Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre (CDC), Sanothimi, Bhaktapur, as an additional material for schools
MODERN GRADED
SCIENCE
Class 9
Jai Prakash Srivastav By
Bijay Shankar Mishra
Khaga Raj Ghimire
Mahendra Bahadur Thapa
VIDYARTHI PUSTAK BHANDAR
Publisher and Distributor
Kamalpokhari/Bhotahity, Kathmandu
MODERN GRADED SCIENCE CLASS 9
Publisher : VIDYARTHI PUSTAK BHANDAR
Kamalpokhari/Bhotahity, Kathmandu
Ph : 01-5327246, 01-5344834, 01-4423333
email: [emailprotected]
www.vidyarthiprakashan.com
Copyright : Authors
Edition : First Edition, 2000
Seventh Revised Edition, 2020
Eighth Revised Edition, 2021
ISBN : 978-99946-1-955-9
Layout : Vidyarthi Desktop
Printed at :
PREFACE
Advancement of science and technology has reached its apex in the 21st century.
Therefore, understanding of facts, concepts and principles of science has become
a dynamic force in our life. Development of science has brought many changes in
the present civilization. So, science has become one of the major disciplines in our
education system.
Modern Graded Science 9 is a part of recently revised and updated edition of
Modern Graded Science series, brought out for the students of Grade 9 in order
to enhance their knowledge, understanding, application and ability as per the
requirements of the current curriculum of science. The immense popularity of the
previous series among our students and teachers has provided us additional energy
and encouragement to make this edition more practical, more user-friendly and
more comprehensive. Based on the latest approaches to teaching science, the series
envisages learner-centred teaching method and activity-based learning techniques.
Heavy emphasis on developing concepts rather than providing mere information,
special efforts to stimulate learners’ creative thinking encouraging them to get
involved in learning activities, presentation of scientific knowledge in a very logical
and sequential manner in simple and lucid language to keep up with the ability of the
young learners are the main features of this series.
Each chapter of the textbook includes Competencies of the chapter to guide
students and teachers towards the right directions, Some Reasonable Facts to enhance
the conceptual knowledge, Things To Know as the summary of the entire lesson and
Things to Do to develop the skills of project work. Similarly, Test Yourself to test the
learning achievement and understanding, Glossary as a list of technical terminologies
and their meanings and Do You Know to impart the knowledge of new facts are
given in the same order. Plenty of exercise questions of different levels of behaviours
including MCQS, are also introduced in the book. Moreover, the book also incorporates
SEE Specification Gird and two sets of Model Questions.
The authors of this series would like to acknowledge Asso. Prof. Yadab Prasad
Adhikari and Vidyarthi Desktop for their contribution in editing and layout designing
respectively.
Further constructive suggestions and recommendations from subject experts,
teachers, students and well-wishers will be highly appreciated.
Authors
Contents
Chapter-1 Measurement 1
8
Chapter-2 Force 31
55
Chapter-3 Machine 72
92
Chapter-4 Work, Energy and Power 112
126
Chapter-5 Sound 151
161
Chapter-6 Light 175
197
Chapter-7 Electricity and Magnetism 211
222
Chapter-8 Classification of Elements 235
244
Chapter-9 Chemical Reaction 280
299
Chapter-10 Solubility 308
328
Chapter-11 Some Gases 340
353
Chapter-12 Metals 369
381
Chapter-13 Carbon and Its Compounds 391
408
Chapter-14 Water 409
Chapter-15 Chemical Fertilizers Used in Agriculture
Chapter-16 Classification of Plants and Animals
Chapter-17 Adaptation of Organisms
Chapter-18 (A) Cell, Tissue and Organ
Chapter-18 (B) Body Systems
Chapter-19 Sense Organs
Chapter-20 Evolution
Chapter-21 Nature and Environment
Chapter-22 Natural Disasters
Chapter-23 Greenhouse
Chapter-24 The Earth in the Universe
Specification Grid (Science)
Model Question Sets
Chapter Measurement
1
Total estimated Pds: 4/ (3T/ 1P)
Competencies
On completion of this chapter, the students will be competent to:
define fundamental and derived units and give an example of each.
explain SI unit with examples.
show the inter-relation between fundamental units and derived units.
Physics deals with the study of natural phenomena. To approach the subject quantitatively, it is
essential to make its measurement. For example, we know that the earth rotates not only about
its own axis but also revolves around the sun. How much time does the earth take to rotate
about its own axis? How much time does the earth take to revolve around the sun? To answer
these questions, we have to take the precise measurements of time.
A physical quantity is the quantity that can be measured directly or indirectly and consists
of a magnitude and a unit. Mass, length, time, density, temperature, etc. are some examples
of physical quantities. Happiness, sadness, etc. cannot be measured. So, they are not physical
quantities. We cannot find the exact value of any physical quantity without measurement.
For the measurement of any physical quantity, we need a fixed quantity as a standard quantity.
The measurement of unknown quantity can be done by comparing it with a standard quantity
of the same kind. Thus, the process of comparison of an unknown quantity with a known
standard quantity is called measurement. For example, 1 kg sugar is weighed with the help of
a standard mass of 1 kg.
Fundamental quantities and derived quantities
Those physical quantities which can neither be derived from nor be further resolved into
other simpler quantities are called fundamental quantities.
For example, length of a body is a fundamental quantity as it cannot be expressed in terms of
other quantities. How many kilograms equal 1m? It does not make any sense. Length is not
dependent on mass. Length, mass, time, electric current, temperature, luminious intensity of
light and amount of substance are the examples of fundamental quantities.
Those physical quantities which depend on two or more fundamental quantities or power of a
fundamental quantity are called derived quantities. Area is a derived quantity.
Measurement 1
Since area = length × breadth, it depends on twice of length (i.e. 2). Similarly, density is also
a derived quantity as density depends on mass (i.e. m) and thrice of length (i.e. 3). Volume,
pressure, force, speed, acceleration, etc. are some other examples of derived quantities.
Unit
A shopkeeper sales different commodities like sugar in kilogram and standard masses using
a pan balance, clothes in metre using a metre rod, oil in litre using a measuring can, etc.
Similarly, the magnitude of the wave of an earthquake is measured in Richter scale. These
kilogram, metre, etc. are called the units of respective physical quantities. Thus, the reference
standard with which we carry out the measurement of any physical quantity of the same kind
is called unit. For example, two metre is 2 times of one metre. So, one metre (m) is the unit of
length. Different physical quantities have different units for their measurement. For example,
kilogram, second and metre are the units of mass, time and length respectively.
The unit selected for measuring a physical quantity must have the following characteristics:
1. The unit should be well-defined and accepted all over the world.
2. The unit should be easily reproducible.
3. The unit should change neither with time nor with physical conditions like temperature,
pressure, etc.
4. The unit should be universally agreed upon so that the results obtained in different
countries are comparable.
5. The unit should be of a suitable size.
For the measurement of a physical quantity, we consider a ‘unit’ and then find the number
which expresses how many times the ‘unit’ is contained in the physical quantity.
Standard system of units
The people from different places of our country still use local units like cubit (Haat) and mana
for the measurement of length and mass. These units are understood by Nepalese people only
and magnitude of the local unit may vary from person to person. In order to maintain the
uniformity in measurement of physical quantities, the following standard systems of units
have been used:
1. MKS system: In this system, length is measured in metre, mass in kilogram and time
in second. It is also called metric system.
2. CGS system: In this system, length is measured in centimetre, mass in gram and time
in second. It is also called French system.
3. FPS system: In this system, length is measured in foot, mass in pound and time in
second. It is also called British system.
2 Modern Graded Science Class 9
4. SI system: The French name for this system is “Sestéme International d’ Unite’s”
(International system of units). This system is,
in fact, the improved and extended version of
MKS system of units. In the year 1960, the The standard one kilogram mass is
Eleventh General Conference of Weights and the mass of platinium-iridium cylin-
Measures in France held among the scientists der kept at 0°C at the International
recommended SI system of units. This system is Bureau of Weights and Measures in
now used all over the world. Thus, the system of France. This mass is equal to mass of
units which is agreed by the international 1 litre of water at 4°C.
convention of scientists held in France in 1960
is called SI unit.
Advantages of SI units
The advantages of SI units over the other systems of units are as follows:
1. It is a rational system of units. That is, this system makes use of only one unit for one
physical quantity. For example, joule is used as the unit of energy for all types of energy.
2. It is a consistent and coherent system of units. That is, in this system, all the derived
units can be easily obtained from fundamental units.
3. It is a metric system. That is, multiples and sub-multiples of SI units can be expressed
as powers of 10. It makes calculation easy and saves time and space.
Fundamental units and derived units
a. Fundamental units
The units of fundamental quantities are Fundamental quantity Unit Symbol
called fundamental units or basic units. Length () Metre m
Fundamental units are independent of Mass (m)
each other and cannot be further resolved Time (t) Kilogram kg
into other simpler units. Units like metre, Temperature (T)
second, kilogram, kelvin, etc. are the Electric current (I) Second s
fundamental units of length, time, mass Luminous intensity (TV)
and temperature respectively. Thus, the Amount of substance (n) Kelvin K
unit which is independent of other units is
called a fundamental unit. Ampere A
Candela cd
Mole mol
The given table shows seven fundamental quantities and their SI units.
