Unit 1: Newtonian Mechanics

1. Elements of Vectors
1.1. Definitions
The physical quantities are mainly classified into two types. They are scalars and vectors.
Scalar: A scalar is a physical quantity which has only magnitude but not direction. Scalars
are specified by (i) Unit, (ii) Number.
Ex: Mass, length, distance, time, etc.
Vector: A vector is a physical quantity which has both magnitude and direction. Vectors are
specified by (i) Unit, (ii) Number, (iii) direction.
Ex: Displacement, Velocity, momentum, etc.
The magnitude of the vector is given by length OA.
Vectors are denoted by ⃗
A.
Resolution of Vector
Consider the case of vector ⃗
A with its tail at the original rectangular
coordinates system as shown in figure. We can resolve this vector
⃗
A into two components. Procedure is as follows.
We draw two perpendiculars from the head of a to X and Y axis
respectively. The intercepts on these axes are called as the scalar
components Ax and Ay. The process is called as resolution of
vectors.
If resolved vectors of ⃗
A are Ax and Ay, then ⃗
A = Ax + Ay
Let i and j be the unit vectors along the axis of X and Y respectively and if Ax and Ay be the
scalar magnitudes of Ax and Ay. Then, Ax = i Ax and Ay = j Ay
Therefore, ⃗
A = i A x + j Ay
If α be the angle made by ⃗
A with positive X axis, then Ax = A cosα and Ay = A sinα
A = √ A x + A y = √ A 2 cos2 α + A 2 sin 2 α
⃗ 2 2

Ax Ax
and tanα = or α = tan–1( )
Ay Ay
In three dimensions, ⃗
A = i Ax + j Ay + k A z
Addition of Vectors:
Let us add two vectors ⃗
A and ⃗
B by component method. The vectors in terms of components
in three dimensions can be expressed as
⃗
A = i Ax + j Ay + k A z
⃗
B = i B x + j B y + k Bz
 ⸫⃗
A+⃗
B = i (Ax + Bx) + j (Ay + By) + k (Az + Bz)
This method can be extended to any number of vectors. The magnitude is given by
| A+ B|=√ ( A x + B x ) + ( A y + B y ) + ( A z + Bz )
2 2 2

Similarly, for subtraction of two vectors, we have

⃗
A–⃗
B = i (Ax – Bx) + j (Ay – By) + k (Az – Bz)
| A−B|= √( A x −B x ) + ( A y −B y ) + ( A z−B z )
2 2 2

