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The Trajectory of a Simple Projectile is a Parabola
For
Prof. Osama Shalabeia
Associate Prof. Ali Ibrahim
2014 -2015
1
Preface
General Physics is written for a one-year course in introductory physics usually
taken by students majoring in physics, chemistry, medical students and
technical fields. The mathematical techniques used in this book include algebra,
geometry, and trigonometry, but not calculus. We have refined the text to better
meet the needs of students and teachers. This textbook, (Chapters 1–7) deal
with Newtonian mechanics.
The main objectives of this introductory textbook are twofold: to provide the
student with a clear and logical presentation of the basic concepts and
principles of physics, and to strengthen an understanding of the concepts and
principles through a broad range of interesting applications to the real world.
To meet those objectives, we have emphasized sound physical arguments and
problem-solving methodology. At the same time we have attempted to
motivate the student through practical examples that demonstrate the role of
physics in other disciplines.
2
Contents
Introduction
7
Chapter 1
Units and Dimensions
1-1
System of Units
11
1-2
Mass Units
13
1-3
Length Units
13
1-4
Time Units
13
1-5
Consistency of Units
14
1-6
Units Conversion
15
2
Dimensions
17
2-1
Dimensional Analysis
17
3-
Exercises
22
Chapter 2
Vectors
2-1
Vectors Properties
29
2-1-1 Adding and Subtracting Vectors Graphically
25
2-1-2 Unit Vectors
27
2-1-3 How to describe a two-dimension vector?
27
2-1-4 Vector Components
27
2-1-5 Properties of Vector Components
28
3
2-1-6 Addition &Subtraction of Vectors by means of Components 28
2-2
Multiplying and Dividing a Vector by a Scalar
31
2-3
The Scalar Product of Vectors (dot product )
32
2-4
The Vector Product (cross product)
33
2-5
Exercises
34
Chapter 3
Motion in One and Two Dimensions
3-1
Motion in One Dimension
45
3-1-1 Displacement
45
3-1-2 Average Velocity
45
3-1-3 Instantaneous Velocity
46
3-1-4 Acceleration
47
3-2
50
One Dimensional Motion with Constant Acceleration
3.2.1
Derivation of Kinematics Equations of Motion
3.2.2
Vertically Thrown Up and Freely Falling Bodies
3.3
Motion in Two Dimensions
50
Motion
56
61
3.3.1 Displacement, Velocity and Acceleration in 2-Dimensions
4
61
3.4
Projectile Motion
65
Horizontal Range and Maximum Height of a Projectile
68
Chapter 4
Newton's Laws of Motion
4.1
Force and Fundamental Forces of Nature
83
4.1.1 Force
83
4.1.2 The fundamental Forces of Nature
87
4.2
Newton's First Law
88
4.3
Newton's Second Law
88
4.4
Newton's Third Law
90
4.5
Frictions
91
4.6
Experimental Facts about Friction
92
4.7
Applications of Newton's Laws
93
4.8
Problem Solving Strategy
93
4.9
Solved Problems
94
4.10 Exercises
108
Chapter 5
Work, Energy and Power
5-1
Work
113
5-2
Kinetic Energy and the Work Energy Theorem
115
5
5-3
Gravitational Potential Energy
118
5-4
Potential Energy Stored in a Spring
118
5-5
Conservation Laws
120
5-5-1 Conservative and Non-Conservative Forces
120
5-5-2 The Conservation of Mechanical Energy
121
5-5-3 Non-Conservative Forces and the Work-Energy Theorem
121
5-6
Power
125
5-7
Solved Examples
125
5-8
Exercises
130
Chapter 6
Linear Momentum, Impulse and Collisions
6-1
Linear Momentum and Impulse
134
6-2
Conservation of Linear Momentum for Two particle system
135
6-3
Collisions and Kinetic Energy
138
6-3-1 Types of Collisions
138
6-3-2 Head on Collisions and Glancing Collisions
140
6.4
Solved Examples
142
6-5
The Center of mass
153
6-6
Exercises
156
6
Chapter 7
Rotation of Rigid Bodies
7-1
Introduction
160
7-2 Angular Velocity and Acceleration
161
7-3 Formulae for Constant Angular Acceleration
162
7-4 Relationship between Linear and Angular Quantities
162
7-5 Centripetal Acceleration
164
7-6 Solved Problems
165
7-7
170
Exercises
7
Introduction
Why PHYSICS is important?
It is clearly important that students need to know why Physics is
important and what careers or other benefits may stem from studying
physics. Without an understanding of the basic physics of the problem,
we would be setting in the dark, so why is Physics important? This is
because none of these today’s amazing technologies (electricity,
computers, X-ray, mobile, internet, artificial satellites, laser,..etc) would
be possible without Physics.
A firm grasp of Physics is needed to be able to even begin thinking
about producing such technologies. Engineers of today design buildings,
bridges, computers, simple robots, aircraft and almost all of the objects
that surround us. While these kinds of engineers will never go away, we
will need to add new fields of engineering to accommodate the
production of advanced technologies. None of these will be possible
unless interested and capable minds pursue an education in physics.
What is Physics?
Indeed, many students find it difficult to even define physics.
Physics is one of the fundamental (or Natural) sciences (Mathematics,
Physics, and chemistry).
It is interested in understanding natural phenomena that occur in
our universe and find out the fundamental laws of such natural
phenomena and their applications. It is often considered the most
fundamental of all the natural sciences and its theories attempt to
describe the behavior of the smallest building blocks of matter(molecules,
atoms, nuclei, light)up to the Universe and everything in between.
8
Physical quantities
A physical quantity is a physical property of a phenomenon, body, or
substances that can be quantified by measuring Physical quantities are divided
into two types:
i.
ii.
Fundamental quantities, such as Length, Mass and Time.
Derived quantities, from fundamental quantities either by multiplication
or division such as velocity, volume, area, .etc.
For any physical quantity, it must have units and dimensions. It must also be
defined either in scalar units (magnitude only) or vector units (magnitude and
direction).The following schematic figure shows the basic properties of any
physical quantity.
Units
Dimensions
Physical Quantities
PhysicalQuantities
Vectors
Scalars
Figure: 1 the properties of physical quantities
The following are some examples which state some of natural phenomena that
Physics is interested in and applied to our daily life:
1234-
Medical- Geo- and Biophysics
Laser physics and its applications
The motion of Earth around its axis ( i.e., day and night)
Revolution of the moon around the Earth
9
56789-
Projectile motion and rockets
Artificial Satellites and communication systems,
Lighting and sparing
Solar Eclipses
Electromagnetism
10
Chapter 1
UNITS AND DIMENSIONS
1-1
System of Units
1-2
Mass Units
1-3
Length Units
1-4
Time Units
1-5
Consistency of Units
1-6
Units Conversion
2
Dimensions
2-1
Dimensional Analysis
2-2
Solved Problems
3
Exercises
11
UNITS AND DIMENSIONS
Objectives
To know the difference between units and dimensions
To be able to convert units
To understand the SI (System Internationale) of units
To know the SI prefixes from nano- to gigaTo understand and apply the concept of dimensional analysis
1-1 System of Units
There are three different systems of units that are most commonly used : Meter- Kilogram- Second or (MKS) or System International (SI)
units.
 The laboratory system Centimeter-Gram- Second or (CGS) system.
 The Foot- Pound- Second or (FPS) British system.
The system of SI is the most commonly used all over the world. This systemis
based on powers of 10. In order to change from one unit to another in the
metric system it is a matter of multiplying or dividing by powers of 10. We can
also, specify various units by the use of prefixes. These prefixes and their
meanings must be memorized. Table 1 shows the SI prefixes:-
12
Table 1: SI prefixes
Factor Name Symbol Factor Name Symbol
1024
Yotta Y
10-1
Deci
D
1021
Zeta
Z
10-2
Centi C
1018
Exa
E
10-3
Milli
1015
Peta
P
10-6
Micro µ
1012
Tera
T
10-9
Nano N
109
Giga
G
10-12
Pico
106
Mega M
10-15
Femto F
103
Kilo
K
10-18
Atto
102
Hector H
10-21
Zepto Z
101
Deka Da
10-24
Yocto Y
M
P
A
Consequently, once you know these prefixes, you can convert from one unit to
another.
1-2 Mass units
It is important to understand the concept of mass. Mass is a measure of the
amount of matter that an object contains. This is not to be confused with
weight, which is a measure of the gravitational pull of the earth on a mass.
There is a related system of units called SI. In SI, the unit of mass is the
Kilogram (103 grams), where the kilogram is defined as the mass of a particular
cylinder of platinum-iridium alloy that is kept under controlled atmospheric
conditions at the International Agency of Weights and Measure in France.
13
1-3 Length units
In the SI system, the unit of length is the meter (m). The definition of length was
originally the one ten-thousandth of the distance from the North Pole to the
Equator through Paris, France. It is currently defined as a given number of
wavelengths of the emitted light from a Cesium atom. This is a more accurate
and reproducible way of defining a unit of length.
1-4 Time units
In the system of SI, the unit of time is the second. The second is the duration of
9 192 631 770 periods of the radiation corresponding to the transition between
the two hyperfine levels of the ground state of the 133Cs atom.
Ampere, unit of electric current
The ampere is that constant current which, if maintained in two straight and
parallel conductors of infinite lengths, of negligible circular cross- section, and
placed 1 meter apart in vacuum, would produce between these conductors a
force equal to 2x10-7Newton per unit length.
Kelvin, unit of thermodynamic temperature
The Kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the
thermodynamic temperature of the triple point of water.
Mole, unit of amount of substance
The mole is the amount of substance of a system that contains as many
elementary entities as there are atoms in 0.012 kilograms of C12. When the mole
is used, the elementary entities must be specified and may be atoms, molecules,
ions, electrons, other particles, or specified groups of such particles.
Candela, unit of luminous intensity
14
The candela is the luminous intensity of a source that emits monochromatic
radiation of frequency 540 x 1012 Hertz and that has a radiant intensity of 1/1687
Watt per steradian.
1-5 Consistency of Units
Principle of consistency of units:
 units on the left side of an equation must be the same as those on
the right side of an equation
 dimensional homogeneity
When using units in Mathematics operations:
 Adding numbers with units, they must have the same units—
also the same for subtraction
 Multiply or dividing numbers with units, also the units
multiply and divide
 Never add or subtract terms with different dimensions
For examples
1- 12 m (length) x 12 m (length) = 144 m2 (area)
2- 3.0 m(Length) / 3 s (time) = 1.0 m/s (velocity)
3- 0.1 g (mass) / 10 s (time) = 0.001 g / s (flow rate)
OK
OK
OK
4- 10 Kg (mass) + 10 m (length)
Wrong
1-6 Units Conversion
We often need to change the units in which a physical quantity is
expressed. We do so by a method called Chain- Link Method of conversion. In
this method, we multiply the original measurement by a conversion factor (a
ratio of units that is equal to one). This method can be used to convert from a
unit to another.
For example, because 1 min and 60 seconds is the same time intervals, this
is not the same as writing 1/60 min. = 1 s or 60=1; the number and its unit must
15
be treated together. Another example 1 meter and 100 centimeter are the same
length but we cannot write 1/100 =1 or 100=1 as just numbers, again dealing
with physical quantities the number and its unit must be treated together.
Because multiplying any quantity by unity (by one) leaves it unchanged, we
can use such conversion factors, wherever we find them useful. In chain-link
conversion method, we use the factors in such a way that the unwanted units
cancel out.
Example 1-1
How many seconds are in 2 minutes?
Solution:
2 min  2 min .
60s
 120s
1 min
Example 1-2
Which of the following is correct?
a. There are exactly 100 m/cm.
b. There are exactly 100 cm/m.
c. Neither a not b.
(Do it yourself)
Example 1-3
How many seconds are in 2.0 years?
Solution:
16
Example 1-4
The density of water is 1g/cm3. What is the density of water in MKS system?
Solution
We know that
1 g = 10-3 Kg
and
1 cm3 = 10-6 m3
 103 Kg / g 
  1g / cm x  6 3
 103 Kg / m 3
3 
 10 m / cm 
3
Homework
i.
ii.
iii.
iv.
v.
How many millimeters are in 20.0 meters?
How many square centimeters are in an area 6 km2?
The volume of a box is 6.30 m3. This is equivalent to how many cubic
nanometers?
120 Km/h =……………….m/s.
A car moves with a constant speed of 90 km/hr. Find this speed in terms of m/s,
km/min, and mi/hr (Hint: 1 km = 3281 ft , while 1 mile = 1. 609 km).
2- Dimensions
The term Dimensional Analysis makes use of the idea that dimensions may
be treated as algebraic quantities. It helps us to check the relationship of
various quantities either correct or non-correct also, and to: i. Verify Physical Equations
ii. Deduce Physical Equations
iii. Deduce the dimensions of Physical Constants
17
2-1 Dimensional Analysis
Dimensions and units can be treated algebraically
Variable from
Eq.
x
M
t
L
M
T
Dimension[ ]
Length
Mass
V=(xf-xi)/t
a=(vf-vi)/t
L/T
L/T2
Time
Base Units of the International System (SI)
Fundamental Dimension
Base Unit
length [L]
meter (m)
mass [M]
kilogram (kg)
time [T]
second (s)
electric current [A]
ampere (A)
absolute temperature [Ө]
kelvin (K)
Supplementary SI Dimensions
Supplementary Dimension
Base Unit
Plane angle
Plane angle
Solid angle
Solid angle
Example 1-5
Check dimensionally (correct or not) the following equation:
18
t  2
l
g
Where t is the periodic time of a simple pendulum of length l and g is the
acceleration due to gravity.
Solution:
The dimensions of the left hand side = [L H S] = [ t ] = [ T ] ………………...(1)
The dimensions of the right hand side = [R H S] = [ l/g ]1/2 = [ L/LT-2 ] = [ T ]…...
(2)
From 1 and 2,
The equation is dimensionally correct
Example 1-6
Do it yourself
Checking equations with dimensional analysis:
(x is the distance, v velocity, a acceleration, and t time)
• Each term must have the same dimension
• Two variables cannot be added if dimensions are different
• Multiplying variables is always fine
• Numbers (e.g. 1/2 or π) are dimensionless
Example 1-7
A body exerts a force of magnitude 8 x 107 dyne to push a body along a flat
surface with a constant speed. Express this force in Newton. ( Hint: 1 dyne = 1
g. cm/ s2).
Solution
19
F = 8 x 107 dyne = 8 x 107 g. cm/s2
But we know that:1 g = 10 -3 Kg and 1 cm = 10-2 m

 
 

F  8x 107 g .cm / s 2 x 103 Kg / g x 102 m / cm  800Kg .m / s 2 (N )
Example 1-8
Consider the following equation:
v2
mM
m
G 2
r
r
Where m and M are masses, r is a radius and v is a velocity. What are the
dimensions of G?
Solution
From the above equation:- then G= v2 r/ M
Then the Si units of G is (m/s)2 m/Kg, so it will be m3/ Kg. s2
As a result of this the dimensions of [G] = [L3. M-1. T-2]
Example 1-9
If“x” has dimensions of distance, “v" has dimensions of velocity, “m” has
dimensions of mass and “g” has dimensions of acceleration. Is the following
equation dimensionally valid?
x 
 4 3 vt

1  2 gt 2 / x
Solution
20

The equation will be valid if and only if the dimensions of the L.H.S are equal
to the dimensions of R. H.S.
Then [L. H.S]= [x] = [L]
While [R. H.S] = [
(1)
 4 3 vt