Measurement 3
b. Derived units
The units of derived quantities are called derived units. That is, those units which depend on
two or more fundamental units or power of a fundamental unit are called derived units. For
example, the unit of speed is metre per second or m/s. It clearly depends on two fundamental
units; unit of length (metre) and unit of time (second). Hence, m/s is a derived unit. Units of
area, velocity, acceleration, etc. are some other examples of derived units.
Some derived units and their relation with fundamental units are given in the following table:
S.No. Derived quantity Related formulae Sumbol of Derived Fundamental
unit name units unit involved
1. Area (A) length×breath (×b) square metre m2 m×m
2. Volume (V) length × breadth× height cubic metre m3 m×m×m
(l×b×h) m/s
ts
3. Velocity (v) displacement/time metre per ms-1
second
4. Acceleration (a) change in vt-u metre per ms-2 m/(s×s)
velocity/time square sec-
ond
5. Force (F) mass × acceleration (m×a) newton N kg×m/(s×s)
mass/volume mv
6. Density kilogram per kg/m3 kg/(m×m×m)
(δ or d) cubic metre or kgm-3
F
7. Pressure (P) force/area A pascal Pa or kg/(m×s×s)
Nm-2
8. Momentum (p) mass × velocity (m×v) Kilogram kg ms-1 kg×m/s
metre per
second
9. Work and Ener- force × displacement (F×s) joule J (kg×m×m)/
10 gy (W/E) watt (s×s)
Power (P) w hertz
work done/time t volt W (kg×m×m)/
(s×s×s)
v
11. Frequency (f) velocity/wavelenth λ Hz cycle/s
12. Potential differ- work done/charge wq V (kg×m×m)/
ence (V) (s×s×s×A)
13. Resistance (R) potential difference/ curvrent ohm Ω (kg×m×m)/
I (s×s×s×A×A)
In this way, the unit which is expressed in terms of two or more fundamental units is called a
derived unit.
4 Modern Graded Science Class 9
Example 1: The unit of power is a derived unit. Show how?
Ans: The unit of power is watt.
We know that,
Power = work done = w = F×s = m×a×s [ F = m.a]
time t t t
Now, substituting the related units,
∴ Watt = kg×m×m
s×s×s
= kg m2s-3
The unit of power (watt) depends on the fundamental units kg, m and s. Thus, it is a derived unit.
Differences between fundamental unit and derived unit
Fundamental unit Derived unit
1. This is the unit of a fundamental quantity 1. This is the unit of a derived quantity.
2. This unit is independent of other units. 2. This unit is obtained from fundamental units.
3. There are only seven fundamental units. They 3. It is formed in many types using seven
are metre, kilogram, second, Kelvin, ampere, fundamental units. They are Newton, Pascal,
candela and mole. Joule, Watt, Hertz, Ohm, etc.
S me Reasonable Facts
1. We use SI units for a scientific purpose all over the world. This is because the results of
experiments conducted in different countries are comparable.
2. Sometimes, a special name is given to the derived units. This is done to honour
a noteworthy physicist. For example, the derived SI unit of force (kg ms-2) is called
Newton in honour of Isaac Newton.
Things To Know
1. The quantity that can be measured and expressed in unit is called a physical quantity.
2. The reference standard with which we carry out measurement of a physical quantity is
called its unit.
3. The units of fundamental quantities are called fundamental units.
4. The units of derived quantities are called derived units.
5. The fundamental units which are independent units, are also called basic units. Metre,
kilogram, second, Kelvin, candela, mole and ampere are seven fundamental units.
Measurement 5
6. The unit which is expressed in terms of two or more fundamental units is called a derived
unit e.g. m2, m/s, etc.
7. SI (Sestéme International d’ Unite’s) is defined as the system of units which is agreed by
the International Convention of Scientists held in France in 1960.
Things To Do
Suspend a small metal bob using a thread as shown in the adjoining
figure. Its initial position is at the point O. Now, displace it slightly
from the point O. It displaces from A to O, O to B, B to O and back to
A. This is one complete oscillation. Count the time for 20 oscillations
using a stop watch. Dividing the time for 20 oscillations by the total
number of oscillations (20), we get the time period of the pendulum.
Test Yourself
1. Multiple Choice Questions (MCQs)
a. The rate of doing work is called ………………….. .
A. power B. energy C. work D. pressure
b. The SI unit of measurement of fo rce is …………………… .
B. Joule B. Hertz C. Newton D. Pascal
c. Which one of the following is a derived unit?
A. Kelvin B. Mole C. Hertz D. Candela
d. The mass per unit volume is called ………………. .
A. weight B. velocity C. work D. density
e. Mass is defined as ………….. .
A. amount of substance in a body B. force per unit area of a body
C. weight per unit volume of a body D. force on the total area of a body
2. Answer the following questions.
a. Define a unit? Write its types.
b. What is the unit of work? Explain whether the unit of work is a fundamental or
derived unit.
c. Name the fundamental units used in physics?
d. Write the importance of International Bureau of Weights and Measures in the
country.
e. What is measurement? Why is it important to have regular supervision of the weights
and measurements in the market?
6 Modern Graded Science Class 9
f. Define fundamental unit with an example. Why is the unit of force called a derived unit?
g. What is a derived unit? Give its two examples.
h. Define SI units?
i. Define:
i. physical quantity ii. unit iii. measurement
j. Write SI units of the following physical quantities:
i. volume ii. potential difference iii. temperature iv. electrical resistance
k. What is the advantage of the SI unit over CGS units?
l. Point out the fundamental and derived units from the units given below:
joule, watt, hertz, second, Pascal, Kelvin, mole, Newton
m. Write the basic units involved in the following derived units:
Joule, Pascal, hertz, Newton
3. Differentiate between: (Two points each)
a. Fundamental quantity and derived quantity b. Fundamental unit and derived unit
c. MKS and CGS systems d. SI unit and derived unit
4. Prove that:
a. The unit of pressure is a derived unit.
b. The unit of work is a derived unit.
Metric system : the system of measurement that uses metre, kilogram and second as
basic units
Luminous intensity : measure of the light emitting ability of a source of light complete
Frequency : the number of waves produced in one second
∆∆∆
Measurement 7
Chapter Force
2
Total estimated Pds: 6/ (5T/ 1P)
Competencies
On completion of this chapter, the students will be competent to:
describe the uses and effects of force on the objects at rest and the objects in motion both.
define inertia of rest and inertia of motion; explain both of them.
introduce uniform speed and non-uniform speed.
describe acceleration and retardation with examples.
explain the equations of motion and solve simple mathematical problems.
explain Newton’s laws of motion with examples.
What do we experience when a moving bus suddenly stops or a bus at rest suddenly starts
moving? It is a rational question to ask why and how this happens. Similarly, when we kick
a ball, it strikes on the wall and gets reflected towards us. In this way, we experience various
events like motion of a body, rest, inertia, etc. due to laws of force in our daily life. In this
chapter, we shall study about these scientific facts and principles.
Rest and motion
Everything in this world is moving and nothing is stationary. All the things or objects are
in motion. For example, a flying bird, a moving bus on the road, a walking man, etc. are in
motion. In reality, our earth is also in motion. Have you ever watched the night sky? You
might have observed that the positions of stars and moon change while we remain stationary
or are standing at the same position. Whenever we say that a body is at rest or in the state of
motion, we are indicating their position with respect to some other bodies which we identify
as the point reference.
A body is said to be at rest if its position does not change with time with respect to an observer
or a reference point. For example, while two passengers are sitting in a moving bus, for each
of them, another passenger is at rest while the bus itself is in the state of motion. Therefore, we
may say that the passengers sitting in the moving bus are at rest with respect to the bus.
A body is said to be in motion if its position changes with time with respect to an observer
or a reference point. For example, the two passengers sitting in a moving bus are viewed
by a person on the ground. The person on the ground sees the bus along with these two
passengers in motion.
8 Modern Graded Science Class 9
“Rest and motion” are relative terms. A body may seem to be at rest with respect to one object,
but may appear to be in motion with respect to another object. The passengers sitting in the
moving bus are at rest with respect to the bus, but they are in motion with respect to the
surroundings objects outside the bus.
Force
Imagine that a book is lying on a
table. It continues to be lying on
the table at the same position until
somebody displaces it to other
position. To move the book, one has
to either push it or pull it. Similarly,
we can move the position of a cart by
pushing or pulling as shown in the
figure. Such push or pull on a body is
called force. Thus, in order to bring Fig. 2.1 (a) pushing (b) pulling
a body into motion from the state of
rest or from motion to rest a force has to act on it.
Thus, force is that push or pull on a body that changes or tends to change the state of rest or of
uniform motion in a straight line.
Force acting on a body can do the following things:
1. It can change the position of the body.
2. It can change the speed of the body.
3. It can change the direction of motion of the body.
4. It can change the shape of the body.
The SI unit of force is newton (N). In CGS system, the unit of force is dyne.
