VECTOR
IMPORTANT DEFINITIONS
Scalar Quantities:
A physical quantity which can be described completely by its magnitude only and does not require a direction
is known as scalar quantity.
Examples: Distance, mass, time, speed, density, volume etc.
Vector Quantities:
A physical quantity which has magnitude and direction and obeys all the laws of vector algebra is called a vector
quantity.
Examples: Displacement, velocity, acceleration, force etc.
Parallel Vectors:
Those vectors which have same direction are called parallel vectors.
Angle between two parallel vectors is always 0°
Equal Vectors:
Vectors which have equal magnitude and same direction are called equal vectors.
→
→
,A = ,B
--- ---- ------ ---------- ------Equal vectors are always parallel. Angle between two parallel vectors is always 0°
Anti-parallel Vectors:
Those vectors which have opposite direction are called anti-parallel vector.
Angle between two anti-parallel vectors is always 180°
Opposite (or Negative) Vectors:
Vectors which have equal magnitude but opposite direction are called opposite vectors.
→
→
Here ,AB & ,BA are opposite vectors
→
→
,AB = – ,BA
Angle between two opposite vectors is always 180°
Collinear Vectors:
The vectors lying in the same line are known as collinear vectors.
Angle between two opposite vectors is either 0° or 180°.
Ex. of collinear vector.
………………….. ………………………
……………….
Coplanar Vectors:
Vectors located in the same plane are called coplanar vectors.
Note: Two vectors are always coplanar.
Null or Zero Vector:
Vector whose magnitude is zero is called a null vector. Its direction is arbitrary and is not specified.
Example: Sum of two vectors is always a vector. Therefore ……………………..
Unit Vector:
A vector whose magnitude is 1, is called unit vector. A unit vector is represented by ………………
Note: A unit vector is used to specify the direction of a vector.
Three Standard Unit Vectors:
In x – y – z co-ordinate frame there are three-unit vectors ………………………………………….
Polar Vectors:
Vectors which have initial point or a point of application are called polar vectors.
Examples: Displacement, force etc.
Axial Vector:
These vectors are used in rotational motion to define rotational effects. Direction of these vectors is always along
the axis of rotation in accordance with right hand screw rule or right hand thumb rule.
Examples: Small acceleration displacement……………………………………………….
Triangle Law of Addition of Two Vectors:
If two vectors are represented by two sides of a triangle in same order then their sum or 'resultant vector' is given
by the third side of the triangle taken in opposite order of the first two vectors.
Process:
i.
Resultant………………………………………
Parallelogram Law of Addition of Two Vectors (Alternate Method) :
If two vectors are represented by two adjacent sides of a parallelogram which are directed away from their
common point then their sum (i.e. resultant vector) is given by the diagonal of the parallelogram passing away
through that common point.
………………………………………………………………
Important Points
1.
Vector addition is commutative, i.e. ………………..
2.
Vector addition is associative, i.e. ………………………..
3.
Resultant of two vectors will be maximum when they are parallel i.e. angle between them is 180
…………………..
4.
Resultant of two vectors of unequal magnitude can never be zero.
5.
If vectors are of unequal magnitude then minimum three coplanar vectors are required for zero resultant.
6.
………………………………………………….
7.
If two vectors have equal magnitude ………………………………………………..
8.
If resultant of two unit vectors is another unit vector then the angle between them ……………………. OR
……….
9.
If the angle between two unit vectors ……………………………..
Addition of More Than Two Vectors (Law if Polygon):
If some vectors are represented by sides of a polygon in same order, then their resultant vector is represented by
the closing side of polygon in the opposite order.
…………………………………………..
Special Cases
1.
In a polygon if all the vectors are in the same order then their resultant is a null vector. e.g. in the given
figure ……………………….
2.
If n vectors of equal magnitude are arranged at equal angles of separation then their resultant is always
zero.
To Find Angle Between Two Vectors (When vector diagram is given)
The angle between two vectors is ……………. if either the heads or the tails of both of the vectors are connected
but if the head of one vector is connected with the tail of another vector, then angle between them is
………………………………..
…………………………………………………………
……………………………………………..
……………………………………………………..
Important Points
1.
The vectors subtraction doesn't follow communicative law i.e. ………………………………..
2.
The vectors subtraction doesn't follow associative law i.e. …………………………….
3.
If two vectors have equal magnitude, i.e. …………………………..
4.
If difference of two unit vectors is another unit vector then the angle between them is
60……………………….
5.
In physics whenever we want to calculate change in a vector quantity, we have to use vector subtraction.
For example, change in velocity………………………
6.
If two vectors are such that their sum and their difference vectors have equal magnitude then angle between
the given vectors …………………………
7.
If ……………………………………..
RESOLUTION OF VECTORS INTO RECTANGULAR COMPONENTS
When a vector is splitted into components which are at right angle to each other then the components are called
rectangular or orthogonal components of that vector.
…………………………………….
Now according to rule of vector addition
……………………………
From Pythagoras theorem ……………………………….
Angle of ………………………..
Angle of ………………………………………………………………
RECTANGULAR COMPONENTS OF A VECTOR IN THREE DIMENSIONS
In terms of x, y and z-components ……………………………………
Magnitude of ………………………………………………..
Important Points
1.
A vector can be resolved into maximum infinite number of components. For example
…………………………………………….
2.
Maximum number of rectangular components of a vector in a plane (2-dimensions) is two. But maximum
number if rectangular components in space (3-dimensions) is three which are along X, Y and Z axes.
3.
A vector is independent of the orientation of axes but the components of that vector depends upon the
orientation of axes.
4.
The component of a vector along its perpendicular direction is always zero.
MULTIPLICATION OF VECTORS
Vectors of different types can be multiplied to generate new physical quantities which may be scalar or a vector. If, in
multiplication of two vectors, the generated physical quantity is a scalar, then their product is called scalar or dot
product and if it is a vector, then their product is called vector or cross product.
SCALAR PRODUCT OF TWO VECTORS
Definition:
The scalar product (or dot product) of two vectors is defined as the product of their magnitudes with cosine of the
angle between them. Thus if there are two vectors …………………………………….. between them then their scalar
product is written as ………………………………………………………………
Important Points
1.
2.
3.
4.
5.
6.
7.
8.
It is always a scalar, which is positive if angle between the vectors is acute (i.e. 90° <  < 180°) and
negative if angle between them is obtuse (……………….)
It is commutative i.e. …………………….
it is distributive i.e. …………………………..
According to definition…………………………………….
The angle between the
vectors…………………..
Scalar product of two vectors will be zero when …………………………………………
……………………………………….. ……… i.e., if the scalar product of two nonzero vectors is zero
then vectors are orthogonal or perpendicular to each other.
In case of orthogonal unit vectors…………..
The scalar product of a vector by itself is termed as self dot product and is give by………………………
In terms of components……………………..
CROSS OR VECTOR PRODUCT
Definition:
The vector product or cross product of two vectors is defined as a vector having magnitude equal to the product of
their magnitudes with the sine of angle between them, and its direction is perpendicular to the plane containing both
the vectors according to right hand screw rule or right hand thumb rule.
Thus, if …………………………………………………………………………………………….
Important Points
1.
Vector product of two vectors is always a vector perpendicular to the plane containing the vectors, i.e.,
orthogonal (perpendicular) to both the vectors ………………………
2.
Vector product of two vectors is not commutative i.e. ………………………………………..
3.
The vector product is distributive when the order of the vectors is strictly maintained, i.e.
4.
According to definition of vector product of two vectors………………………
5.
The vector product of two vectors will be maximum when ……………………..
6.
The vector product of two non-zero vectors is zero vector, then the vectors are collinear.
7.
The self cross product, i.e. cross product of a vector by itself is a zero vector or a null vector.
8.
In case of orthogonal unit vectors……………………………………
9.
In terms of components…………………………..
10. A unit vector ……………………………………………
11. ………………………………………
12. Angle between ……………………….
Examples of Vector Product:
1.
Torque ………………
2.
Angular momentum ………………
3.
Acceleration …………………
GEOMETRICAL MEANING OF VECTOR PRODUCT OF TWO VECTORS
Area of …………………….
Formulae to Find Area:
1.
If ………………………….
2.
If ………………………………
3.
If …………………………
UNITS ……… DIMENSIONS
1.
PHYSICAL QUANTITIES
All the quantities which are used to describe the laws of physics are known as physical quantities.
1.1 Classification: Physical quantities can be classified on the following bases:
I.
Based on their directional properties
1.
Scalars: The physical quantities which have only magnitude but no direction are called scalar quantities.
e.g. mass, density, volume, time. etc.
2.
Vectors: The physical quantities which have both magnitude and direction and obey laws of vector algebra
are called vector quantities. e.g. displacement, force, velocity, etc.
II.
1.
2.
Based on their dependency
Fundamental or base quantities: The quantities which do not depend upon other quantities for their
complete definition are known as fundamental or base quantities. e.g. length, mass, time, etc.
Derived quantities: The quantities which can be expressed in terms of the fundamental quantities are
known as derived quantities. e.g. Speed ( = distance/time), volume, acceleration, force, pressure, etc.
………………………………………………………….
………………………………………………………………..
UNITS OF PHYSICAL QUANTITIES
2.1 System of Units:
1.
FPS or British Engineering system: In this system length, mass and time are taken as fundamental
quantities and their base units are foot (ft), pound (lb) and second (s) respectively.
2.
CGS or Gaussian system : In this system the fundamental quantities are length, mass and time and their
respective units are centimetre (cm), gram (g) and second (s).
3.
MKS system: In this system also the fundamental quantities are length, mass and time but their
fundamental units are metre (m), kilogram (kg) and second (s) respectively.
4.
International system (SI) of units : This system is modification over the MKS system and so it is also
known as Rationalised MKS system. Besides the three base units of MKS system four fundamental and
two supplementary units are also included in this system.
2.2
1.
Classification of Units: The units of physical quantities can be classified as follows
Fundamental or vase units: The units of fundamental quantities are called base units. In SI there are seven
base units.
S.No.
Physical quantity
Unit
Symbol
1.
Length
metre
m
2.
Mass
kilogram
kg
3.
Time
second
s
4.
Temperature
kelvin
K
5.
Electric current
ampere
A
6.
Luminous intensity
candela
cd
7.
Amount of substance
mole
mol
While defining a base unit or standard for a physical quantity the following characteristics must be considered :
(i) Well defined
(ii) Invariability (constancy)
(iii) Accessibility (easy availability)
(iv) Reproducibility (more models can be produced)
(v) Convenience in use
2.
Derived units : The units of derived quantities or the units that can be expressed in terms of the base units are
called derived units.
e.g.
3.
4.
5.
2.3
unit of speed = Ошибка! = Ошибка! = ms–1
Some derived units are named in honour of great scientists.
e.g.
unit of force – newton (N), unit of frequency – hertz (Hz), etc.
Supplementary units : In SI two supplementary units are also defined which are radian (radi) for plane angle and
steradian (sr) for solid angle.
(i) radian : 1 radian is the angle subtended at the centre of a circle by an arc equal in length to the radius of
the circle.
(ii) steradian: 1 steradian is the solid angle subtended at the centre of a sphere, by that surface of the sphere
which is equal in area to the square of the radius of the sphere.
Practical units : Due to the fixed sizes of SI units, some practical units are also defined for both fundamental and
derived quantities, e.g. light year (ly) is a practical unit of distance (a fundamental quantity) and horse power
(hp) is a practical unit of power (a derived quantity).
Practical units may or may not belong to a particular system of units but can be expressed in any system of units.
e.g. 1 mile = 1.6 km = 1.6 × 103 m = 1.6 × 105 cm.
Improper units: These are the units which are not of the same nature as that of the physical quantities for which
they are used.
e.g. kg – wt is an improper unit of weight. Here kg is a unit of mass but it is used measure the weight(force).
Conversion factors:
To convert a physical quantity from one set of units to the other, the required multiplication factor is called
conversion factor.
Magnitude of a physical quantity = numeric value (n) × unit (u)
While conversion from one set of units to the other the magnitude of the quantity must remain same. Therefore
n1u1 = n2u2
or
nu = constant
or
n  Ошибка!
It means that the numeric value of a physical quantity is inversely proportional to the base unit.
e.g.
1m
=
100 cm
=
3.28 ft
=
39.4 in.
(SI)
(CGS)
(FPS)
3.
Dimensions
Dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to
represent that quantity.
3.1 Dimensional formula: The dimensional formula of any physical quantity is that expression which
represents how and which of the base quantities are included in that quantity.
It is written by enclosing the symbols the for the base quantities with appropriate powers in square brackets
i.e. []
e.g. Dim. formula of mass id [M1L0T0] and that of speed (= distance/time) is [M0L1T–1]
3.2 Dimensional equation: The equation obtained by equating a physical quantity with its dimensional
formula is called a dimensional equation. e.g.
[v] = [M0L1T–1]
For example [F] = [MLT–2] is a dimensional equation, [MLT –2] is the dimensional formula of the force
and the dimensions of force are 1 in mass, 1 in length and –2 in time
3.3
1.
Applications of dimensional analysis:
To convert a physical quantity from one system of units to the other:
This is based on a fact that magnitude of a physical quantity remains same whatever system is used for
measurement i.e.
magnitude = numeric value (n) × unit (u) = constant
or n1u1 = n2u2
So if a quantity is represented by [MaLbTc] then
n2 = n1
……………………………….
2.
To check the dimensional correctness of a given physical relation:
If in a given relation, the terms on bot the sides have the same dimensions, then the relation is dimensionally
correct. This is known as the principle of homogeneity of dimensions.
It is not necessary than every dimensionally correct relation physically may be correct.
3.
To derive relationship between different physical quantities:
Using the same principle of homogeneity of dimensions new relations among physical quantities can be derived
if the dependent quantities are known.
Limitations of this method:
i.
This method can be used only if the dependency is of multiplication type. The formulae containing exponential,
trigonometrical and logarithmic functions can't be derived using this method. Formulae containing more than
one term which are added or subtracted like s = ut + Ошибка! also can't be derived.
ii.
iii.
The relation derived from this method gives no information about the dimensionless constants.
The relation derived from this method gives no information about the dimensionless constants.
ERRORS IN MEASUREMENTS
1.
INTRODUCTION
The difference between the measured value and the true value of a quantity is known as the error of measurement.
2.
CLASSIFICATION OF ERRORS
Errors may arise from different sources and are usually classified as follows:
2.1
Systematic or Controllable Errors: Systematic errors are the errors whose causes are known. The can be either
positive or negative. Due to the known causes these errors can be minimized. Systematic errors can further be
classified into three categories:
i.
Instrumental errors: These errors are due to imperfect design or erroneous manufacture or misuse if the
measuring instrument. These can be reduced by using more accurate instruments.
ii. Environmental errors: These errors are due to the changes in external environmental conditions such as
temperature, pressure, humidity, dust, vibrations or magnetic and electrostatic fields.
iii. Observational errors: These errors arise due to improper setting of the apparatus or carelessness in taking
observations.
2.2
RANDOM ERRORS: These errors are due to unknown causes. Therefore they occur irregularly and are
variable in magnitude and sign. Since the causes of these errors are not known precisely they can not be
eliminated completely. For example, when the same person repeats the same observation in the same conditions,
he may get different readings different times.
Random errors can be reduced by repeating the observation of a large number of times and taking the arithmetic
mean of all the observations. This mean value would be very close to the most accurate reading.
Note: If the number of observations is made n times then the random errors reduces to ……………….. times.
Example: If the random error in the arithmetic mean of 100 observations is 'x' then the random error in the
arithmetic mean of 500 observations will be Ошибка!
2.3
Gross Errors: Gross errors arise due to human carelessness and mistakes in reading the instruments or
calculating and recording the measurement results.
For example:
i.
Reading instrument without proper initial settings.
ii. Taking the observations wrongly without taking necessary precautions.
iii. Exhibiting mistakes in recording the observations.
iv. Putting improper values of the observations in calculations.
These errors can be minimized by increasing the sincerity and alertness of the observer.
3.
REPRESENTATION OF ERRORS
Errors can be expressed in the following ways:
Absolute Error (………..): The difference between the true value and the individual measured value of the
quantity is called the absolute error of the measurement.
3.1
3.2
Relative or Fractional Error: It is defined as the ration of the mean absolute error
(…………………………….) to he true value or the mean value (…………………………) of the quantity
measured.
Relative or fractional error = Ошибка! = …………………………..
When the relative error is expressed in percentage, it is known as percentage error.
percentage error = relative error × 100
or,
percentage error = Ошибка! × 100% = ……………………………….
PROPAGATION OF ERRORS IN MATHEMATICAL OPERATIONS
RULE I:
The maximum absolute error is the sum or difference of the two quantities is equal to the sum of the absolute errors
in the individual quantities.
If X = A + B or X = A – B and if ………………………………………… represent the absolute errors in A and B
respectively, then the maximum absolute error in X = …………………
Maximum percentage error
= …………………
The result will be written as …………………………..
RULE II:
The maximum fractional or relative error in the product or division of quantities is equal to the sum of the fractional
or relative errors in the individual quantities.
If ………………………………….
RULE III:
The maximum fractional error in a quantity raised to a power (n) is n times the fractional error in the quantity itself.
i.e.
…………………………………………
Golden Key Points
……………………………………………………………………………………….
KINEMATICS
The study of motion without considering the force (which causes the motion) is known as Kinematics.
If the position of a particle changes with time, then it is in motion. But if possible does not change with time, then it
is at rest.
Motion and Rest both are relative terms because they depend upon the 'frame of reference' (or observer). Motion and
Rest cannot be absolute because they cannot be studied without reference.
Frame of Reference
It is the reference with respect to which position or motion of a particle is studied. Sometimes Cartesian Co-ordinate
System (or XYZ Co-ordinate System) is taken as the frame of Reference. But, if it is not specified, then normally
ground is taken as the frame of Reference.
TYPES OF MOTION
I.
One Dimensional Motion (1-D Motion) or Rectilinear Motion:
If a particle moves along a straight line then its motion is known as one dimensional motion or rectilinear motion.
In this motion, minimum one co-ordinate changes with time.
For example:
i.
A body dropped from a small height (freely falling body).
ii. Motion of a vehicle along a straight road.
iii. Motion of a train along a straight track, etc.
II.
Two Dimensional Motion (2-D Motion or Motion In A Plane):
If a particle moves in a plane then its motion is known as two dimensional motion or motion in plane. In this
motion, minimum two co-ordinates changes with time.
For example:
i.
Motion of queen on carrom board
ii. Projectile motion:
………………………………………………………..
III. Three-Dimensional Motion (3-D Motion) or Motion in Space:
If a particle moves in a space such that all the three co-ordinates changes with time, then its motion is known as
3-D Motion or Motion in space. For example:
i.
Motion of aeroplane.
ii. Motion of birds.
iii. Motion of kites, etc.
ONE DIMENSIONAL MOTION
Study of motion involves the detailed study of following parameters of motion:
1. Distance
2. Displacement
3.Speed
4. Velocity
5. Acceleration
DISTANCE
The length of actual path travelled by a body in a given time interval is known as distance travelled by it.
Important points about Distance:

The dimension of distance is [M0L1T0] and SI unit is meter (m).

Distance is a scalar quantity.

Distance depends upon the actual path which was followed by the particle during its motion. For example:
………………………………………………………….

For a moving body, distance travelled always increase with time. It can never decrease.

For a moving body, distance travelled is always positive. It can never be negative or zero.

A vector which is drawn from the initial position to the final position of the particle is known as displacement.
For example:
…………………………………………………

The magnitude of displacement is always equal to shortest distance between the initial and final position.
Important points about Displacement:

Its dimensions are [M0L1T0] and SI units is metre (m).

It is a vector quantity.

It depends upon the initial and final positions only and does not depend upon the actual path followed by the
body. For example:
……………………………….

For a moving body, displacement can increase or decrease with time. The displacement increases when the body
moves away from the initial position and it decreases if it moves towards the initial position.

For a moving body, displacement can be positive, negative or zero. The displacement becomes zero when the
particle returns back to the initial position.

For the motion between two fixed points, the displacement can have only one unique value. But the distance
travelled can have more than one value depending upon the path.
Relation between distance ……………………… displacement:
……………………………………….
…………………………………………
Average Velocity (………….):
The ratio of displacement to the total time taken is known as Average Velocity.
……………………………………………………………………..

It is a vector quantity whose direction is always along the displacement.

Its dimensions are = [M0L1T–1] and SI units is m/s or ms–1.
Average Speed (……………………):
The ratio of total distance travelled to the time taken is known as Average Speed.
……………………………………..

It is a scalar quantity.

Its dimensions are = [M0L1T–1] and SI unit is m/s or ms-1.

For a moving body, distance travelled can never decrease but average speed may decrease.
Relation between Avg. Velocity (………..) and Avg. Speed (V avg):
We know that,
Total distance travelled ………………………..
Now, dividing this equation by time taken
Ошибка! ………………………….
……………………………………………..
………………………………………………
Instantaneous Velocity (………………….):
The velocity of an object at a particular time instant or moment is known as instantaneous velocity.
If the time interval is made negligibly small means infinitely small(∆t→0), then average velocity is converted into
instantaneous velocity.
…………………………………………..
= Instantaneous rate of change of position vector w.r.t time or rate of displacement.
Note: The direction of average velocity is always along the net displacement but the direction of instantaneous velocity
is always tangential to the path.
For example:
……………………………………………..
Instantaneous Speed (Vinst):
It is always equal to the magnitude of instantaneous velocity because in very-very short interval of time, the
particle will be moving along a straight line in a direction for which distance travelled and …………………………..
will be equal. It means that –
Inst. Speed = ……………………
or
Vinst = …………….
or
V = ………………………..
ACCELERATION
The ratio of change in velocity to the time taken is known as average acceleration.
……………………………………….

It is a vector quantity whose direction is along the direction of change in velocity (……………).

Its dimensions are [M0L1T–2] and its unit is m/s2 or ms–2.
Instantaneous Acceleration (………….):
It is acceleration of a particle at a particular time instant or moment. If the time travel is made negligibly
small or infinitely small, then average acceleration is converted into instantaneous acceleration.
……………………………………………….
Acceleration can be positive, negative or zero.
Zero acceleration means that the velocity is constant on both magnitude and direction.
SOME IMPORTANT POINTS:

If speed is constant, then path of particle may or may not be straight line.

If velocity is constant, then path must be straight line.

If acceleration is constant, then path may or may not be straight line.

It u = 0 and ………………….., then path must be straight line.

If u ≠ 0 and ………………….., then path may or may not be straight line.
I case: ……………………………
II case: …………………………………………..

For circular path, a force perpendicular to velocity should be present.

The direction of motion can't be decided by the directions of displacement or acceleration, but it is decided
by the direction of instantaneous velocity, which is tangential to the path.

The increment or decrement of speed can not be decided by the acceleration alone, but it is decided by the
direction of both the velocity …………….. acceleration.
I Case: ………………….
II Case: ……………….
FOR QUESTIONS BASED ON CALCULUS:
…………………………………………………………………………………
Non-Uniform Motion (Acceleration motion)
……………………………………………………
……………………………………………..
……………………………………………
Equation
In Scalar Form
1.
v = u + at
2.
S = ut + Ошибка! at2
3.
v2 = u2 + 2aS
In vector form
4.

The ratio of distance travelled in 1s, 2s, 3s ……… is the ratio of squares of natural numbers.
means S1 : S2 : S3 …………= 1 : 4 : 9………….

The ratio of distances travelled in successive 1 second is the ratio of successive odd numbers.
means SI : SII : SIII …………= 1 : 3 : 5………….
This ratio of odd successive numbers is also valid for successive equal time intervals.
Note: Both the above ratios are applicable in all those conditions in which u = 0 and ………….. is constant.
VERTICAL MOTION UNDER GRAVITY

Earth attracts every body towards itself by the gravitational force if attraction which is known as gravity.

If a body moves under the effect of the gravitational attraction of earth only then its motion is known as
motion under gravity.