1  2 gt / x
2

] = [LT-1. T+1]/[L. T-2. T+2. L-1]= [L]
(2)
From 1 and 2 , then [L.H.S] = [R. H.S] . So the equation is VALID
Note
•
•
•
•
•
Dimensions: L, T, M, L/T …
Units: m, mm, cm, kg, g, mg, s, hr, years …
When the equation is all algebra: check the dimensions
When the numbers are inserted: check the units
Units obey the same rules as dimensions: Never add terms with different
units
• Angles are dimensionless but have units (degrees or radians)
• In Physics sin () or cos() never occur unless  is dimensionless.
Example 1-10
Find the units of the constants A and B in the following equation:
v  At  Bt 2 , where t is time (s) and v is the speed (m/s).
Solution
In the above equation, each term of the R. H. S must have the same dimensions
of speed as the L. H. S. Therefore,
So that:[At]= [v] = [L.T-1]
21
But [t] = [T], thus
[A] T = [L. T-1]
That implies to [A] = [L. T-2] , then the unit of the constant A is m/s2.
Similarly, for the constant B, we have
[B t3] = [v] = [L. T-1]
So that, [B] T3 = [L. T-1], that implies to [B] = [L T-4], then the unit of the constant
B is m/s4.
Example 1-11
A particle moving in a circle of radius r with constant speed v undergoes an
acceleration a . Express this acceleration in terms of v and r.
Solution
First of all, we begin by expressing the acceleration a in terms of the other
quantities as follows:
a= k vxry
Where k is some dimensionless constant while x and y are to be determined.
Next we insert the dimensions of each quantity where [a] is [L. T-2] , [v] is [L.T-1]
and [r] is [L]:
[RHS] = [LHS]
[L.T-2] = [L.T-1]x . [L]y
By comparing both sides, we get:
For L:
1 = x+y
For T:
2=x
22
These equations are easily solved and yield x = 2 and y = -1. Therefore the
acceleration is defined as:
a = k v2r-1
v2
or a  k
r
23
3-
Exercises
1-The speed of light in a vacuum is 0.300 Gm/s. What is this in meters per
second?
2- A computer can do 2.00 Giga-calculations per second. How many can it
do in a millisecond (ms)?
3- A human hair has a thickness of 70.0 µm. a) What is this in meters (m)?b)
What is this in km?
4- Deduce the dimensions of Newton's gravitational constant G in the
following equation:Fg= G m1 m2 / r2
Where Fg is the force between the any two particles G Universal gravity
constant&m1 and m2 are two particle masses.
5- Part A: See if you can determine the dimensions of the following
quantities:
1.
2.
3.
4.
5.
volume
acceleration (velocity/time)
density (mass/volume)
force (mass × acceleration)
charge (current × time)
Part B: Now find the dimensions of these:
1.
2.
3.
4.
5.
pressure (force/area)
(volume)2
electric field (force/charge)
work (in 1-D, force × distance)
energy (e.g., gravitational potential energy = mgh = mass × gravitational
acceleration × height)
24
6. square root of area
6- Part A: - You have already determined the dimensions of each of the
following quantities. Can you determine the corresponding SI units?
1. density
2. pressure
3. energy
Part B: - Choose the corresponding name of the SI unit.
1. Energy
Kg.m2/s3
2. Power
A.s
3. Frequency
(cycles).s-1
4. Charge
J/s
5. Force
kg.m/s2
Kg.m2/s2
7- Which one of the following quantities are dimensionless?
1.
2.
3.
4.
5.
6.
7.
68
sin (68 )
e
force
6
frequency
log (0.0034)
8- What are the dimensions of the following?
1.
2.
3.
4.
[3]
Given sin (wt), where t is time, [w] =?
[sin (wt)]
[force]
25
5.
6.
7.
8.
9.
[height]
[frequency]
[displacement]
[area × volume]
[0.5 × volume]
9- Determine whether the following equations are dimensionally correct.
x = xo + vo t + (1/2) a t2 where x is the displacement at time t
xo is the displacement at time t = 0
vo is the velocity at time t = 0
a is the constant acceleration
P=
where P is pressure
ρ is density
g is gravitational acceleration and h is the height
10- Choose the correct answer
1. Which symbol represents the Dimension of Length
a) (M)
b) [ M]
c) (m)
d) [m]
e) [L]
2. The SI unit of velocity is
a) Kg m/s2
b) N
c) m/s2
d) m/s
e) a& b
3. 1 Nano-meter = ……. meter
a) 10-9
b) 10-6
c) 10 9
d) 106
e) 10 12
4. 20 Mega-pixel = … pixel
a) 2x10-9
b) 4x10-6
d) 2x 107
e) 4x1012
c) 2x1010
5. If you have a box its dimension 2 meter length and 50 cm width and
height 1 meter , then its volume is :
a) 1 m3
b) 1 m2
c) 2 cm2
d) 2 cm3
e) none
11- Use dimensional analysis to check the validity of the following Equations:
26
2x + vt = a, and v-v0 = at
Where; x is the displacement, vo is the initial velocity, v is the velocity, a is the
acceleration, and t is the time .
12- Answer the following
If the area of the class is 5 m2, this area = …… cm2
The shutter of a camera opens for 1/500 second. Express this time in
hours, minutes, and microseconds.
A cube of edge 80 nm, find its volume in the SI- units.
i.
ii.
iii.
13-
Which of the following quantities are scalars or vectors?
1234-
Volume
Acceleration
Pressure
Work
14-The density of water is 1 gm/cm3. What is the density of water in SI system?
15- Is this equation dimensionally valid?
t   2x / a  2 Where x is distance, a acceleration, and t is time.
1
16-What are the dimensions and units of:a.
b.
c.
d.
e.
f.
g.
h.
i.
Velocity
Acceleration
Force
Pressure
Displacement
Distance
Sin(Ɵ)
Frequency
Density
27
j. Area
17- The gravitational constant has the value G=6.67 X10-11 (N-m2/Kg2). Find the
value of this constant in CGS system.
18- A particle is moving along a straight line with an acceleration a. Express the
distance x covered in a time interval t.
19- Find the unit of the constant b in the following equation
Where t is time, v is speed and F is force.
20If not, write the correct expression. Where t is time, v is speed, m is mass, a is
acceleration and F is force.
21- The micrometer (1 µ m) is often called the micron.
i. How many microns (µ) make up 1.0 km?
ii. What fraction of a centimeter equals 1.0 µm?
iii. How many microns are in 1.0 yd?
22- Earth is approximately a sphere of radius 6.37 x106 m.
What are:i. Its circumference in kilometers
ii. Its surface area in square kilometers, and
iii. Its volume in cubic kilometers?
23- Suppose that, while lying on a beach near the equator watching then Sun set
over a calm ocean, you start a stopwatch just as the top of the Sun disappears.
You then stand, elevating your eyes by a height H = 1.70 m, and stop the watch
when the top of the Sun again disappears. If the elapsed time is t = 11.1 s, what
is the radius r of Earth?
28
24- (a) Assuming that water has a density of exactly 1 g/cm3,find :-The mass of one
cubic meter of water in kilograms. (b) Suppose that it takes 10.0 h to drain a
container of 5700 m3 of water. What is the “mass flow rate,” in kilograms per
second, of water from the container?
25- Iron has a density of 7.87 g/cm3, and the mass of an iron atom is 9.27 x10-26 kg. If
the atoms are spherical and tightly packed,
a. What is the volume of an iron atom and
b. What is the distance between the centers of adjacent atoms?
26- A person on a diet might lose 2.3 kg per week. Express the mass loss rate in
milligrams per second, as if the dieter could sense the second-by-second loss.
29
Chapter 2
30
VECTORS
2.1Vectors properties
2.1.1Adding and subtracting Vectors Graphically
2.1.2 Unit Vectors
2.1.3 How to describe a two-dimension vector?
2.1.4 Vector Components
2.1.5 Properties of vector components
2.1.6 Adding & subtracting vectors by means of components
2.1.7 Multiplying and dividing a vector by a Scalar
2.2
The Scalar Product of Vectors (dot product .)
2.3
The Vector Product of vectors (cross productx)
2.4
Exercises
31
VECTORS
Chapter goals
To learn how vectors are represented and used
Student learning outcomes
123452.1
To understand the basic properties of the vectors
To add and subtract vectors graphically and using components
To understand the components of vectors
To recognize the unit Vectors
To understand how to multiply vectors.
Vectors properties
A vector is a quantity that has both a magnitude (size) and a direction.
Both of these properties must be given in order to specify a vector
completely. In this section, we describe how to write down vectors, how to
add and subtract them, and how to use them in physical problems. An
example of a vector quantity is a displacement. This tells us how far we are
from a fixed point; it also tells us our direction relative to that point. A
quantity with magnitude only, but no direction, is not a vector, it is called a
scalar quantity. An example of scalar is distance. This tells us how far we are
from a fixed point, but does not give us any information about the direction.
Then, the Vector has both magnitude and direction, and both these
properties must be given in order to specify it. Vectors, such as
displacement, have both magnitude and direction (5 m, north) and obey the
rules of vector algebra.
A quantity with magnitude but no direction is the Scalar. Scalars, such as
temperature, have magnitude only. They are specified by a number with a
unit (such as 10°C) and obey the rules of arithmetic and ordinary algebra.
Not all physical quantities involve a direction. Temperature, pressure,
32
energy, mass, and time, for example, does not “point” in the spatial sense.
We call such quantities scalars, and we deal with them by the rules of
ordinary algebra. A single value, with a sign (as in a temperature of 40°F),
specifies a scalar.
2.1.1Adding and subtracting Vectors Graphically
Two vectors
and
may be added geometrically by drawing them to a
common scale and placing those head to tail. The vector connecting the tail
of the first to the head of the second is the vector sum . To subtract
, reverse the direction of
to get - ; then add
from
to - .Vector additionis
commutative and obeys the associative law.
A variety of mathematical operations can be performed with and upon
vectors. One such operation is the addition of vectors. Two vectors can be
added together to determine the result (or resultant). This process of adding
two or more vectors has already been discussed in an earlier unit. Recall in
our discussion of Newton's laws of motion, that the net force experienced by
an object was determined by computing the vector sum of all the individual
forces acting upon that object. That is the net force was the result (or
resultant) of adding up all the force vectors. During that unit, the rules for
summing vectors (such as force vectors) were kept relatively simple.
Observe the following summations of two force vectors:
33
These rules for summing vectors were applied to free-body diagrams in
order to determine the net force (i.e., the vector sum of all the individual
forces). Sample applications are shown in the diagram below.
In this section, the task of summing vectors will be extended to more
complicated cases in which the vectors are directed in directions other than
purely vertical and horizontal directions. For example, a vector directed up
and to the right will be added to a vector directed up and to the left. The
vector sum will be determined for the more complicated cases shown in the
diagrams below.
34
There are a variety of methods for determining the magnitude and
direction of the result of adding two or more vectors.
Example 2-1
Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east.
Determine Eric's resulting displacement.
Solution:This problem asks to determine the result of adding two displacement
vectors that are at right angles to each other. The result (or resultant) of
walking 11 km north and 11 km east is a vector directed northeast as shown
in the diagram to the right. Since the northward displacement and the
eastward displacement are at right angles to each other, the Pythagorean
Theorem can be used to determine the resultant (i.e., the hypotenuse of the
right triangle).
35
The result of adding 11 km, north plus 11 km, east is a vector with a
magnitude of 15.6 km.
The magnitude and direction of the sum of two or more vectors can also
be determined by use of an accurately drawn scaled vector diagram. Using a
scaled diagram, the head-to-tail method is employed to determine the
vector sum or resultant. A common Physics lab involves a vector walk. Either
using centimeter-sized displacements upon a map or meter-sized
displacements in a large open area, a student makes several consecutive
displacements beginning from a designated starting position. Suppose that
you were given a map of your local area and a set of 18 directions to follow.
Starting at home base, these 18 displacement vectors could be added together in
consecutive fashion to determine the result of adding the set of 18 directions.
Perhaps the first vector is measured 5 cm, East. Where this measurement
ended, the next measurement would begin. The process would be repeated
for all 18 directions. Each time one measurement ended, the next
measurement would begin. In essence, you would be using the head-to-tail
method of vector addition.
36
The head-to-tail method involves drawing a vector to scale on a sheet of
paper beginning at a designated starting position. Where the head of this
first vector ends, the tail of the second vector begins (thus, head-to-tail
method). The process is repeated for all vectors that are being added. Once
all the vectors have been added head-to-tail, the resultant is then drawn
from the tail of the first vector to the head of the last vector; i.e., from start to
finish. Once the resultant is drawn, its length can be measured and
converted to real units using the given scale. The direction of the resultant
can be determined by using a protractor and measuring its counterclockwise
angle of rotation from due East.
A step-by-step method for applying the head-to-tail method to determine
the sum of two or more vectors is given below.
1.
Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a
diagram that is as large as possible, yet fits on the sheet of paper.
2.
Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude
and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
3.
Starting from where the head of the first vector ends, draw the second vector to scale in the indicated
direction. Label the magnitude and direction of this vector on the diagram.
4.
Repeat steps 2 and 3 for all vectors that are to be added
5.
Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as
Resultant or simply R.
37
6.
Using a ruler, measure the length of the resultant and determine its magnitude by converting to real
units using the scale (4.4 cm x 20 m/1 cm = 88 m).
7.
Measure the direction of the resultant using the counterclockwise convention discussed earlier in this
chapter.
An example of the use of the head-to-tail method is illustrated below. The
problem involves the addition of three vectors:
20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.
SCALE: 1 cm = 5 m
The head-to-tail method is employed as described above and the resultant is
determined (drawn in red). Its magnitude and direction is labeled on the
diagram.
SCALE: 1 cm = 5 m
38
Interestingly enough, the order in which three vectors are added has no effect
upon either the magnitude or the direction of the resultant. The resultant will
still have the same magnitude and direction. For example, consider the addition
of the same three vectors in a different order.
15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.
SCALE: 1 cm = 5 m
39
When added together in this different order, these same three vectors still
produce a resultant with the same magnitude and direction as before (20. m,
312 degrees). The order in which vectors are added using the head-to-tail
method is insignificant.
SCALE: 1 cm = 5 m
General procedure for adding two vectors graphically:
• On paper, sketch vector to some convenient scale and at the proper
angle.
• Sketch vector to the same scale, with its tail at the head of vector, again
at the proper angle.
• The vector sum
is the vector that extends from the tail of to the head
of . Then R  a  b
40
Examples
Two important properties of vector additions
1- Commutative law:
2- Associative law:
a b  b a
d  a  b  a  (b )
Vectors Subtraction
Check Your Understanding
- Two vectors,
and
resultant vector
:
, are added by means of vector addition to give a
. The magnitudes of
and
are 3 and 8 m, but
they can have any orientation. What is:(a) The maximum possible value for the magnitude of ?
(b) The minimum possible value for the magnitude of ?
2.1.2 Unit Vectors
The unit vectors are dimensionless vectors that point in the direction along a
coordinate axis that is chosen to be positive
41
2.1.3 How to describe a two-dimension vector graphically?
2.1.4 Vector Components:
The projection of a vector on an axis is called its component.
ax  a cos 
a y  a sin 
a  ax  a y  ax i  a y j
2.1.5 Properties of vector components
i.
ii.
The vector components of the vector depend on the orientation of the
axes used as a reference.
A scalar is a mathematical quantity whose value does not depend on the
orientation of a coordinate system. The magnitude of a vector is a true
scalar since it does not change when the coordinate axis is rotated.
However, the components of vector (Ax, Ay) and (Ax′, Ay′), are not scalars.
42
iii.
iv.
It is possible for one of the components of a vector to be zero. However,
this does not mean that the vector itself is zero. For a vector to be zero,
every vector component must individually be zero.
Two vectors are equal if, and only if, they have the same magnitude and
direction
Components of vector
Example 2-2
A displacement vector r has a magnitude of r= 175 m and points at an angle of
50° relative to the x axis as in Figure. Find the x and y components of this
vector.
Solution
x  r cos   175(m ) cos(500 )  112.48m
y  r sin   175(m ) sin(500 )  134.05m
Magnitude and direction of vector
Example 2-3
Find:- the magnitude and direction, if ax= 3 m& ay= 4 m
43
Solution
Magnitude:- a  ax2  b y2  a  32  42  5m
Direction:-
  tan 1
ay
ax
1
   tan
4
 530
3
2.1.6 Addition & subtraction of Vectors by Means of Components
For any two vectors A and B . Then
C  A so,
B
  tan 1 (Cy / Cx )
44
Example 2-4
Find the sum of the following displacement vectors:
= 5.0 m at 370 N of E
= 6.0 m at 450 N of W
= 4.0 m at 300 S of W
= 3.0 m at 600 S of E
N
B
A
Solution:
E
W
C
D
S
a)
Resolve each vector into components:
b)
Ax = 5cos 37o = 3.99 m
Ay = 5sin 370 = 3.01 m
Bx = - 6cos 450 = - 4.24 m
By = 6sin 450 = 4.24 m
Cx = - 4cos 300 = - 3.46 m
Cy = - 4sin 300 = - 2.00 m
Dx = 3cos 600 = 1.50 m
Dy = - 3sin 600= - 2.60 m
Add up all the x-components and all the y-components:
Rx = Ax + Bx + Cx + Dx = - 2.21 m
Ry = Ay + By + Cy + Dy = 2.65 m
c)
Find | | and :
45
R  R x 2  R y 2  3.45m
  tan 1
Ry
Rx
= 500N of W
Homework 1
Three vectors:
A = 5 km, 60° W of N ,B = 5 km, 30° E of S , and C = 4 km, 45° W of S
i- Draw these vectors
ii- Find their components
iii- Add these vectors by components.
Example 2-5
If you are given the following three vectors
Calculate
Solution:
In this operation we directly combine the similar components
Then we get
Example 2-6
Two vectors are defined as
A third vectors
is defined as
Express this vectors in terms of unit
vectors and then find its magnitude and direction.
solution
We know that if any two vectors
and
product of them is zero. Therefore
46
are perpendicular, and then the dot
Or
That implies to
Finally we get
The magnitude of
The angle that
is
makes with the x-axis is
Check Your Understanding
• Two vectors, A and B, have vector components that are shown (to the
same scale) in the first row of drawings. Which vector R in the second
row of drawings is the vector sum of A and B?
Homework 2
Do it your self?
The Component Method of Vector Addition
47
A jogger runs 145 m in a direction 20° east of north (displacement vector A) and
then 105 m in a direction 35° south of east (displacement vector B). Determine
the magnitude and direction of the resultant vector C for these two
displacements.
Homework 3
Do it yourself/
3- A wagon is being pulled by a rope that makes a 25o angle with the
ground. The person is pulling with a force of 103 N along the rope.
Determine the horizontal and vertical components of the vector.
4- A plane flies 34 km [N30oW] and after a brief stopover flies 58 km
[N40oE]. Determine the plane's displacement.
The Zero Vector and its Properties
48
Zero vector or null vector is a vector which has zero magnitude and an
arbitrary direction. It is represented by
result is a zero vector.
. If a vector is multiplied by zero, the
It is important to note that we cannot take the above result to be a number, the
result has to be a vector and here lies the importance of the zero or null vector.
The physical meaning of
2.2
can be understood from the following examples.

The position vector of the origin of the coordinate axes is a zero vector.

The displacement of a stationary particle from time t to time t is zero.

The displacement of a ball thrown up and received back by the thrower is
a zero vector.

The velocity vector of a stationary body is a zero vector.

The acceleration vector of a body in uniform motion is a zero vector.
Multiplying and Dividing a Vector by a Scalar
eV  e(Vx  Vy )  e(Vx i  Vy j )  (eVx )i  (eVy ) j
2-3: The Scalar Product of Vectors (dot product)
49
The dot product is a scalar.
• If the angle  between two vectors is 0°, dot product is maximum
• If the angle  between two vectors is 90°, dot product is zero
One of the most important features of the dot product of vectors is, the
commutative law:
Homework 4
1- What is the angle between the two vectors?
a  3.0i  4.0 j and b  2.0i  3.0k
Note that: i .i  j . j  k .k  1 while
i . j  j .k  k .i  0
50
2.4
The Vector Product (cross product)
For any two vectors AandB , then C  AxB
c  A B sin  nˆ
1- Cross production is a vector
Where n̂ the unit vector is normal to the plane contains the two vectors
(2) Magnitude is
c  A B sin 
(3) Direction is determined by right-hand rule
Example 2-7
Two vectors are defined as
If these two vectors are perpendicular, calculate the
value
of a.
Solution
We know that if any two vectors
and
product of them is zero. Therefore.
Or
That implies to
51
are perpendicular, and then the dot
Example 2-8
Find a vector of length 5 m in the xy plane that is perpendicular to
Solution
We know that if
and
are perpendicular; the dot product of them is zero
therefore
Or
But because the unknown vector
z-axis is zero
lies in the
plane, its component along the
. Therefore the above equation will be as follows
So we get
But the length of the vector
is 5 m, so that
Now substitute the definition of
in the above equation, we shall get
Or
52
Therefore the x-component is
The required vector is
Property of vector cross product
• The order of the vector multiplication is important.
 If two vectors are parallel or anti-parallel,
 If two vectors are perpendicular to each other, the magnitude of
their cross product isa maximum.
ˆ ˆ  kˆ , jxk
ˆ ˆ  iˆ and kxi
ˆ ˆ  jˆ & ixi
ˆ ˆ  jxj
ˆ ˆ  kxk
ˆ ˆ  0̂ (zero vector).
Note that: - 1- ixj
2- For two vectors a  a1iˆ  a2 jˆ  a3kˆ and B  b1iˆ  b2 jˆ  b3kˆ then
If the cross product of a two vectors (
x
= ) is called the zero vector or
Null vector.
Example 2-9
Find a vector of length 5 m that is normal to both vectors
and
Solution
In this case we shall find a unit vector normal to both
and
, then we will
convert the obtained unit vector to a vector of 5 m as length. As we did in the
previous example, we get:
53
= 14 +19 -
This vector is normal to the vectors and ,so that the unit vector
is defined
as:
But
= 23.6
The unit vector is
+
-
Therefore, the vector that is normal to and and has a length of 5 m is:
54
2-5 Exercises
12-
3-
4-
A force F of magnitude 10N makes an angle 300 with the
horizontal axis ( x-axis) . Find its components?
A car moves 5 Km towards East and then 5 Km towards North.
Find its total displacement and the direction of this
displacement?
A particle moves 10 Km towards East and then 20 Km with an
angle 300 towards North East. Find the total displacement
(magnitude and direction).
After an airplane takes off, it travels 0.4 Km West, 8.7 Km
North and 21.3 Km up. How far is the airplane from the take off point?
5-
Find:-
iˆ.iˆ
and
6-
Find :-
iˆ. jˆ
and
78910-
kˆ.kˆ
jˆ.kˆ
ˆˆ
ˆ ˆ
Find :- ixi
and ixj


Let A = 5N and B = 6N are two vectors subtended by an angle
 
 
0
60 . Find I) A.B and II) AxB


If A has 6 units along x-axis and B has 4 units and makes an
 
angle 600 with x-axis. Find: AxB


 


 
D

2
i

3
j

4
k
E

i

2
j  3k . Find:If
and
 
I) D  E
 
II) D  E
 
III) D.E
Also, find the angle between them.
55
and
 
IV) DxE
11- Add graphically the following vectors
12) i- A vector E = 3i- 4 j
11- Find the magnitude and direction of the vector E
i) If
= 2i + j ,
= i- 4j + k,
and
= j + k,
12- Find the magnitude of the vector ;where
13-
=
+
i) If F1 = 3 N and F2 = 4 N are parallel, then the magnitude ofF1 x F2 =?
ii) If
= 2 i - 5j + 7k and =3 i +2j + k, then, find:- . = ? & x =??
14- If you give the following three vectors:
A  3iˆ  4 jˆ  4kˆ
Calculate: (a) A  B  C
B  2iˆ  3 jˆ  7kˆ And C  4iˆ  2 jˆ  5kˆ
(b) A .B
and
(c)
(AxB ).C
15- A vector , when added to the vector = 3.0i + 4.0j, yields a resultant
vector that is in the positive y direction and has a magnitude equal to
that of . What is the magnitude of ?
56
16- Two vectors and have the components, in arbitrary units, ax = 3.2, ay
= 1.6, bx = 0.50, by = 4.5.
Find:(a)The angle between the directions of
(b) The components of a vector
and .
that is perpendicular to
plane, and has a magnitude of 5.0 units
57
, is in the xy
Chapter 3
58
Chapter 3
MOTIONS IN ONE AND TWO DIMENSIONS
3-1
Motion in One Dimension
3-1-1 Displacement
3-1-2 Average Velocity
3-1-3 Instantaneous Velocity
3-1-4 Acceleration
3-2
One Dimensional Motion with Constant Acceleration
3.2.1
Derivation of Kinematics Equations of Motion
3.2.2
Vertically thrown Up and Free Falling Bodies
3.3
Motion in Two Dimensions
3.3.1 Displacement, Velocity and Acceleration in 2-Dimensions
3.4
Projectile Motion
Horizontal Range and Maximum Height of a Projectile
3.5 Exercises
Exe
59
3.1 Motion in One Dimension
In this section, we discuss motion in one dimension. We introduce definitions
for displacement, velocity and acceleration, and derive equations of motion for
bodies moving in one dimension with constant acceleration. We apply these
equations to the situation of a body moving under the influence of gravity
alone.
3.1.1 Displacement
Definition: Displacement is the change in position,
x  x f  x i

Where: x f is the final position and xi is the initial position. The arrow indicates
that displacement is a vector quantity: it has direction and magnitude. In one
dimension; there are only two possible directions which can be specified with
either a plus or a minus sign. Other examples of vectors are velocity,
acceleration and force. In contrast, scalar quantities have only magnitude. Some
examples of scalars are speed, mass, temperature and energy.
3.1.2 Average Velocity
Definition: Average Velocity is the displacement over total time.
Mathematically:
v


x x f  xi change in displaceme nt


t
t f  ti
change in time
(3.1)
the overbar is frequently used to denote an average quantity
t is always > 0 so the sign of depends only on the sign of x .
Graphical interpretation of velocity: Consider 1- D motion, from point P (with
coordinates x1 ,t1 ) to point Q (at x2 , t2 ). We can plot the trajectory on a graph
(see Figure 3.1).
60
Figure 3.1: Graphical interpretation of velocity
P
X1
Q
X2
Then from Eq. (3.1) is just the slope of the line joining P and Q.
3.1.3 Instantaneous Velocity
Definition: Instantaneous velocity is defined mathematically:
x
t 0 t
v  lim
or
(3.2)
Table 2.1:- provides data on the position of a runner on a track at various times.
t(s)
1.00
1.01
1.10
1.20
1.50
2.00
3.00
x(m)
1.00
1.02
1.21
1.44
2.25
4.00
9.00
Find:- the runner's instantaneous velocity at t = 1.00 s?. As a first estimate,
calculate the average velocity for the total observed part of the run?. We have:
v 
x 9.00m  1.00m

 4m / s
t
3.00s  1.00s
(3-3)
61
From the definition of instantaneous velocity Eq.(3.2), we can get a better
approximation by taking a shorter time interval. The best approximation we
can get from this data gives,
v 
x 1.02m  1.00m

 2m / s
t
1.01s  1.00s
(3-4)
We can interpret the instantaneous velocity graphically as follows: Recall that
the average velocity is the slope of the line joining P and Q (from Figure 3.1). To
get the instantaneous velocity, we need to take t 0, or P Q. When P Q,
the line joining P and Q approaches the tangent to the curve at P (or Q), Thus
the slope of the tangent at P is the instantaneous velocity at P. Note that if the
trajectory were a straight line, we would get v = , the same for all t .
Note:


Instantaneous velocity gives more information than average velocity.
The magnitude of the velocity (either average or instantaneous) is
referred to as the speed.
3.1.4 Acceleration
Definition: Average acceleration is the change in velocity over the change in
time:
a
v v f v i

t
tf ti
(3.5)
Definition: Instantaneous acceleration is calculated by taking shorter and
shorter time intervals, i.e. taking t 0:
v
t 0 t
a  lim
or
(3-6)
Note:
62




Acceleration is the rate of change of velocity.
When velocity and acceleration are in the same direction, speed increases
with time. When velocity and acceleration are in opposite directions,
speed decreases with time.
Graphical interpretation of acceleration as shown in the figure below: on a
graph of v versus t , the average acceleration between P and Q is the slope
of the line between P and Q, and the instantaneous acceleration at P is the
tangent to the curve at P as shown in the figure.
From now on ``velocity'' and ``acceleration'' will refer to the
instantaneous quantities.
Q
P
Example 3-1:
The velocity of a particle moving along the x axis varies in time according to the
expression v=(40 - 5t 2) m/s, where t is in seconds. Find:
1) The average acceleration in the time interval t=0 to t=2.0 s.
2) The instantaneous acceleration at t =2.0 s
Solution:
1) The average acceleration is given by:
63
aav 
Δv vf  vi v(t  2s)  v(t  0s) (40  5  2 2 )  (40  5  0)  20




 10m / s 2
Δt t f  ti
20
2
2
2) By differentiating, the instantaneous acceleration
a 
dv d
 ( 40  5t 2 )  10t
dt dt
At t=2s, a(t=2s)=-20m/s2
Thus, the particle is slowing down.
Example 3-2:
The position, x (in m), of an object that moves along a straight line is changing
with time, t (in sec), as follows:
x = 16 − 12t + 2t2
1)Make a plot of the acceleration, a (in m/s2), vs. time
from t= 0 to t= 6.
2) What is the instantaneous velocity at t = 0, +2, and +4?
3)What is the instantaneous acceleration at t = 0, +2, and +4?
4) When is the velocity zero, and what then is the x position of the
object?
5) What is the average velocity between t = 0 and t = +6?
Solution:
1) By differentiating, a(t) = 4 m/s2(see plot)
2) v(0)=-12m/s, v(2)=-4m/s and v(4)=4m/s
3) a(0)=a(2)=a(4)=4m/s2.
64
Figure
4) v(t)=0 = −12 + 4t, to find that t=3s. The position of the object when the
velocity is 0 is x(3) =-2 m. To summarize: x(3) = -2m, v(3)= 0m/s and of
course a(3)=4m/s2.
5) The average velocity between two times is defined as
vav 
Δx
x(6)  x(0) 16  16


 0(m / s)
Δt
6  (0)
6
3.2 One Dimensional Motion with Constant Acceleration
Constant acceleration means velocity increases or decreases at the same rate
throughout the motion. Example: an object falling near the earth's surface
(neglecting air resistance).
3.2.1 Derivation of Kinematics Equations of Motion
Choose ti =0, xi= 0, vi=v0 , and write xf= x , vf= vand tf= t .

a = constant
a = . Then Eq.(3.5)
a = v-v0/tor
v = v0 + at

(3.7)
a = constant v changes uniformly
= 1/2 (v0 + v). From Eq.(3.1) =
x/t . Combining: x = t = ½ (v0 + v)t . Using Eq.(3.7) we get:
(3.8)
x = v0 t + ½ at2
Eq.(3.7)
t = (v - v0)/a . Substitute into Eq.(3.8)
x = (v + v0)(v - v0)/(2a) or,
(3.9)
v 2 = v02 + 2ax

Note that only two of these equations are independent.
65
Example 3-3 :
An automobile initially traveling at 50 (km/h) crashes into a stationary, rigid
barrier. The front end of the automobile crumples, and the passenger
compartment comes to rest after advancing by 0.40 m, assuming constant
acceleration during the crash:
1) What is the value of the acceleration?
2) How long does it take the passenger compartment to stop?
Solution:
The known quantities are the initial velocity vo = 50 km/h just before the
automobile comes in contact with the barrier, the final velocity (v = 0 when the
passenger compartment comes to rest), and the change of position of the
passenger compartment (x-x0=0.4 m).
1) To find the acceleration, we use the equation:
a
v 2  v02
2( x  x0 )
v0  50 1000 /(60  60)  13.9m / s
a
v 2  v02
0 2  (13.9)2

 240m / s2
2( x  x0 )
2(0.4)
This is a large deceleration. A passenger involved in such a crash would suffer
severe injuries, unless well restrained by a snug seat belt or an air bag.
2) To calculate the time the passenger compartment takes to stop, from Eq.
(3.7):
66
v  v0  at  t 
v  v0 0  13.9

 0.058s
a
 240
Example 3-4 :
The minimum distance required to stop a car moving at 35.0 (mi/h) is 40.0 ft.
What is the minimum stopping distance for the same car moving at 70.0 (mi/h) ,
assuming the same rate of acceleration?
Solution:
The known quantities are:
v  35mi / h  35 1.609  56.315km / h  35 1609 /(3600)  15.64m / s
xstop  40.0 ft  40  0.3048m  12.16m
The acceleration of the car is: a 
v 2  v02
0  (15.64)2

 10.06m / s 2
2( x  x0 )
2  12.16
For the same car moving atv = 70 mi/h = 31.28 m/swith the same acceleration, the
2
2
stopping distance is given by: v  v 0  2a(x  x 0 )
v 2  v02  2a( x  x0 )  x  ( x f  x0 ) 
v2  v20
0  ( 31.28)2

 48.63m
2a
 2  (10.06)
Example 3-5 :
A car is approaching a hill at 30.0 m/s when its engine suddenly fails, just at the
bottom of the hill. The car moves with a constant acceleration of a = 2 m/s2
while coasting up the hill:
1) Write equations for the position along the slope and for the velocity as
functions of time, taking x = 0 at the bottom of the hill, where v0 = 30 m/s.
2) Determine the maximum distance the car travels up the hill.
Solution:
67
The known quantities are: v0  30m / s, a  -2m/s2 and x0  0m
1) Along the slope the equation of position and velocity are respectively:
x
1
1 2
at  v0t  x0  x   ( 2)t 2  30t  0  t 2  30t
2
2
v  v0  at  v  30  2t
2) If the car becomes at rest: v  0  30  2t  t rest 
30
 15s
2
Therefore, the maximum distance traveled by the car is:
xmax 
1 2
at  v0t  x0  (15) 2  30  (15)  225m
2
Example 3-6:
A car travelling at a constant speed of 30 m/s passes a police car at rest. The
policeman starts to move at the moment the speeder passes his car and
accelerates at a constant rate of 3.0 m/o s2 until he pulls even with the speeding
car. Find
i.
ii.
The time required for the policeman to catch the speeder,
The distance travelled during the chase.
Solution:
We are given, for the speeder:
V0s = 30 m/s
as = 0
And for the policeman:
v0p
ap
=
=
0 m/s
3.0 m/s2.
68
a)
Distance travelled by the speeder xs = vst = (30)t . Distance travelled by
policeman xp = v0p + 1/2 apt2 = 1/2 (3.0)t2. When the policeman catches the
speeder xs = xp or,
30t = ½ (3.0)t2.
Solving for t we have t = 0 or t = 2/3 (30) = 20 s . The first solution tells us that
the speeder and the policeman started at the same point at t = 0, and the second
one tells us that it takes 20 s for the policeman to catch up to the speeder.
b)
Substituting back in above we find,
xs= 30(20) = 600 m
And,
xp = 1//2 (3.0)(20)2 = 600 m = xs.
Example 3-7:
A car decelerates at 2.0 m/s2 and comes to a stop after travelling 25 m. Find a)
the speed of the car at the start of the deceleration and b) the time required to
come to a stop.
Solution:
We are given:
a
v
x
=
=
=
- 2.0 m/s2
0
25 m
a)
From v2 = v02 + 2ax we have v02 = v2 - 2ax = - 2(- 2.0)(25) = 100 m2/s2 or v0 =
10 m/s.
69
b)
From v = v0 + at we have t = (v - v0)/a=
1
 10  = 5 s.
2.0
Example 3-8:
The current in a river is 1.0 m/s. A man swims 300 m downstream and then
back to his starting point without stopping. If he can swim 2.0 m/s in still water,
find the time of the round trip.
Solution:
We need to find the velocity of the man relative to the shore for each part of the
swim. Let downstream be the positive direction and let vw be the velocity of the
water. ym/wis the velocity of the man relative to the water and vw/s is the velocity
of the man relative to the shore. Then: (i) going downstream vw/s = vw + vm/w= 1.0
+ 2.0 = 3.0 m/s (ii) going upstream vw/s= 1.0 - 2.0 = - 1.0 m/s .
To find the time to go 300 m in each direction use x = v0t + 1/2 at2. With a = 0 we
have t = x/v0 .
This gives (i) downstream:
td 
300m
 100s
3m / s
(ii) Upstream:
tu 
300m
= 300 s
1.00m / s
The total time of the swim is tt = 100 s + 300 s = 400 s .
Problem 3-9:
The man in the previous problem swims across the river to the opposite bank
and back. The river is 300 m wide and he swims perpendicular to the current so
he ends up downstream from where he started. Find the time for the return
trip.
70
Solution:
Since the man swims perpendicular to the current we can define the y-axis as
parallel to the river and treat the x and y motion independently. We are only
interested in the motion in the x-direction. For the first half of his swim we
have:
ax = 0
vx0 = vx = 2.0 (m/s)
x = 300 m.
To find the time to cross the river we use x = vx0t + 1/2 axt2 which gives,
t 
x
vx0

300m
 150s
2.0m / s
Since the motion is symmetric, the time to return is the same as the time to
cross. The total time is tt = 2(150 s ) = 300 s.
3.2.2 Vertically Thrown up and Free Falling Bodies Motion
A freely falling object is an object that moves under the influence of gravity
only. Neglecting air resistance, all objects in free fall in the earth's gravitational
field have a constant acceleration that is directed towards the earth's center, or
perpendicular to the earth's surface, and of magnitude | | g = 9.8 m/s2. If
motion is straight up and down and we choose a coordinate system with the
positive y-axis pointing up and perpendicular to the earth's surface, we
describe the motion with Eq.(3.7), Eq.(3.8), Eq.(3.9) with a - g , x y .
71
Photo of a freely falling ball
Equations of motion for the one dimension (1-D) vertical motion of an object in
up (thrown up motion):
v = v0 - gt
y = v0t –½ gt2
v2 = v02 - 2gy
Note: Since the acceleration due to gravity is the same for any object, a heavy
object does not fall faster than a light object.
In free fall v0= 0, then from equations mentioned above:
v = - gt
y = – ½ gt2
v2 = - 2gy
Minus sign means negative y-axis motion
Example 3-10:
A swimmer amuses jumping from a 36 m-high cliff into the sea.
72
1) How long does he fall?
2) What is his impact velocity on the water?
Solution:
v 0  0m / s , a  -g  -9.8m/s 2 and y0  0m
Take the upward as the positive y direction.
1
2
1
2
2
2
1) y f  y sea   gt  v0 t  y 0  36    9.8  t  t 
2)
72
 2.71s
9.8
v  v0  gt  v  9.8  2.71  26.56m / s downword
Example 3-11:
A golf ball is released from rest from the top of a very tall building. Calculate:
(a) the position and (b) the velocity of the ball after 1.00 s, 2.00 s, and 3.00 s.
Solution:
Choose the origin at (y=0, t=0) at the starting point of the ball and take upward
as positive. Then y0=0, v0=0 and a=-g=-9.8m/s2. The position and the velocity at
time t become:
1
1
1
y f  y0   gt2  v0t  y f   gt2   (9.8)t 2
2
2
2
and
v f  v0   gt  v f  9.8t
a) at t=1s: yf= -4.9m
att=2s: yf=-19.6m
att=3s: yf= -44.1m
b) at t=1s: vf= -9.8m/s
att=2s: vf=-19.6m/s
73
att=3s: vf= -29.4m/s
Example3- 12:
A ball is thrown vertically upward from the ground with an initial speed of 15.0
m/s.
1) How long does it take the ball to reach its maximum altitude?
2) What is its maximum altitude?
3) Determine the velocity and acceleration of the ball at t = 2 s.
The free fall acceleration a=-g=-9.8m/s2.
Solution:
v0  15m / s, a   g  9.8m / s2 and y0  0m ( the ball is thrown upward from the ground)
We choose y- axis positive in the upward direction.
1) The time taken by the ball to reach its maximum altitude is:
v  v0  gt  t 
v0  v 15  0

 1.53s
g
9.8
2) The maximum altitude (v=0m/s) is given by the equation
v  v  2 g ( y  y0 )  ymax
2
2
0
v 2  v02 0  (15) 2


 11 .48m
 2g
 2  9.8
3) The velocity of the ball at t=2s: v  v0  gt  15  9.8  2  4.6m / s
The acceleration of the ball at t=2s is equals a= -g=-9.8m/s2
Example 3-13:
A stone is thrown vertically upward from the edge of a building 20 m high with
initial velocity 15 m/s. The stone just misses the building on the way down.
Find a) the time of flight and b) the velocity of the stone just before it hits the
ground.
74
Solution:
We are given,
v0 = 15 (m/s)
a = - 10 (m/s2)
At the time the stone hits the ground,
x
a)
=
- 20 m
From x = v0t + 1/2 at2 we have,
-20 =
15t+1/2 (-10) t2
Then t2-3t-4=0 by solving this equation:(t-4)(t+1) = 0
The two solutions are t = 4 s and t = - 1 s. The second (negative) solution
gives the time the stone would have left the ground, and is unphysical in
this case. The solution we want is the first one.
b)
We substitute to find v = v0 + at = 14.7 - 9.8(4) = - 24.5 m/s. Note that the
negative velocity correctly shows that the stone is moving down.
Example 3-14:
A rocket moves upward, starting from rest with an acceleration of 29.4 m/s2 for
4 s. At this time, it runs out of fuel and continues to move upward. How high
does it go?
Solution:
For the first stage of the flight we are given:
v0
a
t
=
=
=
0
29.4 m/s2
4 s
75
This gives, for the velocity and position at the end of the first stage of the flight:
v1 = v0 + at = (29.4 m/s2)(4 s ) = 117.6 m/s and x1 = v0t + ½ at2 = ½ (29.4)(4)2 =
235.2 m.
For the second stage of the flight we start with,
v1 = 117.6 m/s
a = - 9.8 m/s 2
And the end up with v2 = 0. We want to find the distance travelled in the second
stage (x2 - x1). We have,
V22 - v12 = 2a(x2– x1)
(x2 - x1) = 1 (v 22 v 12 )
2a
=

1
(117.6) 2
2(9.8)

= 705.6 m.
Therefore, x2 = x1 + 705.6 = 235.2 + 705.6 = 940.8 m.
3.3
Motion in Two Dimensions
In two dimensions, it is necessary to use vector notation to describe physical
quantities with both magnitude and direction. In this section, we define
displacement, velocity and acceleration as vectors in two dimensions. We also
discuss the solution of projectile motion problems in two dimensions.
3.4.1 Displacement, Velocity and Acceleration in 2-Dimensions
Recall: - In 1-dimension, the vector nature of velocity and acceleration is taken
into account by displacement.
In Figure 3.2, an object is at position ri at time ti (point P). Sometime later, tf, the
object is at position rf (point Q). The displacement vector of the object is given
by Eq. (3-10):-
76
(3-10)
Figure 3.2: Displacement
The sign of the physical quantity displacement is (positive or negative).
In 2-dimensions: - motion must use two components to specify a velocity or
acceleration vector.
We define the average velocity of a particle during the time interval t as the

displacement vector  r of the particle divided by that time interval t:


r
v av 
t


r
The instantaneous velocity v is defined as the limit of the average velocity
as
t
t approaches zero as in Figure 3.3:



r dr
v  lim

t 0 t
dt

The magnitude of the instantaneous velocity vector v  v is called the speed,
which is a scalar quantity.
77
y
P
o
Figure 3.3: Instantaneous velocity in 2-D.
x
Interpretation: When t 0, the point Q in the figure gets closer and closer to
the point P and the direction of
approaches the direction of a tangent to the
curve at point P. Thus the instantaneous velocity is parallel to the tangent and
in the same direction as the motion.
Average Acceleration
The average acceleration of a particle as it moves from one position to another

is defined as the change in the instantaneous velocity vector v divided by the


v
time interval t during which that change occurred: aav 
t

Instantaneous Acceleration: The instantaneous acceleration a is defined as the

v
limiting value of the ratio
as (t) approaches zero:
t





v dv d  dr  d 2 r
a  lim

  
t 0 t
dt dt  dt  dt 2
Note:a particle can accelerate in different ways:
78
1. The magnitude of can change in time, while the direction of motion stays
the same.
2. The magnitude of , | | , can stay constant, while the direction of motion
changes. This only happens in more than one dimension.
3. Both | | and the direction of can change.
Example 3-15:
A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as
functions of time are given by the following expressions:
x  (18.0m / s)t
And y  (4.0m / s) t  (4.9m / s 2 ) t 2
(a)Write a vector expression for the ball’s position as a function of time, using
the unit vectors
and ĵ . By taking derivatives, obtain expressions for (b) the


velocity vector v as a function of time and (c) the acceleration vector a as a
î
function of time. Next, use unit-vector notation to write expressions for (d) the
position, (e) the velocity, and (f) the acceleration of the golf ball, all at t = 3.0 s.
Solution:

a) r  18.0tiˆ  (4.0t  4.9t 2 ) ˆj



2
b) v  (18.0m / s)iˆ  4.0m  (9.8m / s )t ˆj

2
c) a  (9.8m / s ) ˆj

d) r (3.0s)  (54m)iˆ  (32.1m) ˆj

e) v (3.0s)  (18.0m / s)iˆ  (25.4m / s) ˆj

2
f) a (3.0s)  (9.8m / s ) ˆj
79
3.4
Projectile Motion
Projectile motion is a particular kind of two - dimensional motion. We make the
following assumptions:



The only force present is the force due to gravity.
The magnitude of the acceleration due to gravity is | | = g = 9.8 m/s2. We
choose a coordinate system in which the positive y-axis points up
perpendicular to the earth's surface. This definition gives, ay = - g and ax =
0.
The rotation of the earth does not affect the motion.
Projectile Photo
Initial Conditions:
We choose the coordinate system so that the particle leaves the origin ( x0 = 0, y0
= 0 ) at time t0 = 0 with an initial velocity of v 0 .
Procedure for Solving Projectile Motion Problems
1. Separate the motion into the x (horizontal) part and y (vertical) part.
80
2. Consider each part separately using the appropriate equations.
The equations of motion become:(a) x motion ( ax =0): vx = vx0 = constant& x = v0xt
(b) y motion ( ay = - g ):
vy = vy0 – gt
y = vy0t – 1/2 at2
vy2 = vy02 – 2gy
3.
Solve the resulting system of equations for the unknown quantities.
Furthermore, let us assume that at t= 0, the projectile leaves the origin
(x0=0, y0=0) with speed v0 , as shown in Figure 3.4. The vector v 0 makes an angle
0 with the horizontal, where 0 is the angle at which the projectile leaves the
origin. Therefore, the initial v 0 x and v 0 y components of velocity are:
v0 x  v0 cos( 0 )
v0 y  v0 sin( 0 )
,
(3.11)
Therefore, the velocity vector at t=0 is:

v0  v0 x iˆ  voy ˆj  v0 cos( 0 )iˆ  v0 sin( 0 ) ˆj
The components of velocity and position at time t are:
vx  v0 x  v0 cos( 0 ) , v y  v0 y  gt  v0 sin( 0 )  gt
x  v0 x t  v0 cos( 0 ) t ,
1
1
y  v0 y t  gt 2  v0 sin( 0 )t  gt 2
2
2
(3.12)
(3.13)
Therefore, the velocity vector and the position vector of the projectile as
functions of time are, respectively:

v  vx iˆ  v y ˆj  v0 cos( 0 )iˆ  (v0 sin( 0 )  gt ) ˆj
81

1
r  xiˆ  yˆj  v0 cos( 0 )tiˆ  (v0 sin( 0 )t  gt 2 ) ˆj
2
When analyzing projectile motion, consider it to be the superposition of two
motions: (1) constant-velocity motion in the horizontal direction and (2) free-fall
motion in the vertical direction.
If we eliminate the time between x and y, we find the path of a projectile, which
we call trajectory, is always a parabola. From equation (3.13) we find
t
x
and substitute this expression for t into the equation of y; this
( v0 cos( 0 ))
gives:

 2
g
 x
y  tan( 0 )x   2
2
2
v
cos
(

)
 0
0 
The equation is of the form y = ax-bx2, which is the equation of a parabola that
passes through the origin.
Figure 3.4
Horizontal Range and Maximum Height of a Projectile:
82
In projectile motion, two points are especially interesting to analyze: the peak
point A,which has Cartesian coordinates (R/2, h), and the point B, which has
coordinates (R, 0). The distance R is called the horizontal range of the projectile,
and the distance h is its maximum height (Figure 3.5). Let us find h and R in
terms of
, i = 0, and g.
Figure 3.5
We can determine h by noting that at the peak, v yA  0 . Therefore, we can use
Equation (3.12) to determine the time tA at which the projectile reaches the
peak(maximum height):
0  v0 y  gt  v0 sin( 0 )  gt A
 v sin( 0 ) 

t A   0
g


(3.14)
Substituting this expression for tA into the y part of Equation (3.13) and
replacing y = yA with h, we find:
v sin(0 ) 1  v0 sin(0 ) 

h  (v0 sin(0 )[ 0
]  g 
g
2 
g

83
2
v02 sin 2 ( 0 )
h
2g
(3.15)
The range R is the horizontal position of the projectile at a time that is twice the
time at which it reaches its peak, that is, at time t
B
= 2tA. Using the x part of
Equation (3.13), noting that vx  vB  v0 x  v0 cos( 0 ) and setting
at t = 2tA, we find that:
R  v0 x t B  v0 x 2t A
2v0 sin( 0 ) v02 2 cos( 0 ) sin( 0 ) v02 sin(2 0 )
R  v0 cos( 0 )


g
g
g
The maximum value of R from Equation (3.16) is Rmax
(3.16)
v02

obtained for
g
0=45°. The next figure illustrates various trajectories for a projectile having a
given initial speed but launched at different angles.
Figure
The figure below is an experiment illustrating the independence of
horizontal and vertical motion. The gun is aimed directly at the target ball and
84
fired at the instant the target is released. In the absence of gravity, the projectile
would hit the target because the target wouldn’t move. However, the projectile
still hits the target in the presence of gravity. That means the projectile is falling
through the same vertical displacement as the target despite its horizontal
motion.
Example 16:
A long-jumper leaves the ground at an angle of 20.0° above the horizontal and
at a speed of 11.0 m/s as shown in figure.
1) How far does he jump in the horizontal direction? (Assume his motion is
equivalent to that of a particle.)
2) What is the maximum height reached?
85
Solution:
v0  11.0m / s,  0  20 
1) We set our origin of coordinates at the takeoff point and label the peak as A
and the landing point as B. The horizontal motion is described by Equation:
R  xB  v0 x t  v0 cos( 0 ) t B  (11.0m / s)(cos(20.0°))tB
The value of xB can be found if the time of landing tB is known. This time is
twice the time tA in which the horizontal component of velocity of projectile is
equals zero (at maximum height). Therefore, we use the equation (3.14), we can
found the time tB:
 v sin( 0 ) 
  2(11.0. sin(20.0  ) /( 9.8)  2  0.384s  0.768s
t B  2t A  2 0
g


Therefore: R  xB  v0 cos( 0 )t B  11.0 cos(20 )(0.768)  7.94m

This is a reasonable distance for a world-class athlete.
The range could be also be found from (3.16) as:
v02 sin( 2 0 ) 11 2 sin( 40)
R

 7.94m
g
9.8
2) We find the maximum height reached by using equation (3.15):
86
h  ymax 
v02 sin 2 ( 0 ) (11sin(20 )) 2

 0.722m
2g
2(9.8)
Example 3-17:
A stone is thrown from the top of a building upward at an angle of 30.0° above
horizontal with an initial speed of 20.0 m/s, as shown in Figure 3.6. If the height
of the building is 45.0 m:
1) How long does it take the stone to reach the ground?
2) What is the speed of the stone just before it strikes the ground?
Solution:
We have indicated the various parameters in Figure 3.6.The initial x and y
components of the stone’s velocity
are: v0 x  v0 cos( 0 )  20 cos(30)  17.3m / s
Figure : 3. 6
87
v0 y  v0 sin( 0 )  20 sin(30)  10.0m / s
1) To find t, we can use the equation:
2)
There is a minus sign on the numerical value of yf because we have chosen the
top of the building as the origin.
Solving the quadratic equation for t gives, for the positive root, t = 4.22 s. Does
the negative root have any physical meaning?
3) We can use Equation: v y  v0 y  gt with t=4.22s to obtain the y component
of the velocity just before the stone strikes the ground:
v yf  v y  10.0  9.8x(4.22)  31.4m / s
The
negative
Because vx f
sign
indicates
 vox  17.3m / s ,
that
the
stone
is
moving
downward.
the magnitude of velocity (speed):
v f  vxf2  v yf2  (17.3)2  (31.4)2  35.9m / s
Example 3-18:
A plane drops a package of emergency rations to a party of explorers. The
plane is traveling horizontally at 40.0 m/s and 100 m above the ground, as
shown in Figure 3. 7:
I.
Where does the package strike the ground relative to the point at which it
is released?
88
II. What are components of the velocity of the package just before it hits the
ground?
III. Where is the plane when the package hits the ground? (Assume that the
plane does not change its speed or course.)
Solution:
To solve this problem, we choose the coordinate system, in which the origin is
at the point of release of the package. The various parameters are:
 0  0  , v0 x  40.0m / s , v0 y  0m / s and a  -g  -9.8m/s 2
i) Consider first the horizontal motion of the package, the distance traveled
along the horizontal direction is given by:
x  v0 x t  v0 cos( 0 ) t  40.0 cos(0  )t  40.0t
Figure3.7: Plane drops a package
89
To find t, we use the equations that describe the vertical motion of the package.
We know that at the instant the package hits the ground, its y coordinate is:
1
1
y F  100  v0 y t  gt 2  0  (9.8)t 2
2
2
t= 4.52s
Substitution of this value for the time of flight into the equation for the x
coordinate gives: x f  40.0t  40.0x4.52  181m
The package hits the ground 181 m to the right of the drop point.
ii) The components of the velocity just before the package hits the ground are
given by the equation:
vx f  v0 x  v0 cos( 0 )  40.0m / s
v y f  v0 y  gt  0  gt  9.8x(4.52)  44.3m / s
iii) Directly over the package. Why?
Example 3-19:
A projectile is fired with an initial speed of 113 m/s at an angle of 600 above the
horizontal from the top of a tower 49 m high (see Figure 3.8). Find:a- The time to reach the maximum height,
b- The maximum height,
c- The total time in the air,
d- The horizontal range and
e- The component of the final velocity just before the projectile hits the
ground.
90
Solution:
Set up the coordinate system.
A
600
49m
B
Figure 3.8: (Problem 3.3)
Consider the x- and y-motion separately. We are given:
Table 3.2: (Problem 3.3)
y-motion
x-motion
yB = - 49 m
x =?
v0x = 113cos 60o v0y = 113sin 60o
v0x = vAx = vBx
vAy = 0
ay = - 9.8 m/s2
ax = 0
a) Find the time to reach the maximum height:
vAy = v0y + aytA
v Ay v 0 y
 tA 
ay
113sin 600
 9.99s
9.8
b) Find the maximum height:
=
(3-15)
91
yA= v0y tA + 1/2 ay t2A
= (113sin 600)(9.99) – 1/2 (9.8)(9.99)2 = 489 m
(3-16)
c) Find the total time tB in the air:
yB = v0ytB + 1/2 ay t B2
- 49 m = (113sin 60)tB–1/2 (9.8) t B2
(3-17)
0 = 4.9t2B - 97.9tB– 49
Solving the quadratic we obtain,
tB =
1
(97.9  (97.9)2  4(4.9)(49)) )
9.8
Which gives, tB = 20.5 s or tB = - 0.49 s. We reject the second solution (it gives the
time the projectile would have left the ground, if it had been thrown from
there).
d) Find the horizontal range ( xB ):
xB = v0xtB + 1/2 ax t B2
= (113cos 60)(20.5) = 1158 m.
(3-18)
e) Findthe components of the final velocity ( vBx , vBy ):vBx = v0x = 113cos( 60) = 56.5 (m/s)
vBy = v0y + aytB
= 113sin
- 9.8(20.5) = - 103 (m/s).
(3-19)
Note that the negative value of vBy correctly gives the direction as down.
92
Example 3-20:
A projectile is fired at a falling target as in figure 3.9. The projectile leaves the
gun at the same instant that the target falls from rest. Assuming that the gun is
initially aimed at the target, show that the bullet will hit the target.
Solution:
Let xB and yB be the x and y positions of the bullet, and let xT and yT be the x and
y positions of the target. We need to show that, yB = yT and xB = xT at some
common time tc , the time at which the bullet will hit the target.
Figure 3.9: (Example 3-20)
The motion of the bullet is described by the two equations:
xB =
yB =
cos(Ө)t
sin(Ө)t –
.
It will take the bullet some time, tc , to arrive at the x position of the target, xT .
At that time, the bullet and target will be at the same x position (the target's x
position does not change since it falls straight down). Thus, we have at time tc :
xT = xB=
At time tc .
93
In order for the bullet to hit the target, the y positions must be equal at tc . The y
position of the target is given by :
yT = y0 – g t c2
And for the bullet:
2
tc – ½ g t c
yB = v0 sin
We can see from the above equations that in order for yB = yT at (tc) , we need to
have y0 = v0sin tc .
To show this, consider Figure 3.9:
y sin 
sin 
 tan   
 y 0  xT
x cos 
cos 
Using the previous result, xT = v0cos
y0 
tc or cos
= (xT)/(v0tc) gives:
x T sin 
 v 0 sin t c
x T / v 0t c
Thus we have shown that at time tc ,xT = xB and yT = yB , and therefore that the
bullet will hit the target.
94
Exercises:
1- In order to equality for the finals in a racing event, a race car must
achieve an average speed of 250 km/h on a track with a total
length of 1600 m. If a particular car covers the first half of the track
at an average speed of 230 km/h, what minimum average speed
must it have in the second half of the event in order to quality ?
2- A certain car has an accelerating rate of 0.60 m/s2 . How long does
it take for this car to go form a speed of 55 mi/h to a speed of 60
mi/h?
3- An aircraft has a take-off speed of 120 km/h.
a. What minimum constant: acceleration does this require if the
aircraft is to be flying after a takeoff run of 240 m ?
b. How long does it take the aircraft to have a steady flying?
4- A train is traveling at 20 m/s when the driver starts to apply the
brakes. As a result, the train decelerates with -1 m/s2 until it stops.
How far does the train move just after breaking to be stationary?
95
5- A rock is dropped into a 200-mdeep well.
a. How long does it take to fall halfway to the bottom?
b. What is its speed at that height?
6- The velocity of a uniformly accelerated body increases from
to
over the interval from
a. Find the body's acceleration.
b. Find the body's initial velocity at t=0.
7- A boy throws his dummy vertically upward with a speed of 17
m/s, and catches it again exactly where it left his hand.
a. How long does the ball remain the air?
b. How high does it go?
8- A certain vehicle requires a distance of 49 m to stop when it is
traveling at 28 m/s. Under these condition,
a. What acceleration is produced by braking?
96
b. If a body is 35 m directly in front of the vehicle when the driver first
applies the brakes, how much time does the body have to get out of
the way?
9- A car accelerates uniformly from rest to a speed of 20 m/s over a 65 time interval, starting at t=0
a. What is the cars average velocity during this time interval?
b. At what instant is the cars instantaneous velocity equal to its
average velocity?
c. What is the total distance traveled at the end of this interval?
10A truck covers 40 m in 8 s while smoothly slowing down to a
final speed of 2.50 m/s.
a. What is the trucks original speed?
b. Find its acceleration.
11-
A ball is shot with speed vo= 30 m/s with an angle 300 with the
x-axis from the origin. Find:
i.
ii.
iii.
iv.
v.
The position and velocity after 2s
The time required to reach the maximum height
The max. height
The range
The flight time
97
12-
A car moves with a horizontal constant velocity v0 = 9 m/s.
Find the position and velocity after 0.5s when the car leaves the
edge of the road.
13A ball is thrown so that its initial vertical and horizontal
components of velocity are 40 m/s and 20 m/s, respectively. Use a
motion diagram to estimate the ball’s total time of flight and the
distance it traverses before hitting the ground. Also, Estimate the
maximum height.
14- A long jumper (Figure below) leaves the ground at an angle of 200 to the
horizontal and at a speed of11.0 m/s. Then:a)
How long does it take for him to reach maximum height?
b)
What is the maximum height?
c)
How far does he jump? (Assume that his motion is equivalent to that of a
particle, disregarding the motion of his arms and legs.)
d)
Find the maximum height he reaches.
98
15- A jet plane traveling horizontally at 1.00 x 102 m/s drops a rocket from a
considerable height. (See Figure) The rocket immediately fires its engines,
accelerating at 20.0 m/s2 in the x-direction while falling under the influence of
gravity in the y-direction. When the rocket has fallen 1.00 km. Find:a)
Its velocity in the y-direction,
b)
Its velocity in the x-direction, and
c) The magnitude and direction of its velocity. Neglect air drag and
aerodynamic lift.
17-
The position of a particle moving in xy plane is given by
Where t is measured in seconds. Calculate the magnitudes and directions of the
particles position, velocity and acceleration at
99
18-
The position of a particle moving along the x axis is given in
centimeters by:-
Where t is in seconds. Consider the time interval t = 2.0s to t = 3.0s, calculate:
a. The average velocity during this interval
b. The instantaneous velocity at t=2.0s.
c. The instantaneous velocity at t=3.0 s.
d. The average acceleration during this interval
19-
A bullet is fired at an angle of 250 below the horizontal with a
velocity of 200 m/s, if the bullet was 25 m above the ground.
Find the total Flight time.
20-
A ball is shot with a velocity of v0 at an angle
above the
horizontal. If the range is R=250 m and the maximum height is
H= 36 m. Findv0 and
100
Chapter 4
101
NEWTON'S LAWS OF MOTION
4-1 Force and Fundamental Forces of Nature
4-2Newton's First Law
4-3Newton's Second Law
4-4 Newton's Third Law
4-5 Frictions
4-6 Experimental Facts about Friction
4-7 Applications of Newton's Laws
4-8 Problem Solving Strategy
4-9 Solved Examples
4-10
Exercises
102
Newton's Laws of Motion
Goals of Chapter 4
4-1
• To understand the meaning of force in Physics
• To view force as a vector and learn how to combine forces
• To understand the behavior of a body on which the forces
“balance”: Newton’s First Law (NFL) of Motion
• To learn the relationship between mass, acceleration, and force:
Newton’s Second Law (NSL) of Motion
• To relate mass and weight
• To see the effect of action-reaction pairs: Newton’s Third Law (NTL)
of Motion
• To learn how to make free-body diagrams
•
Force and The fundamental forces of nature:
4-1-1 Force
In this section, we introduce the concept of force. We discuss Newton's laws,
which describe the way a body responds to a net force. We discuss frictional
forces and the way they can be mathematically represented. We study several
applications of Newton's laws.
Idea: Force is the cause of motion in classical mechanics. Classical mechanics
deals with systems of size 10-10 m (atomic dimensions) and velocity 3.0 x 108 m/s
(the speed of light).
Notes:


Force is a vector.
There are two kinds of forces:
o Contact Forces - involve physical contact between objects.
Examples: the force involved in kicking a ball, pulling a wagon,
compressing a spring, etc.
103
o
Field forces - don't involve physical contact between objects.
Examples: the gravitational force and the electromagnetic force.
What are some properties of a force?
Types of Forces
A force is a push or pulls acting upon an object as a result of its interaction with
another object. There are a variety of types of forces, a variety of force types are
placed into two broad category headings on the basis of whether the force
resulted from the contact or non-contact of the two interacting objects.
Contact Forces
Field (Action-at-a-Distance) Forces
Frictional Force
Gravitational Force
Tension Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
-------
Applied Force
--------
Spring Force
---------
104
What are the magnitudes of common forces?
Drawing force vectors
Use a vector arrow to indicate the magnitude and direction of the force as
shown in Figure (1).
Figure (1)
Case I( pull directed 300 above horizontal), we have: Fnet   F  ma
105
For x-axis, we have:
Fx   Fx  max
, therefore F cos300= m a
If the object moves with a constant speed, the above equation becomes F =0
For y-axis, we have Fy   Fy  may . There is no motion along y-axis, so that ay=
0. Therefore, N+ F sinӨ = mg, or
= mg – F sinӨ.
Case II( push directed 450 above horizontal ).
For x-axis, we have: Fx   Fx  max , therefore FcosӨ= Fcos300= ma
For y-axis, we have Fy   Fy  may . Where ay=0, so N-F sinӨ = mg, or
N=mg+ FsinӨ.
Case III: An object moving along a rough surface with a force F
a)
F parallel to the surface
For x-axis, we have Fx   Fx  max , therefore F-fk = ma wherefkis the kinetic
friction force. If the object moves with a constant speed, then the equation
becomes: F = fk
For y-axis,
Fy   Fy  may
, where ay = 0, therefore
106
= mg.
b) F makes an angle to the surface
For x-axis, we have Fx
  Fx  max
, therefore F cosӨ -fk = max , where fk is the
kinetic friction force. At a motion of constant speed ax=0, then FcosӨ = fk
For y-axis, Fy   Fy  may , where ay = 0, therefore N+ F sin Ө= mg.
4-1-2 The fundamental forces of nature
According to current understanding, all forces are expressions of four
distinct fundamental forces:
1. Gravitational Force
The gravitational force acts between any two bodies of matter in the
universe, since MASS is its source. It is a force of infinite range which
obeys the inverse square law. It is weak, very long ranged, and always
attractive.
2. Electromagnetic Force
It can be attractive or repulsive, and acts only between bodies of matter
carrying electrical charges. It is long ranged, but much weaker that the
strong force. It is a force of infinite range which obeys the inverse square
law, has the same from as the gravitational force , and holds atoms and
molecules together
3. The Strong Interaction Force
107
It is a force which can hold a nucleus together against the enormous
forces of repulsion of the protons. It is not an inverse square force like the
electromagnetic. It has very short range of order 10-11 m and the
strongest of the four fundamental forces.
4. The Weak Interaction Force
This force is crucial for the universe in that the sun would not burn
without it (formation of deuterium and deuterium fusion). It is necessary
for the buildup of heavy nuclei..
4-2 Newton's First Law
Newton's First Law States: an object at rest stays at rest, and an object in motion
stays in motion with a constant velocity, if there is no net external force
between the object and the environment. In equation form we can write:
F  0  a  0
(4.1)
4-3 Newton's Second Law
If
≠0 (i.e. there is a net external force acting on an object) then,
(4.2)
Definition:Inertia is the tendency of an object to resist any attempt to change its
state of motion. Mass is the force required per unit of acceleration produced
and is a measure of inertia. Mass is a scalar and has SI units of kilograms (Kg).
Example: If a bowling ball and a golf ball are hit with a bat, the bowling ball
would be much harder to get moving since it has greater mass and thus greater
inertia.
Note:

is inversely proportional to m . This means that, for the same force, a
smaller mass will have a larger acceleration.
108



Newton's second law is a vector equation which contains three scalar
equations (in three dimensions):
= max ,
= may ,
= maz
The first law is a special case of the second law.
The SI unit of force is the Newton (N). Definition: 1 Newton is the force
that produces an acceleration of 1 m/s2 when acting on a 1 kg mass. In the
CGS system: 1 dyne = 1 g cm/s2= 10-5N. In the British engineering system:
1 pound (lb) = 4.448 N.
Definition: Weight ( ) is the force exerted on an object by a gravitational field.
From Newton's second law,
w = mg
(4.3)
Notes:




Weight is a vector with direction towards the earth's center, or
perpendicular to the earth's surface.
The weight of an object is different on the earth and on the moon since the
strength of the gravitational field is different ( gearth ≠ gmoon ).
The value of g varies with distance from the center of the earth as a
consequence:
o Since the earth isn't a perfect sphere, the weight of an object varies
slightly from place to place on the earth's surface.
o The weight of an object varies slightly with altitude above the
earth's surface.
In comparison, mass is a scalar with a value independent of location.
Notice however that, in the approximation that g is constant, mass is
proportional to the magnitude of the weight and the two quantities can be
used interchangeably. This is called the equivalence principle.
4-4 Newton's Third Law
Idea: Forces in nature always exist in pairs. Newton's third law states: "For
every action, there is an equal and opposite reaction". When two bodies
interact:
109
2 on 1 = -
1 on 2
(4.4)
Where 2 on1 is the force exerted on body 1 by body 2 and 1 on 2 is the force
exerted on body 2 by body 1.
For example:-If you exert a force on a body, the body always exerts a force (the
“reaction”) back upon you.
• Figure 4.2 shows “an action-reaction pair.”
Another Example: When an object falls towards the earth, the earth exerts a
force on it that causes it to accelerate towards the earth. According to Newton's
third law, the object exerts a force on the earth as well, and the earth accelerates
towards the object. Why don't we feel the earth accelerate?
Solution:
2nd Law
3rd law
obj on earth
obj on earth
=-
= me
earth on obj
e
=-
 ae   / me
110
Conclusion: the acceleration of the earth is too small to detect because the mass
of the earth is much larger than the mass of the object.
4-5
Frictions
Frictional forces (i.e.; Interaction or Contact)
A surface can always supply a normal force, perpendicular to the surface.
However, a surface quite often also supplies a fiction force parallel to the plane.
Friction forces always opposite to the motion -- or prevent the motion.
• When a body rests or slides on a surface, the friction force is parallel to
the surface.
Friction between two surfaces arises from interactions between molecules on
the surfaces as shown in Figure (4.3).
Figure (4-3)
Also, the contact force between the surface and the body is normal to the
surface and called normal force “
.”
111
Kinetic and static friction
•
•
•
•
Kinetic friction acts when a body slides over a surface.
Static friction acts when there is no relative motion between bodies.
The kinetic friction force is fk = µk .
The static friction force can vary between zero and its maximum value: fs ≤
µs .
• Static friction followed by kinetic friction.
• Before the box slides, static friction acts.
• But once it starts to slide, kinetic friction actsas in Figure (4-4).
Figure (4.4)
4-6
Experimental facts about friction
1. fs  μs
where μs is the coefficient of static friction and
is the magnitude
of the normal force ( =mg). Equality holds when the object is on the point of
slipping: fs(max) = μs
.
112
2. fk = μk
where μk is the coefficient of kinetic friction and is approximately
constant for any given pair of materials.
3. Values of μs and μk depend on the nature of the surfaces that are in contact,
usually μk<μs.
Examples: rubber on concrete μs = 1.0, μk = 0.8 ; waxed wood on wet snow
μs = 0.14, μk = 0.10.
4. The direction of the force of friction is opposite to the direction the object
wants to move.
5. μk and μs are nearly independent of the area of contact between the two
surfaces.
6. μk is nearly independent of the velocity of the object under consideration.
Table (4-2): lists some approximate values for µs and µk for various pairs of surfaces
Table 4-2
4-7Applications of Newton's Laws
Goals of applying Newton's laws:
• Use Newton’s 1st law for bodies in equilibrium (statics)
• Use Newton’s 2nd law for accelerating bodies (dynamics)
113
• Study types of friction
• Using Newton’s First Law when forces are in equilibrium
• A body is in equilibrium when it is at rest or moving with constant
velocity in an inertial frame of reference.
• We treat objects as point particles (no rotational motion).
• We neglect masses of ropes and springs. One consequence of this
assumption is that the force exerted along a rope is the same at all
points in the rope.
Note: In problems with several bodies, apply Newton's 2nd law to one body at
a time.
4-8 Problem Solving Strategy


