?
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 30 𝑁𝑁 ∙ 𝑠𝑠 to the right
Let’s take positive to the right:
2.3 Data
𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣⃗
𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1
𝑚𝑚
= 10
𝑘𝑘𝑘𝑘⃗
+ 80
= 10𝑣𝑣
𝑣𝑣⃗ =?
Let’s𝑣𝑣⃗ take
= + 8positive
𝑚𝑚 ∙ 𝑚𝑚−1to the right:
𝑝𝑝⃗ =Data
𝑚𝑚𝑣𝑣⃗
2.4
𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘
+
= 10𝑣𝑣⃗
to the
𝑚𝑚𝐵𝐵right
= 80
8 𝑘𝑘𝑘𝑘
𝑠𝑠 −1 to the right
𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 8 𝑚𝑚 ∙ −1
𝑣𝑣⃗ = + 8 𝑚𝑚 ∙ 𝑚𝑚
𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 0
2.4 Data
=?
𝑣𝑣⃗𝐴𝐴𝐴𝐴 Data
2.4
𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘
For
this
problem, we are going to select right as positive.
𝑚𝑚𝐵𝐵 =
8 𝑘𝑘𝑘𝑘
𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 8 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right
𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
∑
𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = 0= 0
𝑣𝑣⃗𝐴𝐴𝐴𝐴 = ?
Commented [U157]: Formatting: pse check use of
and standardise appearance of all eqivalent items.
Commented [U157]: Formatting: pse check use of
and standardise appearance of all eqivalent items.
85
For this problem, we are going to select right as positive.
∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = 0
∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0
⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −
0 𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 0
𝑝𝑝⃗∆𝑝𝑝
𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
85
= 𝑝𝑝⃗𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
− 𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
= 0
𝑝𝑝⃗𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝⃗𝑝𝑝⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
= 𝑝𝑝𝑝𝑝⃗⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 +
𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 𝑝𝑝⃗𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
+
𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
= 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+ 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚
⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
+ 𝑝𝑝
= 𝑝𝑝⃗𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+ 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑝𝑝⃗𝐴𝐴𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
After𝑚𝑚the
collision,+the
trolleys=move
together
as𝐵𝐵 𝑣𝑣one
object; therefore, the velocity is the same:
⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
𝑚𝑚𝐵𝐵two
𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
+ 𝑚𝑚
𝐴𝐴 𝑣𝑣
𝑣𝑣⃗𝑖𝑖𝑖𝑖 collision,
+ 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 the
= 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓
)
𝑚𝑚𝐴𝐴the
𝐴𝐴 + 𝑚𝑚𝐵𝐵move
After
two(𝑚𝑚
trolleys
together as one object; therefore, the velocity is the same:
(10)(+8)
+𝐵𝐵(8)(0)
𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓
+ 𝐵𝐵8)
⃗𝑓𝑓𝑓𝑓𝑓𝑓
𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑣𝑣=
(𝑚𝑚(10
)
𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑖𝑖𝑖𝑖 + 𝑚𝑚
𝐴𝐴 + 𝑚𝑚
+80
= 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓+(18)
(10)(+8)
(8)(0) = 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 (10 + 8)
= 80
𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 (18)
𝑣𝑣⃗+80
𝑓𝑓𝑓𝑓𝑓𝑓 =
80
18
−1
𝑓𝑓𝑓𝑓𝑓𝑓
𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣⃗=
+=
4,44
18 𝑚𝑚 ∙ 𝑠𝑠 .
−1 −1
𝑣𝑣⃗𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓
=
4,44
𝑚𝑚
∙
𝑠𝑠
to .the right.
=
+
4,44
𝑚𝑚
∙
𝑠𝑠
𝑓𝑓𝑓𝑓𝑓𝑓
𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 = 4,44 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right.
2.5
We can apply Newton’s second law in terms of momentum:
∆𝑝𝑝⃗
2.5
apply
Newton’s
second
in terms
of momentum:
cancan
apply
Newton’s
second
lawlaw
in terms
of momentum:
𝐹𝐹⃗2.5 = WeWe
𝑛𝑛𝑛𝑛𝑛𝑛
∆𝑡𝑡
∆𝑝𝑝⃗
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝑡𝑡
5.5
Vertical Projectile Motion
Exercise
5.5 1Vertical Projectile Motion
Exercise 1
1.1
1.1
𝑝𝑝𝑓𝑓 − 𝑝𝑝𝑖𝑖
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =
𝑝𝑝𝑓𝑓∆𝑡𝑡
− 𝑝𝑝𝑖𝑖
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =70 − 80
∆𝑡𝑡
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =
700,2
− 80
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −10
0,2
⃗
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 =
−10
0,2
𝐹𝐹⃗ =
⃗
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛= −50
0,2𝑁𝑁
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −50 𝑁𝑁
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
1
2,2  vi (0,28)  (9,8)(0,28) 2
21
2,2  vi (0,28)  (9,8)(0,28) 2
2
2,2 – 0,38 = vi (0,28)
2,2 – 0,38 = vi (0,28)
vi = 6,5 m·s-1
vi = 6,5 m·s-1
1
y  viAt  gt 2
21
y  viAt  gt 2
2
781.2.
1.2.
v 2f  vi2  2 gy
v 2f  vi2  2 gy
(6,5) 2  02  2(9,8)y
(6,5) 2  02  2(9,8)y
(6,5) 2  02  2(9,8)y
2
(6,=5)2,16
 0m2  2(9,8)y
Δy
Δy = 2,16 m y = 2,16 + 2,2
y = 2,16 + 2,2
86
86
𝑝𝑝𝑓𝑓 − 𝑝𝑝𝑖𝑖
∆𝑡𝑡
70 − 80
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =
0,2
−10
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =
0,2
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −50 𝑁𝑁
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =
7.3 Vertical Projectile Motion
Exercise 1
5.5
Vertical Projectile Motion
Exercise 1
1.1
𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
1.1
1
2,2  vi (0,28)  (9,8)(0,28) 2
2
2,2 – 0,38 = vi (0,28)
vi = 6,5 m·s-1
v 2f  vi2  2 gy
(6,5) 2  02  2(9,8)y
y  viAt 
1
gt 2
2
(6,5) 2  02  2(9,8)y
Δy = 2,16 m
1.2.
y = 2,16 + 2,2
1
∆y = viA ∆t + g∆t 2
2
86
1.2.
y = 2,16 + 2,2
y = 4,3 m downwards
y = 4,3 m downwards
1
2,16  0t  (9,8)t 2
2
t 2 
Commented [G158]: Corrected Afrikaans word remove
2  2,16
9,8
Δt = 0,66 s
Δttotal = Δt1+ Δt2
Δttotal = 0,66 + 0,28
Δttotal = 0,94 s
5
y (m)
4
3
2
1
0
0
0.2
0.4
0.6
0.8
1
t(s)
79
1.4
v f  vi  gt
1.4
1.4
v vf 1f v0i g9t,8  0,66
1
,8 0s,66
v vf 1f 1  06,59m
v f 1  6,5m  s 1
1.4
v vf f22 
0v,944
f 
i  gt
 00
9,89,80v,
944
v1f  0  9,8  0,66
1.4
m  s 1s 1
v vf f22  9v9,f21
,21
 vi m
 gt
v f 1  6,5m  s 1
v f 1  0  9,8  0,66
v f  vi  gt
1
v (m·s-1)  6,5m
v  0 v9f,18 
0,66v f2s  0  9,8  0,944
1.4
f1
v (m·s-1) 1
v f 2  9,21 m  s 1
5)m  s
f 1  6,-1
vv (m·s
v f 22  0  9,8  0,944
Exercise
m  s 1
v fPositive
 0 v9fdownwards
,28 90,,21
944
2
v (m·s-1)
Exercise
Exercise2 2
v f 2.1
 9,21 m  s
2
Commented [U159]: Formatting: pse check use of italics;
standardise with all equivalent items.
1
v (m·s ) Exercise 2
Positive
Positivedownwards
downwards
-1
Option 1
-1
v (m·s
𝑣𝑣⃗𝑓𝑓2 =) 𝑣𝑣⃗𝑖𝑖2 + 2𝑔𝑔
⃗Δ𝑦𝑦⃗ downwards
Positive
2.1 2 Exercise 2
2.1v f  02  (2)(9,8)(150)
2.1
Exercise
2
2 Positive downwards
Option
v f 1 2940
Option
1 Option 1
Comme
standard
Option 2
𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 + 𝑔𝑔⃗𝛥𝛥𝛥𝛥
t 
Commented [U159]: Formatting: pse ch
standardise with all equivalent items.
2y
g
Commented [U159]: Formatting: pse check use of ita
standardise with all equivalent items.
Option
2
2 x150Option
Option 22
 5,53s
𝑣𝑣
⃗
=
𝑣𝑣
⃗
9,8 𝑣𝑣⃗𝑓𝑓 𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 𝑖𝑖 ++𝑔𝑔⃗𝛥𝛥𝛥𝛥𝑔𝑔⃗𝛥𝛥𝛥𝛥
t 
Commented [U159]: Formatting: pse check use of italics;
2.1
Positive
2
2
⃗ 𝑓𝑓 downwards
=𝑣𝑣
𝑚𝑚
𝑠𝑠 −1
𝑣𝑣⃗ 2∙ =
𝑣𝑣⃗downwards
+ 2𝑔𝑔⃗Δ𝑦𝑦⃗
standardise with all equivalent items.
𝑖𝑖 ⃗
𝑣𝑣⃗𝑓𝑓2 𝑣𝑣=
⃗54,22
⃗Δ𝑦𝑦
𝑖𝑖 + 𝑓𝑓 2𝑔𝑔
9,8 x5,53
2y
Option
2
2.1 2 Option 1 v 2  02  (2)(9,8)(150)vf = 0+
f
-t 
y 2 2
2
vf = 54,22
downwards
𝑣𝑣⃗𝑓𝑓 = m.s
𝑣𝑣⃗
𝑔𝑔⃗𝛥𝛥𝛥𝛥 2
g
𝑖𝑖 +
f 𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 +2 2𝑔𝑔⃗Δ𝑦𝑦⃗
t1 
Option 1 2
Option 2
g
2 v f  2940
 ⃗0Δ𝑦𝑦
⃗ (2)(9,8)(150) −1𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 +t 𝑔𝑔⃗𝛥𝛥𝛥𝛥2yt  2 x150
𝑣𝑣⃗𝑓𝑓2 =2 𝑣𝑣⃗𝑖𝑖2 +v f 2𝑔𝑔
 5,53s
g
𝑣𝑣
⃗
=
54,22
𝑚𝑚
∙
𝑠𝑠
downwards
𝑓𝑓
2.22 f
9,8
2y
2 2
2
x
150
v f  0 v f(2)(9,82940
)(150)
t 
= 0+ 9,8 x5,53  5,53s
vf
2 xt150
gt  
 5,53m.s
−1 downwards
⃗ 𝑓𝑓 =1𝑣𝑣⃗ 𝑓𝑓54,22
𝑚𝑚𝑚𝑚∙ ∙𝑠𝑠𝑠𝑠−1
2 𝑣𝑣
54,22
Option
= 54,22
downwards
9,s28 -1 downwards 9v,8f =Option
v When
f  2940
2
x
150
the two objects meet their position is the same: 150 = |ΔyA| +|ΔyB|
150 = the
vf = 0+
9,8
t 
 5v
,53=x5,53
s 0+ 9,8 x5,53
𝑣𝑣⃗ 𝑓𝑓 =yA54,22
= yB 𝑚𝑚 ∙ 𝑠𝑠 −1 downwards
addition
of the absolute
values of
9v,8f = 54,22f m.s
1 downwards
2.2 1
v
=
54,22
m.s
1
downwards
the
displacements
of
A
and
B.
Commented [U160]: Pse check.
1
f
2
2
v
=
0+
9,8
x5,53
y Ai  viAt  gt  y Bi  viBft  gt 1
vf = 54,22
2.2
2
2 m.s 1 downwards
150 = | viAt  gt 2 | + |
Option
1
2Option 2
Level2.2
zero at the
projecting
point of A (reference
When the two objects meet their position is the same:
150 = the
1
point).
2 150 = |ΔyA| +|ΔyB|
viB t Option
gt | 2
Option 1
y
A = yB
2.2
addition
of
the
absolute
values of
2.2
2
1
Option 11
Option
2
0  100t  x9,8 xt 2  1501 0t 2 x9,8 xt 2
the
displacements
of
A
and
Commented [U160]: Pse check.
1
1
2
2
When
the
two
objects
meet
their
position
is
the
When the
same:
150 = the B.
A|t+|Δy
2 two
150 g- v150
= B|
 viAtmeet
 their
2 y Bi isvthe
y objects
gt position
tiB t= |Δy
g
iB t 
1
2
Option 1
Option
2addition
yA = yB Ai
2of
2
2
150
=
|
|
+
|
same:
the
absolute
values
of



v
t
g
t
iA
100t  150
Option
1 objects
Option
2
2
When the two
their
position
is the
same:
150
= |Δy
+|Δy
150of= Athe
1A| displacements
Level meet
zero
the
projecting
point
of1A (reference
2 B|
the
and B.
Commented [U160]: Pse check.
2 𝑠𝑠
2v t 
𝛥𝛥𝛥𝛥1 at
=
1,5

g
t
iA









y
v
t
g
t
y
v
t
g
t
1
yA = ythe
point).
2of|ΔyA| +|ΔyB|
B Ai two
iA objects meetBitheir
iB position
When
is
the
150
=
150 = the
addition
of2same:
the absolute
values
1
t  ggt2t| +| |
2
2
iBt
150 = | viAv
2 B.
the
displacements
of
A
and
Commented [U160]: Pse check.
1
1
1
1
2
2
2
2
2
100
t