1 newton = 105 dynes.
Force is measured by using a spring balance or a top pan balance.
Balanced and unbalanced forces
When a number of forces acting on a body do Fig. 2.2 tug of war 9
not change its state of rest or uniform motion in
a straight line, the forces are said to be balanced Force
forces. For example, in the game of tug of war, two
teams pull a rope in opposite directions. When
the two teams exert equal and opposite forces on
the rope, the rope remains steady and does not move in any direction. It appears as if no force
is acting on the rope. Thus, the two forces acting on the rope in this case are balanced forces.
When a number of forces acting on a body change its state of rest or uniform motion in a
straight line, the forces are said to be unbalanced forces. For example, if we push a toy car
towards east, it moves towards east. If it is pushed towards west, it moves towards west. This
unbalanced force produces motion.
Speed
Suppose a body moves 20 m in 2 seconds, either on a curved path or on a straight path. So, 10
m is the distance travelled by the body in each second which can be written as 10 m/s. This is
called speed of the body. Thus, speed is defined as the distance travelled per unit time. It is the
rate of distance travelled by a body.
Mathematically, Distance travelled
Time taken
Speed =
i.e., v = d
t
In SI System the unit of distance is metre and time is second. Thus, the SI unit of speed is m/s.
Speed is a scalar quantity because it has magnitude only.
Velocity
Velocity is the distance travelled by a body per unit time in a particular direction. Suppose a
body moves 10 m due east on a straight line in 2 seconds. In each second, it moves by 5 m on
the straight line due east. This can be written as 5 m/s. Velocity is defined as displacement per
unit time.
Mathematically,
Displacement (s)
V elocity (v) = Time (t)
i.e. v = st
Displacement is measured in metre and time in second in the SI system. Thus, velocity is
measured in m/s.
Velocity is a vector quantity because it has both magnitude and direction. For example, the
statement "the velocity of a body is 50 m/s", is incomplete because this statement does not tell
about the direction of velocity. However, the statement "the velocity of the body is 50 m/s due
east" is complete.
Velocity and speed have the same meaning only when a body moves on a straight line. If the
body moves on a curved path, it is better to describe its motion by speed rather than by velocity.
10 Modern Graded Science Class 9
a. Uniform velocity
Fig 2.3 uniform velocity
Suppose a boy travels 2 m each second on a straight line as shown in the figure. The velocity of
the boy is then 2 m/s throughout the motion. Such
velocity is called uniform velocity. A body is said
to move with uniform velocity if the body covers
an equal distance in an equal interval of time on The rate of change of distance on a circular
a straight line. The velocity is same throughout path cannot be called velocity. This is
the motion. The uniform velocity is also called because at every point on the circular
constant velocity. path, there is a change in the direction of
the body.
b. Non-uniform Velocity
Suppose, a boy travels 2 m in 1 s i.e. 2m/s, then 1 m/s and again travels at the rate of 4 m/s on
a straight line. The velocity throughout the motion is different. Such velocity is called non-
uniform or variable velocity. A body is said to move with non-uniform velocity if the body
covers unequal distances in equal intervals of time on a straight line.
Fig. 2.4 non-uniform velocity
For a body moving with non-uniform velocity, we calculate average velocity. Average velocity
is defined as the total displacement made by a body per unit time.
Mathematically, Total displacment i.e. A. V. = st
Total time taken
Average velocity =
The velocity of a body can be changed by:
1. Changing the speed of a body and keeping the direction same. Force 11
2. Changing the direction of motion by keeping the speed constant.
3. Changing both the speed and direction of the body.
Acceleration
Usually, moving bodies in our surroundings have no uniform motion. This is because they
cannot cover equal distance in equal time interval. Thus, the velocity of a body changes with
time. This kind of motion of a body is called accelerated motion.
Fig. 2.5 increasing Velocity
For example, a boy starts to move on a straight line from point A. His initial velocity at point A
is zero. After one second, his velocity at point B is 2 m/s; after two seconds his velocity at point
C is 4 m/s; after three seconds his velocity is 6 m/s at point D as shown in the figure. Here, the
velocity of the boy is increasing by 2 m/s in every second. This increase in the velocity of the
boy per second is called acceleration. Thus, acceleration is defined as change in velocity per
unit time.
Mathe mA oarcticA ceacllcelerya,l etrioanct ion i .e(a. ) a === FvCt-ihnuTaainlmgveeeTlitoniamckvieteeynlto(avck)iety-nI(nti)tial velocity (u)
In the above figure,
Final velocity (v) = 6 m/s
Initial velocity (u) = 0 m/s
Time (t) = 3 sec
we have that a = v -t u
6-0
= 3 = 2 m/s2
If the motion of a body is on a straight line, acceleration occurs in the direction of the velocity.
Therefore, acceleration is a vector quantity.
In the SI system, change in velocity is measured in ms-1 and time in second. So, acceleration
is measured in m/s2.
For a body moving with uniform velocity, there is no change in velocity and hence it has
zero acceleration. For example, if a boy is running with a uniform velocity of 2m/s, then he
has zero acceleration.
12 Modern Graded Science Class 9
Example: If a car travelling at a velocity of 4 m/s along a straight line speeds up uniformly to
the velocity of 6 m/s in 2 s, find the acceleration.
Solution
Here, Initial velocity (u) = 4 m/s
Final velocity (v) = 6 m/s
Time taken (t) = 2s
Acceleration (a) = ?
We know that,
a = v-u
t
Therefor e, the ac=cel6e2r-a 4tio=n 2 =1 m/s2 1 m/s2.
2 the car is
of
Retardation
If the brake of a moving bicycle is pressed, its velocity decreases. The bicycle is then said to
have retardation. Retardation is defined as the decrease in velocity per unit time or negative
acceleration.
Thus, the negative acceleration is called retardation or deceleration. For example, if acceleration
of - 5 m/s2, it means the retardation is 5 m/s2. It shows that the body is decreasing its velocity
by 5m/s in each second. Retardation has the same unit as acceleration.
Equations of motion
Equations involving displacement, initial velocity, final velocity, acceleration and time of
motion of a moving body are equations of motion. These equations are derived from the
definitions of average velocity and acceleration.
Consider a body moving on a straight line with uniform acceleration as shown in the figure.
Let,
Displacement = s Fig. 2.6 uniform velocity
Initial velocity = u
Final velocity = v
Acceleration = a
Time taken = t
a. Relation between u, v, a and t or first equation
Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so
Force 13
that the final velocity be 'v' after a time interval 't'. Now, by the definition of acceleration,
we have
a = v t- u
- u
o r at = v
∴ v = u + at ................ (i) Fig. 2.7 uniform velocity
b. Relation between s, u, v, and t or second equation
Let us assume a body is moving with an initial velocity 'u'. Let after time 't' its final velocity
becomes 'v' and the distance travelled by a body is 's'. For uniformly accelerated motion,
average velocity is the mean of initial and final velocities. Therefore, we have
A.v. = u2+ v - ----- --- (1)
Average velocity is also calculated by -
A.v. = ts --------- (2)
From equation (1) and (2)
Fig. 2.8 uniform velocity
u+v s
2 = t
or, 2 s = (u + v) t
∴ s = u + v × t ................. (2)
2
c. Relation between s, u, a and t or third equation
Let us assume a body moving with an initial velocity 'u'. Let its final velocity 'v' after a time
interval 't' and the distance travelled by the body becomes 's' then, we already have,
v = u + at .................... (i)
s = u + v × t ................ (ii)
2
Puttin g the value of v from the equation (i) in the equation (ii), we have
Fig. 2.9 uniform velocity
14 Modern Graded Science Class 9
d- Relation between u, v, a and s or fourth equation
Let us assume a body moving with an initial velocity 'u'. Let its final velocity be 'v' after a time
't' and the distance travelled by the body be 's'. We already have,
Fig. 2.10 uniform velocity
or, 2as = v2-u2
Application of equations of motion in different situations
Case 1: For a body moving with a uniform velocity, initial velocity (u) = final velocity (v).
Hence, acceleration (a) = 0. Therefore, from the equation (iii), we have
s = ut + 21 at2
s = ut (∴ 21 at2 = 0 )
Case 2: For a freely falling body under gravity (a = g and s = h), equations (i), (ii), (iii) and (iv)
are respectively written as,
α Ϊ v = u + gt
α ʹΪ έ h = u + v × t
2
α Ϊ ʹͳ ʹ h = ut + 1 gt2
2
ʹ α ʹ Ϊ ʹ v2 = u2 + 2gh
Case 3: For a body thrown vertically upwards (s=h and a=-g), then equations (i), (ii), (iii) and
(iv) are respectively written as
Force 15
ȋȌ
α – v = u - gt
α ʹΪ έ h = u + v × t
2
α Ϊ ʹͳ ʹ h = ut - 1 gt2
2
ʹ α ʹ Ϊ ʹ v2 = u2 - 2gh
Notice that in case 2 and case 3, a is replaced by g and s by h in equations given above. Here,
a negative sign is assigned for g. This acceleration due to gravity (g) is oppositely directed i.e.
motion is directed vertically upwards but g is directed vertically downwards.