The acceleration produced in a body (freely falling body) due to the gravitational attraction of earth is known
as acceleration due to gravity. It is a constant (for smaller values of height) whose value is:
g = 9.8 m/s2  10m/s2 (In MKS)
g = 980 cm/s2  1000cm/s2 (In CGS)
g = 32 ft/s2 (FPS)

The direction of acceleration due to gravity is always vertically downwards irrespective of upward
(ascending) or downward (descending) motion of body.

If a body is dropped or released or it starts falling from a stationary frame of reference, then its initial velocity
is taken as zero. (u = 0)

When a hydrogen balloon or a rocket starts (begins) its motion, then its initial velocity is also zero. (u = 0).

If a body is projected / thrown / fired, then its initial velocity is non-zero (u≠0).
IMPORTANT CONCLOSIONS:

If the air resistance or air friction is neglected, then ascending time is equal to descending time.

During the ascending & descending motion, speed and acceleration remain constant at a particular point but
the direction of velocity reverses.

During ascending motion speed decreases because ……………………… are in opposite (slowing down)
direction but during descending motion speed increases because ……………. are in same direction (speeding
up).

The speed with which body was projected upwards from the ground is equal to its speed with which it returns
back at the ground. But the direction of velocity reverses.

The speed of body one second before the topmost position and one second after the topmost position are
same (nearly 10m/s) and it is independent by u.

At the topmost position velocity is zero but acceleration is non-zero, means it is possible to have zero velocity
with non-zero acceleration. [NCERT]

xI > xII > xIII. It means that the distance travelled by body in successive one second cannot be equal in one
particular direction (either ascending or descending)

The distances travelled by body in the last second of ascending motion and the first second of descending
motion are equal.

The distances travelled in the last 'n' seconds of descending motion are equal.

For a freely falling body a = g.

Projectile motion is also example of freely falling motion.

Time is an independent variable and it has natural tendency to increase.
GRAPHS IN KINEMATICS

In kinematics, usually time is taken on x-axis and other variables such as position, distance, displacement,
speed, velocity, acceleration, etc. om y-axis.
……………………………………………………….

Slope of (x-t) graph = velocity.

Slope of (v-t) graph = acceleration.

Slope of (a-t) graph = Nothing..
…………………………………………………………..

Area under (a-t) graph = ∆v

Area under (v-t) graph = x

Area under (x-t) graph = Nothing
RELATIVE MOTION
Relative Velocity
Whenever a moving object observes the motion of other objects, it actually experiences the difference in their
velocities, which is known as Relative Velocity.
……………………………………………………………………….
Note: Velocity of an object with respect to itself is always zero because …………………………….
It means that a moving object experience its own motion.
Similarly,
……………………………………………….
……………………………………………….
To Find Relative Speed:
For finding relative speed, first find the relative velocity and then take its magnitude.
means ………………………………
Note: If both the bodies move along same straight line, then relative speed can be obtained without calculating
relative velocity. The following cases are possible –
I Case: If both the bodies mov e in same direction, then the relative speed is equal to the positive difference of both
the given speeds.
………………………………………………………………
……………………………………………………………….
II Case: If the bodies are moving in opposite directions, then the relative speed is sum of both the given speeds.
……………………………………….
………………………………………
III Case: In all other conditions
VRel = |
→
,VRel
|
"RAIN & MAN"
→
→
→
,VRM =
,VR –
,VM
→
→
→
,VRM +
,VM =
,VR
tan α =
……………………………………………………………………………
…………………………………………………
……………………………………………………..
………………………………………………….
….
"SWIMMING IN RIVER"

If the reference for swimming speed is not given then consider w.r.t river or wwater.

Speed of swimming in still water = speed of swimmer w.r.t water or river

Velocity of swimmer w.r.t river does not depend upon the river flow speed
→
→
Means ( ,VSR) in still water = ( ,VSR) is flowing water

Upstream means in the opposite direction of river flow
VNet = (Vb or VM) – VR

Downtown means in the direction of river flow
VNet = (Vb or VM) + VR
"CROSSING OF RIVER"
Minimum distance or shortest path approach
……………………………………………..
………………………………………………
……………………………………………….
…………………………………
…
→
→
→
,VSR =
,VSG - ,VR
→
→
→
,VSR + ,VR =
,VSG
I.
Now,
or

|→,V | = V2,SR – V2,R
|→,V | = V cos
SG
SG
SR
[If VSR & VR are given]
[If VSR &  are given]
To cross the river with minimum displacement swimmer should swim at an angle  from perpendicular of
river flow in the upstream direction or at an angle (90° + ) from the river flow direction.
Time of crossing = Ошибка! = Ошибка! = Ошибка!= Ошибка!
In this condition displacement is minimum but time is not minimum.
For minimum time cos  should be maximum means  = 0°
II.
Minimum time approach
→
→
→
Here
,VSR =
,VSG –
,VR
→
→
→

,VSR +
,VR =
,VSG
Now,
|→,V | =
SG
V2,SR + V2,R
Time of crossing = Ошибка! = Ошибка! = Ошибка! = Ошибка!
Drift = VR × t.
PROJECTILE MOTION

When a body is projected such that velocity of projection is not collinear to the force or acceleration, then
body moves along a curved path. This motion is example of two-dimensional motion. If force (and hence
acceleration) on the body is constant, then the curved path of the body is parabolic. This body is known as
projectile and its motion is known as projectile motion.

The time taken by body to return back at the ground or the time for which it remained in air is known as time
of flight.

The horizontal distance travelled by the projectile during its motion is known as horizontal range.
Let the horizontal direction be along x-axis, vertical direction be along y-axis, point of projection at origin
(0,0) and vertically upward direction as positive.
…………………………………………….
…………………………………………….
…………………………………………..
Important points about projectile motion:

Projectile motion is the example of two dimensional motion with constant acceleration (a net=g).

Projectile motion is the combination of two mutually perpendicular motions which are horizontal motion and
vertical motion. Both these motions are independent from each other
………………………………………………………………
…………………………………………………………….
………………………………………………………………

In the horizontal direction, there is no force on the projectile. Due to it horizontal acceleration is zero, means
the horizontal component of velocity remains constant throughout the projectile motion.
⸪
Fx = 0
⸫
ax = Ошибка! = Ошибка! = –0

Ошибка!(vx) = 0 Ошибка!
⸫
vx = constant.

In the vertical direction, force is not zero. The body experiences gravitational attraction force of earth in the
vertically downward direction during the whole projectile motion and due to it vertically downwards
acceleration due to gravity is experienced by it. This downward 'g' decreases the vertical speed during the
ascending motion and increases the vertical speed during the descending motion.

At the topmost position, vertical velocity is zero but horizontal velocity is not zero. Therefore,
(Speed)top =
|→,v | =
top
v2,x + v2,y =
(u cos)2 + 02 = u cos 

(KE)top = Ошибка!mОшибка! = Ошибка!mu2cos2 = (KE)Gcos2

The direction of motion at any point of projectile motion is along the instantaneous velocity, means tangential
to the path.

At the topmost position, velocity is horizontal but acceleration is in the vertically downward direction.
→
→
Therefore, velocity is perpendicular to the acceleration. It means that angle between ,v and ,a is 90° at
the topmost position.

When the body returns back at the ground, its speed becomes 'u' but direction of velocity changes. The angle
between initial and final velocity is 2.
……………………………………………………..
……………………………………………………..
……………………………………………………….
……………….

During the projectile motion, horizontal component of velocity, acceleration, force and total mechanical
energy (K.E. + P.E.) remains constant.

During the projectile motion, vertical component of velocity, speed, velocity, momentum, kinetic energy and
potential energy do not remain constant.

At the topmost position, kinetic energy, speed, velocity and momentum become minimum but potential
energy becomes maximum.
Change in momentum of projectile
|
Total change in momentum, ∆
→
→
,P = m ∆ ,v = 2mu sin
|
|
|
Time of Flight(T):
At the topmost position,
vy = O
Let upwards direction is positive
Applying I EOM in vertical direction.
vy = uy + ayt.
 0 = uy – gt
 gt = uy
 t = Ошибка! [Ascending time]
⸫ But, Time of Flight = Ascending time + Descending time = 2 × Ascending time
[⸪Ascending time = Descending time]
⸫ Ошибка!
⸫
T α uy
Horizontal Range(R):
In horizontal direction (ax = 0)
Using distance = Speed × Time


R = ux × T

R = ux × Ошибка!

Ошибка!
R = Ошибка! = Ошибка! = Ошибка!
⸫
[⸪sin 2 = 2 sin  cos]
Ошибка!
Note: R will become maximum if sin 2 = 1 (maximum value) means,

2 = 90°
⸫
 = 45°
Rmax = Ошибка! = Ошибка!
[At =  = 45°]
Ошибка!
Maximum Height Attained (H):
Let upwards direction be +ve
At topmost position, vy = 0
Using III EOM in vertical direction
⸫

Ошибка!
(PE)top = mgH = mg Ошибка! = Ошибка! mu2sin2 = (KE)Gsin2
Note: If  = 90°, then sin  = 1, which is the maximum value of sin .
But, if sin  is maximum, then H will also become maximum.
⸫
H = Ошибка! = Ошибка!
Ошибка!
Here,

Ошибка!
[At  = 90°]
Velocity after 't' time (
→
,V)


= Vx ,i + Vy ,i


= ux ,i + (uy – gt) ,j


= (ucos) ,i + (usin – gt) ,j


Speed after 't' time
2
2
=
V ,x + V ,y
=
(ucos)2 + (usin – gt)2
Angle of velocity from hz. after 't' time is given by:
tan α = Ошибка! = Ошибка! = Ошибка!

Co-ordinates after 't' time = (x, y) where x = uxt and y = uyt – Ошибка! gt2


Displacement = x ,i + y ,j
|Displacement| = x2 + y2
Angle of Elevation at max. height:
tan β = Ошибка!
tan β = Ошибка! = Ошибка! = Ошибка!
Here  = Angle of projection
and β = Angle of elevation

If  = 45°, then angle elevation tan β = Ошибка!
β = tan–1 Ошибка!
()
Equation of Trajectory (Path):
The relation between x-displacement & y-displacement is known as equation of trajectory.
After 't' time,
In Hz. direction (ax = 0):
using, displacement = velocity × time
So,

sx = ux × t

x = u cos  t

t = Ошибка!
Ошибка!
⸫
…(1)
Path of projectile is parabola,
y = x tan Ошибка!

y = x tan Ошибка!

y = x tan Ошибка!
⸫
Ошибка!
Ex.
A particle of mass 'm' is projected from the ground at  angle from the horizontal with velocity 'u' when the
particle returns back at the ground, find,
1.
Change in speed
2.
Change in velocity
3.
Change in momentum
Sol.
[Putting 2 sin cos = sin 2]
Ошибка!
…………………………………………………………………………
1.
Change in speed = u – v = 0
→
→
2.
Change in velocity = ,vf - ,vi




= (u cos  ,i – u sin  ,j) – (u cos  ,i + u sin  ,j)

= – 2 u sin  ,j
= 2 u sin , (Vertically downward direction)
3.
→
→
Change in momentum = m ,vf - m ,vi
→
= m(∆ ,v)

= m (–2 u sin  ,j)
= 2mu sin , (Vertically downwards)
Ex.
A particle of mass 'm' is projected from the ground at  angle from horizontal with velocity 'u'. When the
particle reaches maximum height, find1.
Change in speed
2.
Change in velocity
3.
Change in momentum
…………………………………………………………………………….
Sol.
1. Change in speed =
→
→
,vf - ,vi
= (u cos – u)
→
→
2. Change in velocity = ,vf - ,vi



= u cos  ,i – (u cos  ,i – u sin  ,j)

= –u sin  ,j
= u sin , in vertically downward direction
→
3.Change in momentum = m(∆ ,v)

= – m (u sin ) ,j

= – m u sin  ,j
= (mu sin ), vertically downwards
Horizontal Projection
…………………………………fig………………………………….
Let downwards direction be +ve
Here, ux = u, uy = 0
ax = 0, ay = g
sx = R, sy = H
In vertical direction (ay = g):
Using, sy = uyt + Ошибка!ayt2

H = 0 + Ошибка!gT2

T2 = Ошибка!

T = Ошибка!
"Time of flight" Ошибка!
In Horizontal direction (ax = 0):
Using, distance = (Speed) × (Time)

sx = uxt
Ошибка!
Velocity ∆ after 't' time:
………………….fig…………………………………….