150
y
=
y
at
the
projecting
point
Av tzero
B 0 
x9v,iB
8 xt tofA150
addition of2 the absolute values of

t y Bi 
t  0t  x9,8 xt 1
y Ai Level
g100
t 
g(reference
iA
2 = 1,5 𝑠𝑠 1
point). 2
2
2 = | viAv
-|2 +
2
150
v| iB| t  gt =

t  1g150
t𝛥𝛥𝛥𝛥

g
t
iB t 
the
displacements
of A and B.
Comme
2
2 22
Level zero at the projecting
point 2of2 A100
(reference
1
t  150 1
2
0

100

t

x
9
,
8
x

t

150

0

t

x
9
,
8
x

t
1
1
Ai
iA
Bi
iB
point).
1
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠2
viAt1gtg2==
t 2|
viB t  150
gt- 2v| 
2
viAt  gt 2 | + |
iB t  150
2.3
2
1
1
2 2
2
2
100t 0150
0
 100zero
t  at
x9,8the
xtprojecting
 150
tpoint
 x9of
,8 xA
t (reference 1
Level
100t  1502
2
150 - viB vt 
 g1g
t 2t=2
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 2
iA t 2
1𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 point).
2
100t  150
viB t  88gt 2 |
1
2
100t  1502
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠
viAt  gt
2
2
2
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 1
2.3
2
100t  150
v  0  (2)(9,8)(150)
v
 2940
y  v t 
1
1
gt  y  v t  gt
2
2
1
1
x9,8 xt  150  0t  x9,8 xt
2
2
150 - v t  gt =
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 iB
2
100t  150
2.3
0  100t 
2.3
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠
viAt 
1
gt 2
2
88
88
88
100t  150
𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 2.3
88
80
2.3
Option 1
𝑣𝑣⃗𝑓𝑓 = 11𝑣𝑣⃗𝑖𝑖 + 𝑔𝑔⃗𝛥𝛥𝛥𝛥
Option
Option
Option
1
100
=𝑣𝑣⃗𝑣𝑣⃗𝑖𝑖−100
𝑣𝑣⃗𝑣𝑣𝑓𝑓⃗𝑓𝑓 ==
+ 𝑔𝑔⃗𝑔𝑔𝛥𝛥𝛥𝛥
⃗𝛥𝛥𝛥𝛥+ 9,8 𝛥𝛥𝛥𝛥
𝑖𝑖 +
𝛥𝛥𝛥𝛥
=
20,41
100
100 == −100
−100++𝑠𝑠9,8
9,8𝛥𝛥𝛥𝛥
𝛥𝛥𝛥𝛥
𝛥𝛥𝛥𝛥
𝛥𝛥𝛥𝛥 == 20,41
20,41𝑠𝑠𝑠𝑠
Option 2
Option2 2
Option
1
gt 2
2
11 2 2
yy viAv
t  ggtt
iAt
1
22
011  100
t  9,8t 2
2 2
2
0 0  100
tt  9,89
 100
,8tt
Option 2
y  viAt 
22

100
0t
100
t
4,4
9,9
t2 t 2 t  4,9t
0 0 
100
2
2 2
100

t t100
4,49,
t
100
9
tt  4,9t
2
ΔtΔt
==
20,41
s sΔt = 20,41 s
20,41
2.4
2.4
2.4
2.4
Positive
Positiveupwards
upwards
Positive upwards
Positive
downwards
Positive
downwards
Positive downwards
5.6
Work Energy Power
5.6
Work Energy Power
Exercise 1
Exercise 1
1.1
Greater than.
1.1
Greater
than.(non-conservative) force acting; mechanical energy is not conserved.
There
is friction
There is friction (non-conservative) force acting; mechanical energy is not conserved.
(3)
(3) net work done on an object is equal to the change in the kinetic energy of the object.
1.2
The
1.2
The (2)
net work done on an object is equal to the change in the kinetic energy of the object.

5.61.3 Work
Energy Power
 (2)
1.3
1.3.1 Wnet  E f  E i 
Exercise
1.3.1 Wnet1  E f  E i 
1.1
1.2
Commented [U161]: Pse check – only 2
Commented [U161]: Pse check – o
Commented [U162]: Pse complete.
Suggest
cross-referencing
of entire
text is re
Commented
[U162]:
Pse complete
Suggest cross-referencing of entire text
89
Greater than.
There is friction (non-conservative) force acting; mechanical energy is not 89
conserved.
(3)
The net work done on an object is equal to the change in the kinetic energy of the object.
 (2)
Comm
1.3
1.3.1 Wnet  E f  E i 
Comm
Sugges
81
89
7.1 Work Energy Power
Exercise 1
1.1
5.6
Greater than.
Work Energy Power
There
Exercise
1 is friction (non-conservative) force acting;
mechanical energy is not conserved.
(3)
Greater than.
1.2 1.1 The
net work done on an object is equal to the change in the kinetic energy of the object.  (2)
1.3 1.2
1.3
There is friction (non-conservative) force acting; mechanical energy is not conserved.
(3)
The net work done on an object is equal to the change in the kinetic energy of the object.
 (2)
Commented [U161]: Pse check – only 2 ticks.
Commented [U162]: Pse complete.
Suggest cross-referencing of entire text is required.
1.3.1
1.3.1 Wnet  E f  E i  (5)
1
Fnet x cos   1m(v 2f 2 vi2 2) 
Fnet x cos  2 m(v f  vi ) 
0
2
1
2
89
2
 Fnet (10) cos 180 0 1(24,7)(18 2 24,25 2)
 Fnet (10) cos 180 2 (24,7)(18  24,25 )
2
 Fnet  326,12 N  frictional force / wrywingskrag 
 Fnet  326,12 N  frictional force / wrywingskrag 
Commented [U163]: Pse check.
Commented [U163]: Pse check.
(5)
(5)
1.3.2
1.3.2 Wnc  E K  E P 
1.3.2
Wnc  E K  E P 
1
f  x cos   1m(v 2f 2 vi2 2)  mgh 
f  x cos  2 m(v f  vi )  mgh 
2
1
326,12  x cos1800 0  1(24,7)(0  18 2 2)   24,7(9,8)(x sin 50 0 0) 
326,12  x cos180  2 (24,7)(0  18 )   24,7(9,8)(x sin 50 ) 
 326,12 x  4001,4  2185,4287 x
 326,12 x  4001,4  185,4287 x
 x  7,82 m 
 x  7,82 m 
(6)
(6)
(6) [16][16]
[16]
7.5 Doppler
Doppler EffectEffect
7.2
7.5
Doppler Effect
Exercise 1
Exercise 1
1.1.
Exercise
1 Lower than
1.1.
1.1.
1.2
1.2
Lower
As
the than
speed of sound is constant wavelength is inversely proportional to
As the speed
of sound is
is higher.
constant wavelength is inversely proportional to
frequency
and frequency
frequency
and
frequency
is
higher.
Lower
than
OR
OR than
Lower
Lower than
1 is constant wavelength is inversely proportional to frequency
As the speed of sound
When v is constant λ 1f and frequency is higher.
When
v
is
constant
λ
 f and frequency is higher.
and frequency is higher.

OR
OR than
Lower
OR
Lower
thanheard is higher, the wavelength is smaller, because when the speed
As
frequency
frequency
heard is higher,
theand
wavelength
is smaller,
because
when the speed
ofAs
sound
is constant,
wavelength
frequency
are inversely
proportional.
of sound
is constant, wavelength and frequency are inversely proportional.
Lower
than
Doppler
When veffect
is constant λ
Doppler effect
OR
1.3
1.3
1.2
1
α
(4)
(4)
(1)
(1)
f and frequency is higher.
Option 1
Optionthan
1
Lower
(6)
(6)
v
 vlistener 

f l    vsound
sound  vlistener f s 
f l  vsound
 vsource
  f s is higher, the wavelength is smaller, because when the speed of
As frequency
heard
 vsound  vsource 
 0  wavelength and frequency are inversely proportional.
sound is 340
constant,
f l   340  0   x 620

f

340

30
.56   x 620
l

Doppler
effect
 30.56 
 340
f l  681,23 Hz
f l  681,23 Hz
90
90
82
(4)
(1)
(4)
1.2
Doppler effect
(1)
Option 1
1.3
(6)
Option 1
1.3
(6)

v
v
f l   sound listener  f s 
 vsound  vsource 
 340  0 
fl  
  x 620
 340  30.56 
f l  681,23 Hz
v  f 
90
v  f 
340  681,23 
340  681,23 
  0,50 m
  0,50 m
Option 2
v Option
f  2


v
Option
 f s 
f   2 sound

340 l 681
,
23

v
v

 sound
vsound source 
 f s 
f l  
vsource  
 340
sound
vm
  0f,l50

  x 620
 30.56
 340340
fl  
  x 620
340

30.56 

Option 2
f l  681
23 Hz 
v,sound
 f s 
f l  
f l v681
 vHz
sound,23
source 
v  f 
340


vf l f 
  x 620
340 340
681,2330
 .
56 
340  681,23 
  0,50 m
f l 681
,23m
Hz
0,50
 681.23 
v 340
f 
340
.23 
  0,681
50 m
340  681,23 
  0,50 m
7.6
Electrostatics

0,50 m
Exercise 1Solutions
7.6
Electrostatics
7.3Exercise
Electrostatics
1 Solutions
1.1 What is the relationship between the electrostatic force between two charges and the
340  681.23 
distance between those charges?
1.1
the relationship
Exercise
1isSolutions
1.2 What
Electrostatic
 between the electrostatic force between two charges and the
m
  0,50force
distance
between
those
charges?
1.3 Magnitude of the charges

1.2 Electrostatic force 
1.1
What
is
the
relationship
between
the electrostatic force between two charges and the distance between those charges?
1.4
1.3
Magnitude
of
the
charges

7.6
Electrostatics
Exercise 1 Solutions
1.4
2000
1.2 Electrostatic
force
1000
500
1.4
5000
2000
0
0
0
1500
F(N)
1.4
F(N) F(N)
1.1 What is the relationship between the electrostatic force between two charges and the
1500
distance2000
between those charges?
1.2 Magnitude
Electrostatic force
1.3
of thecharges
1500
1000
1.3 Magnitude
of the charges 
2
4
2
4
d(m)
6
8
10
6
8
10
d(m)
1000
500
0
0
2
4
6
d(m)
8
10
91
91
91
83
CRITERIA
CRITERIA
MARKS/PUNTE
MARKS/PUNTE
CRITERIA
CRITERIA
Correct
shape
Correct
shape
MARKS/PUNTE

Correctshape
shape
Correct
Three
or or
more
coordinates
plotted
correctly
Three
more
coordinates
plotted
correctly


Three or more coordinates plotted correctly

Three or more coordinates plotted correctly
Commented
[U164]:
Pse Pse
check.
Commented
[U164]:
check.
Pse Pse
check
all instances
where
another
language
is included
andand
it it
check
all instances
where
another
language
is included
Commented
[U164]:
Pse check.
appears
as though
something
has has
been
copies
andand
pasted
but but
not not
appears
as though
something
been
copies
pasted
Pse
check all instances where
language
is document.
included and it
over-written/corrected
so that
it fits
this this
document.
over-written/corrected
so another
that
it with
fits
with
appears as though something has been copies and pasted but not
over-written/corrected so that it fits with this document.
MARKS/PUNTE


(2)
(2)(2)
(2)
1.5The
Themagnitude
magnitude
ofelectrostatic
the electrostatic
force two
between
two
chargesproportional
is inversely
proportional
to the square of
1.51.5
of the
force
between
charges
is inversely
to to
thethe
square
The magnitude
of the
electrostatic
force
between two
charges
is inversely
proportional
square
the
distance
between
the
charges.
 of
the
distance
between
charges.