Case 4: For a body just released from rest from height (h), u = 0, a = g and s = h, then the
equations of motion become,
α Ϊ v = gt
α Ϊ ͳʹ ʹ
ʹ α ʹ Ϊ ʹ h = 1 gt2
2
v2 = 2gh
Also remember that:
1. When a body is thrown vertically upwards, the final velocity at maximum
height (v) = 0 (Here a = -g).
2. When a body is allowed to fall vertically downwards, the initial velocity u = 0.
(Here a = +g)
3. The value of retardation is always negative (–a)
4. In the problem, if three quantities are given we can calculate the rest quantities using
the equations of motion.
Example 1: A bus starts from rest. If the acceleration of the bus is 0.5 m/s2, what will be its
velocity at the end of 2 minutes and what distance will it cover during that time?
Solution
Here, initial velocity (u) = 0
Acceleration (a) = 0.5 m/s2
Time taken (t) = 2 minutes = 2×60 = 120 s [∴ 1 min. = 60 sec.]
Final velocity (v) = ?
16 Modern Graded Science Class 9
Distance covered (s) ?
We have,
v = u + at
Or, v = 0 + 0.5 × 120 = 60 m/s
Again,
s = u + v × t
2
= 0 + 60 × 120 = 3600 m
2
Therefore, the final velocity of the bus is 60 m/s and it will cover 3600 m in 2 minutes.
Example 2: A ball is thrown vertically upwards with a velocity of 20 m/s. Calculate (i) the time
taken for it to return the initial position. (ii) the maximum height travelled by it.
Solution
Here, Initial velocity (u) = 20 m/s
Acceleration (a) = -10 m/s2 [ the ball is thrown upwards a is taken - g.]
The velocity of the ball goes on decreasing as the ball attains height. At the highest point, final
velocity (v) = 0.
We have,
v = u + at
Or, 0 = 20 + (-10) t
Or, 10t = 20
Or, t = 20 ∴ t = 2s
10
Since 2 s is the time taken by the ball to reach the maximum height, so time taken to return to
the initial position is 2t = 2×2s = 4 s.
α Ϊ ʹͳ ʹ
Again, we have
ǡ α ʹͲ έ ʹ Ϊ ͳʹ ȋǦ ͳͲȌ έȋʹȌʹ
ǡ α ͶͲ Ϊ ͳʹ έ ȋǦͶͲȌ
ǡ α ͶͲ Ǧ ʹͲ α ʹͲ
Therefore, the maximum height attained is 20 m and the time taken for it to return to the
initial position is 4 s.
Force 17
Example 3: A car is travelling at a speed of 90 km/hr. On seeing a baby 20 m ahead on the
road, the driver of the car presses the brakes and the car stops at a distance of 15 m. What is
its retardation and how much time does it take to come at rest?
Solution
Although the driver sees a baby 20 m a head on the road, he stops the car at a distance of 15
m, hence,
Distance covered (s) = 15 m.
Initial velocity (u) = 90km/hr = 906×0 ×106000sm = 25 m/s
Final velocity (v) = 0
Retardation (u) = ?
Time taken (t) = ?
We know,
v2 = u2 + 2as
or, (0)2 = (25)2 + 2a × 15
or, - 625 = 30a
or, a = - 625 = -20.83 m/s2
30
Again, we have
v = u + at
or, 0 = 25 + (-20.83) t
or, - 25 = -20.83 t
or, t = 202.583 = 1.2 s
Therefore, the acceleration of the car is - 20.83 m/s2 and it takes 1.2 s to come at rest.
Inertia
The property of a body due to which it remains or tends to remain in the original state of rest
or uniform motion on a straight line unless an external force acts on it is called inertia. Only
unbalanced force can change the state of rest to the state of motion and vice versa. Because of
this property of the body, it opposes any change in its state of rest or of uniform motion on a
straight line.
18 Modern Graded Science Class 9
Relation between inertia and mass
Activity 2.1
To prove that inertia depends on mass.
1. Take two identical water bottles; one empty and another filled with water.
2. Suspend them separately as shown in the figure.
3. Displace the bottle for a certain
distance and leave them to move at the
same time. We feel that a larger force
is required to set the bottle filled with
water in motion. It keeps on oscillating
for a longer time than the empty bottle.
This means that a heavier body possesses
more inertia than a lighter one.
From the above activity, it follows that the mass of a body is the quantitative measure of its
inertia. The greater the mass of a body, the greater is its inertia.
Inertia may show itself in any of the following three forms: inertia of rest, inertia of motion
and inertia of direction.
a. Inertia of rest
Inertia of rest is the property of a body by virtue of which it remains or tends to remain in the
state of rest unless external forces act on it. For example,
1. When we stand or sit loosely on a bus and the bus suddenly starts to move, we jerk
backwards on the bus.
2. When a blanket is given a sudden jerk, the dust particles fall off.
3. A coin placed on the top of a greeting card covering a tumbler falls into the tumbler
Fig. 2.12 inertia of a coin
Force 19
b. Inertia of motion
Inertia of motion is the property of a body by virtue of which it remains or tends to remain
in the state of uniform motion in a straight line unless external forces act on it. For example,
1. A passenger in a moving bus jerks forward when the bus stops suddenly.
2. A coin thrown vertically upwards above a moving bus comes back in the hands of the
thrower.
3. An athlete runs for some distance before taking a long jump.
c. Inertia of direction
Inertia of direction is the property of a body by virtue of which it maintains or tends to
maintain its direction of motion unless external forces act on it. For example,
1. When a running bus suddenly takes a turn, the passengers experience a jerk in the
outward direction.
2. When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially.
Momentum
Imagine that a car and a truck both have the same speed on the road. Which one can stop
easily? Of course, it is hard to stop the truck relative to the car. The reason is that the car and
the truck have the same speed but different masses. Thus, we can say that to stop a heavier
body is harder than to stop a lighter one even both of them have the same speed.
If two tenis balls are hit towards you in different velocities, it is not equally easier to stop them.
The ball in more velocity is difficult to stop relatively. Thus, we can say that to stop a body in
more velocity is harder than to stop a body with less velocity, eventhough they have the same mass.
The physical quantity that describes the quantity of motion of a body is called momentum.
The momentum of a moving body or linear momentum is defined as the product of mass and
velocity of a moving body.
Mathematically,
Momentum = mass × velocity
i.e. , p = m × v
The SI unit of momentum = kg.m/s. Since velocity is a vector quantity and multiplied with
mass (scalar quantity), momentum becomes a vector quantity. Direction of momentum is the
same as velocity.
20 Modern Graded Science Class 9
Newton’s law of motion
After Galileo, Sir Isaac Newton (1642-1727 AD) of England made a detailed and systematic
study of the motion of bodies and formulated the three laws of motion. These laws are known
after his name as Newton’s laws of motion.
Newton’s first law of motion
Newton’s first law of motion states that ‘everybody continues in its state of rest or uniform
motion along a straight line unless an external force is applied on it’. Interpretation of the
law: According to this law, a body continues to remain at rest if no external force acts on it.
Similarly, the body will remain moving with a uniform motion on a straight line unless an
external force acts on it. This concept is, in fact, the concept of inertia. Hence, Newton’s first
law of motion is also called the law of inertia.
Activity 2.2
To prove Newton’s first law of motion
Take carom coins and make a pile as shown in the figure.
Hit the lowest coin hard with the striker. The lowest
coin moves away but the pile of coins drop down and
will remain in the same position. This is because the
force exerted by the striker on the lowest coin remains Fig. 2.13 carrom board with carrom coins
in a state of motion but the rest of the pile remains at its original state of rest due to inertia.
Now, the force exerted by the lowest coin on the coin above it is for a very short time, which
is not able to move the upper coins to the horizontal direction. This activity proves Newton’s
first law of motion. For example,
1. When we shake a branch of mango tree, the mangoes fall down. It is due to the fact that
shaking brings the branches of the tree in motion while the mangoes tend to be at rest
due to inertia of rest. Consequently, the mangoes get detached and fall down.
2. A man getting out of a moving bus or train falls in the forward direction. When he steps
down from a moving bus or train, his/her feet come at rest while the upper portion of
his/her body is still in motion and he/she falls in the forward direction. To avoid falling
down, a person, getting out of a moving bus or train, should run forward in the same
direction for a distance.
3. A bullet fired on a window pane makes a clear hole on it without causing cracks. This is
because the window pane is in the state of rest and is unable to share the motion of the
bullet. The portion of the window pane shares the motion of the bullet’s size. As a result,
the bullet makes a clear hole on the window pane without causing cracks. 21
Force
4. The blades of a fan continue to rotate for some time when the current is switched off. This
is due to inertia of motion of the blades. After some time, it stops due to friction.
5. Athletes run before taking a long jump. It is because the running changes the inertia of
athletes into motion that helps them to take a long jump.
Newton’s second law of motion
Newton’s second law of motion states that, “acceleration produced in a body is directly
proportional to the force applied to it in the direction of motion and inversely proportional to
its mass”. This Newton's second law of motion describes the relationship between force, mass
and acceleration.