Parallelogram of Vectors:
According to this law, if two vectors acting simultaneously at a
point are represented both in magnitude and direction by the two
adjacent sides of a parallelogram drawn from one point, then
their resultant is represented both in direction and magnitude by
the diagonal of the parallelogram passing through the same
point.
P and ⃗
Thus, let ⃗ Q be two vectors drawn from the same point O and let ⃗
P be represented by
OA and ⃗
Q be represented by OB. Then their resultant ⃗
R is represented both in magnitude and
direction by the diagonal OC of the parallelogram. Let θ be the angle between the two vectors
P and ⃗
⃗ Q . Let α be the angle which the resultant makes with ⃗
P. From C, let CD be the
perpendicular on OA produced.
Magnitude of the resultant:
By the right-angled triangle OCD,
OC2 = OD2 + CD2
OC2 = (OA + AD)2 + CD2 = OA2 + 2 (OA X AD) + AD2 + CD2
But, AD2 + CD2 = AC2
OC2 = OA2 + 2 (OA X AD) + AC2
= OA2 + AC2 +2 (OA X AC cosθ)
Since OB = AC = ⃗
Q
R2 = P2 + Q2 + 2 PQ cosθ
⸫ |R2|=√ P 2+Q 2+2 PQ cosθ …….. (1)
Direction of Resultant:
CD CD Q sinθ
tanα = = =
OD OA + AD P+ Q cosθ
where CD = AC sinθ and AD = AC cosθ
Q sinθ
⸫ tanα = ……... (2)
P+ Q cosθ
Special Cases:
P and ⃗
Case (i): If ⃗ Q are acting in the same direction then the resultant is given by
|⃗R| = |⃗P|+|⃗
Q| [⸪ θ = 0]
P and ⃗
Case (ii): If ⃗ Q are acting in the opposite direction then the resultant is given by
|⃗R| = |⃗P|−|⃗
Q| [⸪ θ = 180]
P and ⃗
Case (iii): If ⃗ Q are acting perpendicular to each other than the resultant is given by
|⃗R| = √ P2+ Q2 [⸪ θ = 90]
P is equal to the magnitude of ⃗
Case (iv): If the magnitude of ⃗ Q then the resultant is given by
θ
|⃗R| = 2 P cos [⸪ ⃗
P=⃗
Q]
2
Scalar product of two vectors:
The scalar product or dot product of two vectors A and B is defined as the product of the
magnitudes of two vectors and the cosine of the angle between them.
⃗
A .⃗
B = | A||B| cosθ, where θ is angle between the two vectors.
A .B
⸫ cosθ =
| A||B|
This product is a scalar quantity. The product may be positive or negative depending upon the
π 3π
angle θ. The product is negative when θ is between to .
2 2
⃗
A .⃗
B = | A||B| cosθ = | A|¿ cosθ) = (| A| cosθ)|B|
Hence, the scalar product of two vectors will be equal to the product of length
of one vector and the length component of the second vector along the direction of
first vector.
Properties of scalar or dot product:
(i) The scalar product is commutative, i.e.,
⃗
A .⃗
B= ⃗
B.⃗
A
(ii) The scalar product follows the distribution law, i.e.,
⃗
A .( ⃗ C) = ⃗
B+ ⃗ A .⃗
B+ ⃗
A .⃗
C
(iii) The scalar product of unit orthogonal vectors, i, j, and k have the following
relations.
i.i = j.j = k.k = 1
i.j = j.k = k.i = 0
Examples:
(i) The work done W by a force ⃗ F . ⃗S
F acting on a body is given by W = ⃗
Where ⃗S is the displacement produced by the force.
 (ii) The Potential difference V between two points is given by V = ∫ E . dl
Where ⃗
E is electric field strength between the two points.
(iii) For a conservative field ⃗
F

∫ ⃗F . ⃗
dl = 0

(iv) Divergence of field is result of dot product of two vectors ∇ and field ⃗
F
Divergence = ∇ .⃗
F
Vector product or cross product:
The vector product or cross product of two vectors is defined as a vector having magnitude
equal to the product of the magnitudes of two vectors and sine of the angle between them.
The direction being perpendicular to the plane containing the two vectors.
If ⃗ B are the two vectors then their vector product is ⃗
A and ⃗ AX⃗
B, which can be expressed as
⃗
AX⃗
B = | A||B| sinθ n^
⃗
A X⃗ B
⸫ sinθ n^ =
| A||B|
Properties of vector product:
(i) The vector product is not commutative, i.e.,
⃗
AX⃗
B≠⃗
B X⃗
A
(ii) The vector product is distributive. i.e.,
⃗
A X (⃗ C) = ⃗
B+⃗ B+⃗
A X⃗ A X⃗
C
(iii) The vector product of unit orthogonal vectors i , j , k having the following relations
ixi=jxj=kxk=0
ixj=–jxi=k
jxk=–kxj=i
kxi=–ixk=j
Examples of vector product:
(i) Let a body is moving along a circular path with angular velocity ω. If V be the linear
vector of the body, then the relation between angular velocity (ω) and linear vector V
is given by
V =r xω
Where r is radius of the circular path.
(ii) In a rotatory motion, the torque acting on a body is given by
τ =r xF
(iii) The angular momentum L and linear momentum p are related as
L=r x p
 Where r is the redial vector.
2. Newton’s laws of motion
2.1. Concept of force and types
Force: Force is that which changes or tends to change the state or rest or of uniform motion
of a body. Units: Newtons
Types of forces: There are two types of basic forces (i) contact force and (ii) non-contact
force
Contact force: Contact forces act on objects when they are in physical contact with each
other.
Ex: Tension
Non-contact force: Non-contact force act on objects without direct physical contact.
Ex: Gravitational force, electromagnetic force
2.2. Newton’s laws:
The relation between force and motion was described by Sir Isaac Newton in 1686. These are
the basic laws of mechanics known as Newton’s laws of motion.
Newton’s first law:
Everybody continues to be in its state of rest or of uniform motion in a straight line unless it
is compiled by an external force to change that state. It gives the concept of force and inertia.
It is also known as the law of inertia.
Examples:
(i) The passenger in a stationary train falls backward when the train starts suddenly.
(ii) A body released from an artificial satellite continues to move in the same orbit at
the same speed.
(iii) The sparks emitted out from the grinding stone are tangential to the rotating stone.
Newton’s second law:
The acceleration of an object is directly proportional to the net force acting on it and is
inversely proportional to the mass of the object. The direction of the acceleration is in the
direction of the net force acting on the object.