,v = vx ,i + vy ,j = u ,i + (gt) ,j = u ,i + ( 2gh) ,j

→

Speed =

tan  = Ошибка! = Ошибка! = Ошибка! [where  = angle of Ошибка! from vertical after 't' time]
|→,v| =
v2,x + v2,y = u2 + 2gh
Equation pf Trajectory:
For
Vertical motion (ay = g)
y = uyt + Ошибка!ayt2
y = Ошибка!gt2
Ошибка!
[⸪ uy = 0]
[using x = ut  t = Ошибка!]
Projection from a moving system:
Ex.
Consider a man who throws a ball from a moving trolley. Let the velocity of ball relative to man be u
→
→
→
,vball, trolley = ,vball – ,vtrolley
…………………………….fig…………………………………..
CIRCULAR MOTION

If a particle moves in a plane in such a way that its distance from a fixed point always remains constant, then
its path is a circle and its motion is known as circular motion. The fixed point is known as center and the
constant distance is known as radius of circle.
SOME BASIC TERMS:
1.
Time Period(T):
Time taken by the particle to complete one revolution is known as time period. The SI unit of time period is
second & the dimensions are [M0L0T1] or [T1]
2.
Frequency(n, f, v):
The number of revolutions made by particle in unit time is known as frequency.
⸪
Number of revolutions in T time = 1
⸫
Number of revolutions in 1 unit time = Ошибка!
⸫
Ошибка!
Units of frequency →
→
Revolutions per second (r.p.s)
Revolutions per minute (r.p.m)
1rps = Ошибка! = 60rpm
So,
1 r.p.s. = 60 r.p.m.
The SI unit of frequency is hertz (Hz) & its dimensions are:
[M0L0T–1]
or
[T–1]
3.
Angular displacement (∆):
The change in angular position of a particle is known as angular displacement. 'OR'
The angle subtended by the position vector at the center of circle is known as angular displacement.
Here, Angular displacement = Change in angular position
means ∆ = 2 – 1
Now,
Angle = Ошибка!

∆ = Ошибка!
Ошибка!
Note: In the above formula, angle must be in 'Radian'.
Important Points About Angular Displacement:

The S.I. unit of angular displacement is radian and other units are degree or revolutions.
1 revolution = 2 rad. = 360°

Dimensions of angular displacement = Ошибка! = Ошибка! = [M0L0T0]
It means that angular displacement is a dimensionless quantity.
Note: A unitless quantity must be dimensionless but a dimensionless quantity may or may not have units.
For example, Angle has units but it is dimensionless.
4.
→
Angular Velocity ( ,):
Angular velocity per unit time or the rate of change of angular displacement w.r.t. time is known as angular
velocity.
Angular velocity is of two types:
i.
Average Angular Velocity:
Ошибка!
ii.
Instantaneous Angular Velocity:
Ошибка!
The SI unit of angular velocity is radian per second and its dimensions are:
[] = Ошибка! = Ошибка! = [M0L0T–1] or [T–1]
Note: The dimensions of angular velocity and frequency are same but their units are different.
Note: Quantities having same units must have same dimensions but the quantities having same dimensions
may have different units.

For 1 complete revolution:
Ошибка!
So
 Ошибка!
i.e.
If T, then 
and
i.e.
If T, then 
n
If n, then 
If n, then 
→
→
Relation Between Linear Velocity ( ,v) and Angular Velocity ( ,):
Using

∆ = Ошибка!
Dividing by ∆t, we get,
Ошибка! = Ошибка! × r

and
v = r
→
,v =
[In scalar form]
→
, ×
→
,r
[In vector form]
∆s = (∆)r
5.
→
Angular Acceleration ( ,α):
The rate of change of angular velocity w.r.t. time is known as angular acceleration.
There are two types of angular acceleration.
i.
Average Angular Acceleration:
Ошибка!
ii.
Instantaneous Angular Acceleration:
Ошибка!
→
→
Relation Between Linear Acceleration ( ,a) & Angular Acceleration ( ,α):
→
→
→
We know, ,v = , × ,r
Differentiating w.r.t. time,
Ошибка!(Ошибка!) = Ошибка!(Ошибка! × Ошибка!)

Ошибка! = Ошибка! + Ошибка!
→
But
,aT 
→
|→,a| =
⸫
,acp
a2,T + a2,cp
………………………………fig…………………………………..
Angle between
→
,v and
→
,a:
Ошибка!
→
Tangential Acceleration ( ,aT):
→
→
→
We know,
,a = ,α × ,r
⸫
|→,a| = |→,α × →,r|
So,
aT = αr sin90°
So,
aT = αr
[⸪sin90° = 1]
………………….fig………………………………..
Since, the tangential acceleration is in the direction of velocity, therefore it can change only the magnitude
of velocity (speed). It cannot change the direction of velocity. It means that if the speed is changing in a
circular motion, then tangential acceleration will be responsible for it.
⸫
aT = Rate of change of speed
Ошибка!
Note: If a particle executes circular motion with uniform (constant) speed, means v = constant, then
aT = Ошибка! = Ошибка! = 0
and,
α = Ошибка! = Ошибка! = 0
This type of circular motion is known as Uniform Circular Motion (U.C.M.). In U.C.M.,
→
,a =
→
,acp
[⸪ aT = 0]
→
Centripetal Acceleration ( ,acp):
⸫
Ошибка!
→
So, Centripetal force ( ,Fcp):
Ошибка!
Some Important Points:

Since, tangential acceleration is along the direction of velocity. Therefore, it can change only
magnitude of velocity but it cannot change the direction of velocity.

Since centripetal acceleration is perpendicular to the direction of velocity. Therefore, it can change
only the direction of velocity but it cannot change magnitude of velocity.

In the circular motion, direction of velocity continuously changes. Therefore, centripetal
acceleration or the centripetal force must be present.

Centripetal force is not a realistic or practical force in itself, but some other practical forces such as
electrostatic force, gravitational force, tension etc. provide the required centripetal force for
executing circular motion.

Work done by the centripetal force is always zero.
→
→
Here,
,Fcp  ,S
So,
Work done by Fcp,
→
→
W = ,Fcp. ,S = FcpS. cos90° = F.S.0 = 0
NON UNIFORM CIRCULAR MOTION

Work done by the centripetal force will be zero but work done by tangential force is not zero.

Total work done
→
W = ,FT.S = FT . S cos 0° = FTS

Power = Ошибка! = Ошибка! = Ошибка!.Ошибка!
= FT.v.cos0° = Ftv

Angle between velocity & acceleration is given by:
Ошибка!
CIRCULAR MOTION IN VERTICAL PLANE:
…………fig……………………………………………………
Case-I:
At rest (v = 0)
So,
T = mg
Case-II:
At lowest position (v ≠ 0)
Net force towards the centre = Fcp.

T – mg = Ошибка!
So,
Ошибка!
Case-III:
At horizontal position
Here, T = Fcp
Ошибка!
Case-IV:
At topmost position
Here, T + mg = Fcp

T + mg = Ошибка!
Ошибка!
Note:
Tension is maximum at the lowest position & it is maximum at the topmost position. It decrease
during the ascending motion & increases during the descending motion.
TL > THz > TTop because Ошибка! > Ошибка! > Ошибка!
CaseV: At  angle from vertical:
Here,
T – mg cos  = Fcp

T = Fcp + mg cos 
Ошибка!
Condition For Completing Vertical Circle:
For completing the vertical circle, tension at the topmost position should be greater than or equal to
zero.
means, TTop ≥ 0

⸫
– mg ≥ 0

Ошибка! ≥ mg 
v2 ≥ rg 
Ошибка!
vTop ≥ rg
CIRCULAR TURNING
I.
Cyclist on a horizontal road:
When a cyclist takes a circular turn, then he bends himself towards the centre of circular tracks.
Due to it, a component of normal reaction is exerted on him in the direction towards the center. The
component provides him the required centripetal force, means,
N sin  = Fcp

N sin  = Ошибка!
…(1)
For vertical equilibrium,
N cos  = mg
…(2)\
Dividing equation (1) by (2), Ошибка! = Ошибка!

Ошибка!
II.
Banking of Tracks:
For high speed driving, the outer edge of tracks is raised slightly above the inner edge. This phenomenon is
known as banking of tracks. On inclining the track one component of normal reaction is directed towards the
canter of track, which provides the required centripetal force and he remaining other vertical component
balances, the weight means,
N sin  = Ошибка!
&
N cos  = mg
…(1)
…(2)
Dividing equation (1) by (2),
Ошибка!
…………………fig………………………………………….
III.
Conical Pendulum:
Tsin = mr2……………(1)
Tcos = mg ……………. (2)
Dividing equation (1) by equation (2)
tan = Ошибка!
Ошибка! = Ошибка!  2 = Ошибка!

Now, Time period (T) = Ошибка!
Substituting value of ,
T = Ошибка! = Ошибка!
⸫
Ошибка!
…………………………..fig…………………………………
Ошибка!(⸪ h = L cos)
IV.
Dearth well:
In it a motor cycle rider drives against a cylindrical at vertical wall and does not fall down due to force of
friction acting upwards. The normal reaction exerted by vertical wall provides the required centripetal force
for executing circular motion.
So,
For vertical equilibrium
N = Ошибка!
Fr = mg
N ≥ mg
[⸪ Flim = N]
Ошибка! ≥ mg Ошибка!
V ≥ Ошибка!
LAWS OF MOTION
FORCE
Any push or pull which either changes or tends to change the state of rest or of uniform motion (constant
velocity) of a body is known as force.
Effect of Resultant force:
A non zero resultant force may produce following effects on a body:
i.
It may change the speed of the body.
ii.
It may change the direction of motion means the direction of instantaneous velocity.
iii.
It may change both the speed and direction of motion.
iv.
It may change the size and the shape of the body.
Units for measurement of force:
Absolute units
Gravitational / Practical units
i. N (M.K.S)
kg – wt or kg – f (kg. force)
ii. dyne (C.G.S)
g – wt or g - f
Relation between above units:
1kg-wt = 9.8 N
1 g-wt = 980 dyne
1 N = 105 dyne
INERTIA
Inertia is the property due to which a body tries to maintain its state of rest or of uninform motion. It means
that inertia is the resistance or opposition to the change of state.
Note:
Inertia of a body is measured by its mass.
Inertia  Mass
It is because a heavier body requires greater force for the change in its state and hence its inertia is more.
Similarly, the inertia for lighter body will be less.
NEWTON'S FIRST LAE OF MOTION
According to Newton, a body continues to be in the state of rest or of uniform motion along a straight line,
unless it is compelled by some external force.
Newton's first law defines force. It tells about the qualitative property of force such that with the help of
force, we can change the state of body.
We define inertia in 3 parts:
1. Inertia of rest.
2. Inertia of motion.
3. Inertia of direction.
1. INERTIA OF REST
A body at rest cannot change its state of rest by itself. This property is known as inertia of rest.
Examples:
i.
When a bus or train starts moving suddenly, then the passenger inside falls backward.
2. INERTIA OF MOTION
A moving body cannot change its state of motion by itself. This property is known as inertia of motion.
Examples:
i.
When a moving bus or train stops suddenly, then the passenger standing inside it falls forward.
ii.
A ball is thrown in the vertical upward direction by a passenger sitting inside a moving train. The
ball will return:
I Case: In the hands of the passenger, if the train is moving with constant velocity.
II Case: Ahead the passenger, if the train is retarding means slowing down.
III Case: Behind the passenger, if the train is accelerated means speeding up.
3. INERTIA OF DIRECTION
A moving body cannot change its direction of motion by itself. This property is known as inertia of direction.
Examples:
i.
When a car takes circular turn, then the person sitting inside it experiences a force away from the
centre of circular path. This force is known as centrifugal force.
(Centripetal force is towards the centre and Centrifugal force is away from the centre)
MOMENTUM(p)
The product of mass and velocity of a body is known as momentum.
→
→
,p = m ,v
It is a vector quantity whose direction is along the velocity.
Unit of momentum = kg-m/.
Dimensions of momentum = [m] [v] = [M1L1T–1]
NEWTON'S SECON LAW OF MOTION
According to Newton, the rate of change of momentum of any system is directly proportional to the applied
external force and this change in momentum takes place in the direction of applied force.
→
,F  Ошибка!
→
,F = k Ошибка!
→
+
→