Let us assume a body of mass 'm' is moved after applying a force 'F'. If 'a' is the acceleration
produced on the body on a straight line, then according to the statement of Newton's second
law of motion.
a ∝ F ........... (i) (when ‘m’ is constant)
a ∝m1 ............ (ii) (when F is constant)
Combining these two equations, we have,
a ∝mF
F ∝ ma ..... (iii)
or, F = k m a ............ (iv) (where k is the proportionality constant)
The value of k depends upon the unit chosen for measuring force. That is, if m = 1 unit, a = 1
unit and F = 1 unit, then on substituting these values in the equation (iv) we get,
1=k×1×1
or, k = 1
The equation (iv) now becomes,
F = ma ........... (v)
This relation between force, mass and acceleration can be regarded as the mathematical
statement of the Newton’s second law of motion.
22 Modern Graded Science Class 9
Some examples of Newton's second law of motion are :
1. A cricket player while catching a ball moves his hands backwards. As we know that the
cricket ball is very fast in motion, the cricketer has to stop the ball by taking to acceleration
=v tc-ruic]k. Ieft
(a) [a the time is more, acceleration will be less. It makes force also less as F = m.a.
Thus, players draw their hands back while catching the ball.
2. A person jumping on a cemented floor is injured more than a person jumping on sandy
floor. When a person jumps on a cemented floor, he/she takes less time to stop. But on
sand, the time taken to stop is more as he/she sinks in the sand. We know that t ∝ 1/a as
aw=hyv,tt-hue.pIefrlseossnriestianrjduarteiadnmisorperoodnutcheedc, elemssenretetadrdflionogr.force is required as F = m.a. That is
3. It is difficult to catch a cricket ball as compared to a tennis ball moving with the same
velocity. In the relation F = m.a. = m×(v-u)/t if, (v-u)/t is constant. Thus, if a player
catches (v = 0) either a tennis ball or a cricket ball moving with velocity u in the same time
interval (t), then F ∝ m the player has to apply more force in the case of a cricket ball than
that for a tennis ball. This is because the cricket ball is heavier than the tennis ball.
4. It is easier to stop a fast running bicycle by taking a long time than to stop it suddenly. It
is because when the bicycle is stopped suddenly, it takes a short period thus retardation is
more as a = (v-u)/t and due to more retardation, more effort is also required as F = m. a.
But when the bicycle is stopped taking a long time, the condition is opposite and it is easier
to stop.
5. It is easier to drag a stone than to kick it. It is because dragging needs longer time than
kicking the body. We also know that, a ∝ 1/t (∴ a = v-u/t) and F a) [∴ F = m.a.]. Due to
more time required for dragging, there is less acceleration and it causes less effort to drag it.
But in the case of kicking, the condition is opposite and it needs more effort to kick the stone.
Activity 2.3
To prove Newton’s second law of motion
Apparatus required: toy trolley, storing balance, string, identical blocks.
Procedure: (a)
1. Take a toy trolley with a block on a table and pull
it by applying on effort.
2. Now pull the trolley applying more effort and (b)
observe the acceleration. You will find
Fig. 2.14
a ∝ F ........................... (i)
Force 23
3. Now make the force constant but add some other blocks on the trolley. What is the change
in acceleration when mass is increased? Now you
will find that,
a ∝m1 (c)
From this activity, it is proved that the acceleration produced on a body is directly proportional
to the applied force keeping the mass constant and inversely proportional to the mass i.e. a
heavier mass has less acceleration than a lighter one keeping the applied force constant.
Example 1: A cricket ball of mass 200 g moving with a velocity of 15 m/s is brought to rest by
a player in 0.05 s. What is the average force applied by the player ?
Solution
Here, Mass (m) = 200g = _200 kg = 0.2 kg
1000
Initial velocity (u) = 15m/s
Final velocity (v) = 0
Time taken (t) = 0.05s
We have,
v = u + at
or, 0 = 15 + a × 0.05
or, 0.05a = 0-15
or, a = 0-.1055 , = - 300m/s2
Again, we have
F = ma
or, F = 0.2 × (- 300) = - 60 N
Therefore, the force applied is 60 N but in opposite direction of motion (-60 N).
Example 2: A force acting on a body of mass 200 g displaces it through 200 cm in 5 s. Find the
magnitude of the force if the initial velocity of the body is zero.
Solution
Mass (m) = 200 g =1200000 kg = 0.2 kg
Distance covered (s) = 200 cm = 2 m
Time taken (t) = 5 s
24 Modern Graded Science Class 9
Initial velocity (u) = 0
ǡ
ǤǤǤ α Ϊ ʹͳ ʹ
ǡ ʹ α Ͳ έ Ϊ ʹͳ ȋͷȌʹ
or, 25 a= α4 ʹͶͷ Ȁʹ
∴
ǡ
ǤǤǤ α
ǡ α ȋͲǤʹȌ έ ʹͶͷ α ͲǤͲ͵ʹ
ǡ ͲǤͲ͵ʹ Ǥ
Newton’s third law of motion
Newton’s third law of motion states that “to every action there is an equal and opposite
reaction”. Action and reaction always act on different bodies. According to this law, whenever
a body exerts a force on another body, the second body also exerts force of equal magnitude
on the first body of but in the opposite direction.
Interpretation of the third law of motion
Consider a body of weight W is resting on a horizontal
surface as shown in the figure. The body exerts a force,
equal to its weight W, on the surface. The surface in turn,
exerts a force R in the upward direction such that R = W.
For example,
1. In order to swim, a man pushes water backwards
with his hands. Due to the reaction offered by water
to the man, the man is pushed forwards. Fig. 2.15
2. When the mouth of an inflated balloon is left open, the balloon flies away. This is because
when the air inside the balloon escapes out (action), the balloon gets pushed (reaction)
in the opposite direction by the air escaping out.
3. When a bullet is fired from a rifle with a certain force (action), there is an equal and
opposite force exerted on the rifle in the backward direction (reaction). That’s why, the
rifle recoils when a bullet is fired from the rifle.
4. In rockets, the burnt gases are exhausted out vertically downward with a great force
(action). These gases exert an equal and opposite force on the rocket in the vertically
upward direction (reaction) and this force causes the rocket to move forward.
Force 25
5. When a man jumps out from a boat, he presses the boat with his feet (action) and the
boat in turn pushes the man (reaction) in
the opposite direction. As a result, the man
moves in the forward direction whereas the Newton’s second law of motion is called
boat gets pushed in the backward direction. the real law of motion because the first
law and the third law of motion can be
Activity 2.4 obtained by it.
To prove Newton’s third law of motion
Apparatus required: two identical spring balances, stand.
Procedure: spring
1. Hang a spring balance on a nail and fasten another spring balance in balance
it as shown in the diagram.
(a)
2. Note the reading shown by them.
3. Now pull the lower spring balance applying a force on it. Note the
readings.
4. Repeat the activity 3 applying more effort. Note the readings. spring
Now analyse the readings obtained. You will find that the readings of both balance
spring balances are same though only a single spring balance is pulled. It (b)
proves that to every action there is an equal and opposite reaction. Fig. 2.16
S me Reasonable Facts
The velocity of an accelerated body may be zero. Suppose a body is thrown vertically upwards
with an initial velocity u, say 20 m/s. Due to the gravity, its velocity decreases gradually as it
moves up. At the maximum height, its velocity is zero but it is still accelerating downwards
with an acceleration of g = 9.8 m/s2 (approximately).
Things To Know
1. Force is the push or pull on a body which changes or tends to change the state of rest
or of uniform motion in a straight line.
2. The property of a body due to which it remains or tends to remain in the state of
rest or uniform motion in a straight line unless external unbalanced forces act on it
is called inertia.
3. Speed is defined as the distance travelled per unit time.
4. Velocity is defined as displacement per unit time.
5. Acceleration is defined as change in velocity per unit time.
26 Modern Graded Science Class 9
6. The physical quantity that describes the quantity of motion in a body is called
momentum.
7. The equations of motion are:
v = u+at ............................ (i)
s = u+v ×t ............................ (ii)
2
s = ut + 21 at2 ............................ (iii)
v2 = u2 + 2as ............................ (iv)
8. Newton’s first law of motion states that “everybody continues to be in its state of rest or
of uniform motion along a straight line unless an external force acts on it”.
9. Newton’s second law of motion states that “acceleration produced in a body is directly
proportional to the force applied to it in the direction of motion and inversely
proportional to its mass”.
10. Newton’s third law of motion states that “to every action there is an equal and opposite
reaction”.
Things To Do
Take a straw and pass a thread inside it. Tie the
both ends of the thread as shown in the diagram.
Now inflat a balloon and paste it on the straw using
tapes. Now leave the mouth of the balloon open.
Observe the result and discuss the actionreaction
on the balloon.
Test Yourself
1. Multiple choice questions (MCQs)
a. The rate of change of ………. is velocity.
A. distance B. displacement C. wavelength D. mass
b. Inertia depends on ...............
A. acceleration B. weight C. force D. mass
c. The product of mass and velocity of a moving body is called ……………….. .