Draw a picture of the situation and a force diagram of all the forces for
each body (a free body diagram).
o In the force diagram for each object, include only the forces acting
on that object.
o The force exerted by a rope is called the tension and usually
denoted .
o The contact force exerted by a surface is called the normal force and
always acts perpendicular to the surface.
Set up a coordinate system and apply Newton's second law:
F
x

4-9
 max ,  Fy  may
If necessary, use the kinematic equations of motion to solve for the
desired quantities.
Solved Examples
One-dimensional equilibrium: Tension in a massless rope
Example 4-1
114
A gymnast with mass of mG= 50 Kg is suspended from a hanging rope of
negligible mass. Find
1- The gymnast's weight,
2- The force the rope exerts on the gymnast
3- The tension at the top of the rope
Solution:
Fig. (4.5)
There are two bodies at equilibrium; the gymnast G and the rope R.
Applying Newton's First Law:
For G
F
y
0
Then TR on G - wG =0
So TR on G= wG = 50 x 9.8 = 490 N
For R
T C on R- T G on R = 0
115
T C on R =T G on R = 490 N
One-dimensional equilibrium: Tension in a rope with mass
Exercise: What is the tension in the previous example if the rope has mass?
Fig. (4.6)
Example 4-2
A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a
rope attached to the box. Find:
a. The acceleration of the box, and
b. The vertical velocity of the box after 2 seconds.
Solution:
116
Figure 4.7: example 4.2
a)From Newton's 2nd Law:ma = T – mg
= 68N  9.8(m / s 2 )  3.8(m / s 2 )
5Kg
The direction of the acceleration is that of the tension force
b) Since a is constant
v = v0 + at
= 0 + 3.8(2) = 7.6 m/s.
(4.5)
(4.6)
Example 4-3
Straight-line motion with constant force Newton's Second Law NSL.
An Iceboat is at rest. A wind is blowing after 4s from its release. So it attains a
velocity of 6 m/s. The wind exerts a constant horizontal force on the boat. Find:
-the horizontal force Fw which exerts on the boat if it’s mass = 200 kg, as shown
in Figure (4.8).
117
Figure (4-8) (Example 4.3)
Solution
From the given information, we have:
v0 = 0 m/s, after t= 4 s, the iceboat velocity is v = 6 m/s
Then from the equation:
v = v0 + a x t
In one dimensional motion
So, 6 = 0+ax (4) and so ax = 6/4 = 1.5 m/s2
From Newton's Second Law, we have
In this problem,
F
y
 F  ma
 0 &  Fx  max  Fw and so, Fw  200(Kg )(1.5m / s 2 )  300N
Straight-line motion with constant force
Example 4.4
An Iceboat is moving as considered in the previous example.If the position of
the boat as a function of time is given by:
x  1.2t 2  0.2t 3
118
Find:-Fw after t = 3s . Also, find the time at which Fw=0, Fw> 0 and Fw<0.
Solution
d 2x
From Second Law: we have  F  ma  m 2 Fx  Fw
dt
dx
 2.4t  0.6t 2
dt
Then
dv
a
 2.4  1.2t
dt
v 
After t= 3s, so v = 2.4(3) - 0.6(3)2 = 7.2 – 5.4 = 1.8 m/s
Also, a= 2.4 – 1.2(3) = -1.2 m/s2 , minus sign means only negative x –axis motion.
Then after t = 3s
Fw = max = 200(-1.2) = -240 N, minus sign means only
negative x –axis motion.
Fw = 0 N when ax = 0 then, 2.4 – 1.2 t= 0 , so t = 2.4/1.2= 2 s.
Fw>0 at t< 2s while Fw<0 at t>2s
Example 4-5(Frictions)
A hockey puck of mass 0.145 kg travelling at 10 m/s slows to 2.0 m/s over a
distance of 80 m. Find:f- The frictional force acting on the puck and
g- The coefficient of kinetic friction between the puck and the surface.
119
Solution:
Figure 4.9: Example 4.5
First we find the acceleration of the puck from the kinematics equations of
motion. We have, v0 = 10 m/s, vf = 2 m/s and x = 80 m. The third equation of
motion gives,
v2 = v + 2ax
2
0
v 2 v 0 2 4  100

 0.6m / s 2
Then  a 
2x
2(80)
From the Newton's Second Law:
In the x-direction,
i- fk = ma =0.145(- 0.6) = - 0.087 N.
(4.8)
ii- Use fk = -μk
. From the y-component of the Newton 2nd Law
(fk = ma) & - mg = 0 . Combining these equations….
120
- μk mg = m a
 μk = - a/g = 0.061< 1
(4.9)
Example 4-6
A student of mass 50 kg tests Newton's laws by standing on a bathroom scale
in an elevator. Assume that the scale reads in Newton's. Find:- the scale reading
when the elevator is:a) Accelerating upward at 0.5 m/s2,
b) Going up at a constant speed of 3.0 m/s and
c) Going up but decelerating at 1.0 m/s2.
Solution:
Figure 4.10: Example 4.6
From the Newton's 2nd Law:
Fs - mg = ma  Fs = m(g + a)
(4.10)
121
This gives:
a)
Fs = 50(9.8 + 0.5) = 515 N
b)
Fs = 50(9.8 + 0) = 490 N
c)
Fs = 50(9.8 - 1.0) = 440 N
Example 4-7
A wooden plank is raised at one end to an angle of 300. A 2.0 kg box is placed
on the incline 1.0 m from the lower end and given a slight tap to overcome
static friction. The coefficient of kinetic friction between the box and the plank is
μk = 0.20. Find:a) The rate of acceleration of the box and
b) The speed of the box at the bottom. Assume that the initial speed of the box
is zero.
Solution:
Figure 4.11: Example 4.7
y
x
a) Find the components of the weight of the object:
122
wx = - mg sin
wy = - mg cos
Write out the two components of Newton's 2nd Law:
x : - mg sin + fk = - ma
y : – mg cos = 0.
Using fk = μk
a
(4.11)
we get,
ma = -μk (mg cos ) + mgsin
= g(sin - μk cos )
= 9.8(sin 30 - 0.2cos 30) = 3.20 m/s2
(4.12)
b) Since a is constant and v2 = v02 + 2ax . With x = 1 m , v0 = 0, we have:
v = 2(3.2)(1)  2.53 m/s .
Example 4-8
A 10 kg box is attached to a 7 kg box which rests on a 300 incline. The coefficient
of kinetic friction between each box and the surface is μk = 0.10. Find:a) The rate of acceleration of the system.
b) The tension in the rope.
Solution:
Figure 4.12: Example 4.8
123
We apply the 2nd law separately to each box.
For the 10 kg box:
y direction:
– m2 g = 0
&
x direction:
T - fk2 = m2 a ,
T-μk
= m2 g
= m2 a Then T-μk m2 g = m2a
(4-13)
For the 7 kg box:
y direction:
= m1gcos ,
x direction:
m1gsin
m1gsin
m1gsin
- T - fk1 =
=
- T - µk
m1a
m1a
=
m1 a
- T –μkm1gcos
(4.14)
We have a system of two equations and two unknowns: a and T. We can solve
as follows.
a) From equation (4.1), T = m2a + μk m2g . Substituting into equation (4.2) gives,
m1a = m1g sin -μk m1gcos - m2a - μkm2g
m1a + m2a = m1g sin - μk m2g - μk m1g cos
(4.15)
124
 a = (1/m1+m2)[m1gsin
- μkm2g - μkm1gcosƟ]gμk]
= (1/17){[7(9.8) sin(30) - (0.1)(10)(9.8) - (0.1)(7)(9.8) cos(30)] }= 1.1 m/s2
(4.16)
b) Then substituting into the first equation gives,
T = m2(a + gμk)
= 10(9.8(.1) + 1.1) = 20.8 N.
Example 4.9
A traffic light weighing 125 N hangs from a cable tied to two other cables fastened
to a support. The upper cables make angles of 370 and 530 with the horizontal.
Find the tension in the three cables.
Solution:
370
T3
530
T2
T1
T3
a
T1
T2
370
b
530
T3
W
Figure


 0
 F
Applying this equation at point a, considering the traffic light as a point, we find:
T3 = W = 125 N
Applying the same equation at point b, we get two equations:
Since the traffic light is at equilibrium, then:
F
x
 0 T2
cos 53 - T1 cos 37  0
125
ext
F
y
 0 T1 sin
37  T2 sin 53 - 125  0
Quiz: using the last two equations to find T1 and T2 , prove that:
T1  75.1 N
T2  99.9 N
Problem 4-10:
A crate of mass 5 kg is placed on a frictionless inclined plane of an angle θ= 300.
Find:i) The acceleration of the crate after it is released.
ii) The normal force on the crate.
y
n
ax
mg
sinθθ
θ
mg cos θ
θ
x
θ mg
Figure
Solution:
Now we apply Newton’s second law in component form, noting that ay = 0:
 Fx  m a x mg sin 
F
y
 0 n - mg cos 
 m ax
0
From the first equation, we get:
ax = g sin θ = 4.9 m/s2
126
From the 2nd equation, we find:
n = mg cos θ =42.43 N
Example4-11:
Two blocks are connected by a light string over a frictionless pulley as shown in
Figure. The coefficient of sliding friction between m1 and the surface is ().
Find:- the acceleration of the two blocks and the tension in the string.
Figure
Solution:
Consider the motion of m1. Since, its motion to the right, then T>f.
Fx = T - f = m1a , Fy = N - m1g = 0  N = m1g
Since f = N = m1g, then:
T = m1(a+g)
For m2, the motion is downward, therefore m2g>T. Note that T is uniform
through the rope. That is the force which acts on the right is also the force
which keeps m2 from free falling. The equation of motion for m2 is:
Fy = T - m2g = - m2a

T = m2(g-a)
Solving the above two equations for T together, we find:
m1(a+g) - m2(g-a) = 0
127
And the tension T is
Example 4-12:
A 3kg block starts from rest at the top of 30o incline and slides a distance of 2m
down the incline in 1.5s. Find:a) The acceleration of the block,
b) The friction force acting on the block,
c) The coefficient of kinetic friction between the block and
the plane, and
d) The speed of the block after it has slid 2m.
Solution
Given m = 3kg,  = 30o, x = 2m, t = 1.5s
x = 1/2at 2  2 = 1/2a (1.5)2
a) a = 1.78m/s2
mg sin30 - f = ma  f = m (g sin30 -a)
Figure
b) f = 9.37N
N–mgcos30 = 0 
 N=
 c)
N = mg cos30
25.46N
k = f / N = 0.368
128
v2 = vo2 + 2a (x-xo )
v2 = 0 + 2(1.78)(2) = 7.11
 d)
v= 2.67 (m/s)
129
4-10:- Exercises
1- A car engine with weight w hangs from a chain that is linked at ring O to
two other chains, one fastened to the ceiling and the other to the wall.
Find the tension in each of the three chains, assuming that w is the given
weight and the rings and chains have no significant weight as in the
Figure.
2- If a net horizontal force of 130 N is applied to a person with mass 60.0 Kg
who is resting on the edge of a swimming pool, what horizontal
acceleration is produced?
3- Find the magnitude of net force required giving a 135-Kg refrigerator an
acceleration of magnitude 1.40 m/s2?
4- A box rests on a frozen pond, which serves as a frictionless horizontal
surface. If a fisherman applies a horizontal force with magnitude 48.0 N
to the box and produces an acceleration of magnitude 3.00 m/s2, what is
the mass of the box?
5- A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on
by a net horizontal force of 140 N.
iii) What acceleration is produced?
130
iv) How far does the crate travel in 10.0 s?
v) What is its speed at the end of 10.0 s?
6- At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81
m/s2. A watermelon weights 44.0 N at the surface of the earth.
i) What is the mass on the earth's surface?
ii) Calculate its mass and weight on the surface of Io?
7- Complete the following:
a) The coefficient of kinetic friction µk=..../... and its SI unit is
……..
8- Choose the correct answer:


Two perpendicular forces F1 and F2 are applied to an object, with
magnitudes F1 = 20 N and F2 = 15 N. The resultant force
magnitude in Newton's is:
A) 25
B) 50
C) 75
D) 100
i)
ii)
The first condition of equilibrium is
A)  Fext  0
iii)
A)
2 v 02
g
B)  Fext  0
C)

F
ext

0
D)

F
ext

0
The maximum range of a projectile is given by:
B)
v02
g
C)
2g
v02
D)
g
2v02
A body of mass 10 kg is moving along an incline of angle 300 with
the horizontal axis. The normal force on this body is……
A) 98 N
B) 86.6 N
C) 49 N
D) 29 N
iv)
131
In the following problem: F is the applied force; fs and fk are the static and
kinetic friction forces.
9- A body is moving with constant speed, then……..
A) F = fs
B) F = fk
C) F >fs
D) F >fk
10- A body of weight 500 N is pulled with a horizontal force 230 N. When
the body started to move, it acquired a constant velocity with a force 200
N. Find the coefficients of kinetic and static friction.
11- An object, weighing 490 N, initially at rest, is acted upon by a net force
of 100 N. How fast is the object moving after 5 seconds?
12- A car having a mass of 800 kg is initially traveling with a speed of 90
km/hr. It has a constant acceleration while braking. it takes 5 seconds to
come to a stop.
a.What is its acceleration?
b. What is the net force on the car shile its breaking?
c. What is the distance the car travels before it stops?
13- A 2 kg ball is dropped from the top of a cliff and falls with a constant
acceleration due to gravity.
a.How fast is the ball moving after 4 seconds?
b. How far has the ball gone after 4 seconds?
c. What is the force on the ball?
14- A car slows down with a constant acceleration when the brakes are
applied. The net force that causes the car to stop is applied on it by the
132
road. If that force has a magnitude of 2500 N and it takes 4 seconds for the
car come to a complete stop when it is initially moving with 60 km/hr,
what is the cars mass?
15- An unknown mass and a 7.3 kg mass are tied to a light string and hung
over a frictionless pulley. If the tension in the string is 63.95 N, what is the
unknown mass?
16- Ali is pulling a 55-kg cart with a force of of 2050 N. Neglecting friction,
what is the acceleration of the cart?
17-
A 6.6 kg mass and a 14.4 kg mass are tied to a light string and hung
over a frictionless pulley. What is their common acceleration?
18-
A 0.6 kg mass and a 4.4 kg mass are tied to a light string and hung
over a frictionless pulley. What is the tension in the sting?
19-
A 7.6 kg mass and a 13.8 kg mass are tied to a light string and hung
over a frictionless pulley. How fast is the heavier mass falling at t=0.50 s if
they are released from rest?
20-
A box is resting on a plane inclined at 200 to the horizontal. If the
normal force on the box from the surface is 90 N, what is the mass of the
box?
133
Chapter 5
134
WORK, ENERGY AND POWER
Main contents
5-1
Work
5-2
Kinetic Energy and the Work Energy Theorem
5-3
Gravitational Potential Energy
5-4
Potential Energy Stored in a Spring
5-5
Conservation Laws
5-5-1
Conservative and Non-Conservative Forces
5-5-2
The Conservation of Mechanical Energy
5-5-3
Non-Conservative Forces and the Work-Energy Theorem
5-6
Power
5-7
Solved Problems
5-8 Exercises
135
Work, Energy and Power
In this chapter, we introduce the concepts of work, energy and power. We
define kinetic energy, gravitational potential energy, and the potential energy
stored in a compressed or stretched spring. If all forces are conservative, the
mechanical energy of an isolated system is constant. If non-conservative forces
are present, we use the work-energy theorem to equate the work done by the
non-conservative forces and the change in mechanical energy.
5-1 Work
Definition: The work done by an agent exerting a constant force ( ) and
causing a displacement ( ) equals the magnitude of the displacement, ,
times the component of along the direction of . In Figure 5.1, the work done
by is:
W=F
cos ( ).
Figure 5.1: Work
Note:


If
= 0
W = 0. (i.e.: no work is done when holding a heavy box, or
pushing against a wall).
W = 0 if
(i.e.: no work is done by carrying a bucket of water
horizontally).
136




The sign of W depends on the direction of relative to : W > 0 when
component of along is in the same direction as , and W < 0 when it
is in the opposite direction. This sign is given automatically if we write (θ)
as the angle between and
and write W = FΔr cos .
If acts along the direction of then W = F , since cos = cos 0 = 1.
Work is a scalar.
The SI units of work are Joules (J) (1 Joule = 1 Newton meter). In CGS
units are: 1 erg = 1 dyne cm.
Example 5-1:
A man cleaning a floor by pulling a vacuum cleaner with a force of
magnitude F=50 N at an angle of 30° with the horizontal (Figure. 2a). Calculate
the work done by the force on the vacuum cleaner as the vacuum cleaner is
displaced 3.00 m to the right.
Figure 5.2
137
Solution:
Fd cos= 50 (3) cos (30°) =130 N.m =130 J
The normal force n , the force of gravity w = mg, and the upward component of
the applied force (50.0 N) (sin 30.0°) do no work on the vacuum cleaner because
these forces are perpendicular to its displacement.
Example 5-2:
A particle moving in the x-y plane undergoes a displacement
a constant force
as
acts on the particle.
a) Calculate the magnitude of the displacement and that of the force.
b) Calculate the work done by
.
Solution:
a) d = 3.6m , F = 5.4 N
b) W=
The angle between
and
is equals 35°
5-2 Kinetic Energy and the Work Energy Theorem
Idea: Force is a vector, work and energy are scalars. Thus, it is often easier to
solve problems using energy considerations instead of using Newton's laws (i.e.
it is easier to work with scalars than vectors).
Definition: The kinetic energy (K.E) of an object of mass m that is moving with
velocity v is:
138
K.E = 1/2 mv2
(5-1)
Note:


Kinetic energy is a scalar.
The units are the same as for work (i.e. Joules, J).
Relation between KE and W: The work done on an object by a net force equals
the
change
in
kinetic
energy
of
the
object:
W = K.Ef – K.Ei
(5-2)
This relationship is called the Work-Energy Theorem.
Proof (for parallel to ):
1. W = FΔr W = (ma)Δr (by Newton's second law).
2. From the third equation of motion: aΔr = ( - )/2
K.Ef–K.Ei .
W = ½ m(
-
)=
Note:



If the speed of an object increases ( v> v0 ) W > 0.
If W < 0 then the object is doing work on the agent exerting the net force.
Interpretation of Eq.(5.2): We can think of K.E as the work an object can
do in coming to rest.
Example 5-3:
A 6.0-kg block initially at rest is pulled to the right along a horizontal,
frictionless surface by a constant horizontal force of 12 N (figure 3.a). Find:-a)
the speed of the block after it has moved 3.0 m.
139
Figure 5.3
The work done by this force is:
Using the work – kinetic energy theorem:
vf=3.5 m/s
b)Find the final speed of the block described in (figure5.3.b), if the surface
is not frictionless but instead has a coefficient of kinetic friction of 0.15.
Applyingthe work – kinetic energy theorem in the case of kinetic friction:
vf = 1.8m /s
140
Example: 5-4
To move a car of mass 1000 Kg along a level road you need to apply a force of
magnitude of 200 N, in the direction of motion to overcome the friction force
and get going. If the frictional force is 100N and you push the car 20 m. Find :the speed of the car?
Solution
The net force on the car is
= 200-100 = 100 N,so if a is the constant
acceleration ,Newton's second law gives:
100 = ma
Then 100 = 1000 a
Giving a = 100/1000 = 0.1 m/s2
Since the car is initially at rest, if v is its speed after being pushed 20 m is:
So, vf = 2m/s
5-3 Gravitational Potential Energy
Definition: Gravitational Potential Energy ( P.Eg ) is given by:
(5-3)
P.Eg = mgy
Where m is the mass of an object, g is the acceleration due to gravity, and y is
the distance the object is above some reference level.
141
The term ``energy'' is motivated by the fact that potential energy and kinetic
energy are different aspects of the same thing (mechanical energy).
For Example: When an object is dropped from rest at some height above the
earth's surface, it starts with some P.Eg but no K.E. As the object falls towards
the Earth, it loses (P.Eg) and gains K.E. Just before the object hits the ground, it
has lost its entire initial P.Eg but gained an equal amount of K.E.
Proof: Find the work done by the force of gravity (conservative force) when an
object falls from rest at position yi to yf= 0 . We have W = Fs , F = |m | = mg and
s = (yi - yf) = yi . This gives, W = mgyi .
Combining with Eq.(5.2) gives m(v f2 - 0) = mgyi or PEi = KEf .
5-4 Potential Energy Stored in a Spring
Definition: The spring constant, k , is a measure of the stiffness of a spring
(large k stiff spring, small k soft spring).
To compress a spring by a distance x we must apply a force Fext = k x . By
Newton's 3rd law, if we hold a spring in a compressed position, the spring
exerts a force Fs = - k x . This is called a linear restoring force because the force
is always in the opposite direction from the displacement.
Note