Ошибка! = Ошибка! [value of k = 1by experiment]
,F = k Ошибка! = Ошибка!(mОшибка!)
,vОшибка! (general-form)
………………………………………………………….
or
Ошибка! = m Ошибка!
IMPULSE
When a large force is applied on a body for a very short interval of time, then the product of force and time
interval is known as impulse.
→
→
→
d ,I = ,Fdt = d ,p
Ошибка!
I Case: If this force is working from t1 to t2, then integrating the above equation, we get→
,I = Ошибка!Ошибка!dt = Ошибка!dОшибка!
II Case: If a constant or average force works on body, then:
→
→
Ошибка!
=
,FavgОшибка! dt = Ошибка!dОшибка!
→
1

,I = Ошибка!Ошибка!avg dt = Ошибка!dОшибка!
,I =
→
t
,Favg [t] 2, =
t1
→
,Favg [→,p]
→,p
2
,→
,p1

→
,I =
→
→
→
,Favg (t2 – t1) = ( ,p2 - ,p
)
→
,I =
→
→
,Favg∆t = ∆ ,p
Impulse Momentum Theorem:
The impulse of force is equal to change in momentum of body. This relation is known as impulse momentum
theorem.
Units of impulse = N – s or kg-m/s
Dimensions of impulse = [F] [t] = [m] [a] [t] = [M 1L1T–2T1] = [M1L1T–1]
Important points about Newton's Second Law of Motion:
1.
Newton's first law of motion defines force and second law of motion measures force. It gives the units,
dimensions and magnitude of the force.
Unit of force = (Unit of mass) × (unit of acceleration) = 1kg × 1 m/s 2 = 1N
1N = 1 kg-m/s2
1 dyne = 1 g-cm/s2
1N = 105 dyne
Dimensions of force:
2.
[m] [a] = [M1] [L1T–2]
= M1L1T–2
If a particle moves uniformly, means velocity = constant
→
So,
→
,a = Ошибка! = Ошибка! = Ошибка!
→
→
→
,F = m ,a = m× ,0 = ,0
It means that, in the absence of external force, a particle moves uniformly. This is Newton's first law of
motion. It means that, with the help of Newton's second law of motion (NSLOM), we can derive Newton's
first law of motion (NFLOM)
3.
The slope of momentum-time graph is equal to force on the particle
…………………..fig…………………………………..
At point A,
4.
Ошибка!
The area under the force – time graph represents impulse or change in momentum.
………………………..fig…………………………………..
Note:
Here,
Ошибка!
Since
Favg = Ошибка!
Therefore, for a particular (constant) momentum change if time interval is increased, then the average force
exerted on body will decrease.
Collision of ball with wall:
A ball of mass m strikes a surface with velocity 'u' at angle '' from the normal of surface and rebounds with
the same speed in time 't'. Find the change in momentum and the force exerted on ball due to the surface.
Here


→
,Pi = mu cos( ,i) – mu sin( ,j)
→


,Pf = mu cos( ,i) – mu sin( ,j)
Now,
∆
→
,P =
→
,Pf –
→
,Pi




= –mu cos( ,i) – mu sin( ,j) – mu cos( ,i) + mu sin( ,j)

= -2mu cos( ,i)
It means that momentum changes only along the normal to the surface but not along the surface.
|∆→,P| = 2mu cos 
→
,F = Ошибка! = Ошибка!(Ошибка!)
Ошибка!
Two special cases:
Case (I)
If  = 0°, means ball is thrown perpendicular to the surface, then

→
∆ ,P = –2mu cos0°( ,i)

→
∆ ,P = –2mu( ,i)
→
,F = Ошибка! = Ошибка!(Ошибка!)
NEWTON'S THIRD LAW OF MOTION
According to Newton's third law, to every action, there is always an equal (in magnitude) and opposite (in
direction) reaction.
This law is also known as action-reaction law. Here
equal in magnitude and opposite in direction to
→
→
→
,F12 (means force on first body due to second body) is
,F21 (means force on second body due to first body).
→
,F12 = – ,F21
…………………fig……………………………….
Important points about Action – Reaction law:
i.
Action - reaction law is applicable to all the interaction forces eg. gravitational force, electrostatic force,
electromagnetic force, tension, friction, viscous force, etc.
ii.
Action and reaction always act on two different bodies.
iii.
Action and reaction never cancel each other because they are exerted on two different bodies.
ROCKET PROPULSION
I Case: If rocket is accelerating upwards, thenNet upwards force on rocket = ma
Ошибка!
II Case: If rocket is moving with constant velocity, then a = 0
⸫ vОшибка! – mg = m × 0
Ошибка!
NORMAL REACTION
When a stationary body is placed on a surface then, that surface exerts a contact force on that body which is
perpendicular to the surface and towards the body. This force is known as Normal Reaction
e.g.
A block of mass "m" is placed on a horizontal table. Then the forces exerted on it are:
1.  Gravitational force of attraction on body due to earth, means weight of the block (=mg).
2.  Normal reaction exerted by the surface of table on the block.
Since the body is in equilibrium, therefore net force on it is zero.
⸫
→
,N +
→
→
,N = –
,W =
→
→
,O
,W
N=W
⸫ N = mg
[⸪
|→,N| = |–→,W|]
[⸪ W = mg]
EFFECTIVE OF APPARENT WEIGHT OF A MAN IN LIFT
I Case: If the lift is at rest or moving uniformly (a = 0), then
N = mg
So, Wapp = Wacual
………………………………………………………………….
II Case: If the lift is accelerated upwards, then
Net upward force on man = ma
N – mg = ma
N = mg + ma
⸫N = m (g + a)
or N = m (g + a)
So, Wapp > Wactual
…………………………………………..
III Case: If the lift is retarding upwards, then
N – mg = m (–a)
N = mg – ma
⸫
Wapp or N = m (g – a)
So, Wapp < Wactual
IV Case: If the lift is accelerated downwards, then
mg – N = ma
N = mg – ma
⸫
Wapp or N = m (g – a)
So, Wapp < Wactual
V Case: If the lift is retarding downwards, then
mg – N = m (–a)
mg – N = –ma
N = ma + mg
Wapp or N = m (g + a)
So, Wapp > Wactual
Two Special Cases of downward acceleration (IV Case):
I Special Case: If the lift is falling freely, means its acceleration in the vertically downward direction is
equal to acceleration due to gravity. i.e. a = g, then
Wapp = m (g – a) = m ( g – g) = m × 0 = 0
⸫
Wapp = 0
Means the man will feel weightless. This condition is known as Condition of Weightlessness.
The apparent wt. of any freely falling body is always zero (0).
II Special Case: If the lift is accelerating downwards with an acceleration which is greater than 'g', then
the man will move up with respect to lift and he will stick to the ceiling.
MOTION OF BODIES IN CONTACT (CONTACT FORCE)
Two bodies in contact:
F – f = m1a
…1
f = m2a
…2
………………fig…………………..
______________________________
On adding above equations
F = m1a + m2a
F = a (m1 + m2)

Ошибка!
…3
OR
Ошибка!
putting value 'a' from equation (3) into (2), we get:
Ошибка!
(here f = contact force)
Find acceleration (a) of blocks and contact force (f) between them

Ошибка!
…4
Putting value of 'a' from equation (4) into (3), we get
Ошибка!
Putting value of 'a' from (4) to (1):

Ошибка!
SYSTEM OF MASSES TIED BY STRINGS
Tension in a String:
It is an intermolecular force between the molecules of a string, which acts or reacts when the string is
stretched.
Important points about the tension in a string:
a.
i.
String is massless and frictionless so that tension throughout the string remains same.
ii.
If the string is massless but not frictionless, at every contact tension changes.
iii.
If the string is not light, tension at each point will be different depending on
acceleration of the string.
MOTION OF CONNECTED BODIES:
Two Connected Bodies:
……………..fig…………………………………………..
Ошибка!
Ошибка!
Three Connected Bodies:
OR
Ошибка!
the
…………………………………..fig……………………………………
F = (m1 + m2 + m3)a
⸫
Ошибка!
Ошибка!
Ошибка!
or
Ошибка!
TENSION IN ROD:
Given
Mass of Rod
=
M
Length of Rod
=
L
F – T = ma
T = F – ma
T = F – m Ошибка! Ошибка!
⸪ Mass of length 'L' = M
⸫ Mass of unit length = Ошибка!
⸫ Mass (m) of length 'x' = Ошибка! x
Put this value of 'm' in equation (1)
T = F – Ошибка! × Ошибка!
T = F – Ошибка!
Ошибка!
BODIES HANGED VERTICALLY
Since the bodies are in equilibrium, therefore net force on all the bodies is zero
T1 = m1g
…(1)
T2 = T1 + m2g
T2 = m1g + m2g
[From equation (1)]
T2 = (m1 + m2)g …(2)
T3 = T2 + m3g
T3 = (m1 + m2)g + m3g
T3 = (m1 + m2 + m3)g
[From equation (2)]
FIXED PULLEY
SOME CASES OF PULLEY
Case I
Case II
…………………………..
m1 = m2 = m
Tension in the string
T = mg
Acceleration 'a' = zero
Thrust/Reaction at the suspension/clamp of the pulley
T = 2T = 2 mg
Case III:
For mass m1 : T1 – m1g = m1a
For mass m2 : m2g + T2 – T1 = m2a
For mass m3 : m3g – T2 = m3a
a = Ошибка!g
we can calculate tensions T1 and T2 from above equations
……………………………fig…………………………………………..
Case IV:
For mass m1:
T = m1a
For mass m2:
m2g – T = m2a
Solving:
a = Ошибка! and T = Ошибка!g
Case V:
Mass suspended over a pulley from one another on an inclined plane.
For mass m1: m1g – T = m1a
For mass m2: T – m2g sin = m2a
acceleration a = Ошибка!g
T = Ошибка!g
Case VI:
Masses m1 and m2 are connected by a string passing over a pulley (m1 > m2)
Acceleration a = Ошибка!g
Tension
T = Ошибка!g
INERTIAL AND GRAVITATIONAL MASS
1.
The ratio of force applied on a particle to its acceleration is known as inertial mass Ошибка!
2.
The ratio of gravitational force to gravitational acceleration is known as gravitational mass
Ошибка!
It is experimentally proved that both masses are equal means mi = mg
Ошибка!
I. Note: A bird is sitting at the base of an air tight cage. Now if the bird starts flying, thenI Case: Weight will not change, if the bird flies with constant velocity.
II Case: Weight will increase, if the bird flies with upward acceleration.
III Case: Weight will decrease, if the bird flies with downward acceleration.
II. Note: A bird is sitting at the base of wire cage. Now if the birds flies upward then weight will decrease in all the
cases.
FRAME OF REFERENCE
A system with respect to which position or motion of a particle is described as frame of reference. We can
classify frame of reference in two categories:
i.
Inertial Frame of Reference:
The frame for which law of inertial is applicable is known as inertial frame of reference.
All the frame which are at rest or moving uniformly with respect to an inertial frame,
inertial frame.
ii.
are also
Non-Inertial Frame of Reference:
The frame for which law of inertial is not applicable is known as non-inertial frame of reference.
All the frames which are accelerating or rotating with respect to an inertial frame will be non
inertial frames.
Note:
In case of rotating frames, a pseudo force [i.e. centrifugal force] acts on the body away from the centre of
the circular path. Its magnitude is Ошибка!.
Pseudo force does not follow action-reaction law.
…………………………………………………………………….fig
FRICTION
The force which opposes tendency of motion or relative motion between surfaces is known as Friction.
Theories about the origin of friction:

Acc. to it every surface in the nature is rough. So, when we place one surface over other, the projections of
one surface get interlocked with the depressions of other surface. This interlocking opposes the tendency of
motion or relative motion between surfaces and the force of opposition is known as friction.