A. weight B. speed C. momentum D. force.
Force 27
d. …………. that produces 1 m/s2 acceleration on a body of 1 kg mass.
A. 1 J work B. 1 watt power C. 1 N force D. 1 J energy
e. The rate of change of velocity which expressed in negative is called ………...
A. speed B. retardation C. uniform acceleration D. acceleration
2. Answer the following questions.
a. What is the relation between inertia and mass ?
b. State Newton’s first law of motion.
c. State Newton’s second law of motion.
d. State Newton’s third law of motion.
e. What happens when an air filled balloon is left with its mouth downwards and allowed
the air to escape?
f. Action and reaction are equal in magnitude and opposite in direction. Then, why do
they not balance each other?
g. Name the principle in which a rocket works.
h. State Newton’s second law of motion. Prove that F = ma.
i. State Newton’s first law of motion. Explain in brief this law is also called the law of
inertia.
j. What is the definition of force according to Newton’s first law of motion?
k. What is retardation? What will be the acceleration of a body moving with uniform
velocity?
l. What is the relation of acceleration of a body to its mass and applied force.
m. State Newton’s third law of motion. How does this explain the action of a bullet fired
from a gun.
n. Prove that: 21
i. s = ut + at2.
ii. v2 - u2 = 2as
iii. v = u+at
o. Define:
i. average velocity ii. inertia iii. momentum
vi. balanced force
iv. acceleration v. one newton force
vii. equation of motion
p. What is retardation?
q. What is the SI unit of acceleration and force?
r. What is uniform acceleration?
28 Modern Graded Science Class 9
3. Distinguish between:
a. Speed and velocity b. Uniform and non-uniform velocity
d. Acceleration and momentum
c. Acceleration and retardation f. Balanced force and unbalanced force
e. Scalar and vector
4. Give reasons.
a. If we shake the branches of a tree, the fruits fall.
b. A swimmer pushes water backwards.
c. A cricketer moves his hands backwards when holding a catch.
d. A person has to run in the direction of the bus over some distance after getting down
from a moving bus.
e. Passengers of a bus jerk forward when it stops suddenly.
f. The gun recoils when a bullet is fired from it.
g. A moving truck takes a much longer time to stop than that taken by a car when brakes
are pressed at the same time.
h. When we jump on a concrete surface, the feet are more seriously hurt than while
jumping on sand.
i. An athlete runs some distance before taking a long jump.
j. We beat a blanket with a stick to remove dust particles.
Numericals Problems
[Assume g = 10m/s2]
a. If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her
acceleration. (Ans: 0)
b. A car travelling at a velocity of 15 ms-1 due north speeds up uniformly to a velocity of
30 ms-1 in 3 seconds. Calculate its acceleration. (Ans: 5m/s2 )
c. A bus is travelling at a speed of 60 km/hour. On seeing a boy 11 m ahead on the road,
the driver of the bus applies brakes and the bus stops at a distance of 10 m. What is its
retardation and how long does it take to come at rest? (Ans: 13.89 ms/2, 1.2 s)
d. A jeep starts from the state of rest. If its velocity becomes 60 km/hr in 5 minutes, (i)
What is the acceleration of the jeep? (ii) What is the distance covered by the jeep?
(Ans: 0.056 m/s2, 2500 m)
e. A car is moving with a velocity of 45 km/h. The driver presses the brakes and the
car comes to rest in 3 seconds. What is its retardation? How far does it move before
coming to rest? (Ans: 4.17 m/s2, 18.74 m)
Force 29
f. A stone is thrown upwards with a velocity of 72 km/h. If air resistance is
neglected, find the maximum height travelled by it and the time taken for it to
return to the initial position. (Ans: 20 m, 4s)
g. A bus is travelling with a velocity of 60 km/h. Pressing the brakes suddenly, it is
retarded to 0.5 m/s2. How far will it travel before coming to rest? (Ans: 277.89 m)
h. A bus is running with a velocity of 72 km/h. On seeing a boy on the road 65 m ahead, the
driver jammed on the brakes and the bus stopped with uniform retardation 5 m/s2. Calculate
the distance travelled by the bus and discuss whether the accident took place or not?
(Ans: Any accident does not take place as the bus can be stopped at a distance of 40m
after applying breaks)
i. What acceleration is produced on a mass of 200 g, when a force of 10 N is exerted on it ?
(Ans: 50 m/s2)
j. A constant retarding force of 10 N is exered to a body of mass 20 kg moving initially
with a speed of 10 m/s. How long does the body take to stop? (Ans: 20 s)
k. A constant force acting on a body of mass 4 kg changes its velocity from 2 m/s to 4 m/s
in 4 s. What is the force? (Ans: 2 N)
l. The driver of a three-wheeler moving with a velocity of 40 km h-1 sees a child standing
in the middle of the road and brings his vehicle to rest in 5 seconds just in time to save
the child. What is the average retarding force on the vehicle? The mass of the three-
wheeler is 500 kg and the mass of the driver is 70 kg. (Ans: 1266.67 N)
m. A car of mass 500 kg travelling at 36 km/hr is brought to rest over a distance of 40 m.
Find the average retardation and average braking force. (Ans: 1.25 m/s2, 625N)
Oscillate : move repeatedly and regularly from one position to another and
back again
Gravity
: the force with which the attracts objects towards the center of the
Dyne earth
Absolute
Relative : the unit of force in CGS system
: something that does not depend on anything else
: something that is dependent on other things
∆∆∆
30 Modern Graded Science Class 9
Chapter Machine
3
Total estimated Pds: 7 (5T/2P
Competencies
On completion of this chapter, the students will be competent to:
describe mechanical advantage, velocity ratio and efficiency of a simple machine
(lever, pulley, wheel and axle and inclined plane).
solve numerical problems related to mechanical advantage, velocity ratio and efficiency
of the simple machines mentioned above.
describe the law of moment in lever with examples.
We do different types of work in our daily life. In some cases we do not use any tools to perform
our work but in some cases we use various types of tools. The tools or simple devices which are
used to make our work easier, faster and more convenient are called simple machines. Such
simple machines do not do our work themselves but we have to use force on them to work
with them. Thus, Simple machines are those tools or devices which make our work easier and
convenient to do in the direction of the force.
Purposes of using simple machines
Simple machines are used to make our work easier in the following ways:
1. By multiplying force
2. By changing the direction of force
3. By transfering force from one point to other
4. By increasing the rate of doing work
Some terms related to machines
A. Mechanical advantage (MA)
We know that all the simple machines require force to do work. The resistive force to be
overcome in a simple machine is called load and the force applied to overcome the load
is called effort. The ratio of the load to the effort in a simple machine is called mechanical
advantage or actual mechanical advantage (AMA) of that machine.
Load (N) L
or, Mechanical advantage = Effort (N) = E
MA = L
E
Machine 31
Mechanical advantage has no unit as it is a simple ratio of two forces. Mechanical advantage
of a machine may be one, lesser than one or greater than one. The value of actual MA is not as
given by the formula, because some of the applied effort is wasted by friction in the machine.
Hence, mechanical advantage is affected by friction.
B. Velocity ratio (VR)
Velocity ratio of a machine is the ratio of
velocity of effort applied in the machine
to the velocity of load overcome by it.
If the load overcome by the machine is
‘L’ and the distance travelled by the load
is ‘Ld’. Similarly, the effort applied in the
machine is ‘E’ and the distance travelled
by effort is ‘Ed’, then,
Velocity of effort = distance travelled by effort = Ed and
time t
Velocity of load = distance travelled by load = Ld
time t
Now, Velocity of effort Ed/t = ELdd
Velocity of load Ld/t
Velocity of effort (VR) = =
Velocity of effort = distance travelled by effort in the given time (Ed)
distance travelled by load in the given time (Ld)
Thus, velocity ratio of a simple machine is the ratio of distance travelled by effort to the distance
travelled by load in that machine.
distance travelled by effort (m) Ed
or, Velocity ratio (VR) = distance travelled by load (m) = Ld
As velocity ratio is a simple ratio of two distances, it does not have any unit. Velocity ratio is
also termed as ideal mechanical advantage (IMA) because friction is not involved in it.
C. Efficiency
If a machine overcomes a load ‘L’ and the distance travelled by the load is ‘Ld’, the work done
by the load is L × Ld. It is also called output work or useful work.
Therefore, Output work = L × Ld
Likewise, the effort applied to overcome the load is ‘E’ and the distance covered by effort is ‘Ed’,
the work done by effort is E × Ed. It is also called input work.
32 Modern Graded Science Class 9
Therefore, Input work = E × Ed
The efficiency of a simple machine is defined as the ratio of useful work done by the machine
(output work) to the total work put into the machine (input work).
Therefore, Efficiency (h) = OInuptpuut tWWoorkrk × 100%
Efficiency is expressed in percentage. It is a ratio of two works. Example
Solved Numerical: Fig. 3.2
Calculate the efficiency of the lever given in the diagram.
Consider, a lever is used for lifting a load as shown in
the diagram.