The sign of Fs shows that the spring resists attempts to compress or
stretch it; therefore Fs is a restoring force.
For Example: In Figure (5.4a) x = xf - xi = - 5 which gives Fs = - k(- 5) = 5k
. This force is positive and therefore directed to the right. This means that
the spring resists the compression. In Figure (5.4b) x = xf - xi = 3 which
gives Fs = - 3k . The negative sign indicates that the force is to the left and
that the spring resists the stretching.
142
Figure (5.4): a) Compressed spring b) Stretched spring


The farther we compress or stretch the spring, the greater the restoring
force.
We usually define xi = 0 and xf = x which gives Fs = - kx. This is called
Hooke's law.
To find the potential energy stored in a compressed (or stretched) spring, we
calculate the work to compress (or stretch) the spring: the force to compress a
spring varies from Fext = F(0) = 0 (at xi = 0 ), to Fext = Fx = kx (at xf = x ). Since force
increases linearly with x , the average force that must be applied is
= ½ (F0 + Fx) = ½ kx
The work done by external force
in the spring as potential energy:
is W = Fext .x = ½ kx2. This work is stored
P.Es = ½ kx2
(5-4)
Note:



5-5
P.Es = 0 when x = 0 (at equilibrium).
P.Es always > 0 when the spring is not in equilibrium.
P.Es is the same if x =  xf (same P.Es for equal expansion or compression).
Conservation Laws
There are many forms of energy - mechanical, chemical, electrostatic, heat,
nuclear. In any isolated system, energy can be transformed from one kind to
143
another, but the total amount of energy is constant (conserved). Example: a
battery contains chemical energy and can be used to produce mechanical
energy. Example: when a block slides over a rough surface, the force of friction
gives rise to the heating of block and surface. As a result, mechanical energy is
transformed into heat energy, but total energy is constant. In this section, we
are interested in two kinds of mechanical energy. They are motion energy
(K.Es)(i.e., this is the energy something has because it is moving), and stored
mechanical energy (P.Es) (i.e., this is energy something has stored in it because
of its height above the ground or because it is stretched or bent).
5-5-1 Conservative and Non-Conservative Forces
Idea: It is not always true that the work done by an external force is stored as
some form of potential energy. This is only true if the force is conservative:
Wc = Fcext(xB - xA) = P.EA– P.EB
(5-5)
Definition: The work done by a conservative force on an object in moving it
from A to B is path independent - it depends only on the end points of the
motion this is called conservative force. Examples such as: the force of gravity
and the spring force are conservative forces. For a non-conservative (or
dissipative) force, the work done in going from A to B depends on the path
taken. Examples such as: friction and air resistance forces.
5-5-2 The Conservation of Mechanical Energy
Definition: Mechanical energy is the kinetic energy plus all of the kinds of
potential energy that are present. In the absence of non-conservative forces,
mechanical energy is conserved. Example: if gravitational and spring forces are
present:
K.Ei + P.Egi + P.Esi = K.Ef + P.Egf + P.Esf.
144
(5-6)
Notice that while the total amount of energy is conserved, the distribution of
energy may change. For example, there may be more K.E in the initial state and
more P.E in the final state (or the other way around).
5-5-3 Non-Conservative Forces and the Work-Energy Theorem
Idea: If there are non-conservative forces then mechanical energy is not
conserved. We write,
W = Wnc + Wc = K.Ef – K.Ei
(5-7)
Where Wc is the work done by conservative forces and Wnc is the work done by
non-conservative forces. Since Wc = PEi - PEf we have,
Wnc = (K.Ef– K.Ei) + (P.Ef– P.Ei)
(5-8)
The work done by non-conservative forces is equal to the change in mechanical
energy.
Example 5-5:
A ball of mass m is dropped from a height h above the ground, as shown in
Figure 5.5. Neglecting air resistance, determine the speed of the ball when it is
at a height y above the ground.
145
Figure 5.5
Solution
Applying the conservation of Mechanical energy:
+mgy
As example if h = 100 cm and y = 30 cm, find the final velocity?. Do it yourself!
Example 5-6:
A block having a mass of 0.80 kg is given an initial velocity vA=1.2 m/s to the
right and collides with a spring of negligible mass and force constant k=50 N/m,
as shown in Figure 5.6.
146
(a) Assuming the surface to be frictionless, calculate the maximum
compression of the spring after the collision.
(b) Suppose a constant force of kinetic friction acts between the block and the
surface, with μk =0.5. If the speed of the block at the moment it collides with
the spring is vA = 1.2 m/s, what is the maximum compression in the spring.
Figure 5.6
Solution:
a)Because mechanical energy is conserved(conservative force), the kinetic
energy of the block before the collision must equal the maximum potential
energy stored in the fully compressed spring:
147
b) In this case, mechanical energy is not conserved because a frictional force
acts on the block. The magnitude of the frictional force is:
Applying the mechanical energy theorem:
The physically meaningful root is xB = 0.092 m
The negative root does not apply to this situation because the block must
be to the right of the origin (positive value of x) when it comes to rest. Note that
0.092 m is less than the distance obtained in the frictionless case of part (a). This
result is what we expect because friction retards the motion of thesystem.
5-6
Power
Definition: Power is the time rate of doing work or, the amount of work done
per second.
In everyday life, the amount of work an apparatus can do is not always
important. In general it is more important to know the time within which a
certain amount of work can be done. For example: the explosive effect of
dynamite is based on its capability to release large amounts of energy in a very
short time. The same amount of work could have been done using a small space
heater (and having it run for a long time) but the space heater would cause no
explosion. The quantity of interest is power. The power tells us something
148
about the rate of doing work. If an amount of work W is carried out in a time
interval Δt, the average power for this time-interval is:Average Power:
=
(5-9)
Where t is the time interval in which the work is done.
P = Fv cosθ, so in a circular motion the power is zero (i.e., F is perpendicular to
v).
Instantaneous Power:
Notes:



5-7
Power is a scalar.
SI Units: 1 Watt (W ) = 1 Joule/sec = 1 kg m2/ s3
British Engineering Units: 1 horsepower (hp) = 746 W= 3.6x106 J.
Solved Examples
Example 5-7
A 2000 kg car is travelling 50 miles per hour. Find:i) The kinetic energy in Joules.
ii) The same car is lifted vertically upward and then dropped from rest.
Find the height from which it is dropped if it strikes the ground at 50 miles
per hour (neglect air resistance).
Solution:
149
a)
KE = 1/2 mv2
1
m
1 hr 
 mi
2x 103 Kg 50
x 1609
x
= 2
mi 3600 s 
 hr
= 4.99 x 105 J


2
(5-10)
b)
PEi = KEf
mgh = 1/2 mv2
v2
h = 2g
2
(5-11)
1
 50(1609) 
 25.5m


=  3600  2(9.8)
Example 5-8
An object of mass 1 Kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is µk = 0.10. Use the work - energy theorem to find
the distance the object travels before coming to rest.
Solution:
Figure 5.7
The work- energy theorem gives W = K.E. We have W = - fkd = - μkNd = μkmgd and K.E. = 1/2 mv2f – 1/2 mv2i = - 1/2 mv2i. Combining:
- μk mgd = - 1/2 mv2i
150
1
d = 2 g
=
v i2
25
 13m
2(0.1)(9.8)
(5-12)
Example 5-9
A 30 kg child enters the final section of a waterslide travelling at 2.0 m/s. The
final section is 5.0 m long and has a vertical drop of 3.0 m. The force of friction
opposing the child's motion is 50 N. Find:
a) The loss of potential energy,
b) The work done by friction in the final section and
c) The child's velocity at the end of the section (using energy considerations).
Solution:
Figure 5.8
a)
P.E = mg(hf - hi)
=30(9.8)(0 - 3) = - 882 J
(5-13)
b)
W = - fkx = - 50(5) = - 250 J
c)
Wnc =
K.E +
P.E
151
(5-14)
- 250 = 1 (30)(v 2 )  1 (30)(2.0)(2)  882
f
2
2
v f2 = 2 250  882  1  30  2 2 


30 

2
vf = 6.8 m/s
(15)
Example 5-10
A 2.0 Kg wood block is on a level board and held against a spring of spring
constant k=100 N/m which has been compressed 0.1 m. The block is released
and pushed horizontally across the board. If the coefficient of friction between
the block and the board is μk = 0.20. Find:a) The velocity of the block just as it leaves the spring and
b) The distance the block travels after it leaves the spring.
Solution:
a)
The work - energy theorem gives:
Wnc = KE + PEs
- fkx = (1/2 mv2f - 0) + (0 – 1/2 kx2)
- μkmgx = 1/2 mv2f – 1/2 kx2
b)
vf = 0.33 m/s.
The Work -Energy theorem gives,
(5-16)
- μkmgd =
1
d
=
2k g
 0.33  0.028m

2  0.2  9.8 
2
v
2
i
152
(5-17)
Example 5-11
A man pushes a 100 kg box across a level floor at a constant speed of 2.0 m/s for
10 s. If the coefficient of friction between the box and the floor is μk = 0.20. Find:a) The instantaneous power output by the man.
b) The displacement done by the man on the box.
c) The friction force fk, and
d) The Work done by the man
Solution:
Figure (5-9)
a) Since the acceleration of the box is zero, the force exerted by the man is
obtained from Fman - fk = 0 Fman = fk = μkmg . Then
P = W/Δt = F = µkmg = 0.2(100)(9.8)(2) = 392 W.
(5-18)
b)-The final displacement: - x = v t = 2.0 (m/s) . 10( s )= 20 m
c)- The friction force fk is given by μkmg = 0.2(100 Kg) (9.8 m/s2) = 196 N
d)- The work done by the man is fkd = 196(20 m) = 3920 J
Example 5-12
A motor raises a mass of 3 kg to a height h at a constant speed of 0.05 m/s. The
battery (not shown) which provides energy to the motor originally stores 4 J of
energy, all of which can be used to lift the mass.
153
(a) What is the power developed in the motor?
(b) To what maximum height can the motor lift the mass using its stored
Motor
energy?
Solution
(a)
P
W Fh mg h


 mg v  3kg10 m / s 2 0.05 m / s   1.5 Watts
t
t
t
h
v
Figure
m
(b)
W  mgh
h
4J
W

 0.13 m
mg 3kg  10 m / s 2


Example 5-13
A car moves along a horizontal road against a resistance of 400 N. What is the
greatest speed in (km/h) the car reach if the engine has a maximum power of 16
kW.
Solution
16000 W = 400(N) v
v = 40 m/s
Example 5-14
An elevator has a mass of 1100 kg and carries a maximum load of 774 kg. A
constant frictional force of 3280 N retards its motion upward. What must be the
154
minimum power delivered by the motor to lift the elevator at a constant speed
of 4.3 m/s.
Solution
Total mass = me + mload = 1100 +774 = 1874 kg
Friction force upward = 3200 N
v = 4.3 m/s
F- fk = weight = mg
F = mg + fk = 1874x9.8 + 3200 = 21565.2 N
Pmin. = F.v = 21565.2x4.3 = 92 730.36 Watt
Example 5-15
A 1.00x103-kg elevator carries a maximum load of 8.00x102 kg. A constant
frictional force of 4.00x103 N retards its motion upward, as in Figure below.
What minimum power, in kilowatts and in horsepower, must the motor deliver
to lift the fully loaded elevator at a constant speed of 3.00 m/s?
Strategy:- To solve this problem, we need to determine the force the elevator’s
motor must deliver through the force of tension in the cable, T. Substituting this
force together with the given speed v into P = Fv gives the desired power. The
tension in the cable, T, can be found with Newton’s second law.
Solution
Apply Newton’s second law to the elevator:
155
The velocity is constant, so the acceleration is zero. Theforces acting on the
elevator are the force of tension in the cable, T , the friction f, and gravity , Mg
where M is the mass of the elevator.
Write the equation in terms of its components: T- f-Mg = 0
Solve this equation for the tension T and evaluate it:
T = f +Mg
T = 4.00x103 N + (1.80x103 kg)(9.80 m/s2) = 2.16x104 N
Substitute this value of T for F in the power equation:
P = Fv = (2.16 x104 N)(3.00 m/s) = 6.48 x104 W
Remarks The friction force acts to retard the motion, requiring more power. For
a descending elevator, the friction force can actually reduce the power
requirement.
156
5-8
a.
Exercises
A futon of mass 40 kg is moved 2.5m through a dogleg path of 2 and 1.5 m
long because the presence of a heavy coffee table. How much more work
must be done to push the futon if
b.
=0.2?
If a Saturn V rocket with an Apollo spacecraft attached has a combined mass
of 2.9 × 105 kg and is to reach a speed of 11.2 km/s , how much kinetic
energy will it then have?.
c.
( 1.8x 1013 J)
If an electron (mass m = 9.11 × 10−31 kg) in copper near the lowest possible
temperature has a kinetic energy of 6.7×10−19 J, what is the speed of the
electron?
d.
( 1.21x106 m/s)
A floating ice block is pushed through a displacement of d = (15m)i − (12m)j
along a straight embankment by rushing water, which exerts a force F =
(210N)i−(150N)j on the block. How much work does the force do on the
block during the displacement?
e.
( W= 4950 J)
A 40 kg box initially at rest is pushed 5.0m along a rough horizontal floor
with a constant applied horizontal force of 130N. If the coefficient of friction
between the box and floor is 0.30. Find:i.
The work done by the applied force,
157
f.
ii.
The energy lost due to friction,
iii.
The change in kinetic energy of the box, and
iv.
The final speed of the box.
A force of 50 N acting on a body displaces it through 10m at an angle of 30 0,
calculate the work done by the force.
g.
When a conservative force does positive work on a body, what happens to
the potential energy of the body?
2- Choose the correct answer:
i)
A)
ii)
A)
iii)
The kinetic energy of particle has a velocity v and a mass m is:
B)
D)
The gravitational potential energy of an object at a height h and has
a mass m is:
B)
C)
D)
The elastic potential energy of spring stretched a distance x and has
a constant spring k is:
B)
iv)
C)
C)
D)
Which one of the following expressions has the same units as work?
A)
B)
C)
D)
v)
A ball is projected vertically upwards and then returns to the
ground. Ignore all frictional forces. Which of the following
158
statements about the kinetic energy and potential energy of the ball
is true?
A)
B)
C)
D)
3- Choose the correct answer
1- Which of the following is NOT a correct unit for work?
a. erg
b. ft·lb
c. Watt
d. Newton·meter
e. Joule
2- Which of the following groups does NOT contain a scalar quantity?
a. velocity, force, power
b. displacement, acceleration, force
c. acceleration, speed, work
d. energy, work, distance
e. pressure, weight, time
3- A boy holds a 40-N weight at arm’s length for 10 s. His arm is 1.5m above the
ground. The work done by the force of the boy on the weight
while he is holding it is:
a. 0
b. 6.1J
c. 40 J
d. 60 J
e. 90 J
4- An object of mass 1 g is whirled in a horizontal circle of radius 0.5m at a
constant speed of 2m/s. The work done on the object during one
revolution is:
159
a. 0
b. 1 J
c. 2 J
d. 4 J
e. 16 J
5- The work done by gravity during the descent of a projectile:
a. is positive
b. is negative
c. is zero
d. depends for its sign on the direction of the y axis
e. depends for its sign on the direction of both the x
and y axis.
6- A body of mass m=50 g is dropped from rest at a height h=12m
above the Earth's surface. What will its speed be just before it
strikes the ground?
7- A block of mass m=3.63 kg slides on a horizontal frictionless table
with speed of v=1.22 m/s. It is brought to rest by compressing a
spring in its path By how is the spring compressed if its force
constant k is 120 N/m?
8- The kinetic energy of an electron is 6.71 x 10-19 J. Find its velocity if
you know that the mass of the electron is 9.11x10-31 kg.
9- A truck has a weight of 2.94x105 N. It takes 4 min to go from a
height of 300 m to a height of 4000 m. Fund the work done due to
the gravity.
10-
If a force
displaces a 0.25-kg object by
. Find the work done and the speed of the object?
160
11A 5-kg body is initially at a height 20 m above the ground. It is
dropped freely. What is its speed 1.5 s later? What is the average
power delivered by the body at this time?
12A 125g block is attached to a horizontal spring with a force
constant 0.5 N/m. The spring with the block is compressed 0.06 m
and then released. What is the maximum velocity of the block?
13A rock is dropped from a height of 15.2 m. What is its speed
(in m/s) when it is half way to the ground?
14A rock is dropped into a 200-m-deep well. What is its speed at
that depth?
15A 50 g ball is thrown from a window with an initial velocity of
8 m/s at an angle of 300 above the horizontal. Using energy
methods, determine:a. The kinetic energy of the ball at the maximum height, and
b. Its speed when it is 3.0 m below the window.
16A ball is kicked off with initial speed of 19.5 m/s at an angle of
450 with the horizontal. Find the maximum height reached by the
ball.
17A ball is thrown vertically upward with initial velocity of 20
m/s. Find the maximum height the ball can reach.
18A block of mass 1 kg is released from rest at the top of a 1-m
high incline plane making an angle of 300 with the horizontal. The
161
coefficient of kinetic friction is 0.3. Find the speed of the block
after it has traveled I m down the incline.
19A 1 kg block collides with a horizontal weightless spring of
force constant 2 N/m. The block compresses the spring 5 m from
its equilibrium position. The coefficient of Kinetic friction is 0.25.
What was the speed of the block at the time of collision?
20A 125 g block attached to a horizontal spring with a spring
constant 0.5 N/m, rests on on a horizontal frictionless surface. The
spring is compressed 0.06 m and the block is released. What is the
maximum speed of the block?
162
Chapter 6
163
LINEAR MOMENTUM, IMPULSE AND COLLISIONS
Main contents
6-1
Linear Momentum and Impulse
6-2
Conservation of Linear Momentum for Two particle system
6-3
Collisions and Kinetic Energy
6-3-1 Types of collisions
6-3-2 Head on Collisions and Glancing Collisions
6.4
Solved Examples
6-5
The Center of Mass
6-6
Exercises
164
LINEAR MOMENTUM, IMPLUSE AND COLLISIONS
In
this chapter, we define momentum and impulse. We discuss the
conservation of momentum in collisions and the conservation of kinetic energy
in the context of three different types of collisions.
6.1
Linear Momentum and Impulse
Definition: Linear momentum

The linear momentum of a particle of mass m moving with a velocity v is
defined as:


P  mv
(6-1)
Note:


Momentum is a vector.
The units of linear momentum are: (kg. m/s) (N.s).
We can rewrite Newton's Second Law in terms of momentum:
 F ma  m
v p

t
t
(6-2)
Interpretation: Force can be expressed as:
Definition: Impulse ( )
The impulse of any body is equal to the change in its momentum this is called
(Impulse – momentum theory), and is given by the following equation:-
165
I  F t  p
(6-3)
Note: Usually the force is a strongly dependent function of time. In this case, we
need to use the average force in the above equation.
Example 6-1: Breakup of a Kaon at Rest
One type of nuclear particles, called the neutral kaon (K), breaks up into a pair of
other particles called pions that are oppositely charged but equal in mass, as
illustrated in Figure 6.1Assuming the kaon is initially at rest, prove that the two
pions must have momenta that are equal in magnitude and opposite in
direction.
Figure 6.1 A kaon at rest breaks up spontaneously into a pair of oppositely charged pions. The pions
move apart with momenta that are equal in magnitude but opposite in direction.
Solution: The breakup of the kaon can be
Written as K0   -   
Because the kaon is at rest before the breakup, then the initial momentum is: Pi
=0
From the momentum conservation: Pi = Pf
 Pf  P   P   0  P   - P 
166
Where P+ is the momentum of the positive pion and P- is the momentum of the
negative pion
6.2
Conservation of Linear Momentum for Two particle system
Momentum is conserved in any collision if the effect of any external forces
present is negligible relative to the effect of the collision. Consider a collision of
two particles A and B in one dimension as shown in Figure (6.2). Note that, the
particles after collision may have the same direction or vice versa.
Figure 6.2: 1-D Collision
Apply
the
impulse-momentum
theorem
F1t =
p1 = m1v1f – m1v1i
F2 t =
p2 = m2v2f – m2v2i
to
m1
and
m2
separately,
Where F1 = the average force of m2 on m1 , and F2 = the average force of m1 on
m2 . By Newton's third law F1(t) = - F2(t) which gives F1  F2 and so,
 F  F  t  m v
1
2
1 1f
 m1v 1i  m 2v 2f  m 2v 2i
p1f + p2f = p1i + p2i
This is the statement of the conservation of momentum.
167
(6-4)
Note:


The system must be isolated: the effect of all external forces acting on m1
and m2 must be negligible.
The conservation of momentum holds for a collision involving any
number of objects:
p

  p ni
nf
(6-5)
Momentum is a vector, and each component is conserved separately. The
equation for conservation of momentum really contains three equations,
one for each dimension.
Example 6- 2:
In
a
particular
crash
test,
an
automobile of mass 1500 kg collides
with a wall, as shown in Figure 6.3 The
initial and final velocities of the

automobile are
vi
= -15.0 m/s

and v f = 2.6 m/s, respectively. If the
collision lasts for 0.150 s. Find: - the
impulse caused by the collision and
6.3(a) this car’s momentum changes as a result of its
the average force exerted on the automobile. Figure
collision with the wall. (b) In a crash test, much of the car’s initial
kinetic energy is transformed into energy used to damage the car.
Solution:
The initial and final momenta of the automobile are