When we place one surface over other, then at the area oof actual contact, molecules of one surface get
interlocked with molecules of other surface due to intermolecular forces. This interlocking of molecules
opposes the tendency of motion or relative motion between the surfaces. This opposition is known as friction.
…………………fig…………………………………………………..
We can classify friction in two categories:
1.
Internal friction:
The friction between layers of a fluid is known as internal friction or viscosity.
2.
External friction:
The friction between solid surfaces is known as external friction.
External friction can be divided in two parts.
STATIC FRICTION (fs)
The force which opposes tendency of motion of surfaces is known as static friction.
It is a self adjusting force and its numeric value is equal to external force applied on the body which causes
tendency of motion.
The maximum value of static friction is known as limiting friction (fL).
DYNAMIC/ KINETIC FRICTION (fK)
The force which opposes relative motion between surfaces is known as kinetic friction.
Its value does not depends on the types of motion of body such as accelerated motion, retarded motion or
moving with constant velocity because it is a constant friction.
Graph Between Friction and Applied External Force:
………………fig…………………………………………….
Limiting friction is more than kinetic friction because initially we have to rovide some extra force
to overcome the inertia of rest and in the morning state, molecules will have less time to get interlocked again
with the same strength.
LAWS OF LIMITING FRICTION
1.
Limiting friction is directly proportional to normal reaction, i.e. fL = N
…………………………fig………………………………..
2.
Limiting friction always work in the direction opposite to that of the tendency of motion.
3.
For given surfaces limiting friction does not depend upon apparent constant area of surfaces, as long as
normal remains same.
4.
Limiting friction depends upon the degree of smoothness and material of surfaces.
COEFFICIENT OF FRICTION []
It is of two types:
1.
Coefficient of static friction [s]
2.
Coefficient of kinetic friction [k]
⸪
fL  N
⸫
fL = s × N
So,
s = Ошибка!
Here, s is known as coefficient of static friction
s is unitless & dimensionless.
Coefficient of kinetic friction (k) = Ошибка!
⸪
fL > fk
⸫
s > k
Coefficient of friction depends upon degree of smoothness, material and temperature of surface.
Note: The component of contact force perpendicular to the surface is normal reaction and the ecomponent parallel to
the surface is friction
………………….fig……………………………………………….
PULLING IS EASIER THAN PUSHING
Case of pulling:
Force F is applied to pull a block of weight W.
F can be resolved into two rectangular components: F cos and F sin.
the normal reaction
N = W – F sin [⸪N + F sin = W]
Force of kinetic friction
fk = kN
fk = k (W – F sin)
(i)
Case of pushing:
Force F is applied to push a block
normal reaction N' = W + F sin
Force of kinetic friction
f'k = kN'
f'k = k (w + f sin)
or
(ii)
By (i) and (ii) f'k > fk
The frictional force is more in the case of pushing. Hence it is easier to pull a body than to push it.
SOME SPECIAL CASES OF FRICTION
Case-I: When force F applied on m1 and there is no friction between m1 and surface and the coefficient of
friction between m1 and m2 is . What should be the maximum value of F so that both blocks move together?
from fig. (A),
………..fig………………………………
for mass m
F – f = m1a

F = f + m1a
…i
from fig. (B),
………..fig………………………………
f = m2a
…ii
f = N2 = m2g
…iii
from (ii) and (iii)
m2a = m2g

a = g
…iv
putting values 'f' and 'a' from equation (iii) and (iv) in (i)
F = m2g + m1g
F = (m1 + m2)g
Case-II
When Both Surfaces are rough.
from fig (A)
for mass m1
F – f1 – f2 = m1a
F = f1 + f2 + m1a
from fig (B)
…(i)
…(ii)
f2 = m2a
f1 = 1N1
f1 = 1 (m1 + m2)g
[⸪ N1 = (m1 + m2)g]
…(iii)
from fig (B)
f2 = 2N2
(N2 = m2g)
f2 = 2m2g
…(iv)
from (ii) and (iv)
2m2g = m2a
a = 2g
…(v)
F = 1(m1 + m2)g + 2m2g + 2m1g [by (i), (iii), (iv) & (v)]
F = (m1 + m2) (1 + 2)g
Case – III
When force F is applied on m2 and there is no friction between m and surface.
For mass m2
F – f = m2a
F = f + m2a
…(i)
From fig (B)
f = N2
f = m2g (⸪ N2 = m2g)
…(ii)
from fig (A)
f = m1a
…(iii)
from (ii) and (iii)
m2g = m1a
a = Ошибка!
…(iv)
from (i), (ii) and (iv)
F = m2g + m2Ошибка!
F = (m1 + m2) Ошибка!
Case – IV:
When both surfaces are rough:
For mass m2
F – f2 = m2a
…(i)
F = f2 + m2a
from fig (B)
f2 = 2N2
f2 = 2m2g
…(ii)
from fig (A)
f1 = 1N1
f1 = 1(m1 + m2)g
[⸪ N1 = (m1 + m2)g]
for mass m1
from (ii), (iii) and (iv)
a = Ошибка!
…(v)
…(iii)
from (i), (ii) and (v)
F = Ошибка! + 2m2g
F = Ошибка!
F = Ошибка!
F = Ошибка!
Ошибка!
ANGLE OF FRICTION ():
The angle made by the resultant of limiting friction and normal reaction, from the normal reaction is known
as angle of friction.
tan() = Ошибка! = Ошибка! = s
where

tan() = s

 = tan–1 (s)
s = coefficient of static friction
 = angle of friction
………………….fig……………………………………………..
ANGLE OF SLIDING OR ANGLE OF REPOSE:
The maximum angle of inclination of a plane from horizontal at which a body placed on it remains at rest.
N = mg cos
…1
fL = mg sin
…2
Dividing equation (2) by (1), we get
Ошибка! = Ошибка!
Ошибка! = Ошибка![⸪fL = sN]

s = tan 

 = tan–1 (s)
Angle of friction = Angle of repose = tan–1 (s)
Case – I:
If angle of inclination is less than the angle of repose then the body will not move on the inclined plane and
force of friction acting on the body is equal to "mgsin".
Further if we want to move the body down the incline plane then a minimum downwards force "mgcos –
mgsin" has to be applied on the body.
Case – II:
If the angle of inclination is more than the angle of repose then the body slides down the inclined plane with
net force "mgsin – mgcos" and with net acceleration gsin – g cos 
i.
If we want to prevent downward sliding of body then the minimum required force will be "mgsin
–mgcos"
ii.
If we want to move the block up the inclined plane then the minimum force required is "mgsin +
mgcos"
iii.
If we project the body up the inclined plane with some initial velocity (by providing KE) then
retardation acting on the body will be "gsin + gcos"
CYCLING
In the cycling, rear wheel move by the force communicated to it by pedalling, while front wheel moves by
itself, therefore, like in walking force of friction on rear wheel is in forward direction. As front wheel moves
by itself, force of friction on front wheel is in the backward direction.
When pedalling is stopped, both the wheels moves by themselves so the force of friction on both the wheels
is in backwards direction.
WORK, ENENRGY AND POWER
Work:
Dot or scalar product of force & displacement is known as work.
W=
→
→
,F. ,d
W = Fdcos
Product of component of displacement in the direction of force & force is known as work.
W = (dcos).F
………………………………………………………………………….
………………………………………………….
…
Work is a scalar quantity and its unit
M.K.S. = Joule
C.G.S. = erg.
1J = 107 erg
In atomic and nuclear physics unit of work is
1eV = 1.6 × 10–19J
1MeV = 1.6 × 10–13 J
1kWH = 3.6 × 106 J
Dimension [M1L1T–2][L1] = [M1L2T–2]
Nature of work done:
Positive work: If angle  is acute ( < 90°). Then work is +ve. It means that the external force favours the
motion of body.
Ex.
When a body falls freely under the action of gravity ( = 0°). The W.D. by gravity is positive.
Negative work: If angle is obtuse ( > 90°) the work is -ve. It means force is such that it opposes the motion
of the body.
Ex.
W.D. by kinetic frictional force is -ve when it opposes the motion.
Zero work: W = Fdcos
If F = 0 or d = 0 or  = 90°
Ex.
1.
Body moving with uniform velocity W.D. by nete force = 0
2.
Net force on the particle is zero then W.D. on the body is also zero.
3.
Work done by coolie against gravity is zero.
Golden key points:
Work is defined for an interval or displacement, there is no term like instantaneous work similar to
instantaneous velocity.
For a particular displacement work is independent of time, work will be same for same displacement whether
the time taken is small or large.
When several forces act, work done by a force for a particular displacement is independent of other forces.
Displacement depends on references frame so work done by a force depends on frame of reference so work
done by a force can be different in different reference frames.
Effect of work is change in kinetic energy.
Work done by variable force:
For a variable force work is calculated for infinitely small displacement and for this displacement force is
assumed to be constant.
dW =
→
→
,F.d ,r
Total work done
W = Ошибка!dW = Ошибка!Ошибка!.dОшибка!
let
and
d
→
→



,F = Fx ,i + Fx ,j + Fx ,k
→



,rA = (x1 ,i + y1 ,j + z1 ,k)
→



,rB = (x2 ,i + y2 ,j + z2 ,k)



,r = dx ,i + dy ,j + dz ,k
(⸪
→



→



,r = x ,i + y ,j + z ,k ⸫ d ,r = dx ,i + dy ,j + dz ,k )
dW =
→
→






,F.d ,r = (Fx ,i + Fx ,j + Fx ,k) . (dx ,i + dy ,j + dz ,k)
W = Ошибка!dW = Ошибка!Fxdx + Ошибка!Fydy + Ошибка!Fzdz
Graphical method to calculate work done:
If force displacement (F – x) curve is given then net area under F – x curve is equal to work done.
W = W1 + (–W2)
…………………………….fig……………………………………………………
Energy: Energy is defined as internal capacity of doing work.
When we say that a body has energy, we mean that it can do work. It is a scalar quantity.
Unit & dimensions are same as work.
Mechanical energy: Sum of kinetic & potential energy.
Kinetic energy: Energy due to motion of the particle is known as kinetic energy.

Kinetic energy is always positive.

Kinetic energy is a scalar quantity.

Kinetic energy is a relative quantity & t depends on a frame of reference.
………………………………fig…………………………………….

Kinetic energy of block w.r.t. A is zero.

Kinetic energy of block w.r.t. B is Ошибка! mv2.
E = Ошибка!mv2 = Ошибка!mv2 × Ошибка!  Ошибка!Ошибка! = Ошибка! Ошибка!
E  P2 (P = Linear Momentum)
Graph
………………………………………………………………
………………………………………………………………………………
…………………………………………………………………….
…………………………………………………………………..
Questions based on % change:
Case-I: change ≤ 5%
(A)
K.E. = Ошибка!  K.E.  P2  Ошибка!
(B)
P = 2m K.E.  P  K.E.  P  (K.E.)1/2  Ошибка!
More then 5% (≥ 5%):
E  p2  Ошибка! = Ошибка! × 100  Ошибка!
P  E  Ошибка! = Ошибка! × 100  Ошибка! = Ошибка! × 100
Work energy theorem:
Work done by all forces (Net force) = change in K.E.
W = ∆K.E.
Proof:
i.
For constant force
………………….fig……………………..
v2 = u2 + 2as
v2 – u2 = 2Ошибка!s
s = Ошибка! = Ошибка!
F.S. = ∆K.E.
W = ∆K.E.
ii.
For variable force
W = Ошибка!F dx = Ошибка!ma dx = Ошибка!mОшибка!dx
W = Ошибка!mvdv
W = mОшибка!
W = Ошибка![v2 – u2]
W = ∆K.E.
Golden Key Points:

If there is no change in the speed of a particle, there is no change in kinetic energy. So work done by the
resultant force is zero.

If K.E. of the body decreases then work done is negative i.e. the force opposes the motion of the body.

If K.E. of the body increases then work done is positive.

In above discussion, we have assumed that the work done by the force is effective only in the kinetic energy
of the body. It should however be remembered that work done on a body may also be stored in the body in
the form of potential energy.
Potential energy:
The energy possessed by any body or a system is due to change in its position, shape or configuration is
called as potential energy.

It is a scalar quantity.

It can be positive or negative or zero.

It is a relative quantity & it depends on frame of reference but change in potential energy is independent of
frame of reference.

Potential energy is always defined only for conservative force fields.