Here, Load (L) = 300 N
Effort (E) = 200 N
Distance travelled by load (Ld) = 2 m
Distance travelled by effort (Ed) = 6 m
Now,
Output work = L × Ld
= 300 × 2 = 600 g
Rube Goldberg (1883-1790) was
a famous cartoonist. He spent his
life by creating arts and sculptures
and his most famous work was
“Inventions”. They were a series
of simple machines put together to
accomplish something very simple
but it took many steps to get there.
For ideal or perfect machines, work output is equal to the work input. Ideal or perfect
machines are those imaginary machines which are frictionless. In practice, the work output of
a machine is always less than the work input due to the effect of friction. Since, there is always
some loss of energy in the machine, if the frictional force in machines increases, efficiency
decreases. Because machines are not frictionless in practice, efficiency of a machine is never
100% or greater.
Machine 33
Relation between MA, VR and η
We know that,
Thus, the relation shows that efficiency is the ratio of mechanical advantage to velocity ratio
in percentage.
Effect of friction on MA, VR and η
The force that resists the motion of one surface relative to another with which it is in contact
is called frictional force or friction. The value of friction depends on the nature of surface
(rough or smooth), nature of material, and nature of motion (rolling or sliding), etc. Friction
wastes the kinetic energy of a machine in the form of heat energy. Practically, machines are
not frictionless. Therefore, friction reduces MA and efficiency of machines.
VR of a machine is not affected by friction because it is the ratio of two distances. Due to this
reason, the value of MA is always found lesser than that of VR.
To increase MA and η of a simple machine, we have to reduce friction. We can reduce friction
by the following methods:
i) By using lubricants like oil, grease, etc.
ii) By using ball-bearings, wheels and rollers.
iii) By making the sliding surface smooth.
iv) By rolling instead of sliding.
Principle of simple machines
The principle of simple machines states that “if there is no friction in a simple machine, output
work and input work are found equal in that machine.”
Mathematically,
Output work = Input work
i.e., L × Ld = E × Ed (If there is no friction).
34 Modern Graded Science Class 9
Types of simple machines
Simple machines are classified into the following six types:
1. Lever 2. Pulley 3. Inclined plane
4. Wheel and axle 5. Screw 6. Wedge
1. Lever Fig. 3.3 a crowbar
A lever is a rigid, straight or bent bar which is capable
of rotating about a fixed axis called a fulcrum. In a lever,
effort distance and load distance are measured from
the fulcrum. The distance between the fulcrum and the
load is load distance (Ld) and the distance between the
fulcrum and the effort is effort distance (Ed).
Activity 3.1
To measure MA, VR and effeciency of a first class lever.
Apparatus required: a model lever, a spring balance, a piece of stone/brick tied with a string,
a stand to hang the lever.
Procedure:
1. Take the model lever and hang it in the
stand.
2. Take the weight of the stone or brick and
hang at one side of it. Measure the load
distance. Consider it is 15 cm.
3. Now, using a spring balance, pull the Fig. 3.4 a model lever
load from the other side and note the effort distance and effort applied.
4. Repeat the activity 4 times by applying the effort at 10 cm, 15 cm, 20 cm and 25 cm.
5. Fill the following table and calculate MA, VR and η.
Observation table and calculation:
S.N. Load Ld Ed E MA = L VR= Ed η = ...M.V.........R..A....×%×110000%%
1. 10 cm E Ld η =
10 ...... =
MA = 160 VR = 15 =
.....
15 η = .......... × 100%
2. 15 cm MA = 160 VR = 15 = ...... = .........%
3. 160 g 15 cm 20 cm .....
VR = 2105 = ...... η = .......... × 100%
MA = 160 = .........%
.....
25 η = .......... × 100%
4. 25 cm MA = 160 VR = 15 = ...... = .........%
.....
Machine 35
Example 1: Calculate MA, VR and η of the crowbar shown in the diagram.
Here, Load (L) = 600 N
Effort (E) = 100 N
Load distance (Ld) = 0.20 m
Fig. 3.5 first class lever
Effort distance (Ed) = length of lever – load distance = 1.8 – 0.2 m = 1.6 m
Now, according to the formula,
L 600
MA = E = 100 =6
VR = Ed = 1.60 =3
Ld 0.20
MA
η= VR × 100%
= 6 × 100% = 3 × 100% = 75%
8 4
∴ MA of the crowbar is 6, VR is 3 and η is 75%.
Example 2: Study the diagram and calculate MA, VR and η.
Solution:
Here, Load (L) = 400 N
Effort (E) = 160 N 30 cm 100 N
Load distance (Ld) = 30cm
90 cm
Effort distance (Ed) = 90cm
Fig. 3.6
i. MA = ?
ii. VR = ?
iii. η= ?
36 Modern Graded Science Class 9
2. Pulley
A pulley is a metallic or wooden disc with a grooved rim. The rim rotates about a horizontal
axis passing through its centre.
A pulley can be used as a single fixed pulley as shown in fig ‘a’;as a single movable pulley as
load
load
load
shown in fig ‘b’ or in a combined form (block and tackle) as shown in fig ‘c’. A single fixed
pulley makes our work easier by changing the direction of force only. Mechanical advantage is
not gained by usinlgoadthis pulley because the value of effort distance and load distance is equal
in it. Therefore, the value of effort and lolaodadis also equal in this type of pulley.
load
But when a single pulley is used as a movable pulley, mechanical advantage is gained. In this
condition, the direction of force is not changed but velocity ratio is doubled. When two or
more pulleys are used in the form of block and tackle, the direction of force is changed and
mechanical advantage is gained.
In a block and tackle, MA and VR are directly proportional to the number of pulleys used in it.
Fig. 3.8 block and tackles on work
Machine 37
From the study of the given figures, we find that if the number of pulleys increases in a block
and tackle, effort distance and the value of VR also increase. In the same way, the value of MA
increases but not exactly as given by the formula because some of the applied effort is wasted
due to friction. In a pulley, velocity ratio is equal to the number of pulleys or number of rope
segments that supports the load.
VR in pulleys = No. of pulleys used (except in single movable pulley) or number of rope
segments that supports the load.
Activity 3.2
To measure MA VR and η of the given pulleys.
Apparatus required: a single movable pulley model, a combined double pulley system model,
string, a stone tied in a string, a spring balance.
Procedure:
1. Hang a single pulley to use as a fixed pulley.
Take the weight of the stone.
2. Tie the stone at one end of the string passed
through the pulley and adjust the spring
balance at the other end of the string.
3. Now, pull the load up to 80 cm high. Measure Fig. 3.9 (a) single
the effort distance also. Note the load, load fixed pully
distance, effort distance and effort in the (b) single (c) 3 pully system
movable pully
table.
4. Now, adjust the pully as a movable pulley. Again pull the load to the same height i.e. 80
cm. Measure the effort and the effort distance again and note them.
5. Then adjust the pulleys as block and tackle. Again pull the load high up to 80 cm. Measure
the effort and the effort distance again. Fill them in the given table. Finally calculate MA, VR
and η of the pulleys.
Observation table:
S.N. Load Ld Ed Effort MA = L VR= Ed η= MA ×100%
1. 160 g E Ld VR
..... η = .......... × 100%
2. 160 g 80 cm ....... ....... MA==..1.....6..0. VR = .8..0.. = .........%
3. 160 g =
80 cm ....... ....... MA = 1..6...0 VR==...8....0... η = .......... × 100%
80 cm ....... ....... = ..... ..... = .........%
VR = .8..0..
MA==..1.....6..0. = η = .......... × 100%
= .........%
38 Modern Graded Science Class 9
A load of 650 N is lifted by the five-pulley system up to 4 metres by applying an effort of 300
N. Calculate:
i. MA ii. VR iii. Output work iv. Input work v. Efficiency
Solution
Here, Load (L) = 650 N
Load distance (Ld) = 4 m
No. of pulleys = 5
Effort (E) = 300 N
(i) MA = ?
(ii) VR = ?
(iii) Output work = ?
(iv) Efficiency = ?
According to the formula,
L 650 65 13
i . MA = E = 300 = 30 = 6
ii. VR = No. of pulleys used = 5 For effort distance,
iii. Output work = L×Ld= 650 × 4 = 2600 J Ed
iv. Input work = E × Ed = 300 × 20 = 6000J VR = Ld
v. h = Output work = 100% 5 = E4d
Input work
= 2600 J = 100% ∴ Ed = 20m
6000 J
= 26 J × 100% = 43.33%
60 J
Hence, MA is 13 , VR is 5, work output is 2600 J, wok input is 6000 J and m is 43.33%.
6
3. Inclined plane
Slanted surface is called an inclined
plane and it is used to lift a heavy load by
applying less effort. Winding roads on
hills, an inclined plank used for loading
onto a truck, staircases, etc. are some
examples of this type of simple machine.
In this device, the length of the slope
(t) is effort distance and the height of
the slope (h) is load distance. You must
notice that in an inclined plane, the value of the length of the slope is always greater than the
Machine 39
value of the height of the slope; therefore the value of VR
is always greater than 1 in it. In an inclined plane:
Input work = E × Ed = E × (∴ = length of slope)
Output work = L × L.d = L × h (∴ h = height of slope)
VR =
h
In an inclined plane, if the length of a slope increases by Fig. 3.11 an inclined plane
keeping the height of the slope constant, the values of
both MA and VR increase. Similarly, if the height of a slope increases by keeping the length of
the slope constant, both the values of MA and VR decrease.