1.5of
The
magnitude
of thethe
electrostatic
force
the
distance
between
the
charges.
 between two charges is inversely proportional to the square
of the distance between the charges. 
1.6.1
1.6.1
1.6.1
1.6.1
CrITerIA
CrITerIA
Shape
Shape
CrITerIA
Direction
of of
thethe
arrows
Direction
arrows
Shape
Direction of the arrows
Marks
Marks


Marks



(2)(2)
1.6.2
1.6.2
1.6.2
1.6.2
2
kQ 2
F F 2 kQ

r r2 2
kQ
F 2

r
9 9 2 2
(9 
(910
 10)Q
)Q
 1600
= =

 1600
92 2 2
(
1
)
(9  10(1))Q
 1600 =

(1) 2
-4
Q Q= 4.22
x 10
= 4.22
x 10-4C C 
Q = 4.22 x 10
-4
C
(4)(4)
1.6.3
1.6.3
kQkQ
E E 2

r r2
kQ
E 2

r
4
9
 10
9 910
 109 4,22
 10 4
4,22

E =E =
9 (0,25) 2 2 4
,25) 10
9  10  4(0,22
E=

(0,25) 2
[19]
(3)(3)
(3)
E =E 6,07
x 10
= 6,07
x 10
N·C-1 -1
[19]
[19]
E = 6,07 x 10
N·C 
[19]
-11 -11
-11
N·C 
-1
92 92
92
84
Commented
[U165]:
Formatting:
pse pse
alignalign
items
on right-hand
Commented
[U165]:
Formatting:
items
on right-hand
margin.
Repetitive
problems
seen.
margin.
Repetitive
problems
seen.
Commented [U165]: Formatting: pse align items on right-hand
margin. Repetitive problems seen.
(4)

1.6.3
1.6.3
(2)
(4)
7.4 Electric Circuits
Exercise 1
1.1
Differences between emf and potential difference:
Electromotive force
1.
Electromotive force transmits current both inside and outside the cell.
2.
Electromotive force emf is the cause.
3.
Electromotive force is always greater than potential difference.
4.
Electromotive force creates potential difference entire the circuit.
5.
Electromotive force does not depend on the resistance of the circuit.
6.
Electromotive force remains constant.
7.
The part of the circuit where electrical energy is created from any other energy then that part contains the source of Electromotive force.
Potential difference:
1.
Potential difference current transfers between any two points in the circuit.
2.
Potential difference is the result.
3.
Potential difference is always less than electromotive force.
4.
Potential difference takes place between any two points in the circuit.
5.
Potential difference of two points depends on the resistance of those points.
6.
It does not remain constant.
7.
Potential difference exists in the part of the circuit where electrical potential energy is transformed
into another form of energy.
1.2
The potential difference across a conductor is directly proportional to the current in the conductor at
constant temperature.
85
1.2
The potential difference across a conductor is directly proportional to the current in the conductor at
constant temperature.
1.3
1.3
First we must identify the type of connection; for this purpose we must indicate the direction of the
First we must identify the type of connection; for this purpose we must indicate the direction of the
current.
current.
12 V
S2
A
+
R1
-
R2
+
R3
93
S1
To
current,
we apply
mustOhm’s
applylawOhm’s law
To calculate
calculate current,
we must
Commented [U168]: Pse complete (theory?)
Commented [U168]: Pse complete (theory?)
To calculate
current, we must apply Ohm’s law
𝑉𝑉2

𝐼𝐼To
2 =
Commented [U168]: Pse complete (theory?)
calculate
current, we must apply Ohm’s law
𝑅𝑅22
𝑉𝑉
𝐼𝐼2 = 𝑅𝑅 
𝑉𝑉22

𝐼𝐼The
2 = resistors
are connected
parallel:
𝑅𝑅2
Commented [U168]: Pse complete (theory?)
To
calculate
current,
we mustinapply
Ohm’s law
The
resistors
connected
in parallel:
The resistors
are are
connected
in parallel:
𝑉𝑉
1 = 21 1
1
𝐼𝐼The
resistors
are
connected
in
parallel:
2 =
+ + 
2
𝑅𝑅1𝑒𝑒𝑒𝑒 𝑅𝑅𝑅𝑅
𝑅𝑅
𝑅𝑅
11
12
13
= + + 
𝑅𝑅𝑒𝑒𝑒𝑒
𝑅𝑅11
𝑅𝑅12
𝑅𝑅13
1
=
+
+

The resistors
are
𝑅𝑅
𝑅𝑅
𝑅𝑅3 connected in parallel:
𝑒𝑒𝑒𝑒 =𝑅𝑅1 𝑅𝑅2
𝑅𝑅
𝑒𝑒𝑒𝑒
𝑅𝑅3
𝑅𝑅1𝑒𝑒𝑒𝑒 = 31  1
1
== 𝑅𝑅
+ + 
𝑅𝑅
𝑅𝑅
𝑅𝑅1three
𝑅𝑅2 resistors
𝑅𝑅3
𝑒𝑒𝑒𝑒the
𝑒𝑒𝑒𝑒
As
are in parallel, the potential difference is the same:
3
As the
three resistors are in parallel, the potential difference is the same:
As
the
𝑅𝑅 three resistors are in parallel, the potential difference is the same:
= 𝑉𝑉1=resistors
V
𝑉𝑉1𝑒𝑒𝑒𝑒=
As
the
are in parallel, the potential difference is the same:
𝑅𝑅
=𝑉𝑉23three

𝑉𝑉1 = 𝑉𝑉2 = 𝑉𝑉1= V
Commented [U168]: Pse complete (theory?)
To
calculate
current,
must apply
Ohm’s the
law current through each one is the same.
As
the
the
resistance,
𝑉𝑉2resistors
= 𝑉𝑉1=resistors
V havewe
𝑉𝑉1 =
As
the
three
are same
in parallel,
the potential
difference is the same:
As the
𝑉𝑉2 resistors have the same resistance, the current through each one is the same.

𝐼𝐼Solving:
2 =the
Commented [U169]: Perhaps solution (i.e.noun – solving is a
𝑅𝑅2 resistors have the same resistance, the current through each one is the same.
As
𝑉𝑉2 = 𝑉𝑉1= V
𝑉𝑉1 =
verb).
Commented
[U169]: Perhaps solution (i.e.noun – solving is a
Solving:
As
the
resistors
have
the
same
resistance,
the
current
through
each
one
is
the
same.
𝑅𝑅
verb).
The
resistors
are
connected
in
parallel:
Commented [U169]: Perhaps solution (i.e.noun – solving is a
Solving:
𝑉𝑉
=
𝐼𝐼𝐼𝐼
=
𝐼𝐼

As
have
the
same
resistance,
the
current
through
each
one
is
the
same.
2 the resistors
𝑒𝑒𝑒𝑒
3
𝑅𝑅
verb).
𝑉𝑉21 = 𝐼𝐼𝐼𝐼1𝑒𝑒𝑒𝑒 =1 𝐼𝐼 1
3𝑅𝑅
𝑅𝑅 + = 𝐼𝐼+  
=
𝑉𝑉
=
𝐼𝐼𝐼𝐼
Commented [U169]: Perhaps solution (i.e.noun – solving is a
Solving:
2
𝑒𝑒𝑒𝑒
𝐼𝐼
Solving:
𝑅𝑅
𝑅𝑅
𝐼𝐼 𝑅𝑅 3 𝑅𝑅
verb).
𝐼𝐼2𝑒𝑒𝑒𝑒= 𝐼𝐼𝑅𝑅𝑅𝑅3 1= 3 2  3
𝐼𝐼
3
𝑅𝑅
𝑅𝑅𝑅𝑅
= 3= 
𝐼𝐼𝑉𝑉2 == 𝑅𝑅𝐼𝐼𝐼𝐼𝐼𝐼
𝐼𝐼

𝐼𝐼
2
𝑒𝑒𝑒𝑒
3
 3
𝐼𝐼𝑅𝑅2𝑒𝑒𝑒𝑒==3𝑅𝑅 3=
(6)
𝐼𝐼2 = 3 = 13 𝐴𝐴
3
(6)
𝐼𝐼2 = 3𝐼𝐼𝑅𝑅= 1𝐼𝐼𝐴𝐴
As
the
33 three resistors are in parallel, the potential difference is the same:

𝐼𝐼22 = 𝑉𝑉𝑅𝑅 ==1=
𝐴𝐴
(6)
2.4.
3𝑒𝑒𝑒𝑒𝑒𝑒 3 𝜀𝜀 − 𝑉𝑉𝑖𝑖 
=
𝜀𝜀
− 𝑉𝑉 
2.4.
𝑉𝑉
𝑒𝑒𝑒𝑒𝑒𝑒
𝑉𝑉1 = 3𝑉𝑉
2 = 𝑉𝑉1 = V 𝑖𝑖
=𝑒𝑒𝑒𝑒𝑒𝑒
𝑉𝑉
2.4.
= 𝑉𝑉
=12
1=−
𝐴𝐴𝜀𝜀3−×𝑉𝑉0,4
(6)
𝐼𝐼2𝑒𝑒𝑒𝑒𝑒𝑒
𝑖𝑖 
3
= 12
− 3 × 0,4
𝑉𝑉
𝑒𝑒𝑒𝑒𝑒𝑒the
As
resistors
have the same resistance, the current through each one is the same.
𝑉𝑉
10,8
(3)
= 12
−𝑉𝑉3 
× 0,4
𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉
2.4.
𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 
𝑉𝑉
(3)
Commented [U169]: Perhaps solution (i.e.noun – solving is a
Solving:
𝑒𝑒𝑒𝑒𝑒𝑒 = 10,8 𝑉𝑉 
verb).
2.5
𝑉𝑉
= 10,8 𝑉𝑉 
(3)
𝑒𝑒𝑒𝑒𝑒𝑒Increases
𝑉𝑉
𝑒𝑒𝑒𝑒𝑒𝑒 = 12 − 3 × 0,4
𝑅𝑅
2.5
𝑉𝑉2 =Increases
𝐼𝐼𝐼𝐼𝑒𝑒𝑒𝑒 = 𝐼𝐼 
3
TheIncreases
resistors are
now connected in series then total resistance increases, emf is constant then(3)
current
2.5
𝑉𝑉
𝑒𝑒𝑒𝑒𝑒𝑒 = 10,8 𝑉𝑉 
The
resistors
are
connected
in series
total resistance
increases,
emf=is𝜀𝜀constant
then current
decreases,
dropnow
of potential
in the
batterythen
decreases
and according
to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
− 𝑉𝑉𝑖𝑖 terminal
𝑅𝑅
𝐼𝐼
𝐼𝐼
resistors
are now
connected
in series
total resistance
increases,
emf
then(4)
current
3
decreases,
of potential
in the
batterythen
decreases
and according
to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒
= is𝜀𝜀 constant
− 𝑉𝑉𝑖𝑖 terminal
Commented [U170]: Pse check / clarify.
potential
2.5
= increases.
drop
𝐼𝐼The
2 =Increases
𝑅𝑅
3
(6) [U170]: Pse check / clarify.
decreases,
drop of potential in the battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal (4)
Commented
potential increases.
Exercise
2 are now connected in series then total resistance increases, emf is constant then(4)
3
The
resistors
current
Commented [U170]: Pse check / clarify.
potential
increases.
(6)
𝐼𝐼2 = 3 = 1 𝐴𝐴
Exercise
2 drop of potential in the battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal
2.4.
decreases,
3.1. What is the
relationship between potential difference and the strength of the electric current in a circuit
Exercise
2
Commented [U170]: Pse check / clarify.
potential
increases.
3.1.
What
is 𝜀𝜀the
relationship
between potential difference and the strength of the electric current in(4)
a circuit
=
−with
𝑉𝑉
2.4.
𝑖𝑖  constant resistance?
with 𝑉𝑉
a𝑒𝑒𝑒𝑒𝑒𝑒
resistor
3.1. What is the relationship between potential difference and the strength of the electric current in a circuit
with
a resistor
with constant resistance?
Exercise
3.2.
When
increases, potential difference also increases.
𝑉𝑉
12 2−current
3×
0,4
𝑒𝑒𝑒𝑒𝑒𝑒 =
with
a resistor
with
constant resistance?
3.2.
When
current
increases, potentialpotential
difference
also increases.
3.1.
is the relationship
difference
and the strength of the electric current in a circuit
3.3. What
Independent
variable – between
current
3.2.
When
current
increases,
potential difference also increases.
𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 Independent
= 10,8 𝑉𝑉  variable – current
(3)
3.3.
with aDependent
resistor with
constant
resistance?
variable – potential difference
3.3. Independent variable – current
Dependent
variable
– potential
difference
2.5
3.2.
When
current
increases,
potential
difference also increases.
3.4.Increases
Dependent variable – potential difference
3.4.
3.3. Independent
– current
(3)
The
resistors are variable
now connected
in series then total resistance increases, emf is constant then current
3.4.
decreases,
drop
of potential
in the difference
battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal
Dependent
variable
– potential
Commented [U170]: Pse check / clarify.
potential
increases.
(4)
3.4.
2.5 Increases
Exercise 2
3.1. What
is the relationship
potential
difference
and total
the strength
of the electric
current in aemf
circuit
The
resistors
are now between
connected
in series
then
resistance
increases,
is constant then current
94
with
a
resistor
with
constant
resistance?
decreases, drop of potential in the battery decreases and according to terminal
94 potential increases. 3.2. When current increases, potential difference also increases.
(4)
3.3. Independent variable – current
Dependent variable – potential difference
94
94
3.4.
86
94
Exercise 2
3.1.
What is the relationship between potential difference and the strength of the electric current in a circuit
with a resistor with constant resistance?
3.2.
When current increases, potential difference also increases.
3.3.
Independent variable – current
Dependent variable – potential difference
V (V)
V (V)
3.4.
I (A)
I (A)
2.5
∆YY
∆X
Gradient
==
2.5
gradient
∆