Whenever work is done by the conservative forces potential energy decreases & whenever work is done
against conservative forces, potential energy increases.

It is function of position doesn't depend on path.
Difference between conservative & non-conservative force
……………………………………………………………………………………………………………..
……………………………………………………………………………………………………………
……………………………………………………………………………………………
……………………………………….
Example of conservative force and non-conservative force
…………………………………………………………………………………………………………
………………………………………………………………………………………………………..
………………………………………………………………………………………………
.
Central force:
The forces acting along the line joining the particles are called as central force.
→

,Fcentral = C.F(r). ,r

[C = constant, F(r) = function of position, ,r = direction]
All central forces are conservative forces but all conservative forces are not central forces.
Ex.
All constant forces are conservative forces but they are not central force.
Ex.
Gravitational force, electrostatic force, spring force or restoring force, elastic force, magnetic force produced
by magnetic poles.
Power:
Rate of work done by man or machine is known as power.
Average power Pavg = Ошибка! = Ошибка!
Instantaneous power P = Ошибка! = Ошибка!
P=
→
→
,F. ,v
Unit = Ошибка! = watt
1 HP = 746 watt.
Dimension = [M1L2T–3]
……………………………fig…………………………………………………

Slope of work-time curve represents instantaneous power.
Area under power-time graph gives the work done
Area under power-time graph
= Area ABCD
= Ошибка!Pdt = W
Efficiency:
 = Ошибка! = Ошибка!
or it is defined as the ratio of output work done by a body and input energy supplied to it.
 = Ошибка!
Conservation of linear momentum law:
If net external force on the system is zero then linear momentum of the system remains constant.
According to newtons II law.
→
If
,Fext = Ошибка!
→
,Fext =
→
,0

Ошибка! = Ошибка!
→
→
,P1 +
→
,P2 +
→
,P3 +
→
,Pin =
→
,P4 + ….. +
Ошибка! COLM
,Pfinal
→
,Pn = constant
→
→
→
→
∆( ,P1 + ,P2+ ….. + ,Pn) = ,0
→
→
→
→
∆ ,P1 + ∆ ,P2+ ….. + ∆ ,Pn = ,0
Ex. GUN-BULLET system:
M = mass of gun
⸫
→
→
m = mass of bullet
Total momentum before firing = Total momentum after firing
,Pi =
→
,Psys =
Ошибка!
,Pf
→
,Pb +
→
,PG
= Ошибка!
= Ошибка!
→
→
→
,0 = m ,vb + M ,vG
Ошибка!
Ошибка! > 1
as
so
Ошибка!
After firing bullet the change in K.E. of system
∆KE = KEf – KEi
Ошибка!

This  in K.E. due to the conversion of chemical energy of gun powder into of K.E. of Gun-Bullet system.

If the bullet is the system the force exerted by trigger will be external and the momentum of bullet will change
from 0 to mV.

If the bullet and Gun is the system the force exerted by trigger will be internal.

Conservation of linear momentum is equivalent to Newton's III law of motion for a system of two particles
in absence of external force, by law of linear momentum.
→
,P1 +
→
,P2 = constant
→
→
m1 ,v1 + m2 ,v2 = constant
Differentiating above with respect to time
m1Ошибка! + m2Ошибка! = 0
→
→
m1 ,a1 + m2 ,a2 = 0
→
→
,F1 +
→
,F2 = 0
[as m is constant]
Ошибка!
[as
→
→
,F = m ,a]
→
,F1 = – ,F2
i.e. for every action there is equal and opposite reaction which Newton's III law of motion.
Law of conservation of Mechanical energy:
Under the effect of "conservation force" mechanical energy of the system always remains constant. This is
called law of conservation of mechanical energy.
M.E. = K.E. + U = constant
∆K + ∆U = 0
∆K = –∆U
Ex.
M.E. of a particle remains constant in gravitation field. Prove ME (1) = ME(2) = ME(3)
………………………………fig………………………………..
Special cases:
1.
A particle of mass m is freely released from height h on a spring, spring is compressed by x then prove that:
Ошибка!
………………………………………………..figures………………………………………….
……………………..
…………
………
….
Special notes:

When spring is compressed or elongated from its natural length by same amount then the W.D. by spring
force = –Ошибка! .

When spring is elongated or compressed from its natural length by same amount, the W.D. by external force
= Ошибка! .
COLLISION
Collision: It is an isolated event in which there is mutual interaction between 2 or more then 2 particles
and due
to that there will be change in momentum at least one of the colliding particles. It is not necessary to have physical
contact between bodies. Change in momentum is an essential
characteristic of collision.
In simple language collision is an event in which an impulsive force acts between 2 bodies for a short duration
which results in change of their velocity.
For a system of colliding bodies, total linear momentum can't change as there is no net external force to
change it.
Collision on the basis of direction or dimension:
1.
1-D/head on/direct collision/linear collision: Here colliding particles travels along a straight line before &
after collision.
2.
2-D/oblique collision: Here colliding particles travel in a plane after collision. e.g.: Carrom Board (collision
of striker & coin), Billiards table (collision between ball & ball)
3.
3-D collision (collision in space): Study of 3-D collision is impossible because no. of known equation is less
in comparison to unknown parameters.
Special points about collision:
1.
In a collision we consider the duration of collision negligible as compare to the time for which event is
observed.
2.
In a collision effect of external non-impulsive forces (eg. gravity force mg etc.) can be neglected because
they are very small compare to mutual interaction forces (responsible for collision).
3.
In case of collision if the external impulsive forces are not acting in direction of collision the total momentum
of system in that direction remains conserved.
According to conservation laws collision can be classified into:
Elastic collision Inelastic collision Perfectly inelastic collision
i.
Conservation of Linear momentum
ii.
Conservation of Total energy
Law if collision:
i.
Conservation of linear momentum: Valid for all type of collision.
Total momentum before collision = Total momentum after collision.
…………………………………..fig………………………………………………..
→
→
→
→
m1 ,u1 + m2 ,u2 = m1 ,v1 + m2 ,v2
Law of K.E.:
Valid only for elastic collision
Total kinetic energy before collision = Total kinetic energy after collision
Ошибка!
Newton's law of collision: Valid for all types of collision.
(Relative velocity after collision) = –e(Relative velocity before collision)
→
→
→
→
( ,v2 – ,v1) = –e( ,u2 – ,u1)
Ошибка! = Ошибка!
e = coefficient of restitution
0≤e≤1
e=1
for elastic collision
e=0
perfectly inelastic collision
0 < e < 1 Inelastic collision
Practically value of e can be > 1
Head on elastic collision
→
→
→
→
m1 ,u1 + m2 ,u2 = m1 ,v1 + m2 ,v2
…1
→
→
→
→
( ,v2 – ,v1) = e ( ,u2 – ,u1)
…2
for elastic collision e = 1
By solving equation (1) & (2)
…………………………………………………………..
…………………………………………………………..
Special cases:
a.
If the two bodies are of equal masses:
m1 = m2 = m
→
,v1 =
→
,u2 and
→
,v2 =
→
,u1
Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the
bodies will exchange their velocities and kinetic energy.
b.
If two bodies are of equal masses and second body is at rest:
m1 = m2 and initial velocity of second body
→
→
,v1 = 0,
,v2 =
→
→
,u2 = 0
,u1
When body A collides, against body B of equal mass at rest, the body A comes to rest and the body B moves
on with the velocity of the body A. In this case transfer of energy is hundred percent.
Eg. Biliard's ball, Nuclear moderation
c.
If mass of a body is negligible as compared to other:
i.
If m1 >> m2
…………………………………………………………………………
…………………………………………………………………………..
…………………………………………………………………………..
…………………….
Head on inelastic collision:
→
→
→
→
m1 ,u1 + m2 ,u2 = m1 ,v1 + m2 ,v2
→
→
→
,v2 –
→
→
→
,v1 = e ( ,u1 – ,u2)
,v1 = Ошибка!Ошибка! + Ошибка!Ошибка!
,v2 = Ошибка!Ошибка! + Ошибка!Ошибка!
Less in K.E. = Ошибка! – Ошибка!
Ошибка!
Block bullet system:
……………………..fig……………………….
→
,Pi =
→
,Pf
→
m ,vb + 0 = (M + m)gh
Ошибка!
Ошибка! (M + m)vОшибка! = (M + m)gh
Ошибка!
Loss in K.E. = Ошибка!mvОшибка! – (M + m)vОшибка!
Ошибка!
Ex.
Sol.
A ball is dropped freely from height 'h'. This ball is continuously rebounding, then find out:
i.
velocity of ball after 'n' rebound
ii.
height attained by the ball after 'n' rebound
iii.
time taken by the ball in nth rebound
iv.
total distance covered by the ball before it stops rebounding.
i.
Velocity after 'n' rebound:
………………….fig…………………………….
vn = en 2gh
ii.
Height attained by the ball after 'n' rebound:
hn = e2nh
iii.
Time taken in nth rebound:
Ошибка!
Ошибка!
Special Case:
T = t + 2t Ошибка! = tОшибка! = Ошибка!Ошибка!
iv.
Distance covered by the ball before it stops:
S = hОшибка!
CENTRE OF MASS
Definition: The centre of mass of a body is a point where total mass of the body is supposed to be
concentrated.
Centre of gravity: Centre of gravity of a body is that point where it is assumed that gravitational force of earth on
body i.e. weight of body acts on it.
In normal cases if the acceleration due to gravity remains the same throughout the mass distribution then
centre of gravity coincides with centre of mass and both also coincides with geometrical centre of body.

For very high mountains or multistory buildings (in both cases value of g decreases with height)
⸫
Centre of gravity lies below its centre of mass.

At a place where g = 0 centre of gravity does not exist but centre of mass, exists.

There may or may not be any mass present physically at centre of mass (see figure).
…………………………………………….fig……………………

It may be inside or outside of the body.

Its position depends on the shape of the body.

For a given shape it depends on the distribution of mass of within the body & is closer so massive part.

For symmetrical bodies having homogenous distribution of mass it coincides with centre of symmetry of
geometrical centre.

If we know the centre of mass of parts of the system & their masses, we can get the combined centre of mass
by treating the parts as point particles placed at their respective centre of masses.
→
Position vector of C.O.M. for n-particle system ( ,rcm):
………………………………..fig……………………………………….
→

,rcom = Ошибка!
For two particles system
→
,rcm = Ошибка!
If C.O.M. coincides with origin then
→
,rcm =
→
,0
→
→
→
m1 ,r1 + m2 ,r2 = ,0
→
→
→
→
m1 ,r1 = – m2 ,r2 m1 ,r1 = – m2 ,r2
|
| |
|
m1r1 = m2r2
Position co-ordinates of c entre of mass for a particle system:
xcm = Ошибка!
ycm = Ошибка!
zcm = Ошибка!
GRAVITATION
2.
Newtons universal law of Gravitation:
According to it, every body in the universe attracts every other body by a force which is directly proportional to
the product of their masses and inversely proportional to the square of the distance between them.
So,
F  m1m2 ………… (1)
and
F  Ошибка! ……………… (2)
Figure Page 283
from equation (1) and (2)
F  Ошибка!
Where, 'G' is called "universal gravitational constant"
Important points about 'G' :
1.
It is a scale.
2.
Units and dimensions,
⸪
F = Ошибка!
⸪
G = Ошибка!
SI unit = N m2 / kg2
CGS unit – dyne cm2 / g2
Relation between SI & CGS unit of 'G' :
So,
Ошибка! = Ошибка! = Ошибка!
Dimensional formula of G = Ошибка! = Ошибка! = Ошибка! = [M–1 L3T–3]
So,
3.
4.
5.
[G] = [M–1L3T–2]
Value of 'G' is = 6.67 × 10–11 Nm2 / Kg2 [SI or MKS value]
and 'G' = 6.67 × 10–5 dyne cm2 / g2 [CGS value]
It is a universal constant because its value is same throughout the universe and it does not depend upon the
bodies and medium between them.
In the laboratory, its first successful determination was made by "Henry cavendish" by using torsional
balance."
Important point about gravitational force:
1.
In vector form

,F = Ошибка! Ошибка!
2.

Ошибка!