Activity 3.3
To measure MA, VR and η:
Apparatus required: a model inclined plane/ an inclined plank, a load tied in a string, a
spring balance.
effort
Procedure:
1. Take a model of the inclined plane. Adjust load
it at its maximum height.
2. Weigh the stone tied in the string and place
on the inclind plane.
Fig. 3.12 model inclined plan
3. Pull the stone using a spring balance. Measure Ed, Ld and effort and note them.
4. Repeat the activity 3 for three times by changing the inclination of the slanted surface.
Observation table:
S.N. Load Ld (height) Ed (length) Effort MA = L VR= Ed η= MA ×100%
E
1. 160 g Ld VR
2. 160 g
3. 160 g 80 cm ....... ....... MA = 160 VR = ............... η = .......... × 100%
= .......... = = .........%
4. 160 g VR = ..........
80 cm ....... ....... MA = 1..6..0. η = .......... × 100%
= ..... = .........%
= .....
VR = .......... η = .......... × 100%
80 cm ....... ....... MA = 160 = .........%
= ..........
= .....
VR = ..........
80 cm ....... ....... MA = 160 η = .......... × 100%
= .......... = .........%
= .....
40 Modern Graded Science Class 9
Example: By studying the diagram, calculate: 550N 8m 5m
i. Output work
ii. Input work 750N
iii. Mechanical advantage (MA)
iv. Velocity ratio (VR)
v. Efficiency (η)
Here,
Load (L) = 750 N Fig. 3.13
Effort (E) = 550 N
Length of slope (l) = 8 m
Height of slope (h) = 5 m
i. Output work = ?
ii. Input work = ?
iii. M.A. = ?
iv. V.R. = ?
v. η= ?
According to the formula,
i. Output work = L × L.d (h) = 750 × 5 = 3750 J
ii. Input work = E × E.d (l) = 550 × 8 = 4400 J
iii. MA= L = 750N = 1.36
E 550N
iv. VR = h1 = 58mm = 1.36
v. = M.A. × 100%
V.R.
=185//511 × 100%
= 15 × 5 × 100%
11 8
= 75 × 100% = 85.23%
88
Machine 41
Alternative method
h = OIuntppuutt wwoorrkk = 100%
= 3750 = 100%
4400
= 375 = 100%
440
= 75 = 100% = 85.23%
88
Hence, in this inclined plane, output work is 3750 J, input work is 4400J, MA is 15/11, VR is
8/5 and η is 85.23%.
4. Wheel and axle
Wheel and axle consists of two coaxial cylinders
of different diameters. Various examples of
wheel and axle are used in our daily life. Some
examples of such wheel and axles are string-
roller, screwdriver, MADANI, ROTE PING,
spanners, knobs of a door, steering of vehicles
etc. Most of them are used for magnifying
effort but some are also used to gain speed like
MADANI and ROTE PING. In wheel and axles,
usually effort is applied on big cylinder called
wheel and load is overcome by a small cylinder
called axle. Due to this reason, circumference of
the big cylinder (wheel) is considered as effort
distance and circumference of the small cylinder
(axle) is considered as load distance. Therefore, the calculation of VR in wheel and axle can be
done by the formula-
42 Modern Graded Science Class 9
Activity 3.4
To measure MA,VR. and η of a wheel and axle
Apparatus required: a string roller, a stone tied in a string, a spring balance, a stand and string
string roller
Procedure:
1. Weigh the stone using a spring blance and adjust stand
the string roller on a stand as shown in the diagram.
load
2. Wind the strings on the axle and on the wheel part
in opposite directions.
3. Adjust the stone at the axle and a spring balance at spring balance
the wheel as shown in the diagram. Fig. 3.15 model wheel and axle
4. Now, pull the spring balance to lift the stone up to a certain distance. Note load, effort,
Ed and Ld for calculation. Repeat the activity for three times. Calculate MA, VR, output,
input and η. Find the average of L, E, Ed, and Ld before calculation.
Wheel and axle is also called a continuous lever
In the given diagram, a load is lifted by using a wheel and (wheel)
axle. The load is suspended at point ‘L’ at the circumference (axle)
of the axle and the effort is applied at point ‘E’ at the
circumference of the wheel. Wheel and axle both are load spring balance
pivoted at point ‘F’ i.e., axis. Because there is load, effort Fig. 3.16 continuous lever
and fulcrum, it works as a lever. When the wheel and axle
is in use, the points of load and effort vary continuously for
360° on the circumference of the wheel and axle, hence, it
is considered as a continuous lever.
Example: In a wheel and axle, the radius of the wheel is 24 cm and that of the axle is 5 cm. If
a load of 1200 N is overcome by using an effort of 300 N on it, calculate MA, VR and η of the
machine.
Solution
Here,
Radius of wheel (R) = 24 cm = 120m40m= 0=.005.24
Radius of axle (r) = 5 cm =1050
Load (L) = 1200 N
Effort (E) = 300 N
i. Mechanical advantage (MA) = ?
ii. Velocity ratio (VR) = ?
Machine 43
iii. Efficiency (η) = ?
Hence, in that wheel and axle MA is 4, VR is 4.8 and η is 83.33%.
5. Screw
In screw, effort distance is the circumference
of the screw (2πR) and load distance is pitch effort
(P). Thus, VR in screw is, 2πR .
P
Mathematically, VR = 2PπR
threads
6. Wedge pitch [P]
In wedge, effort distance is the length () load
of the wedge and load distance is thickness (t) Fig. 3.17 (a) screw nail (b) a wage
of the wedge. Thus, VR = t
Note: Screw and wedge are not in our syllabus.
Moment of force
When force is exerted on a body, it results either
in displacement of the body or its rotation about
a point or axis i.e. the rotation or turning effect of
force. The turning effect of force acting on a body
about an axis is called moment of the force. It may
be clockwise or anti-clockwise. The given figure (a) anticlockwise moment (b) clockwise moment
shows the turning effect of the applied force on the
body. In figure (a) the rotation is anti-clockwise while in (b) it is clockwise.
Activity 3.5
1. Apply force in the handle of a door to open it.
2. Now, close the door.
3. Then apply force near the hinge to open it.
44 Modern Graded Science Class 9
Do you find any difference in the force applied while opening the door? We find that less force
is needed to open the door when it is applied at its handle.
From this activity it is clear that moment of a body depends on the perpendicular distance of
the line of action of the applied force from the axis. Thus, the moment of force is equal to the
product of the magnitude of the force and the perpendicular distance of the line of action of
the force from the axis of rotation or fulcrum.
Moment = F × s Here, F = force applied
s = perpendicular distance of the line of action of the force from the
fulcrum
The SI unit of moment is Nm but Ncm is also in use.
Consider, the perpendicular distance between the
fulcrum ‘A’ and the pedal ‘B’ is 20 cm. The line of
action of the force is XY and the applied force is 40 N
and in the figure.
In this condition, Fig. 3.19 perpendicular force on a pedal
Now, Moment = f × s = 40 × 20 = 800 N cm
If the force is oblique to the line of action of force the
value of moment decreases.
In the figure (3.19), the distance between the line of
action of force and the fulcrum is not perpendicular
but it is forming an angle of θ.
AB1 is perpendicular distance between the fulcrum Fig. 3.20 oblique force on pedal
and the line of action of the force.
α
ǡ α ͳ
THorhe,unAsc,Bef1,o=trhAteBhpeseicrnapθlecnudlaitciuolnarodf imstoamnceenAt iBn1 = AB sinθ we will use,
moment=F × AB sin θ the above condition,
moment = F × s sin θ
Machine 45
or,
If F = 40 N, s = 20 cm and θ = 30°
moment = F × s sin θ
= 40 × 20 × sin 30°
= 40 × 20 × 0.5 = 400 N cm
∴The moment will be 400N cm.
In figure (c) the force is parallel with the pedal. In this condition, the
value of θ is 0° hence sin θ also will be zero. It will make the value of
moment zero.
Law of moment Fig. 3.21
Activity 3.6
1. Suspend a metre scale horizontally from its mid-point as the fulcrum.
2. Now, suspend some weights on both
sides of the mid-point and adjust their
distances so that the scale again becomes
horizontal.
When the products of effort and effort distance, Fig. 3.22 a first class balancing lever
and that of load and load distance are calculated,
it is found that both are equal.
Mathematically,
It concludes that:
In the equilibrium condition of a lever, the sum of the anti-clockwise moment is equal to the
sum of the clockwise moment. This is the law of moment.
Example: In the given figure, a lever is shown. Three weights of 20 N, 40 N and 10 N are
suspended on it. Now calculate:
i. Clock wise moment Fig. 3.23
ii. Anti-clockwise moment
iii. Can the lever be balanced in this situation?
iv. Find the location of 10 N weight by keeping
other loads unchanged to balance the lever?
46 Modern Graded Science Class 9