a=
∑ F or ∑ F=ma
m
The rate of change of momentum of a body is directly proportional to the force applied on it
and this change in momentum takes place in the direction of the applied force. Newtons
second law of motion introduces the concept of momentum.
dp d (mv) dv
F ext = = =m =ma
dt dt dt
 ⸫ F ext =m a
If F ext =0, a=0

Newton’s third law:

For every action there is always an equal and opposite reaction. Let F 12 be a force acting on a
body 1 due to body 2 and F 21be the force on body 2 due to body 1, then
F 12=−F 21
Example:
(i) If you push against the edge of a desk, the desk pushes back against your hand.
(ii) Rocket motion
3. Law of universal gravitation:
Statement: Every particle attracts every other particle in the universe with a force that
is proportional to the product of their masses and inversely proportional to the square of the
distance between them. This force acting along the line joining the two particles.
The magnitude of the force can be expressed as
Fα m❑1 m2
1
Fα 2
r
m1 m2
Fα 2
r
m 1 m2
F=G 2
r
Where m1 & m2 are two massive particles and G is the universal constant.
Relation between g and G:
The acceleration due to gravity g on a planet or earth of mass M and radius R is related with
G as follows.
Fα mM
1
Fα 2
R
mM
Fα 2
R
mM
F=G 2 ---- (1)
R
The force of gravity due to the earth is the weight of the object on earth, which is given as
F=mg ---- (2)
By equating eqs (1) and (2), we have
mM
mg=G 2
R
GM
g= 2 ---- (3)
R
2
gR
G= ---- (4)
M
Note: g changes from planet to planet, it depends on M and R of the planet.
Differences between g and G:
Acceleration due to Universal gravitational
gravity (g) constant (G)
Definition An acceleration produced Force of gravitation
on a freely falling body due between two bodies of
to the gravitational force of unit mass kept at unit
earth distance.
Equation GM F−R
2
g= 2 G=
R mM
Stands for ‘g’ represents the ‘G’ represents the
acceleration due to gravity gravitational constant
value g = 9.81 m/s2 G = 6.67 x 10–11 N-m2/kg2
Constant or variable? ‘g’ changes planet to planet G is universal constant
Depends on ‘g’ depends on mass (M) ‘G’ does not depend on
and any factor or physical
Radius (R) of the planet quantity

4. Work-Energy Principle
Work: Whenever a force acting on a body displaces the body in the direction of the force.
Work is said to be done by the product of the force and displacement of the body in the
direction of force.