VYV
=
gradient
X = I
2.5
∆XI

24  6
24  6
0
,
08
Gradient== 0,02
Gradient
V
I
Gradient =
0,08  0,02
Gradient = 300
Gradient==of
300
2.6.
Resistance
the resistor.
Gradient
300
2.7. The
is correct:
the potential difference is directly proportional to the strength of the
2.6.hypothesis
Resistance
of the resistor.
electric
when
resistance
is constant.
2.7.current
The hypothesis
is correct:
the potential difference is directly proportional to the strength of the
2.6. Resistance
of
thethe
resistor.
electric current when the resistance is constant.
2.7. The
is correct: the potential difference is directly proportional to the strength of the electric
7.8 hypothesis
Electrodynamics
current
when1 the resistance is constant.
Exercise
7.8
Electrodynamics
7.5 Electrodynamics
Exercise 1
1.1
DC generator
Exercise 1 Mechanical energy to electrical energy
1.1
1.2
1.1
(2)
DC generator
To make the direction of the (induced) current to be the same in every half cycle/half
DC generator
turn
Commented [U171]: Pse check – not understood.
Mechanical energy to electrical energy
(2)
OR
Mechanical
energy to electrical energy
1.2
(2) cycle/half
To make the direction of the (induced) current to be the same in every half
keepturn
the (induced)
unidirectional.
To To
make
the
directioncurrent
of the
(induced) current to be the same in every half cycle/half turn
1.2
1.3
Commented [U171]
Graph A
OR
OR
(2)
The DC generator becomes an AC generator.
keep
the (induced)
current
unidirectional.
To keepTo
the
(induced)
current
unidirectional.
1.3
(2)
Voltage changes the polarity with every half cycle. 
Graph
A A
1.3 Graph
(2)
(3)
The DC
generator
becomes
an AC
generator.
The
DC generator
becomes
an AC
generator.
(3) (3)
Voltage
changes
the polarity
every
halfcycle.
cycle.
Voltage
changes
the polarity
withwith
every
half
95
95
87
1.4
1.4
1.4
Pav g/ gem 
V 2 rms / wgk

R
Pav g/ gem 
Commented [U172]: Pse c
V 2 rms / wgk

R
Commented [U172]: Pse check items (languages).
7.9
Commented [U172]: Pse check items (languages).
Photoelectric Effect
IrmsEffect
Total/wgk totaal = 2x(0,71) 
Photoelectric
Exercise 1
Exercise 1
1.1
Alternative
solution:
Alternative solution:
V 2 maks 2
rms / wgk )
( Vmax/

Pav g/ gem 
R2
V
Pavg/gem 
Vmax/ maks 2
( max/ maks ) 2
)
(
R
2
2
Pavg 
Vmax/ maks 2
Pavg/gem 
)
(
R
R V 2 rms / wgk
solution:
12 Alternative
( )2
2
1.4

Pav g/ gem 
2 Pavg
R
R2 rms / wgk
6
12
12 2
V
( )2
( )
R Pav g/ gem 

2 
2
R
V
6
6

12
( max/ maks ) 2 R= 12 Ω
( )2
R
R
2
2 
V
Pavg/gem 
6
( max/ maks ) 2
R
R= 12 Ω
R= 12 Ω
Vrms / wgk R
2
Pavg 
Irms / wgk 
R 12 Ω
Rrms/wgk = IrmsTot/wgk tot RT
12
V
R=
Vrms / wgk
( ) 2
Irms / wgk 
2
6R

12 V
max/ maks tot2RT
Vrms/wgk =( V
IrmsTot/wgk
R
)
( ) 2 rms / wgk  I
R
)
rms Total/wgk tot  (
2
2
2
2

Irms/wgk
V maks 2
6 R

Vrms / wgk
)
( max/
R
R= 12 Ω
 Irms Total/wgk tot  ( )R
2
2
2
Irms/wgk 
12
12
R
R= 12 Ω
 Irms / wgk  ( ) 
12 2
Vrms / wgk
)
(
2
 12
Irms / wgk 
2
12
2 V
R
  Irms
rms/wgk
tot RT
 ( = )IrmsTot/wgk
12 2

/ wgk
( )
2
2 12
Irms Total/wgk tot = 1,42 A 
Vmax/ maks 2
Vrms / wgk
 2
R
)
(
Irms/wgk
= 0,71tot = 1,42 A
Irms Total/wgk tot  ( )
12
Irms Total/wgk
2
2
2
Irms/wgk 
R
Irms/wgk = 0,71
Irms Total/wgk totaal = 2x(0,71) 
12
12
 Irms / wgk  ( ) 
12
Irms Total/wgk totaal = 2x(0,71)
( )2 
=1,42 A2
2
 2
=1,42 A  12
Irms Total/wgk tot = 1,42 A 
Irms/wgk = 0,71
7.9
V 2 rms / wgk

Pav g/ gem 
R
Alternative solution:
=1,42 A 
1.1
Minimum energy needed to eject an electron from a metal surface (substance).
7.6
Photoelectric
Effect
Minimum
energy needed1.2
to ejectThe
an electron
from
a metal of
surface
(substance).
extraction
(ejection)
electrons
from a metal surface by radiation (when light strikes its
surface).
Photoelectric
1.2
The7.9extraction
(ejection)Effect
of electrons from a metal surface by radiation (when light strikes its
Exercise
1
surface).
1.3.1
𝐸𝐸 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘
Exercise 1
1.3.1
1.1
1.1
= 𝑊𝑊𝑜𝑜an
+ 𝐸𝐸electron
𝐸𝐸 = 𝑊𝑊𝑜𝑜 +
𝐸𝐸𝑘𝑘
𝑘𝑘
Minimum
energy
needed toℎ𝑓𝑓eject
from a metal surface (substance).
Minimum energy needed to eject an electron from a metal surface (substance).
(6,63 × 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19
ℎ𝑓𝑓 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘
The extraction
(ejection)
of electrons
from a metal
by surface
radiation (when
light strikes
its light strikes its surface).
1.21.2 The
extraction
(ejection)
of electrons
fromsurface
a metal
by radiation
(when
(6,63
× 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19 𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻
surface).
1.3.1
1.3.1
𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻
𝐸𝐸 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘
ℎ𝑓𝑓 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘
96
96
(6,63 × 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19
𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻
1.3.2
1.3.2
𝑐𝑐 = 𝑓𝑓𝑓𝑓
3 × 108 = (1,16 × 1015 )𝜆𝜆
𝜆𝜆 = 2,59 × 10−7 𝑚𝑚
88
96
chemistry
1.
How to use this book
This book serves as a guide to understanding the Grade 12 Physical Sciences. However, it
does not replace your textbook. The book focuses more on the challenges that have been
observed from learners’ responses in the Grades 12 National Examinations over the past
few years.
The authors have used their experience to focus their attention on the areas that learners
seem to struggle with.
2.
Key subject concepts
The two areas of study in Chemistry are: 1. materials 2. chemical change. In demonstrating
concepts, laws and theories in this book, examples are used; this however does not mean that
the laws, concepts and principles applies only to those examples used. It is therefore important
to understand the concept itself and where and when it applies.
Some theories/laws/concepts apply throughout the content in both organic and inorganic
chemistry e.g. atomic theory, The Kinetic Molecular Theory, the mole concept, stoichiometry,
rules of oxidation, inter-molecular forces etc. These theories and concepts are applied
throughout the chemistry content. There are aslo laws that apply only to a particular group or
class of compounds; for example gas laws apply only to gases and not to solids or aqueous
substances. It is important to always know where the laws/ rules can be applied.
3.
The Representation model
This model employs a strategy of three stages in understanding Chemistry Education. These
stages should be applied almost simultaneously when studying chemistry: 1. The macro
level: This is what you see as an observer of either a material or chemical change (state,
colour, size, etc.) 2. The Sub-micro level: This is what happens at the sub-atomic level, e.g.
with electrons, protons in atoms or molecules. 3. The Symbolic level: This is when you
represent what happens at the sub-atomic level and macro level using symbols, e.g. instead
of saying ice, you write H2O (s). Learners of Chemistry must be able to move through these
three levels with ease when they discuss chemistry. When you deal with a chemistry question,
always think about it in terms of these three levels.
Macroscopic
Symbolic
Sub-micro
representation model
89
4. Tips on specific topics

Organic chemistry: Study how to name organic molecules for all
groups. always remember that the carbon atom always takes a
maximum of 4 bonds in all organic compounds, not more, not less.

Electrochemistry: Pay attention to half reaction and their respective
oxidation states. use the Table of Reduction Potentials as discussed in
the text. Read about the similarities and differences between the two
types of cells: Galvanic and Electrolytic!