P i  m v i  1500 x ( - 15 i )  - 2.25 x 10 4 i kg.m/s

P f  m v f  1500 x ( 2.6 i )  0.39 x 10 4 i kg.m/s
The impulse is given as:
168



 I   P  2.64 x 104 i kg.m/s
The average force exerted on the automobile is


P
5
 F av 
 1.76 x 10 i N
t

6.3
Collisions and Kinetic Energy
Idea: Momentum is conserved for any isolated collision, but kinetic energy is
usually not. Kinetic energy can be converted into thermal energy and internal
elastic potential energy (because of deformations).
6.3.1 Types of collisions

1. Inelastic collision: P conserved, but not K.Es. Example - rubber ball on a
hard surface (ball deforms internal elastic P.E).
m1v1i + m2v2i = m1v1f + m2v2f
1
1
1
1
( m1v 12i  m 2v 22i )  ( m1v 12f  m 2v 22f )
2
2
2
2
2. Perfectly inelastic collision: Two objects stick together


. P conserved, but not K.E. Conservation of P gives,
m1v1i + m2v2i = (m1 + m2)vf
1
1
1
( m1v 12i  m 2v 22i )  (m1  m 2 )v f2
2
2
2
(6-6a)
v1f = v2f vf
(6-6b)
Example- two lumps of clay or two cars as in the figure:-
169

3. Elastic collision: P and K.Es are conserved. Example - two billiard balls (no
deformations). We have,
m1v1i + m2v2i = m1v1f + m2v2f
(6-7)
1
1
1
1
m1v 12i  m 2v 22i  m1v 12f  m 2v 22f
2
2
2
2
(6-8)
By combining these two equations we obtain a third (dependent) equation that
tells us that the relative velocity before collision is the negative of the relative
velocity after collision:
v1i – v2i = - (v1f - v2f)
(6-9)
Figure: Schematic representation of an
Figure: Schematic representation of a
elastic head-on collision between two
particles: (a) before collision and (b) after
collision.
perfectly inelastic head-on collision
between two particles :(a)before collision
and (b)after collision.
The following Table describes the comparison between the different types of
collisions:
170
Collision type
(E.C)
( I.E.C)
Momentum
Kinetic energy
Examples
Conserved
Conserved
Two billiard balls
m1v1i + m2v2i = m1v1f + m2v2f
1
1
1
1
m1v 12i  m 2v 22i  m1v 12f  m 2v 22f
2
2
2
2
Conserved
Non- conserved
Rubber ball on a
hard surface
m1v1i + m2v2i = m1v1f + m2v2f ( 1 m v 2  1 m v 2 )  ( 1 m v 2  1 m v 2 )
1 1i
2 2i
1 1f
2 2f
2
P.I.E.C
Conserved
2
2
2
Non- Conserved
1
1
1
( m1v 12i  m 2v 22i )  (m1  m 2 )v f2
2
2
2
6.3.2 Head on Collisions and Glancing Collisions
m1v1i + m2v2i = (m1 + m2)vf
Two cars
Twolumps of clay
Head on Collisions: objects rebound on straight line paths that coincide with
original direction of motion. These collisions can be treated one dimensionally
events.
Glancing Collisions:
The total linear momentum of a system is conserved when the system is
isolated (that is, when no external forces act on the system). For a general
collision of two objects in three-dimensional space, the conservation of
momentum principle implies that the total momentum of the system in each
direction is conserved. However, an important subset of collisions takes place
in a plane. The game of billiards is a familiar example involving multiple
collisions of objects moving on a two-dimensional surface. We restrict our
attention to a single two-dimensional collision between two objects that takes
place in a plane, and ignore any possible rotation. For such collisions, we obtain
two component equations for the conservation of momentum:
m1v1ix+m2v2ix= m1v1f x+m2v 2f x
171
m1v1iy+m2v2iy= m1v1f y+m2v 2f y
We must use three subscripts in this general equation, to represent,
respectively, (1) the object in question, and (2) the initial and final values of the
components of velocity. Now, consider a two-dimensional problem in which an
object of mass m1 collides with an object of mass m2 that is initially at rest, as in
Active Figure below. After the collision, object 1 move at an angle θ with
respect to the horizontal, and object 2 moves at an angle Φ with respect to the
horizontal. This is called a glancing collision.
Figure 6.4: 2-D Glancing collision
Applying the law of conservation of momentum in component form, and noting that the
initial y-component of momentum is zero, we have:-
x-component:
y -component:
m1v1i + 0 = m1v1fcosθ+ m2v2fcosΦ
0 +0 =m1v1f sin θ - m2v2fsin Φ
If the collision is elastic, we can write a third equation, for conservation of energy, in
the form:(*)
If we know the initial velocity vii and the masses, we are left with four unknowns (v1f,
v2f, θ, and Φ). Because we have only three equations, one of the four remaining
quantities must be given in order to determine the motion after the collision from
172
conservation principles alone. If the collision is inelastic, the kinetic energy of the
system is not conserved, and Equation* does not apply.
In glancing collisions, the objects do not smash straight into each other;
they hit each other at an angle. They might push each other up and down in
addition to forward and back (like one dimensional collisions).
You may have seen a glancing collision on a billiard (pool) table.
Momentum is still conserved, but we must work in 2 dimensions:-
(6-10)
Or,
and
(6-11)
(6-12)
Momentum is ALWAYS conserved. It does not matter if the collision is elastic
or inelastic.
Example 6-3:
A car of mass 1800 kg stopped at a traffic light is struck from the rear by a900kg car, and the two become entangled. If the smaller car was moving at 20.0 m/s
before the collision, what is the velocity of the entangled cars after the collision?
Solution:
Pi = m1v1i + 0 = 1.8 x 104kg.m/s
 Pi  Pf

Pf  1.8 x 104 kg.m/s
 Pf  ( m1  m2 ) v f

v f  6.67 m/s
The direction of the final velocity is the same as the velocity of the initially
moving car.
173
Example 6-4: Proton –Proton Collision
Proton 1 collides elastically with proton 2 that is initially at rest. Proton 1 has an
initial speed of 3.5x105 m/s and makes a glancing collision with proton 2, as
shown in Figure 6.5 After the collision, proton 1 moves at an angle of 30° to the
horizontal axis, and proton 2 deflects at an angle 600 to the same axis. Find the
final speeds of the two protons.
Figure 6.5 An elastic glancing
collision between two protons.
Solution:
m1 = m2= m





Momentum conservation: P i  m v1i  m v 2 i  m v1 f  m v 2 f  P f

The y-component of the momentum equation is:
0  v1 f sin 30  v 2 f sin 60  v1 f sin 30  v 2 f sin 60  v1 f  1.73 v 2 f
T
he x-component of the momentum equation is:
v1i  0  v1 f cos 30  v 2 f cos 60  3.5x10 5  v1 f cos 30  v 2 f cos 60
 3.5x10 5  0.87 v1 f  0.5 v 2 f
Hence we have two equations in two unknowns, solving for the unknowns we
find:
v1 f  1.75x10 5 m/s
v 2 f  3.03x10 5 m/s
&
174
6.4
Solved Examples
Example 6-5:
A 0.2 kg baseball is travelling at 40 m/s. After being hit by a bat, the ball's
velocity is 50 m/s in the opposite direction. Find:a)
b)
The impulse
The average force exerted by the bat if the ball and bat are in
contact for 0.002 s.
Solution:
2. Choose the ball's initial direction of motion as the positive xdirection.
I  p
= 0.2(- 50) – 0.2(40) = - 18 kg. m/s
3.
= I/
t = - 18/.002 = - 9000 N
(6-13)
(6-14)
Example 6-6:
A 1000 kg car travelling at 22- m/s (about 50 mi/hr) hits a concrete bridge
support and comes to a stop in 0.5 s. Find:a) The average force acting on the car and
b) If the bridge support had been cushioned so that the stopping time was
increased to 3 s, what would have been the average force?
Solution:
a)
175
F
p
v
m
t
t
 0  22 
4
 1000 
  4.4x 10 N
 0.5 
(6-15)
b)
 0  22 
3
  7.3x 10 N
 3 
(6-16)
= 1000 
Example 6-7:
A 1.0 kg object travelling at 1.0 m/s collides head on with a 2.0 kg object initially
at rest. Find the velocity of each object after impact if the collision is perfectly
elastic.
Solution:
Given: m1 = 1 kg , m2 = 2 kg , v1i = 1 m/s and v2i = 0 m/s, we need to find
and
We use the conservation of momentum to write:
m1v1i = m1v1f + m2v2f
(1)(1) = (1)v1f + (2)v2f(6-17a)
Conservation of kinetic energy gives us:
1
1
1
1
m1v 12i  m 2v 22i  m1v 12f  m 2v 22f
2
2
2
2
2
2
(1)(12) = (1)v 1f  (2)v 2f (6-17b)
We must solve two equations for two unknowns. The first equation gives v2f =
1/2 – ½ v1f. Substituting it into the second equation gives,
176
1 v 1f 
1 = v  2 2 


1  2v 1f  v 1f 2 
2
v  2
 1
0 = 1f
4


2
2
1f
=
2v 12f  1  2v 1f v 12f  2
=
3v 12f  2v 1f  1
0 = (3v1f + 1)(v1f - 1)
(6-18)
The solutions are v1f = 1 m/s andv1f = - 1/3 m/s.The second solution corresponds
to the situation where the two objects miss each other, and the first continues
on with it's speed unchanged. The second solution is the one we want.
Substituting it back into the first equation gives,
v2f = 1/2 (1 - v1f)
= 1/2 (1 + 1/3) = 2/3 m/s.
(6-19)
Example 6-8:
Suppose the collision in the previous problem was completely inelastic. Find:a) The velocity of the objects after impact
b) The fraction of the kinetic energy lost during the collision.
Solution:
a)
The conservation of momentum gives,
177
m1v1i + m2 v2i= (m1 + m2)vf
vf= 1/3 m/s
(6-20)
b)
Then K.Ei = 1/2 (1)(12) = 1/2J and K.Ef = 1/2 (1 + 2)(1/3)2 = 1/6J . The
percentage of K.E that is lost is:-
K.Elost%
 KE i  KE f 
 x 100%
= 
KE i


= 1 2 1 6
12
x 100%  66%
(6-21)
Example 6-9:
In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired
horizontally into a wooden block of mass 0.2 kg. The wooden block is
suspended from the ceiling by a long string as shown in figure (6.6) the
diagram. The collision is perfectly inelastic and after impact the bullet and the
block swing together until the block is 0.12 m above its initial position. Find:a) The velocity of the bullet and block just after impact
b) The velocity of the bullet just before impact. See Figure (6.7).
Solution:
Figure (6.6):-(Example 9)
178
Note: In Figure 6.7, point A is just before the bullet strikes the block, point B is
just after the bullet strikes the block and point C is when the bullet and block
reach their maximum height.
We want to find vB , the velocity of the bullet and block combination just after
impact, and vA , the velocity of the bullet just before impact.
a) Energy is conserved between points B and C:
PEC = KEB
(m1 + m2)gh = 1/2 (m1 + m2)
vB = 2gh
= 2(9.8)(0.12) = 1.53 m/s.
(6-22)
b) Momentum is conserved between A and B:
m1vA + 0 = (m1 + m2)vB
 m1  m 2 
 0.2  0.06 
v



B

1.53  6.6m / s
vA =
m
0.06


1


(6-23)
Example 6-10
Two discs of equal mass are involved in a perfectly elastic glancing collision.
The second disc is initially at rest and is struck by the first disc moving to the
right at 5.0 m/s. After the collision, the first disc moves in a direction that makes
an angle of 370 with its initial direction, and the second disc moves
perpendicularly to the first, as in Figure 6.7. Find the speed of each disc after
the collision.
179
Solution:
Figure 6.7 : Example 10
After Collision
Before Collision
We apply the conservation of momentum in the x and y directions.
x : mv1i = mv1fcos 37o+ mv2fcos 53o
y : 0 = mv1fsin 37o- mv2fsin 53o
The y-equation gives v1f = 1.33v2f . Substituting into the x -equation gives,
5 = (cos 53o+ 1.33cos 37o)v2f
v2f = 3.0 m/s
(6-24)
And,
v1f = 1.33v2f = 4.0 m/s.
(6-25)
Example 6-11:
A car weighing 1500 kg collides with a van weighing 2500 Kg at right angles in
the center of an intersection. A detective arrives at the scene and finds that the
car and van stuck together and skidded 15 m at Ө=530as shown in Figure (6.8).
180
Figure 6.8: Example11
vt
car
van
After pulling out and doing some calculations, he charges the owner of the car
with speeding (in a 60 km/hr zone), as well as running through a stop sign.
How fast was the car going before going through the collision? Take the
coefficient of kinetic friction between the tires and the road to be μk= 0.8 .
Solution:
From the length of the skid mark, he can calculate the speed of the car and the
van immediately after the collision. First, find the acceleration of the pair of
vehicles due to the force of friction.
Ffr = - (mc + mv)a
Ffr

a = m c  mv
=

k g  mc  mv 
mc  mv
  k g  7.84m / s 2
(6-26)
Then, using v = 0 (at the end of the skid), and d = 15 m , he can obtain the
speed immediately after the collision from,
v2 = v02 + 2ad
v0
= 2ad  2(7.84)(15) = 15.33 m/s.
(6-27)
He can use the conservation of momentum to find the initial velocities.
The x -component give :mcvci = (mv + mc)vfcos 53o
vci = ( mv  mc )v cos530  14.76m / s
mc
f
= 53.2 km/hr
(6-28)
181
and the y -component gives,
mvvvi = (mv + mc)vf sin 53o
vvi = mv  mc
(
)v f sin 530  32.7m / s
mv
= 117.6 km/hr> 60 km/hr
Example 6-12
A soccer ball has a mass of 0.40 kg. Initially, it is moving to the left, but then it is
kicked. After the kick it is moving at 45°upward and to the right with speed
(as in Figure 6.9). Find:- the impulse of the net force and the average net force,
assuming a collision time Δt = 0.010 s?.
SOLUTION
Figure (6.9):-(a) Kicking a soccer ball. (b) Finding the average force on the ball from its components.
The ball moves in two dimensions, so
wemust treat momentum and
impulse as vector quantities. We take the x-axis to be horizontally to the right
and the y-axis to be vertically upward. Our target variables are the components
of the net impulse on the ball,
and the components of the averagenet
force on the ball, and we’ll find them using the impulse–momentum theorem in
its component form,
182
EXECUTE: Using cos(450) = sin(450)= 0.707, we find the ball’s velocity
components before and after the kick:
v 1x  20m / s  v 1y  0
v 2 x  v 2 y  (30m / s )(0.707)  21.2 m/ s
The impulse components are
J x  p 2 x  p1x  m (v 2 x v 1x )
 (0.40kg )[21.2 m/ s  (20 m/ s)]  16.5 kg.m/ s
J y  p 2 y  p1 y  m (v 2 y v 1 y )
 (0.40 kg)(21.2 m/ s  0)  8.5 kg.m/ s
The average net force components are
(Fave )x 
Jy
Jx
 1650N  (Fave ) y 
 850N
t
t
The magnitude and direction of the average net force are
Fave  (1650)2  (850)2  1.9x 103 N
   arctan
850N
 270
1650N
The ball was not initially at rest, so its final velocity does not have the same
direction as the average force that acted on it.
Example 6-13:
Two gliders with different masses move toward each other on a frictionless air
track (Figure a). After they collide (Figure b), glider B has a final velocity of
+2.0 m/s (Figure c).
1- What is the final velocity of glider A?
2- How do the changes in momentum and in velocity
compare?
183
SOLUTION
IDENTIFY and SET UP: The total vertical force on each glider is zero, and the
net force on each individual glider is the horizontal force exerted on it by the
other glider. The net external force on the system of two gliders is zero, so their
total momentum is conserved. We are given the masses and initial velocities of
both gliders and the final velocity of glider B. Our target variables are
, the
final x-component of velocity of glider A, and the changes in momentum and in
velocity of the two gliders (the value after the collision minus the value before
the collision).
EXECUTE: The x-component of total momentum before the collision is
p x  m Av A 1x  m Bv B 1x
 (0.50kg )(2.0m / s )  (0.30kg )( 2.0m / s )
 0.40kg .m / s
Figure: - Two gliders collide on an air track.
This is positive (to the right in Figure above) because A has a greater magnitude
of momentum than B. The x-component of total momentum has the same value
after the collision, so
p x  m Av A 2x  m Bv B 2x
184
We solve for:-
v A 2x 
p x  m Bv B 2 x
0.40kg .m / s  (0.30 kg)(2.0 m/ s)

mA
0.50kg
 0.40m / s
The changes in the x-momenta are:
m Bv A 2x  m Av A 1x  (0.50kg )(0.40m / s )  (0.50kg )(2.0m / s )  1.2kg .m / s
m Bv B 2x  m Bv B 1x  (0.30kg )(2.0m / s )  (0.30kg )(2.0m / s )  1.2kg .m / s
The changes in x-velocities are:
v A 2 x v A 1x  (0.40m / s )  2.0m / s  2.4m / s
v B 2 x v B 1x  2.0m / s  (2.0m / s )  4.0m / s
EVALUATE: The gliders were subjected to equal and opposite interaction
forces for the same time during their collision. By the impulse–momentum
theorem, they experienced equal and opposite impulses and therefore equal
and opposite changes in momentum. But by Newton’s second law, the less
massive glider B had a greater magnitude of acceleration and hence a greater
velocity change.
6. 5
The Center of Mass:
In this section, we describe the overall motion of a mechanical system in terms
of a special point called the center of mass of the system. The mechanical
system can be either a system of particles, such as a collection of atoms in a
container, or an extended object, such as a gymnast leaping through the air.
185
The system moves as if the resultant external force were applied to a single
particle of mass M 
m
located at the center of mass.
i
i
The center of mass of the pair of particles described in Figure 6.10 is located on
the x axis and lies somewhere between the particles closer to the more massive
particle.
Its x coordinate is:
m x  m2 x 2
x CM  1 1
m1  m 2
Figure 6.10 The center of mass of two
particles of unequal mass on the x axis is
located at xCM, a point between the particles,
closer to the one having the larger mass.
CM
If the two masses are equal, the center of mass lies midway between the
particles. For a system of many particles in three dimensions, the x- coordinate
of the center of mass is:
x CM 
m
i
xi
i
M
The y and z coordinates of the center of mass are given similarly by:
y CM 
m
i
yi
i
M
& z CM 
m
i
zi
i
M

The center of mass can also be located by its position vector, r CM

r CM 
m

i
ri
i
M
186
Example 6-14:
The Center of Mass of Three Particles
A system consists of three particles located as shown in Figure 6.11a. With m1 =
m2 = 1kg and m3 = 2kg. Find the center of mass of the system.
Figure 6.11 (a) Two 1-kg masses and a single 2-kg mass
are located as shown. The vector indicates the location of
the system’s center of mass. (b) The vector sum of m
Solution:
Using the basic defining equations for the coordinates of the center of mass, we
obtain:
 x CM 
m
i
xi
i
yi
i
M

m1 x1  m 2 x 2  m3 x 3 (1x1)  (1x2)  (2x0)

m1  m 2  m3
11 2

m1 y1  m 2 y 2  m3 y 3 (1x0)  (1x0)  (2x2)

m1  m 2  m3
11 2
 x CM  0.75 m
And:
 y CM 
m
i
M
 y CM  1.0 m
Noting that zCM = 0, then the position vector of the center of mass is





r CM  x CM i  y CM j  0.75 i  1.0 j
187
6-6 Exercises:
1. A 3.00-Kg particle has a velocity of