Ошибка!
Figure
Figure
Radially outwards = Ошибка! Radially inwards = –Ошибка!
,F = Ошибка! Ошибка!
It is the force of interaction which is always produced in the form of action – reaction pair. Hence, it follows
Newton's third law of motion.
Here,

,F12 = Force on 1st body due to 2nd body.

,F21 = Force on 2nd body due to 1st body.
Figure
⸪
⸪
3.
⸪
⸪
|,F12| = |,F21| =

Ошибка! and their direction are opposite.

,F12 = ,F21






,F12 =
,F12 +
,F21
,F21 = 0
means, ,Fnet + ,0
Since, the net force on system is zero. Therefore, linear momentum of the system will be conserved but
the conservation of linear momentum is not applicable on individual bodies because net force on individual
bodies is not zero.
4.
Gravitational force is a central force
Standard form of central force :

,F = Cons. f(r) (,r or –,r)
For example,
1.
Gravitational force

,F = Ошибка! Ошибка!
here, Gm1m2 = constant, Ошибка! = f(r) and (– Ошибка!) = direction
2.
Electrostatic force

,F = Ошибка! (Ошибка! or –Ошибка!)


here, kq1q2 = f(r) and ( ,r or – ,r) = direction
5.
Gravitational force is a conservative force because the work done against it doesn't depend upon the path
but it only depends upon the initial and final positions.
e.g.
Figure
6.
Gravitational force between two bodies does not depend upon the medium between them. For example if
the gravitational force between two bodies in air if 'F' in water, then it provided will remain 'F' if they are
kept at the same separation in water.
7.
8.
9.
It is a very weak force.
Gravitational force is a wide range force.
If there are more than two particles in a system, then the net gravitational force on any particle is the vector
sum of all the gravitational attractions exerted on it due to remaining particles. This is known as "Principle
of super position of forces."
eg.
Figure





,FNet = ,F1 + ,F2 + ,F3 + … + ,Fn
10.
11.
The attraction between any two massive bodies is known as gravitational attraction but the attraction of
the earth on any other body is known as gravity force.
If a particle executes circular motion around another particles under the influence of gravitational attraction
force, then the work done by gravitational force ill be zero.
eg.
Figure


W = ,Fg . ,S = Fg S cos 90° = 0
Here, F = Fg  Ошибка! = Ошибка!
3.
Gravitational Field:

The space or region around a mass particle in which its gravitational effects are experienced is known as
gravitational field.

Gravitational Field Intensity ( ,I):

The gravitational field intensity at a point in some gravitational field is defined as the gravitational force
exerted on unit mass placed at the point.
Ошибка!



Gravitational field intensity is a vector quantity whose direction is same as that of gravitational force.
Its SI unit is 'N/kg'
Dimension's of intensity = Ошибка! = Ошибка! = [M0L1T–2]
Gravitational field intensity = gravitational force exerted on unit mass


,I = Ошибка! (–Ошибка!)
Ошибка!
where 'M' is the mass of that particle due to which we have to find intensity.
Gravitational field intensity due to spherical mass distribution
Note: If the observation point is located on the surface or outside the surface then the spherical mass can be taken as
a mass particle which is situated at the centre of sphere. i.e. point mass.
For solid sphere
Figure
Let 'M' be the mass of sphere, 'R' be the radius of sphere and 'r' the distance of observation point from the centre
of sphere.

Case I :
When r > R, i.e. Outside the sphere then Ошибка!
Figure

Case II :
When r = R, i.e. At the surface then Ошибка!
Figure

Case III : When r < R, i.e. in side the sphere then

,Iin = Ошибка! …………. (1)
Now, we know,
Ошибка! = Ошибка! = Ошибка! = Ошибка!  M' = Ошибка!
Putting value of M' in eq. (1). we get

,Iin = Ошибка! (– Ошибка!)
⸪
Ошибка!
⸪
Ошибка!
⸪
Ошибка!
Important conclusions:
1.
⸪
Iout = Ошибка!
2.
⸪
Isur = Ошибка!
3.
⸪
Iin = Ошибка!
4.
So,
Ошибка!
5.
Icentre = Ошибка! = 0
⸪

Imin = Icentre = 0
Graph between 'I' and 'r' for a solid sphere :
Imax = Isur = Ошибка!
GRAVITATIONAL FIELD INTENSITY DUE TO HOLLOW SPHERE OR SPHERICAL SHELL
Case I :
If r > R1 means outside the shell then Ошибка!
Case II :
If r = R, means of the surface then Ошибка!
Case III :
If r < R, means inside the sheet then Iin = Ошибка!
But inside mass (M') = 0
Ошибка!
→
ACCELERATION DUE TO GRAVITY ( ,g):
4.

Earth attracts other bodies by the gravitational force of attraction which is known as gravity force and the
acceleration produced due to this force in a freely falling body is known as acceleration due to gravity.
We know
Fg = m × a g
 Ошибка! = m × ag
 ag = Ошибка!
For earth's surface (h = 0):
Ошибка!
where,
……………………………………………………
In terms of density of earth, ():
g = Ошибка! = Ошибка!
Note:
⸪
I = Ошибка! and
⸪
I = a = Ошибка!
Ошибка!
a = Ошибка!
It means that gravitational field intensity is equal to acceleration due to gravity.
gmax = gsur = Ошибка!
'g' v/s 'r' graph
………………….fig……………………………
Acceleration due to gravity is zero (minimum value) at the centre of earth and it is 9.8 m/s 2 (maximum value) at the
surface of earth.
For problem based upto 5% variation: Of mass(M) and radius(R) of a planet, if small change occurs in (M) and
(R) then,
By
g = Ошибка!
Ошибка!

If M is constant, then
⸪ g  Ошибка! so Ошибка!

If R is constant then
⸪ g  M so Ошибка!

We know that g = Ошибка! RPG

If density () is constant, then g × R so, Ошибка! = Ошибка!

If R is constant, then g   so, Ошибка! = Ошибка!
Q.1
If radius of earth shrink by 2%, (mass of earth constant), then what will be change in acceleration – due to
gravity?
Q.2
Ans.
+4% (increased by 4%)
If the mass of the earth is increased by 2% and radius is decreased by 1% then find % change in g.
Ans.
+4% (increased by 4%)
FACTORS AFFECTING ACCELERATION DUE TO GRAVITY (g):
1.
Effects of Altitude (height):
Ошибка!

Decrement in 'g' with 'h' height: [(∆g)h]
(∆g)h = g – gh

(∆g)h = g – gОшибка!

(∆g)h = g – g + Ошибка!

Ошибка! (valid upto 5%)
2
Effect of Depth(d):
Acceleration due to gravity in form of density (at earth surface)
So, g = Ошибка! RG
gd = Ошибка! (R – d)G
…1
…2
Eq. (2)  (1)
Ошибка! = Ошибка!
Ошибка!
(Valid for all depths or 100% depth)
Decrement in 'g' with 'd' depth (∆g)d:
We know, (∆g)d = g – gd
 Ошибка!
 (∆g)d = g – g + Ошибка!
 Ошибка!
4.
(Valid for all depths)
Rotation of earth:
N = mg' = mg – m2r cos
mg' = – m2Rcos2

Important points:
1.
At the equator ( = 0°):
geq = g – 2Rcos2(0°)
2.
At the pole ( = 90°):
geq = g – 2 Rcos2(90°) = g – 2R(0)
3.
gmin or geq = g – 2R
gmax or gpole = g
It means that acceleration due to gravity at the poles does not depend upon angular velocity or rotation of
earth.
Difference in 'g' at equator and poles due to rotation of earth:
eg.
(∆g) = g – (g – 2R)

(∆g) = g – g + 2R

(∆g) = 2R
(∆g)eq = 2R = 0.03m/s2
4.
5.
6.
7.
Total difference in 'g' at poles and equator due to shape as well as rotation of earth:
…………………………………………………………….
If the earth suddenly stops rotating about its axis, then apparent weight of bodies or effective acceleration
due to gravity will increase at all the places except poles.
If the angular velocity of rotation of earth is gradually increased, then apparent weight of bodies will decrease
(except at the poles). At a particular '', the apparent weight will become zero. This condition is known as
condition of weightlessness.
'' corresponding to condition of weightlessness:
mg' = mg – m2Rcos2
But for weightlessness, Wapp = 0
 (')2 = Ошибка!
 0 = mg – m(')2R cos2
 Ошибка!
Special case:
At equator ( = 0°)
So, ' = Ошибка!Ошибка! = Ошибка! rad/s  = 0.00125 rad/s = 1.25 × 10–3 rad/s
'  17 × actual
Time period (T):
T = Ошибка! = Ошибка! = 2Ошибка!
So,
T = Ошибка! = Ошибка! = 1600 s = 5063 s  84.6 min  1.4 hrs
5.
GRAVITATIONAL POTENTIAL ENERGY (U):

The gravitational potential energy of mass particle situated at a point in some gravitational field is defined as
the amount of work done required to bring that particle from infinity to that point.
⸫
Gravitation potential energy = (Gravitational Potential) × (Mass of particle)

U = m × V  U = Ошибка!m

U = –Ошибка! or Ошибка!

1.
2.
3.
4.
Important points:
It is a scalar quantity.
It's SI unit is Joule and Dimension are [M1L2T–2]
It is a –ve quantity whose maximum value is zero at infinite separation.
Relation between force & Gr. prt. energy is
Ошибка!
6. The gravitational potential energy of a mass particle of mass ''m' placed at the surface of earth of mass 'M'
and radius 'R' is given by:
Ошибка!
7.To find the work done to raise a particle of mass 'm' to 'h' height above the surface of earth.
W. = ∆U = Uf – Ui
 W. = –Ошибка! – Ошибка!
 W. = GMmОшибка!
 W. = GMm Ошибка!
 W. = gR2m Ошибка!
[⸪ GM = gR2]
Ошибка!
Special cases:
i.
If h <<< R, then Ошибка! = 0
⸫ W = Ошибка! = mgh
ii.
8.
if h = R, then W = Ошибка! = Ошибка!
To find the velocity required at the surface of earth to project a particle to 'h' height from the surface of earth.
Applying 'COME' on the surface and at 'h' height.
(K.E. + U)final

Ошибка!mv2 – Ошибка! = 0 + Ошибка!

Ошибка!mv2 – Ошибка! = Ошибка!

Ошибка!mv2 = Ошибка!
[From 7th point]

v2 = Ошибка!

Ошибка!
Note: If a body is released from 'h' height above the surface of earth, then its velocity on reaching the earth's surface
is also given by:
Ошибка!
9.
surface of earth.
To find the maximum height attained by a body when it is projected with velocity 'v' from the
Using 8th point,
…………………….fig……………………………….
v2 = Ошибка!
 v2 + Ошибка! = 2gh
 v2 = 2gh – Ошибка!
 v2 = hОшибка!
 h = Ошибка! = Ошибка!
Ошибка!
Note: For questions based on working capacity, throwing capacity or jumping capacity, use:
Ошибка!
Q.
A man can jump upto a height of 3m on the earth surface. How much height can be jump at a
surface of a planet having g = 1.96 ms–2
Ans. (h' = 15 m)
6.
GRAVITATIONAL POTENTIAL (V):

Gravitational potential at a point is defined as the amount of work done required to bring a unit mass particle
from infinity to that point in gravitational field.
Ошибка!
* Gravitational potential is a scalar quantity.
* It's SI unit is 'J/kg'
* Dimension of Gravitational potential.
= Ошибка! = Ошибка! = [L2T–2] = [M0L2T–2]
GRAVITATIONAL POTENTIAL DUE TO A MASS PARTICLE (point mass):
Ошибка!
*
Here, negative sign means that this potential is due to an attractive force.
Note: The gravitational potential is maximum at infinite and its maximum value is 'zero'.
POTENTIAL GRADIENT:
The rate of change of potential with respect to distance is known as potential gradient, means
Potential gradient = Ошибка!
Note(1):
F = –Ошибка! and I = –Ошибка! are valid for all conservative force.
Note(2):
For hollow sphere,
⸪ Iin = 0
⸫ –Ошибка!(Vin) = 0 Ошибка!
⸫ Vin should be 'CONSTANT'
* The gravitational potential inside a hollow sphere remains constant and its value is same as that of surface.
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