Fertilizers: The processes involved in the manufacture of fertilisers is clearly
indicated in the text. Study the molecules and their reactions, including their
reaction conditions (these would include temperature, pressure,
concentrations of reacting species and in some cases catalysts).
5. Study and examination tips
6. message to Grade 12 learners from the writers
As you prepare to write your examination, it is important to carefully understand the rules
governing certain aspects of your work, and the laws and concepts. Ensure you understand
these rules/laws/ concepts properly. Understand what they mean, where they apply and
when they apply - and also when and where they do not apply.
Always:
1.
2.
3.
4.
5.
Read the question that you are working on carefully.
Understand what it says and what is required of you.
Write down the information that you have.
Write down the information that you do not have.
Use the information you have to derive the information you need to answer the
question.
6. Check your work by going through these steps again!
90
7. The mole concept
The Mole: The quantity of a substance that contains 6,02 x 1023 particles. This
number is also called Avogadro’s number.
A mole is a very, very
BIG
number and if 12g of Carbon C12
contains this number of Carbon atoms, you can imagine how small atoms are.
Imagine this many atoms contained in only 12g!
This also demonstrates how small atoms are!
All
Chemistry calculations require an understanding of the mole concept. When
we do calculations on reactions in chemistry we always use the moles that reacted NEVER the mass that reacted. If you are given the mass, you must always work out
the number of moles contained in the mass!
The masses that reacts in a chemical reaction will almost always have one or more
of the reactant(s) having a larger than necessary number of moles or fewer that
necessary number of moles and the reactions will almost always leave some
unreacted reactant(s) that we call an excess reactant and the reactant(s) that gets
used up while there is still an excess reactant is called a limiting reactant(s).
8. Stoichiometry
91
8.1
Excess and Limiting Reagents
See how many tyres you are able to use in this example. The remaining tyres are
in excess. The cars are said to be limiting.
Which of the two reactants is in excess and which is the limiting reagent?
Look at the reactants in the jar as they form products in the jar on the right of the
arrow. Study this reaction and decide which of the two molecules that are involved in
this reaction is limiting H2 or N2 is the limiting reagent?
92
9. Practice activities from Grade 10
Reminder: One MOLE has 6.022 x 1023 items or there are 6.022 x 1023 items/mole.
Show your work!
1. How many atoms of potassium make up one MOLE? _____
2. How many atoms of potassium make up 2 MOLES? _____
3. How many formula units of salt make up 10 MOLES? _____
4. How many molecules of water make up 1 MOLE? _____
5. How many molecules of water make up 5 MOLES? _____
6. How many moles are 6.022 x 1023 atoms of sodium? _____
7. How many moles are 12.04 x 1023 atoms of carbon? _____
8. How many moles are 18.06 x 1023 atoms of sodium? _____
9. How many moles are 60.22 x 1023 atoms of sodium? _____
10. How many moles are 6.022 x 1023 molecules of water? _____
11. How many moles are 12.04 x 1023 molecules of water? _____
12. How many moles are 30.10 x 1023 molecules of water? _____
10. dynamic chemical Equilibrium Graphs
93
11.
cONcENTRaTION – TImE GRaPHS
STudy TIP 1: Concentration, pressure and temperature changes made to a gaseous chemical
equilibrium are identified as follows:
 concentration change: The graph of one of the substances spikes upward or downward
 Pressure change:
The graph spikes upward or downward for each substance in the equation.
 Temperature change: The graph does not spike but shows gradual upward and downward
changes that are opposite for the forward and reverse reactions
STudy TIP 2: Upward spiking and gradual upward changes show increase in concentration as well
as reaction rate. Downward spiking and gradual downward changes show decrease in concentration
as well as reaction rate.
STudy TIP 3: Pressure is increased by decreasing the volume of the container containing the
reaction mixture. Conversely, pressure is decreased by increasing the volume of the closed
container. Therefore pressure changes are really concentration changes.
STudy TIP 4: Once the change is made Le Chatelier’s Principle (LCP) can be used to explain the
graph shape.
ExamPLE
In the Haber Process, the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
closed container.
ΔH < 0 takes place in a
Changes are made to the reaction at times t1, t2 and t3. The graphs below show the situation.
1. What do the shapes of the graphs between the times 0 and t1 indicate about the reaction?
2. State the change that was made to the reaction at:
2.1 t1
2.2 t2 2.3 t3. Give a reason for each answer.
3. Use Le Chatteliers Principle (LCP) to explain the shapes of the graphs between:
3.1 t1 and t2 3.2 t2 and t3
aNSwERS
1. The reaction is in equilibrium
2.1 [N2] was increased
94
Reason: There is an upward spike only for N2
ANSWERS
1.
The reaction is in equilibrium.
2.1 [N2] was increased.
Reason: There is an upward spike only for N2.
2.2 Pressure was increased.
Reason: There is an upward spike for N2, H2 and NH3.
2.3 Temperature was increased.
Reason: All changes in the graph line shapes are gradual. The
changes are upward for the reverse reaction and downward for the
forward reaction.
3.1 STEP 1: Change made: [N2] increases.
STEP 2: According to LCP, the forward reaction is favoured (i.e. the reaction that reduces
the [N2] is favoured).
STEP 3: Therefore, the [H2] decreases and the [NH3] increases until a new equilibrium is
established.
3.2 STEP 1: Change made: Pressure was increased.
STEP 2: According to LCP, the forward reaction is favoured (i.e. the reaction that
produces less molecules is favoured).
STEP 3: Therefore the [NH3] increases, but the [N2] and [H2] decrease until a new
equilibrium is established.
95
FERTILIzERS
12.
FERTILISERS
NH4NO3 and (NH4)2SO4
3 Industrial processes
Fractional distillation
of liquid air gives N2.
Ostwald Process
4NH3 + 5O2
2NO + O2
4NO + 6H2O
2NO2
Haber Process
Contact Process
N2 + 3h2
2NH3
catalyst: Iron Oxide
+
SO2
2SO2 + O2
2SO3
SO3 + H2SO4
H2S2O7 (Oleum)
Catalyst in Step2:V2O5
3NO2 + H2O
2HNO3 + NO
Catalyst in step 1: Pt:
HNO3
S + O2
NH3
+
H2SO4
FERTILISERS
(ONLY 2 Fertilisers to study)
NH4NO3 Ammonium Nitrate
96
(NH4) 2SO4 Ammonium Sulphate
Primary nutrients (N P K)
Essential Nutrients (c H and O)
NPK ratio
TIP 1: Make a CHART showing the OSTWALD, HABER and CONTACT processes
and place it in your ROOM where you can always see it. Practice writing out the
equations on your own.
TIP 2: Play this game: Prepare the following cards for your game:
4 cards marked O2; 2 cards marked NO; 2 cards marked NO2; 2 cards marked
SO2;
2 cards marked SO3; 2 cards marked H2O; 1 card for each of the following - S,
H2SO4, H2S2O7 and HNO3.
Use THese CArDs To DePICT THe reACTIoNs For THe TWo
ProCesses oN A WALL. ComPeTe WITH YoUr CLAssmATe To see WHo
FINIsHes FAsTer.
activities
QuESTION 1
The flow diagram below represents two industrial processes that are used
to make fertilizer Z. The letters (a) to (d) represent steps in one of the
industrial processes. The letter J represents the other industrial process.
S + O2
X (a
(b)
SO3
(c)
Gas P + H2
R
J
(d)
Y
NH3
Fertilizer Z
1.1
Write down the NAME or FORMULA of:
1.1.1
Product X formed in step (a).
(1)
1.1.2
Product Y formed in step (d).
(1)
1.1.3
Process J, by which NH3 is produced.
97
1.2
Write down a balanced chemical equation for the formation of
fertilizer Z.
1.3
A farmer has two labelled bags of fertilizer - x and y -as shown
below.
(3)
X
50 kg
20 kg
6:2:1 (20)
3:5:2 (30)
She mixes the contents of x and y together in another bag - z.
Calculate the total mass of phosphorus that z now contains.
(7)
1.3 The bag of fertiliser shown below was prepared by mixing 25 kg of pure
ammonium sulphate (NH4)2SO4 with potassium salts, phosphates and sand.
3:5:1 (30)
x kg
Calculate the mass of the bag of fertiliser.
(4)
(ASSUME THAT AMMONIUM SULPHATE IS THE ONLY SOURCE OF NITROGEN.
QuESTION 2
2.1 The flow diagram below shows the processes involved in industrial preparation of
fertiliser y.
Process
a:
Proces A
H2 + Gas X
NH3
Fertiliser Y
Process B:
Gas B
+ O2
Step 1
98
NO + H2O
+ O2
Step 2
Gas C
H2O
Step 3
HNO3
2.1.1 Write down the NAME of:
a. The RAW material for gas x in process a.
(1)
b. Process a.
(1)
c. Process B, represented by Steps 1 to 3.
(1)
2.1.2 Write down the balanced equation for:
a. Process a
(3)
b. Step 1
(3)
c. Step 2
(3)
d. The reaction that leads to the formation of fertiliser y.
(3)
2.2 Two 50 kg bags contain fertiliser P and Q.
Fertiliser P: 5:2:3 (25)
Fertiliser Q: 1:3:4 (20)
2.2.1 What do the numbers 25 and 20 represent?
(1)
2.2.2 Do a calculation to determine which fertiliser (P or Q) contains the greater
mass of potassium.
(4)
SOLuTIONS
QuESTION 1
1.1.1 Sulphur dioxide (or SO2) - give the NAME or FORMULA, not both.
1.1.2 Sulphuric acid (or H2SO4)
1.1.3 Haber process
1.2 2NH3 + H2SO4
(NH4)2SO4
1.3 Approach:
Calculate the mass of phosphorus in each bag and add these figures together.
Bag X; m Phosphorus = 2/9 x 30/100 x 50 = 2.22 kg
Bag Y: m Phosphorus = 5 /10 x 30 x 20 = 3
mTOTAL = 3 + 2.22 = 5.22 kg
99
1.4 Approach:
First find mass N in (NH4)2SO4
m Nitrogen = % N in (NH4)2SO4 x 25k
= M nitrogen/M (NH4)2 SO4 X 25 Kg
= (2 x 14 ) / (14 x 2) + (2 X 4 X 1)+ 32 + (4 x 16) x 25kg
= 28+3 +32+64+25
= 28/132 x 25 Kg = 5.3 Kg
Mass nitrogen in bag = 5.3 = 3/9 x 30/100 x
5.3
= 0.1 x
53 Kg = x
QuESTION 2
2.1.1 a. Air
b. Haber c. Ostwald
2.1.2 a. N2 + 3H2
b. 4 NH3 + 5 O2
c. 2 NO + O2
d. NH3 + HNO3
2 NH3
4 NO + 6 H2O
2NO2
NH4NO3
2.2.1 Percentage of fertiliser in the bag.
2.2.2
P
mK = 3/10 x 25/100 x 50 = 3.75 kg
Q mK = 4/8 x 20 /100 x 50 = 5
100
Q contains more potassium
13. ORGaNIc cHEmISTRy
ORGaNIc cHEmISTRy
 BaSIc ORGaNIc cHEmISTRy
 PHySIcaL PROPERTIES
 ORGaNIc REacTIONS
13.1 BASIC ORGANIC CHEMISTRY
Structural
Formulae
Condensed
Molecular
Functional groups 9
Empirical
General formulae
Hydrocarbons
Basic
IuPac naming
Organic chemistry
saturated/Unsaturated
Terminology
Homologous series
Functional group
Isomers
Chain
Structural isomers
Positional
Functional
101
13.2 NOMENCLATURE OF ORGANIC COMPOUNDS
NOTE: organic compounds in these notes are named according to the 1993 IUPAC RULES.
STRaIGHT cHaIN aLKaNES
IuPac NOmENcLaTuRE
The IUPAC names of the first eight are:
Name
condensed structural
formula
Name
condensed structural formula
Methane
CH4
Pentane
CH3CH2CH2CH2CH3
Ethane
CH3CH3
Hexane
CH3CH2CH2CH2CH2CH3
Propane
CH3CH2CH3
Heptane
CH3CH2CH2CH2CH2CH2CH3
Butane
CH3CH2CH2CH3
Octane
CH3CH2CH2CH2CH2CH2CH2CH3
IuPac NOmENcLaTuRE OF STRaIGHT cHaIN aLKaNES cONTaINING aN aLKyL
In organic chemistry, an alkyl substituent is an alkane missing one hydrogen. The
term alkyl is intentionally unspecific to include many possible substitutions. ... The
smallest alkyl group is methyl, with the formula CH3−
In general, alkyl substituents are named by changing the ending “ane” in an alkane’s
name to “yl”.
EXAMPLE: CH3 is methyl; it is derived from CH4 (methane) by changing the “ane” ending in
methane to “yl”.
IUPAC RULES should be followed when naming these alkanes:
The following compound will be used to illustrate the IUPAC Rules:
STEP 1:
102
Select the longest continuous chain for the parent name. In this case it has six
carbon atoms and is thus hexane.
STEP 2:
Number the chain from either end, so that the substituents are attached at the
lower numbers. In the example, numbering from left to right gives substituents at
carbon 2 and 4, whereas numbering from right to left gives substituents at carbon
3 and 5. Thus the lower numbers occur when numbering from left to right.
STEP 3:
Substituent groups are assigned the number of the carbon to which they are
attached. In the example, the two substituents are called the 2-methyl and 4-ethyl
substituent groups. Note that the number is written in front of the substituent
group and a hyphen (-) separates the number and the alkyl group’s name.
STEP 4:
The name of the compound is now composed of the parent name preceded by
the names of the substituent groups in alphabetical order. The correct IUPAC
name for the example is 4-ethyl-2-methylhexane.
NOTES:
In IuPac nomenclature, a hyphen (-) separates numbers from words, and a
comma separates numbers from numbers. Otherwise there is no space
between the words that make up the name.
STEP 5:
If the same substituent occurs more than once in the molecule, the prefixes “di”,
“tri”, “tetra” and so on, are used to indicate how many times it occurs. If the
substituent occurs more than once on the same carbon, the number is repeated.
The following examples illustrate this step.
NOTES:
In IUPAC nomenclature, the prefixes “di”, “tri”, “tetra” and so on, are not
alphabetised.
EXAMPLES:
2,3 - dimethylpentane
3,3 – dimethylhexane
NOTE: The IUPAC nomenclature of the following homologous series is similar to that of
alkanes. The only differences in the IUPAC nomenclature of these series is with precedence,
position of the functional group and parent name.
103
HOmOLOGOuS SERIES: aLKENES IuPac NOmENcLaTuRE
Precedence
Position
Parent name
The longest chain is
numbered so that the C=C
has the lowest number in the
chain.
The first carbon in C=C is
used to locate the double
bond.
Change the “ane” ending in
the alkane to “ene” for the
alkene.
EXAMPLES
(a) pent-2-ene
(b) 4, 4-dimethylpent-2-ene
H
H
C
H
H
H
H
C
C
C
H
H
H
C
H
C
C
H
H
H
H
HOmOLOGOuS SERIES: aLKyNES
IuPac NOmENcLaTuRE
Precedence
Position
The longest chain is numbered so
that the C=C has the lowest
number in the chain.
The first carbon in C=C is used to
locate the triple bond.
EXAMPLES
Parent name
Change the “ane” ending in the
alkane to “yne” for the alkyne.
(a) 4,4-dimethylpent-2-yne
(b) 2-methylhex-3-yne
H
H
H
H
C
H
C
C
C
C
H
C
H
104
H
H
C
H
H
H
H
H
H
C
C
H H
C
H
H
C
H
C
H
C
C
H
H
H
HOmOLOGOuS SERIES: HaLOaLKaNES (aLKyL HaLIdES) IuPac NOmENcLaTuRE