( 3i  4 j ) m/s.
i.
Find its x and y components of momentum.
ii.
Find the magnitude and direction of its momentum.
2. A 0.100- Kg ball is thrown straight up into the air with an initial speed of 15.0
m/s. Find the momentum of the ball:a) At its maximum height,
b) Halfway up to its maximum height.
3. A water molecule consists of an oxygen atom with two hydrogen atoms
bound to it (Figure below). The angle between the two bonds is 1060. If the
bonds are 0.1 nm long, where is the center of mass of the molecule?
Consider: mH = 1.67 x 10-27 Kg and mO = 26.55 x 10-27 Kg
Figure
4. A particle of mass 2 Kg moves along the x – axis with initial velocity of 3 m/s
. A force F = 6 N is applied for 3s. Find:a. The impulse,
b. The final velocity of the particle after that time.
188
5.
A body of mass 5 Kg initially at rest, then it is moving with a velocity of 2
m/s.Find:a) The linear impulse of the body.
b) If the force acting on the body is 50N, what is the impact
time?
6.
In splitting logs with a hammer and wedge, is a heavy hammer more
effective than a lighter hammer? If so?
7.
Suppose you catch a baseball and then someone invites you to catch a
bowling ball with either the same momentum or the same kinetic energy as the
baseball. Which would you choose? Explain.
8.
When rain falls from the sky, what happens to its momentum as it hits the
ground? Is your answer also valid for Newton’s famous apple?
9.
A car has the same kinetic energy when it is traveling South at 30 m/s as
when it is traveling Northwest at 30 m/s. Is the momentum of the car the same
in both cases? Explain.
10. A truck is accelerating as it speeds down the highway. One inertial frame
of reference is attached to the ground with its origin at a fence post. A second
frame of reference is attached to a police car that is traveling down the highway
at constant velocity. Is the momentum of the truck the same in these two
reference frames? Explain. Is the rate of change of the truck’s momentum the
same in these two frames? Explain.
11. A 4-Kg medicine ball is thrown at 15 m/s and hits a 55-kg boy who is on
his skateboard. If the medicine ball continues with a speed of 4 m/s, how fast
will the boy and skateboard be going after the collision?
(0.8 m/s)
12. A bullet that has mass of 0.1-kg has a velocity of 420-m/s just before it hits
the 1.5-kg target. If the bullet continues with a speed of 300 m/s, how fast will
the target be going after the collision?
(8 m/s)
189
13. A boy, mass 70.0 kg, riding a skateboard, mass 2.0 kg, is traveling 3.0 m/s
east when he attempts to jump forward from his skateboard. If his velocity
immediately after leaving the skateboard is 3.1 m/s [E], what is the velocity of
the skateboard?
–(0.5 m/s)
14. A rifle has a mass of 7-kg and the bullet inside has a mass of 0.7-kg. If the
velocity of the bullet is 350-m/s after the rifle is fired, what is the recoil velocity
of the rifle?
–35 m/s
15.
A white pool ball (m1 = 0.3 kg) moving at a speed of vo1 = +3 m/s collides
head-on with a red pool ball (m2 = 0.4 kg) initially moving at a speed of vo2 = - 2
m/s. Neglecting friction and assuming the collision is perfectly elastic, what is
the velocity of each ball after the collision?
16.
The diagram below shows a collision between a white pool ball (m1 = 0.3
kg) moving at a speed v01 = 5 m/s in the +x direction and a blue pool ball (m2 =
0.6 kg) which is initially at rest. The collision is not head-on, so the balls bounce
off of each other at the angles shown. Find the final speed of each ball after the
collision.
Before
After
+y
+y
vf2
v01
65˚
+x
white
blue
+x
25˚
vf1
190
MCQ
Q 1: Two objects (m1& m2) have equal kinetic energies. How do the magnitudes
of their momenta compare?
(a)
m12
m22
(b)
m1
m
(c) 1
m2
m2
(d) not enough information to tell.
Q 2. Consider two carts, of masses m and 2m, at rest on an air track. If you push
first one cart for 3 s and then the other for the same time, exerting equal force
on each, the momentum of the light cart is
1. four times
2. twice
3. equal to
4. one-half
The momentum of the heavy cart.
Q3.A cart moving at speed v collides with an identical stationary cart on an air
track, and the two stick together after the collision. What is their velocity after
colliding?
1. v
2. 0.5 v
3. zero
4. Need more information
Q4. Underline the correct alternative:
i- The rate of change of total momentum of a many-particle system is
proportional to the external force/sum of the internal forces on the
system.
191
ii- In an inelastic collision of two bodies, the quantities which do not
change after the collision are the total kinetic energy/total linear
momentum/total energy of the system of two bodies.
192
Chapter 7
193
ROTATION OF RIGID BODIES
Main contents
7-1
Introduction
7-2
Angular Velocity and Acceleration
7-3
Formulae for Constant Angular Acceleration
7-4
Relationship between Linear and Angular Quantities
7-5
Centripetal Acceleration
7-6
Solved Examples
7-7
Exercises
194
ROTATION OF RIGID BODIES
In
this Chapter, the quantities needed to describe circular motion will be
defined. These include angular velocity, angular acceleration, tangential
velocity and acceleration and, centripetal acceleration.
7.1
Introduction
Definition: Rigid Body
A rigid body is one that does not deform during its motion. The distance
between any two points in the rigid body remains fixed.
Angles Describe Rotation of Rigid Body
In order to completely describe the rotation of a rigid body about a fixed axis,
O, it is sufficient to give the angle, θ, between the position vector of any point in
the rigid body (for example the point P in the Figure (7.1), and some arbitrary,
fixed reference line.
p
p
o
Figure (7.1): Rotation of a Rigid Body about the axis O.
Definition: Radian
The angle expressed in radians is defined as the ratio of the arc length, s ,
swept out by the angle, to the radius, r , of the corresponding circle:
Ө (radians) = s/r
(7-1)
195
Note:
For a complete revolution, s = 2 r is the circumference, so that the conversions
between revolutions, radians and degrees are given by:
1 revolution = 2 radians = 360 degrees
(7.2)
1 radian = 360/2 degrees = 57.3 degrees.
(7.3)
Thus
7.2
Angular Velocity and Acceleration
Definition: Angular Velocity
Angular velocity wof a rigid body is the rate of change of its angular position.
Thus, if θ=θ1,at t = t1, and θ= θ2, at t = t2, then the average angular velocity ( ) of
the body over the time interval t = t2 - t1 is:

 2  1
t 2  t1


t
(7.4)
For linear motion, the instantaneous velocity is obtained by making the time
interval very small:
ω = dӨ/dt
(7-5)
The units of angular velocity are most conveniently given in rads/sec, but can
also be expressed in revolutions/sec or degrees/sec using the conversions given
above.
Definition:Angular Acceleration:Angular Acceleration is the rate of change of angular velocity with time. The
average angular acceleration (
) of a rigid body over a time interval t = t2 t1 is:
196
(7.6)
The instantaneous angular acceleration is obtained by taking a very small time
interval:
(7.7)
The units of angular acceleration are normally radians/sec2.
7.3
Formulae for Constant Angular Acceleration
In analogy with linear motion, when the angular acceleration, , is constant, we
have:
(7-8)
Ө - Ө0 = t + 1/2 t2
(7-9)
 2  02  2 (  0 )
(7-10)
In the above equations, Ө0and 0 are the angular position and angular velocity
of the rigid body at t = 0. These can be compared to the analogous formulae for
linear motion.
7.4
Relationship between Linear and Angular Quantities
Figure 7.2: Circular Motion
197
Consider an object that moves from point P to P' along a circular trajectory of
radius r , as shown in Figure (7.2).
Definition: Tangential Speed
The average tangential speed of such an object is defined to be the length of arc,
s , travelled divided by the time interval, t :
(7.11)
vt = 0
The instantaneous tangential speed is obtained by taking
t to zero:
s
t 0 t
v t  lim
(7.12)
Using the fact that:
s=r
(7.13)
We obtain the relationship between the angular velocity of an object in circular
motion and its tangential velocity:

 r
t 0 t
v t  r lim
(7.14)
This relation holds for both average and instantaneous speeds.
Note:

The instantaneous tangential velocity vector is always perpendicular to
the radius vector for circular motion.
Definition: Tangential Acceleration
The rate of change of tangential velocity of an object traveling in a circular orbit
or path is known as Tangential acceleration. It is directed towards tangent to
the path of a body as shown in the figure.
198
The average tangential acceleration is:
at 
v t

r
 r
t
t
(7-15)
Where is the average angular acceleration?
Also, it can be given by:
. Where
is the change in angular velocity,
dt is the change in time.
Tangential acceleration in terms of distance is:The instantaneous tangential acceleration is given by:
v t
 r
t 0 t
 at  lim
(7-16)
where is the instantaneous angular acceleration.
Note:

The above formula is only valid if the angular velocity is expressed in
radians per second.
199

The direction of the tangential acceleration vector is always parallel to the
tangential velocity, and perpendicular to the radius vector of the circular
motion.
Tangential acceleration formula is used to calculate the tangential acceleration
and also the parameters related to it. It is expressed in meter per Second Square
(m/s2).
Example 7-1
A car accelerates uniformly on a circular path from a speed of 30 m/s to 60m/s
in 10s. What is its tangential acceleration?
Solution:
Given: Initial velocity vi = 30 m/s,
Final velocity vf= 60 m/s,
Change in velocity dv = vf - vi= 60m/s - 30 m/s = 30 m/s
Time taken dt = tf - ti= 10s - 0s = 10s
The tangential acceleration is given by at = dv/dt = 30/10 = 3 m/s2.
Example 7-2
A runner starts from rest and accelerates at a uniform rate up to 15m/s in the
time interval of 8s, moving on a circular track of radius 40m. Calculate the
tangential acceleration.
Solution:
Given: Initial velocity vi = 0,
Final velocity vf= 15 m/s,
Change in velocity dv = vf - vi = 15m/s - 0 = 15 m/s
Time taken dt = 8s
200
The tangential acceleration is given by at = dv/dt = 1.875 m/s.
7.5 Centripetal Acceleration
Consider an object moving in a circle of radius r with constant angular velocity.
The tangential speed is constant, but the direction of the tangential velocity
vector changes as the object rotates.
Definition: Centripetal Acceleration
Centripetal acceleration is the rate of change of tangential velocity:
v t
t 0 t
ac  lim
(7.17)
Note:


The direction of the centripetal acceleration is always inwards along the
radius vector of the circular motion.
The magnitude of the centripetal acceleration is related to the tangential
speed and angular velocity as follows:
vt2
ac 
 r 2
r

(7.18)
In general, a particle moving in a circle experiences both angular
acceleration and centripetal acceleration. Since the two are always
perpendicular, by definition, the magnitude of the net acceleration attotal
is:
atotal  ac 2  at 2  r 2 4  r 2 2
201
(7.19)
Definition: Centripetal Force
Centripetal force is the net force causing the centripetal acceleration of an object
in circular motion. By Newton's Second Law:
(7.20)
Its direction is always inward along the radius vector, and its magnitude is
given by:
vt2
Fc  mac  m
 m 2r
r
7.6
(7.21)
SolvedExamples
Example 7.3
A bicycle wheel of radius r = 1.5 m starts from rest and rolls 100 m without
slipping in 30 s. Calculate:a) The number of revolutions the wheel makes,
b) The number of radians through which it turns,
c) The average angular velocity.
Solution:
a) If there is no slipping, the arc length through which a point of the rim moves
is equal to the distance travelled, so that the number of revolutions is:
n
b)
c)
100m
 10.6
2 (1.5m )
= 2 n = s/r = 100/1.5= 66.7 radians.
Average angular velocity:
202


= 66.7 radians/ 30 s= 2.22 rads/s.
t
Example 7. 4
Assuming that, the angular acceleration
above(perviousexample) was constant. Find:
of
the
wheel
given
a) The angular acceleration,
b) The final angular velocity
c) The tangential velocity and tangential acceleration of a point on the rim after
one revolution.
Solution:
a) For constant angular acceleration:
Using
= 0 and solving for

b)
c)
gives:
2
2(66.7)

 0.15 rads/s2.
(t )2
(30)2
w= ω0+
t = 0 + (0.15)(30) = 4.5 rad/s.
After one revolution,
= 2 . Using :
 2  02  2
Weget:-
  2 (2 ) = 1.37 rad./s.
The tangential velocity and acceleration are:
-vt = r = (1.5 m )(1.37 rad/s ) = 2.06 m/s
-a t = r
= (1.5 m )(0.15 rads/s 2) =0 .225 m/s2.
203
Homework(to be solved by the student)
A point on a bicycle wheel moves a distance s= 1.0 m. If this point is 0.4
m from the wheel axis, through what angle has the wheel rotated in a)
radians and b) degrees.
Example 7.5
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the
angular speed of the wheel is 2.00 rad/s at
ti  0 .
a)
Through what angle does the wheel rotate in 2.00 s?
b)
What is the angular speed at t= 2.00 s?
Solution:
a)
according to Eq. (7-9),
1
2
 f  i  i t   t 2   2 rad s  2 s  


1
2
2
3.5 rad s  2 s 
2
 11.0 rad  11 rad   57.3 rad   630  1.75 rev
b)
Because the angular acceleration and the angular speed are both
positive, we can be sure that our answer must be greater than 2 rad/s.
 f  i   t  2 rad s   3.5 rad s2   2 s   9 rad s
Example 7.6
204
a) Find the angular speed of the disc in revolutions per minute when
information is being read from the innermost first track (r = 23mm) and
the outermost final track (r = 58mm), if the constant speed of the surface
at the point of the laser–lens system is 1.3 m/s.
b) The maximum playing time of a standard music CD is 74 minutes and 33
seconds. How many revolutions does the disc make during that time?
c) What total length of track moves past the objective lens during this time?
d) What is the angular acceleration of the CD over the 4473-s time interval?
Assume that  is constant.
Solution:
a)
Using Eq. (7-10)
 2  02  2 (  0 ) ,we can find the angular speed; this
will give us the required linear speed at the position of the inner track,
v
1.3 m s

 56.5 rad s
ri 2.3 10-2 m
i 
For the outer track,
f 
v
1.3 m s

 22.4 rad s
rf 5.8 10-2 m
 2.1102 rev min
b)
According to Eq.  
 f  i
t f  ti


, with replacing the average speed 
t
with its mathematical equivalent i   f  2 :
205
 f  i 
 0
1
i   f  t
2
1
 540 rev min  210 rev min 1 min 60 s  4473 s   2.8 104 rev
2
c) Because we know the (constant) linear velocity and the time interval, this
is a straightforward calculation:
x f  vit  1.3 m s  4473 s   5.8 103 m
d) Because we know the (constant) linear velocity and the time interval, this
is a straightforward calculation:
x f  vit  1.3 m s  4473 s   5.8 103 m
d)  
 f  i
t

22.4 rad s  56.5 rad s
 7.6 103 rad s 2
4473 s
206
7-7
Exercises:-
1- A disk is rotating about its central axis. The angular position is given by
  1  0.6t  0.2t 2
Where:t in seconds θin radians, and zero angular position. Find its
angular velocity (ω) and acceleration (α) after timet= 2 s.
2- The angular position  of a flywheel of diameter 0.36m is given by

  2 rad
s3
t
3
Find:i.
 in radians and degrees at t1=2s, t2= 5s.
ii.
The distance covered at that time interval
iii.
The instantaneous velocity at t=5s.
3- A grind stone rotates at constant angular acceleration α =0.3 rad/s2. At
time t=0, it has an angular velocity ωo= -4.5 rad/s and a reference line on
its horizontal, at  o =0.
Find the angular position at t= 0 and t = 2 s.
4- A particle moves with constant angular speed of 2 rad/s.
Find the angular displacement at t = 3 s.
5- A body rotates in a circle with angular speed of 3 rad/s.
Findits periodic time.
207
6- A particle moves with constant angular acceleration of 2 rad/s2.If the
particle starts from rest. Find :
a) The time needed to make an angular displacement of 30 rad and ,
b) The angular velocity at the end of the 30 rad trip.
7- You have :-
  3tiˆ   4t 2  1 jˆ (rad / s )
Find: the magnitude of the angular velocity after 1 sec, the magnitude of
angular acceleration at t = 2 sec.
8- The angular position of a 0.36-m-diameter flywheel is given by:Ө = 12.0 (rad/s3)t3
Find
iii- θin radians and in degrees, at t1 = 2.0 s and t2 = 5.0 s.
iv- The distance that a particle on the flywheel rim moves overthe
time interval from t1 = 2.0 s to t2 = 5.0 s.
v- The average angular velocity, inrad/s and in rev/min, over that
interval.
vi- The instantaneous angular velocities at t1 = 2.0 s and t2 = 5.0 s.
vii- The average angular acceleration att1 = 2.0 s and t2 = 5.0 s. and,
finally,
viiiThe instantaneous angular acceleration at t1 = 2.0s and t2 =
5.0s.
9- Is that correct:a)1 rad/s is about 10 rpm.
b)A rigid body’s average angular velocity(as shown in figure) and
instantaneousangularvelocity can be positive or negative.
208
c)
10- You have finished watching a movie on Blu-ray and the disc is slowing to
a stop. The disc's angular velocity at t =0 is 27.5 rad/s, and its angular
deceleration is a constant -10.0 rad/s2. A line PQ on the disc's surface lies
along the +x – axis at t = 0 s( as shown in figure) .
(a) What is the disc's angular velocity at t = 0.300 s?
(b) What angle does the line PQ make with the + x-axis at this time?.
209
h- Multiple choice questions
1- A radian is about:
A) 250
B) 370
C) 450
D) 570
E) 900
2. One revolution is about the same as:
A) 1 rad
B) 57 rad
C)
/2 rad
D)
rad
E) 2 rad
3. One revolution per minute is about:
A) 0.0524 rad/s
B) 0.105 rad/s
C) 0.95 rad/s
D) 1.57 rad/s
E) 6.28 rad/s
4. If a wheel turns with constant angular speed then:
A) each point on its rim moves with constant velocity
B) each point on its rim moves with constant acceleration
C) the wheel turns through equal angles in equal times
D) the angle through which the wheel turns in each second increases as time goes on
E) the angle through which the wheel turns in each second decreases as time goes on
210
5. If a wheel is turning at 3.0 rad/s, the time it takes to complete one revolution is about:
A) 0.33 s
B) 0.67 s
C) 1.0 s
D) 1.3 s
E) 2.1 s
6. If a wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:
A) 0.31 rad/s
B) 0.63 rad/s
C) 10 rad/s
D) 31 rad/s
E) 63 rad/s
7. The angular speed in rad/s of the second hand of a watch is:
A)
/1800
B)
/60
C)
/30
D) 2
E) 60
8. The angular speed in rad/s of the minute hand of a watch is:
A) 60/
B) 1800/
C)
211
D)
/1800
E)
60
9. A flywheel, initially at rest, has a constant angular acceleration. After 9 s the flywheel has
rotated 450 rad. Its angular acceleration in rad/s2 is:
A) 100
B) 1.77
C) 50
D) 11.1
E) 15.9
10. Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. Its average
angular acceleration is:
A) 3.14 rad/s2
B) 30 rad/s2
C) 30 rev/s2
D) 50 rev/min2
E) 1800 rev/s2
2. During the time interval from
11.
t1 to t2
t2
velocity in rad/s at time t1 is:
A) zero
B) 1
C)
D)  2
212
E) 2
12. A flywheel rotating at 12 rev/s is brought to rest in 6 s. The magnitude of the average angular
acceleration in rad/s2 of the wheel during this process is:
A) 1/
B) 2
C) 4
D) 4
E) 72
13. A phonograph turntable, rotating at 0.75 rev/s, slows down and stops in 30 s. The magnitude of
its average angular acceleration in rad/s2 for this process is:
A) 1.5
B) 1.5
C)
/40
D)
/20
E) 0.75
14. The angular velocity of a rotating wheel increases 2 rev/s every minute. The angular
acceleration, in rad/s2 of this wheel is:
A) 4 2
B) 2
C) 1/30
D) 2 /30
E) 4
213
15. A wheel initially has an angular velocity of 18 rad/s but it is slowing at a rate of 2.0 rad/s2. The
time it takes to stop is:
A) 3.0 s
B) 6.0 s
C) 9.0 s
D) 12 s
E) never stops
16. A wheel initially has an angular velocity of 36 rad/s but after 6.0s its angular velocity is 24
rad/s. If its angular acceleration is constant the value is:
A) 2.0 rad/s2
B) –2.0 rad/s2
C) 3.0 rad/s2
D) –3.0 rad/s2
E) 6.0 rad/s2
17. A wheel initially has an angular velocity of –36 rad/s but after 6.0 s its angular velocity is –24
rad/s. If its angular acceleration is constant the value is:
A) 2.0 rad/s2
B) –2.0 rad/s2
C) 3.0 rad/s2
D) –3.0 rad/s2
E) –6.0 rad/s2
18. A wheel initially has an angular velocity of 18 rad/s but it is slowing at a rate of 2.0 rad/s2. By
214
the time it stops it will have turned through:
A) 13 rev
B) 26 rev
C) 39 rev
D) 52 rev
E) 65 rev
19. A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev
its angular velocity is:
A) 16 rad/s
B) 22 rad/s
C) 32 rad/s
D) 250 rad/s
E) 500 rad/s
20. A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. The time it takes to
make 10 revolutions is:
A) 0.50 s
B) 0.71 s
C) 2.2 s
D) 2.8 s
E) 5.6 s
Appendix
215
Quantities & Units
216
Prefixes
Values of Common angles
217
Important Formula
218
219
220
Newtonian mechanics
The One Dimensional Equations of Motion for Constant Acceleration
traditional name
1st Equation
equation
Relationship
v = v0 + at
velocity – time
2nd Equation
x = x0 + v0t + ½at2
3rd Equation
v2 = v02 + 2a(x − x0)
velocity – displacement
v = ½(v + v0)
average velocity
merton rule
displacement – time
Related Concepts of Dynamics
I
II
1st law
inertia
m
momentum
p=mv
2nd law
force law
F=ma
impulse-momentum theorem
J = Δp
3rd law
action-reaction
F1 = −F2
conservation of momentum
∑ p = ∑ p0
221
The kinematic equations for a simple projectile are those of an object traveling
with constant horizontal velocity and constant vertical acceleration.
Equation
Horizontal
vertical
Acceleration
ax = 0
ay = −g
velocity-time
vx = v0x
vy = v0y − gt
displacement-time x = x0 + v0xt
y = y0 + v0yt − ½gt2
vy2 = v0y2 − 2g(y − y0)
Velocity-displacement
222
Translational and Rotational Quantities Compared
concept
base quantities
Translation
connection
s, r
s= θ×r
x=
y=
r2 =
θ=
Coordinate systems r = x i + y j
Rotation
Θ
r cos θ
r sin θ
r= rr+θθ
x2 + y2
tan−1 (y/x)
velocity
v = Dr/dt
acceleration
a = dv/dt = d2r/dt2 a = α × r − ω2r α = dω/dt = d2θ/dt2
v= ω×r
v = v0 + at
Equations of motion x = x0 + v0t + ½at2
v2 = v02 + 2a(x − x0)
223
ω = dθ/dt
ω = ω0 + αt
θ = θ0 + ω0t + ½αt2
ω2 = ω02 + 2α(θ − θ0)
References:
1-
David Halliday, Robert Resnick, Jearl Walker, "Fundamentals
of Physics", 9th Edition, 2011.
2-
Hugh D. Young, Roger A. Freedman, "University Physics with
Modern Physics", 13th Edition, ,Lewis Ford, 2012.
3-
R. Serway and R. Beichner. "Physics for Scientists and
Engineerswith Modern Physics". 5thEditionHarcourtCollege
Publishers, 2000.
4-
D. Giancole, "Physics: Principles with Applications", 6th
Edition, Pearson Edition, 2004.
5-
R. Knight, "Physics for Scientists and Engineerswith Modern
Physics", 1st Edition, Pearson Education, 200.
224
225