The halogen substituents that occur on the alkane chains are named as follows:
cl – chloro
Br- bromo
Precedence
Position
The longest chain is numbered to give
the halogen substituents the lowest
numbers in the chain.
The carbon atoms to
which halogen
substituents are
bonded are used to
locate their positions.
Examples: (a) 1-bromo-4-methylpentane
Br
H
Cl
H
H
C
C
C
C
C
H
H
H
2
3
4
H
C
Parent name
There is no change
made to the parent
alkane chain. It
remains an alkane.
(b) 2-bromo-1-chloropropane
H
1
I – iodo
5
Cl
H
H
H
H
HOmOLOGOuS SERIES: aLcOHOLS
C
H
H
H
H
C
C
Br
H
H
IuPac NOmENcLaTuRE
NOTES: Just like there are primary, secondary and tertiary alcohols, there are primary,
secondary and tertiary alkyl halides.
PRImaRy aLcOHOL: The carbon atom attached to the –OH group is attached to one other
carbon atom in the molecule.
Example: CH3CH2CH2CH2OH (butan-1-ol).
SEcONdaRy aLcOHOL: The carbon atom attached to the –OH group is attached to two
other carbon atoms in the molecule.
Example:
TERTIaRy aLcOHOL: The carbon atom attached to the –OH group is attached to three
other carbon atoms in the molecule.
Example:
105
IuPac NOmENcLaTuRE OF aLcOHOLS
Precedence
Position
Parent name
Longest chain is numbered so that the carbon atom
bonded to the OH group has the lowest number in the
chain.
The carbon atom to which the
OH group is bonded is used to
locate it.
Change the “e” ending in the
alkane to “ol” for the alcohol.
Examples
2-methylbutan-1-ol
The table below provides information on the structural formula, condensed structural formula, molecular formula,
isomer(s) and functional group of some alcohols when their IUPAC name is given.
IuPac
Name
Structural Formula
H
Pentan-1-ol
condensed Structural
Formula
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
molecular
Formula
Functional /
Structural
Isomers
Pentan-2-ol
O
H
CH3CH2CH2CH2CH2OH
C5H12O
Pentan-3-ol
3-methylbutan-2ol
Butan-1-ol
106
H
H
H
H
H
C
C
C
C
H
H
H
H
Functional
Group
(structural
formula)
-O-H
Butan-2-ol
O
H
CH3CH2CH2CH2OH
C4H10O
2-methylpropan2-ol
-O-H
HOmOLOGOuS SERIES: caRBOxyLIc acIdS
Precedence
IuPac NOmENcLaTuRE
Position
The longest chain is numbered so that
the carbon atom in the COOH group
has the lowest number in the chain. It
will always be 1, but 1 is omitted in the
IUPAC name.
Parent name
The carbon atom in
COOH is used to
locate the COOH
group.
Examples:
Change the “e” ending
in the alkane name to
“oic acid” for the
carboxylic acid.
CH3
(a) HCOOH
methanoic acid
(b) CH3COOH
(c) CH3CH2COOH
(d) CH3CH2CH3CH2CH2C1OOH
ethanoic acid
propanoic acid
4-methylhexanoic acid
Draw all the molecules above (a-d).
The table below provides information on the structural formula, condensed structural
formula, molecular formula, isomer and functional group of some carboxylic acids when their
IUPAC name is given.
IuPac Name
Structural
Formula
H
Ethanoic acid
H
condensed
Structural
Formula
molecular
Formula
Functional /
Structural
Isomers
O
O
C
C
O
H
CH3COOH
C2H4O2
Methyl
methanoate
H
H
Propanoic
acid
H
H
C
C
H
H
Functional Group
(structural
formula)
Methyl
ethanoate
O
C
- C –O-H
O
O
H
CH3CH2COOH
C3H6O2
- C –O-H
Ethyl
methanoate
107
In general, an isomer of a carboxylic acid is usually an ester, and vice versa.
HOmOLOGOuS SERIES: ESTERS
Precedence
The alkyl group from the alcohol is
named first, followed by the name of
the acid, with the “ic” changed to
“ate”.
Examples:
A) butyl ethanoate
B) ethyl butanoate
C) ethyl ethanoate
108
Position
Not applicable.
Parent name
Not applicable.
D)
ethyl methanoate
HOmOLOGOuS SERIES: aLdEHydES
IuPac NOmENcLaTuRE
Precedence
Position
Parent name
The longest chain is numbered so that
the carbon atom in the CHO group has
the lowest number in the chain. It will
always be 1, but 1 is omitted in the
IUPAC name.
The carbon atom in
CHO is used to locate
the CHO group.
Change the “e” ending
in the alkane to “al” for
the aldehyde.
a)
B)
(a)
HCHO
(b) CH3CH2CHO
methanal
propanal
HOmOLOGOuS SERIES: KETONES
IuPac NOmENcLaTuRE
Precedence
Position
Parent name
The longest chain is numbered so that
the carbon atom in the C=O group has
the lowest number in the chain.
The carbon atom of
the C=O group is used
to locate C=O.
Change the “e” ending
in the alkane name to
“one” for the ketone.
In general, the functional isomer of a ketone is an aldehyde, and vice versa.
Examples:
CH3
(a)
HCOH
methanone
(b) CH3COCH3
Propan-2-one
(c) CH3CH2CH2CH2COCH3
5-methylhexan-2-one
Draw structural formulae for all the molecules above.
109
PHySIcaL PROPERTIES OF ORGaNIc mOLEcuLES
definitions (REFER TO yOuR Exam GuIdELINES)
- Boiling point
- Vapour pressure
- Melting point
Types of inter-molecular forces (NOTE: ALL inter-molecular forces are
CALLED as Van der Waals forces; therefore, Hydrogen bonds are also Van
der Waals forces, according to caPS)
REmEmBER THE FOLLOwING:
 The relationship between inter-molecular forces and boiling point: As the
strength of inter-molecular forces increases, boiling point increases.
 The relationship between inter-molecular forces and vapour pressure: As the
strength of inter-molecular forces increases, vapour pressure decreases.
 What type of inter-molecular forces are present between molecules of
HydROcaRBONS
Alkanes
Alkenes Only London Forces
Alkynes
(Haloalkanes/aldehydes/Ketones contain BOTH London Forces and dipoledipole forces (organic molecules where there is a halogen atom or oxygen atom
(more electro-negative atoms than C there are dipole dipole forces)
 dipole–dipole forces are stronger than London Forces.
In Haloalkanes/aldehydes/Ketones, the stronger force is the Dipole- dipole force.
(Note: London Forces are also present.)
SummaRy
Alkanes
Hydrocarbons
Alkenes
have ONLy London forces between molecules
Alkynes
Haloalkane/aldehydes and Ketones have:
110
BOTH London forces and dipole-dipole forces (i.e. a compound with Br or cl or O
will have dipole–dipole forces).
dipole–dipole forces are stronger than London Forces.
alcohols and carboxylic acids have:
London Forces, Dipole-dipole forces and Hydrogen bonds (the strongest of the
inter-molecular forces) between their molecules.
SummaRy OF TyPES OF BONdS IN ORGaNIc mOLEcuLES
ONLy London
Forces
Alkane
Alkene
BOTH
London forces
and
dipole –dipole forces
Haloalkane
Aldehyde
Alkyne
Ketone
aLL THREE
 London forces
 dipole-dipole forces
 Hydrogen bonding
Alcohol
Carboxylic acid
Summary of bonds
Hydrogen Bonds
dipole dipole
Hydrogen bonding
occurs
in
molecules containing the highly
electronegative elements F, O, or N
that are directly bound to a hydrogen
atom in a molecule. Since H has an
electronegativity of 2.2 the bonds are
not as polarized as purely ionic bonds
and possess some covalent character.
However, the bond to hydrogen will
still be polarized and possess a dipole.
Van der waals dipole-dipole
interactions.
Other groups of molecules
beside
hydrogen
can
be
involved in polar covalent
bonding
with
strongly
electronegative atoms due to
their electro negativities. The
unequal distribution of charge
gives the molecule a dipole
nature. For instance, each of
these molecules contains a
dipole:
The dipole of one molecule can align
with the dipole from another molecule,
leading to an attractive interaction that
we call hydrogen bonding. Owing to
rapid molecular motion in solution,
these bonds are temporary (shortlived) but have significant bond
strengths ranging from (9 kJ/mol (2
kcal/mol) (for NH) to about 30 kJ/mol
These dipoles can interact with
each other in an attractive
fashion, which will also increase
the boiling point. However since
the electronegativity difference
between
carbon
(electronegativity = 2.5) and the
dispersion Forces/
London Forces
The weakest intermolecular forces
of all are called dispersion
forces or London forces. These
represent
the
attraction
between instantaneous dipoles in
a molecule. If you cool a gas like
Argon to –186 °C, you can actually
condense it into liquid argon. The
fact that it forms a liquid it means
that something is holding it
together. That “something” is
dispersion forces. Think about. on
average the electrons in the
valence
shell,
are
evenly
dispersed. But at any given instant,
there might be a mismatch
between how many electrons are
on one side and how many are on
the other, which can lead to
an instantaneous difference in
charge.
111
(7 kcal) and higher for HF. As you
might expect, the strength of the bond
increases as the electronegativity of
the group bound to hydrogen is
increased.
So in a sense, HO, and NH are
“sticky” – molecules containing these
functional groups will tend to have
higher boiling points than you would
expect based on their molecular
weight.
electronegative atom (such as
oxygen or nitrogen) is not as
large as it is for hydrogen
(electronegativity = 2.2), the
polar interaction is not as strong.
So on average these forces tend
to be weaker than in hydrogen
bonding.
For hydrocarbons and other nonpolar molecules which lack strong
dipoles, these dispersion forces
are really the only attractive
forces between molecules. The
dipoles are weak and transient,
they depend on contact between
molecules – which means that the
forces increase with surface area.
A small molecule like methane
has very weak intermolecular
forces, and has a low boiling
point. However, as molecular
weight increases, boiling point
also goes up. That’s because the
surface over which these forces
can operate has
increased. Therefore, dispersion
forces increase with increasing
molecular weight. Individually,
each interaction isn’t worth much,
but if collectively, these forces can
be extremely significant.
COMPARE THE STRENGTH OF INTER-MOLECULAR FORCES
London Forces – WEAKEST
Dipole-dipole - STRONGER than London Forces, but WEAKER than Hydrogen bonds
Hydrogen bonds - STRONGEST
ExamPLES
1.1
Consider the three compounds: a: CH3OH B:CH3CH2OH and
c:CH3CH2CH2OH
1.1.1 Which of the three compounds has the HIGHEST boiling point?
1.1.2 Explain your answer to QUESTION 1.1.1 above
Answer:
Approach: Using the summary above, check where the molecules in question belong
and check the forces that can be found in that group.
The compounds belong to the same homologous series.
Mention the trend with regard to chain length (surface area).
Mention the trend with regard to London Forces.
112
More energy is required to break forces.
Generally: If the compounds belong to the same homologous series, the
explanation must mention London Forces, as they are the oNLY forces affected by
an increase/decrease in surface area.
A common ERROR is using Hydrogen bonds in the answer, since we are dealing
with alcohols - but hydrogen bonds/dipole-dipole forces are not affected by chain
length.
aNSwER:
1.1.1 c
1.1.2 From a to c
Chain length (or surface area or molecular size) increases.
The strength of London forces increases.
More energy is needed to break inter-molecular forces.
(DO NOT SAY “break bonds”. SAY “Break inter-molecular forces”.)
CH3
1.2
Consider the TWO compounds, d: CH3COCH3 and E:CH3CH2CH3
1.2.1 Which compound has the HIGHEST boiling point?
1.2.2 Explain your answer to QUESTION 1.1 above.
Approach:
The two compounds belong to different homologous series
Step 1: Identify the type of inter-molecular force in each
Step2: Compare the strength of the forces.
Step 3: More energy is needed to break inter-molecular forces.
Answer:
1.2.1 d
1.2.2 STEP 1: Between the molecules of d there are dipole–dipole forces.
STEP 2: Between the molecules of E, there are London forces.
Dipole-dipole forces are stronger than London forces.
STEP 3: More energy is required to break inter-molecular forces in d.
113
ExERcISE : PHySIcaL PROPERTIES
Consider compounds a to c in the table below:
compound
a
B
c
IuPac name
ethane
propane
butane
1.1 To what homologous series do compounds a to c belong?
1.2
Is compound a SATURATED or UNSATURATED? Give a reason.
1.3
Which ONE of the compounds has the HIGHEST vapour pressure?
1.4
Explain your answer to QUESTION 1.3 ABOVE.
1.5
Which compound has the highest boiling point?
ExcERcISE 2: PHySIcaL PROPERTIES
2.1 The table shows five organic compounds represented by the letters A to D.
a
B
c
d
E
CH3CH2OH
CH3CO2H
CH3CHO
CH3CH2CH3
CH3CH3
2.1.1 Which compound in the table is an aldehyde?
Consider the boiling points of compounds a to E, given in random order below,
and use them where applicable to answer the questions that follow.
118 oC
-89 oC
78 oC
2.1.2 Write down the boiling point of:
a. Compound B b. Compound c
2.1.3 Explain the difference in the boiling point of:
a. a and B
114
-42oC
21 oC
b. B and c
c. c and d
d. D and E
2.1.4 Describe the trend in vapour pressure from c to E
2.2 The relationship between boiling point and the number of carbon atoms in
straight chain molecules of alkanes, carboxylic acids and aldehydes is
investigated. Curves P, Q and R are obtained.
GRAPH OF BOILING POINT VERSUS NUMBER OF C ATOMS
500
Boiling point (K)
P
Q
300
R
100
2
3
4
5
6
7
8
Number of C atoms
2.2.1 Define the following terms:
a. Boiling point
b. Vapour pressure
(2)
(2)
2.2.2 For curves a, P, Q, R write down a conclusion that can be drawn from these
results. (2)
115
2.2.3 Identify the curve that represents each of the following (a ,B or c):
a. Aldehydes
(1)
b. Carboxylic acids
(1)
2.2.4 Explain your answer to question 2.2.3 above.
(5)
SOLuTIONS: PHySIcaL PROPERTIES
ExcERcISE 1: PHySIcaL PROPERTIES
1.1 Alkanes
1.2 Saturated - it contains only single bonds.
1.3 Ethane
1.4 From a to c:
Surface area (chain length) increases.
Strength of London forces increases.
More energy is needed to break the inter-molecular forces.
1.5 C
ExcERcISE 2: PHySIcaL PROPERTIES
2.1.1. c
2.1.2. a.118 oC
b. 21 oC
2.1.3 a. Both compound A and B have Hydrogen bonds.
Compound B has more sites for Hydrogen bonds than A (therefore the bonds
are stronger in B).
More energy is needed to break inter-molecular forces in B.
b. B has strong Hydrogen bonds.
C has weak dipole–dipole forces.
More energy is needed to break the bonds in B.
c. c has dipole dipole forces.
d has weak London forces.
c needs more energy to break forces.
116
d. From E to d
Surface area (chain length) increases.
The strength of London forces increases.
More energy is needed to break inter-molecular forces.
GENERaLLy, when explaining trends in boiling point and vapour pressure:

IF the compounds differ in chain length:
 Mention the trend in surface area.
 Mention the trend in strength of London forces. (Remember ALL
compounds have London forces, including alcohols and
carboxylic acids.)

 Lastly, more energy is needed to break forces in the compound
where forces are strongest.
IF the compounds belong to different homologous series:
Identify the type of force in each compound.
Compare the strength of the forces - mention which one is stronger.
More energy is needed to break inter-molecular forces in the
compound where forces are stronger.
(NOTE: IT IS NOT ACCEPTABLE TO WRITE, “TO BREAK BONDS, AS THESE
ARE INTER-MOLECULAR FORCES, NOT INTER-ATOMIC BONDS, LIKE
COVALENT BONDS, ETC.”)
117
ORGaNIc REacTIONS
 addITION, SuBSTITuTION, ELImINaTION
 ESTERIFIcaTION, cOmBuSTION, POLymERISaTION
Draw contrasts between substitution and elimination
addition and elimination
SUBSTITUTION
ELIMINATION
(NO alkene in products or reactants)
Alkene is a Product
Reaction Conditions
Reaction conditions
-mild heat for in all Substitution
Strong Heat in all Elimination
For reaction of Haloalkane to
For Hydrohalogenation
Produce Alcohol (hydrolysis)
- Use concentrated strong base KOH
- Use H2O OR
or NaOH
–Dilute strong base KOH or NaOH
ORGANIC REACTIONS
ADDITION
ELIMINATION
Alkene is a reactant (On left of arrow)
Alkene is a product
 Hydrogenation
Alkene to Alkane: Add H2 (Catalyst Pt or Pd)
 Hydrohalogenation: Add HBr or HCl
Alkene to Haloalkane
(Product will have ONE halogen atom)
 Halogenation (Add Br2 or Cl2)
Alkene to haloalkane
Two halogen atoms on Product (Bromine
test)
 Hydration (Add H2O)
Catalyst: H2SO4
.Alkene to alcohol (add H2O catalyst
H2SO4)
118
Reaction Conditions
Strong Heat for all Elimination reactions
 Cracking
e.g. C15H32
2C2H4 + C3H6 + C8H18
Catalytic cracking Low Temp/Low Pressure
Thermal cracking High Temp/High Pressure
 Dehydrohalogenation
(Removal of HBr or HCl)
Conditions: Strong heat/Conc KOH or NaOH
 Dehydration (Removal of H2O)
Conditions: Strong heat and excess conc H2SO4)
TIPS
In addition, reactions involving compounds where there is no symmetry across
the double bond remember H “likes many Hs”)
OH in substitution in addition reaction and substitution reactions.
Think of water H
E.g. Write down the structural formula of the Major product using structural formulae.
1
2
3
CH3CH=CH2 +
H2O
CH3
Approach:
Think of H2O as H
OH
HO
H
H
Answer:
H
C
comes here
C
H
H
C
H
H
C
H
H
Look at the Carbon atoms at the double bond: Carbon 2 has ONE H; carbon 3 has 2.
Therefore, there is NO symmetry across double bond H will add where there are
many Hs, at Carbon 3 and OH- adds on Carbon 2 on to form the major product
H
Answer:
H
H
O
H
C
C
C
H
H
H
H
C
H
H
For substitution the negative part (written second in formula undergoes substitution
e.g. when using dilute KOH it is the OH- that substitutes, not the K+
ExERcISE 1: ORGaNIc REacTIONS
119
The flow diagram below shows how 2-bromopropane can be used to prepare other
organic compounds.
Alkene
a
(Major Product)
B
2-bromobutane
c
Alcohol
d
1.1 Classify 2-bromobutane as a PRIMARY, SECONDARY or TERTIARY
haloalkane.
(1)
1.2 Write down the STRUCTURAL FORMULA of a positional isomer of
2-bromobutane.
(1)
1.3 Write down the type of reaction represented by:
1.3.1 a
1.3.2 c
1.3.3 d
(1x3)
1.4 Write down the:
1.4.1 NAME or FORMULA of the inorganic reagent in
a. reaction a
b. reaction c
c. reaction d
(1x 3)
1.5 Write down the NAME or FORMULA of the inorganic product(s) in
a. reaction a, if KOH is used as a reagent
(1)
b. reaction c, if KOH is used as a reagent
(1)
c. reaction d
(1)
1.6 Write down the FORMULA of the catalyst used in reaction B.
(1)
ExERcISE 2: ORGaNIc REacTIONS
Consider the following compounds: a: CH4 B: Ethanol C: Propanoic acid
2.1 Compound a reacts with unknown compound x in the presence of sunlight to
produce bromomethane.
Write down a balanced equation for the reaction between:
120
2.1.1 compound a and x
(3)
2.1.2 compound a with excess oxygen
(3)
2.2 Compound B and c react in the presence of an inorganic acid.
Write down the:
2.2.1 name of the inorganic acid
(1)
2.2.2 the equation for the reaction between B and c, using structural
formulae for the organic reagents
(1)
2.2.3 reaction condition
(1)
ExERcISE 3: ORGaNIc REacTIONS
3.1 A mixture of 4 grams of ethanol (C2H5OH) and 25 dm3 of air is ignited at
25 ºC and standard pressure. A reaction occurs according to
the balanced equation:
C2H5OH(ℓ) + 3O2(g) → 2 CO2(g) + 3 H2O(g)
3.1.1Determine, with the aid of a calculation, whether all of Compound d will be
used up at the end of the reaction, if the molar volume of a gas at 25 ºC
and standard pressure is 24,47 dm3.
(ASSUME THAT AIR CONTAINS 21% OXYGEN (O2).)
3.1.2 If the ethanol is not used up, calculate the mass in grams of ethanol that is not
used. (3)
(5)
SOLuTIONS
ExERcISE 1: ORGaNIc REacTIONS
1.1 Secondary
H
Carbon 2
H
H
Br
H
c
c
c
c
H
Look at carbon 2 (middle one)
bonded to ONE H; therefore
H
H
H
H
SEcONdaRy
 If there were TwO Hs on the carbon bonded to Br (carbon 2)
then PRImaRy.
 If there were NO Hs on carbon 2, then TERTIaRy.
1.2 Change position of Br. (Change position of functional group, halogen or side
chain to make a positional isomer.)
121
H
H
H
H
Br
c
c
c
c
H
H
H
H
H
1.3.1 A Elimination OR Dehydrohalogenation
1.3.2 c Substitution or Hydrolysis
1.3.3 d Substitution
1.4.1 a. concentrated KOH
b. KOH or H2O
c. HBr
1.5 a. NaBr + H2O
b. KBr
c. H2O
B H2SO4
1.6
ExERcISE 2: ORGaNIc REacTIONS
H
2.1.1
H
C
H
H
H
+ Br2
C
Br
+ HBr
O
H
H
C
C
C
H
H
O
H
H
C
C
H
H
H
H
H
2.1.2 CH4 + 2 O2
2CO2
+ H2O
2.2.1 Sulphuric acid
2.2.2
H
H
H
C
C
H
H
H
122
O
H
+
H
H
C
C
H
H
H
O
O
C
H
+ H2O
2.2.3 Heat mildly over a water bath.
ExERcISE 3: ORGaNIc REacTIONS
3.1 n ethanol = m/M = 4/46 = 0.087 mol
Vo2 in 25 dm3 of air = 21/100 x 25 = 5.25 dm3
n O2 = V/Vm = 5.25/24.47 = 0,215 mol
n ethanol = m/M = 4 /46 = 0.087
Ratio from equation Ethanol: Oxygen
1: 3
0.087 mol ethanol too much for 0.215 oxygen (Ratio 0.087:0.215 = 1:2,5
required ratio 1:3)
O2 is the limiting reagent; Ethanol is in excess
3.2
n ethanol reacting = 1/3 x (0.215) = 0.072 mol
n ethanol in Excess = n initial – n final = 0.087 - .072 = 0.015 mol
m ethanol = nM= 0.015 x 46 = 0.69 g
123
14. ELEcTROcHEmISTRy
ELEcTROcHEmISTRy
E
ELECTROLYTIC CELL
GALVANIC CELL
ELEcTROLyTIc cELLS (There are 5 electrolytic cells that you must understand.)



124
Convert electrical energy to chemical energy.
(Note there is NO mention of mEcHaNIcaL energy in chemistry.)
Reactions at electrodes are ENdOTHERmIc (reactions not self –
sustaining.)
In the diagram for electrolytic cell, there will be ONE CONTAINER and
a power source or battery.
ELECTROLYSIS OF
EELECTROLYSIS OF
saturated NaCl (aq)
concentrated CuCl2(aq) Cℓ ℓ ℓ
EXTRACTION OF ALUMINIUM
Need for cryolite
ELECTROLYTIC CELL
(To lower melting poit saving on
operational costs)
Reactions at electrodes
At CATHODE:
Al3+ + 3e
ELECTROPLATING
Al
At ANODE: C + O2
CO2
The substance that is plated forms
the CATHODE.
The electrolyte must have ions of
the metal that are plated.
ELECTROREFINING of Copper
(Or Purification)
The substance that is refined forms
the ANODE.
The electrolyte must contain ions of
the metal that is refined.
ELEcTROLySIS
ELEcTROLySIS OF cucl2 and Nacl solutions
“Second part of formula” goes to anode for oxidation.
“First part” may go to cathode for reduction if it is a STRONGER OXIDISING agent.
Positive ion “First part of formula” may NOT be reduced at ALL but H2O will undergo
reduction
125
if Positive ion (Metal ion) “First part of formula” is a weaker oxidising agent than H2O
e.g. in NaCl electrolysis Na+ dOES NOT undergo REducTION
Take CuCl2 (Cu2+
Cℓ- Cℓ- ). The chloride ion Cl- “Second part” always migrates
towards +ve ANODE (Positive attracted to negative) to undergo OXIDATION (Cℓ - is
the reducing agent)
Compare oxidising ability of Cu2+ the “first part” of the formula with that of water
H2O in order to decide which species will undergo reduction (H2O or Cu2+).
Cu2+ is a stronger oxidising agent than H2O therefore Cu2+ will undergo reduction
Consider electrolysis of NaCl (aq)
5.1 Write down the following:
5.1.1 oxidation half reaction
5.1.2 formula of the reducing agent
5.1.3 reduction half reactio5.1.4 formula of the oxidising agent
approach:
5.1.1 Negative part of formula (second part) Cl- goes to anode for oxidation
Therefore, Oxidation half reaction 2 Clreversed)
Cl2 + 2 e (From table and
5.1.2 The reducing agent is found in the oxidation half reaction on left of arrow)
2 Cl-
Cl2 + 2 e
Therefore Cl-
5.1.3 Compare the oxidising ability of Na+ with that of water H2O.
H2O is a stronger oxidising agent than Na+, therefore H2O undergoes
reduction.
Therefore: H2O(l) + 2e
H2(g) + OH-
5.1.4 The oxidising agent is at the left in the reduction half reaction
H2O(l) + 2e
H2(g) + OH-
Therefore: H2O is the oxidising agent.
126
TIPS:











You are NOT permitted to put double arrows in equations - always use ONE
arrow.
Reducing agent found on the left of arrow in OXIDATION half reaction
The oxidising agent is ALWAYS on the left of the REDUCTION half reaction.
There will be NO REACTION involving Na+ but H2O in electrolytic cells
Know the properties of ANODE then know those of CATHODE as the
opposite of anode
Reverse the OXIDATION half reaction when taking it from the table
Oxidising agents are bbon the FAR LEFT on TABLE of REDUCTION
POTENTIALS
Reducing agents immediately on the right of arrow when reading from left to
right
Smaller EO value for ANODE
Write the formula as it is given DATA SHEET e.g. Eocell = Eocathode - Eoanode
(No abbreviations are acceptable, e.g. E = Eocat - Eoan - NO marks for this.)
To get marks for the formula, you must substitute values, even if they are
INCORRECT.
In Chemistry, there is no MECHANICAL energy - only CHEMICAL energy
QuESTION 1
Consider the electrolytic cells a and B given below.
X, Y, W and Z are carbon electrodes.
w
X
X
z
Y
Y
CuCl2(aq)
NaCl(aq)
cELL a
cELL B
1.1 Write down the:
a. energy conversion in these cells
b. the reason why a DC power source is used in place of an AC power source
127
1.2 Write down THREE observable changes in cell B.
1.3 Which electrode is the ANODE in:
1.3.1 Cell a
1.3.2 Cell B
1.4 Write down the half reaction taking place at electrode:
1.4.1 x and w
1.4.2 y
1.4.3 z
1.5 Write down the FORMULA and NAME of the reducing agent in cell:
1.5.1 a
1.5.2 B
1.6 Write down the FORMULA and NAME of the oxidising agent in cell:
1.6.1 a
1.6.2 B
1.7 Write down the name of the gas formed at:
1.7.1 electrode x
1.7.2 electrode y
1.8 In cell a, the Na+ ions do not form sodium Na. Explain be referring to the strength
of oxidising agents involved.
QuESTION 2
A chemist wants to determine the PERCENTAGE of copper by mass in an IMPURE
copper sample containing platinum and silver as the only impurities.
He first sets up an electrolytic cell to purify copper.
Power
Source
ImPuRE
cOPPER
P
Q
P
cuSO4(aq)
2.1 Is electrode P the POSITIVE or NEGATIVE electrode?
(1)
2.2 Which electrode (P or Q) will show an increase in mass?
(1)
2.3 Write down the following:
128
PuRE cOPPER
2.3.1 The SYMBOL or NAME of the product formed at electrode P.
(1)
2.3.2 The equation for the half reaction taking place at electrode Q.
(2)
2.4 Use relative strengths of reducing agents involved to explain why the metals
platinum and silver do not form ions during this process.
(2)
[12]
QuESTION 3
The electrolytic cell below is used for electro refining (purification) an IMPURE
copper sample.
Electrode Q contains copper, which contains 3% IMPURITIES by mass.
Power
Source
Pure Copper
IMPURE Copper
P
Q
CuCℓ2(aq)
3.1 Write down the observable changes at the:
3.1.1 Anode
3.1.2 Cathode
3.2 Write down the half reaction taking place at the:
3.2.1 Anode
3.2.2 Cathode
3.3 CuCl2 solution is light green.
How will the intensity of the colour of the solution change as the cell is working?
Give a reason.
3.4 Calculate the mass of the IMPURE copper sample if 2 moles of electrons
are transferred when ALL the copper is extracted from the IMPURE sample.
129
3.5 The IMPURE copper sample contains metals like silver. Explain why silver
does undergo oxidation in this cell.
3.6 What happens to the electrolyte when the cell is in operation? Explain.
GaLVaNIc cELLS
Questions on galvanic cells may include:



dIaGRam OF a GaLVaNIc cELL
REacTION EQuaTIONS (increase in oxidation number for aNOdE)
cELL NOTaTION (anode is on the left)
KNOw THE PROPERTIES OF aN aNOdE.
aNOdE
OXIDATION OCCURS (ELECTRONS MOVE AWAY FROM THE ANODE)
SMALLER EO-values
NEGATIVE ELECTRODE (CONNECTED TO NEGATIVE TERMINAL OF
VOLTMETER)
DECREASE IN MASS
[POSITIVE IONS] INCREASE
ANIONS MOVE TOWARDS ANODE
OXIDATION NUMBER INCREASES
(Based on the notes above, you can deduce what happens at the
caTHOdE.
No need to memorise both.)
cOmmON mIScONcEPTION
The CATHODE DOES NOT undergo REDUCTION.
130
What undergoes reduction (i.e. oxidising agent) is an ION in solution
at the CATHODE half cell.
QuESTION 4
STRucTuREd QuESTIONS
The diagram below shows a galvanic cell operating under standard conditions. The
cell reaction taking place when the cell is functioning is:
6Cℓ - (aq) + 2Au3+(aq) → 3Cℓ2(g) + 2Au(s)
With switch S OPEN, the initial reading on the voltmeter is 0,14 V.
Write down the following:
4.1 The NAME or FORMULA of the oxidising agent. (1)
4.2 The half-reaction that takes place at the anode. (2)
4.3 The cell notation for this cell. (3)
4.4 Calculate the standard reduction potential of Au.
(4)
4.5 The concentration of the Au3+ is increased. How will this change affect the initial
EMF of the cell. Write down INCREASES, DECREASES or REMAINS THE
SAME.
(1)
Switch S is now closed and the bulb lights up.
131
4.5 How will the reading on the voltmeter now compare to the INITIAL reading
of 0,14 V? Write down only LARGER THAN, SMALLER THAN or EQUAL TO.
Give a reason for your answer.
QuESTION 5
A learner sets up a galvanic cell as shown below. The cell functions under
standard conditions. Nitrates are used as electrolytes.
5.1 Define the following terms:
5.1.1 Electrolyte
(2)
5.1.2 Galvanic cell
(2)
5.2 Write down the initial concentration of the Copper (II) nitrate (Cu (NO3)2
Solution.
(1)
5.3 Calculate the mass of Copper (II) nitrate needed to prepare the solution if the
volume of the solution is 250 cm3.
(4)
5.4 In which direction (a or B) will ANIONS move in the salt bridge?
(1)
5.5 Calculate the emf of the above cell under standard conditions.
(4)
5.6 Write down the balanced equation for the net cell reaction that takes place in
this cell.
132
(3)
5.7 Write down the cell notation for this cell.
(3)
QuESTION 6
A standard electrochemical cell is set up using a standard hydrogen half-cell and
a standard X|X2+ half-cell, as shown below. The standard reduction potential of the
X|X2+ half-cell is - 0.74 V.
6.1 Write down the following:
6.1.1 The function of component Q.
6.1.2 The initial reading on the voltmeter.
(1)
(1)
6.1.3 The half-reaction that takes place at the cathode of this cell.
(2)
6. 2 The hydrogen half-cell is now replaced by a m|m2+ half-cell. The cell notation of
this cell is: X(s)| X2+(aq) || m2+(aq) | m(s). The initial reading on the voltmeter is
now 0.61 V.
6.2.1 Identify metal m. Show how you arrived at the answer.
(5)
6.2.2 Is the cell reaction EXOTHERMIC or ENDOTHERMIC?
(1)
6.3 The reading on the voltmeter becomes zero after using this cell for several
hours. Give a reason for this reading by referring to the cell reaction.
(1)
133
QuESTION 7
Consider the following reaction:
Ce3+ (aq) + Ti(s)
Ti3+ (aq) +
Ce(s)
Standard electrode potentials are given in the table below.
Redox pair
Standard Electrode Potential
Ti3+/Ti
-1.63 V
Ce3+/ Ce
-2.48 V
7.1 Determine by calculation if the reaction is spontaneous.
7.2 According to the data in the given table, which substance is the strongest
reducing agent?
SOLuTIONS
ELEcTROcHEmISTRy: GaLVaNIc aNd ELEcTROLyTIc cELLS
QuESTION 1
1.1 a. Chemical energy to electrical energy.
b. To ensure the polarities of the anode and cathode do not change.
1.2 Bubbles of chlorine gas at Z (Anode)
Brown coating on Y (Copper)
Intensity of blue colour decreases. (The concentration of Cu2+ decreases as Cu2+
reduced to Cu)
1.3.1 x
1.4.1 2Cl-
134
Cl2 + 2e
1.3.2
z
1.4.2
2 H2O(l) + 2e
1.5.1 Cl
1.5.2 Cl-
1.6.1 H2O
1.6.2 Cu2+
1.7.1 Chlorine
1.7.2 hydrogen
H2O + OH-
1.8 H2O is a stronger oxidising agent than Na+. Therefore, H2O is reduced.
QuESTION 2
2.1 POSITIVE
2.2 Q
2.3.1 Cu2+ or Copper (II) ions
2.3.2 Cu2+ + 2e
(From oxidation Cu
Cu2+ + 2e )
Cu
2.4 Cu is a stronger reducing agent than both silver and platinum.
QuESTION 3
3.1.1 Mass decreases (ANODE)
3.2.1 Cu
Cu2+ + 2e
3.3 nCu = ½ x 2 = 1 mol
3.1.2 Mass increases (CATHODE)
3.2.2 Cu2+ + 2e
Cu
(Ratio of mol Cu : mol electrons 1:2)
M pure Cu = n .M = 1 x 63.5 = 63.5 g
Therefore: 97/100 x IMPURE Copper = Pure Copper (3% impurities; therefore
100% - 3% = 97% is pure copper)
3.4 Cu is a stronger reducing agent than Ag.
QuESTION 4
4.1 Au3+
(From oxidation numbers Oxidation number of Au DECREASES from +3 to 0)
4.2 2 Cl-
Cl2
+ 2e
4.3
Pt Cl- Cl2
Au3+ Au
4.4 INCREASE
4.5 SMALLER THAN
Lost volts due to internal resistance (when current flows through the cell).
QuESTION 5
5.1.1 Electrolyte – a solution or liquid that conducts electricity because it contains
ions.
135
5.1.2 Galvanic cell: converts chemical energy to electrical energy.
5.2
c = 1 mol.dm-3 (standard conditions)
5.3
mCu = ?
nCu = c.V = 1 x 250/1000 = 0,25 mol
mCu = n.M= 0.25 x 63.5 = 15.88 g
(15.875 g)
5.4 a
5.5 Eocell = Eocathode - Eo anode
= 0.80
(Formula)
– 0.34
(Substitution)
= 0.46 V
(Answer with UNIT)
Cu2+ + 2e
5.6 Cu
Ag+ + e
electrons.)
Ag
2 Ag+ + 2 e
5.7
(Multiply by 2 to balance the number of
2 Ag
Cu/Cu2+// Ag+/Ag
QuESTION 6
6.1.1
Completes the circuit/ensures electrical neutrality of the solutions.
6.1.2
0,74 V ( an use Eocell = Eocathode - Eo anode = 0 – (- 0.74) = 0.74 V
Remember: Eocell cannot be negative
6.1.3
2 H+ + 2e
H2(g)
6.2.1 m is the cathode (M is on the right-hand side in cell notation.)
Eocell = Eocathode - Eo anode
0.61
= Eocathode - (- 0.74)
0.61
= Eocathode + 0.74
0.61 – 0.74 = Eocathode
- 0.31 V = Eocathode m is Pb
136
QuESTION 7
7.1 According to the equation, Ce is the CATHODE
(Reason: Oxidation number decreases from +3 to 0; OR
Ti is the ANODE - Oxidation number increases from 0 to +3.)
Approach:
Calculate EMF. If EmF is POSITIVE it means Reaction is SPONTaNEOuS
Eocell = Eocathode - Eo anode
Eocell = -2.48 – (- 1.63)
= - 0.85 V
Reaction NOT SPONTANEOUS, Eocell is NEGATIVE
7.2 Ce
INdIREcT TRaNSFER OF ELEcTRONS
e. g. A zinc rod is placed in a copper(II) sulphate solution, as shown below.
Zn
CuSO4(aq)
ExamPLE 1
Write down THREE observable changes that will occur.
1.1 Write down the:
1.1.1 oxidation half reaction
1.1.2 reduction half reaction
1.1.3 name of the oxidising agent
1.1.4 name of the reducing agent
1.1 Oxidation half:
Zn
Zn2+ + 2e
137
1.2 Reduction half reaction: Cu2+ + 2e
Cu
1.3 Oxidising agent (on the left of the Reduction half reaction)
Cu2+ + 2e
Cu
Oxidising agent: Cu2+
1.4 Reducing agent (on the left of the Oxidation half reaction)
Zn
Zn2+ + 2e
Zn is the Reducing agent
NOTE: If the reducing ability of the METAL is lower than that of the metal of the
positive ion in solution, no reaction will occur.
EXAMPLE 2: A piece of silver is placed in magnesium sulphate, e.g.:
Ag
MgSO4
2.1 Will a spontaneous reaction occur?
2.2 Can we store a magnesium sulphate solution in a container made of
silver, without the container leaking? Explain.
138
Approach: Consult the Table of Reduction potentials.
Extract form Table 4B
OXIDISING AGENTS
+ e ⇌
Li
Li+
………………………
………………………
Ba
Ba2+ + 2e ⇌
………………………
……………………….
Mg2+ + 2e ⇌
Mg
………………………
………………………
………………………
Cu2+ + 2e ⇌
Cu
…………………….
…………………….
+ e ⇌
Ag
Ag+
……………………..
Cl2 (g) + 2e ⇌
2ClF2(g) + 2e ⇌
Li - strongest reducing agent
REDUCING AGENTS
2F-
F2 - strongest oxidising agent
RULE: Top right reduces bottom left.
Consider example 1 above:
Zn + CuSO4
Find Zn and Cu2+ in the table.
Remember Zn is the reducing agent in Example 1.
Zn2+ +
2e ⇌
Cu2+ + 2e ⇌
Zn
Zn is a stronger reducing agent than Cu.
Cu
Zn reduces Cu2+ to Cu.
139
Example 2:
If a reaction occurs, Ag will be the reducing agent. (It is the METAL.)
Mg2+
Ag+
+
+
2e ⇌
Mg
e ⇌
Ag
THIS CAN NOT TAKE PLACE SPONTANEOUSLY.
In Example 2:
2.1 There will be NO reaction.
Reason: Ag is a weaker reducing agent than Mg.
Ag cannot reduce Mg2+ to Mg
2.2 To be able to store a solution in a container, the SOLUTION AND THE METAL
USED IN THE CONTAINER MUST NOT REACT.
Yes, we can store MgSO4 in Ag container.
Reason:
Reaction will not occur spontaneously, as Ag is a weaker reducing agent than
Mg. Ag cannot reduce Mg2+ to Mg
The table of reduction potentials only gives values of redox reactions under
standard conditions. This means that if the experiment is carried out when the
conditions are not standard, we cannot rely on the usual equation to give us the.
We will need a different type of equation called the Nernst Equation. The Nernst
equation however is not prescribed in our current curriculum. Our calculations
should therefore confine themselves to Standard Conditions only.
140
PRACTICE QUESTIONS
QuESTION 1
An aluminium rod is placed in a copper (II) sulphate solution as shown below.
Al
CuSO4 (aq)
ExamPLE 1
1.1 Write down THREE observable changes that will occur.
1.2 Write down the following:
1.2.1 oxidation half reaction
1.2.2 reduction half reaction
1.2.3 name of the oxidising agent
1.2.4 name of the reducing agent
ExamPLE 2
2.1 Can we store:
2.1.1 CuSO4 in a container made of silver metal? Explain.
2.1.2 MgSO4 in a container made of zinc? Explain.
2.1.3 AgNO3 in a container made of copper?
.
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