Case (i): IF the force and the displacement of the body are in the same direction, the work
done is given by
W =⃗
F . ⃗S ----- (1)
Case (ii): If the force and displacement are in different directions and ‘θ’ be the angle
between them, the work done is given by
W =|F||S| cosθ ----- (2)
Where W is a scalar quantity
Units: Joule
Note: Work done is independent of the path and time taken to do the work.
Energy: Energy the capacity of doing work. It is a scalar quantity.
Units: Joule
Potential Energy (P.E):
The energy possessed by a body by virtue of its position is called potential energy.
Example:
(i) The work done in lifting a body through a height h from the ground is stored in the body
in the form of potential energy. It is also known as gravitational potential energy and is
given by
mgRh
P . E=
(R+ h)
Where ‘R’ is the radius of the earth.
If h << R, then P . E=mgh
(ii) The work done in stretching a spring is stored in it in the form of
elastic potential energy.
1 2
i.e., P . E= K x
2
Kinetic Energy (K.E):
The energy possessed by a body by virtue of its motion is called kinetic energy. It is given by
1 2
K . E= m v
2
Expression for K.E:
Let a constant force ‘F’ be applied on a body at rest till it starts moving with velocity ‘v’ in
the direction of force. Suppose ‘S’ be the displacement of the body during this time. The
work done by the force is given by
W =⃗
F . ⃗S =|F||S|cosθ=F S ----- (1) [⸪θ = 0]
If ‘a’ be the acceleration produced in the body, then according to newtons second law of
motion, we have
F=ma ----- (2)
¿ eq . (1) & (2), we get
W =( ma ) S ------ (3)
To obtain the value of a, we use the formula,
2 2
V =u +2 as
2
V =2 as [⸪ u = 0]
 2
V
a= ----- (4)
2s
By substituting eq. (4) in (3), we get

( )
2
V
W =m S
2S
1 2
W= mv
2
Hence, the work done is equal to the K.E. of the body because work done on the body sets it
in motion.
1 2
Therefore, K . E= m v
2
Work – Energy Theorem:
According to the work energy theorem, the work
done by a force on a particle is always equal to the
change in the kinetic energy of the particle.
Consider that a body of mass ‘m’ is acted upon by a resultant acceleration force F
along the x-axis. Suppose the body moves from a position x1 to position x2 along x-axis. Let
the velocity of the body increases from v1 to v2. The work done by the force in the
displacement is

x2

W =∫ Fdx ----- (2)

x1

But according to Newton's second law

dv dv dx dv
F=ma=m =m =mv ----- (3)
dt dx dt dx
By substitute eq. (3) in (2), we get
x2
dv
W =∫ mv dx
x1
dx
v2

W =∫ mv dv
v1

[ ]
v2
v2
W =m
2 v1

1 2 1 2
W = m v 2− m v 1
2 2
 W =K 2−K 1
1 2
Where the quantity m v is defined as the kinetic energy of the body. K2 and K1 are the final
2
and initial kinetic energies of the body. If ∆K be the change in kinetic energy,
∆ K =K 2 – K 1
⸫ W =∆ K ----- (4)
Hence whenever a body is acted upon by a number of forces such that resultant force is not
zero then the work done by the resultant force is always equal to the change in the kinetic
energy of the body. This is known as work energy theorem.

5. Conservation Laws:
5.1. Conservation of energy – Freely falling body:
The sum of potential energy and kinetic energy is called as total energy according to law of
conservation of energy,
“The energy can neither be created nor destroyed”, But it can be
transformed from one form to another form.
Let us consider the case of a body of mass ‘m’ at a height ‘h’ above the ground.
velocity at height ‘h’ is u = 0
velocity on reaching the ground = v = v

At position A:
kinetic energy (K.E) of the body
1 2
= mu = 0 [⸪ u = 0]
2
potential energy (P.E) = mgh.
Total Energy (T.E) = K.E + P.E = 0 + mgh = mgh
At position C (Ground):
kinetic energy (K.E) of the body
1 2 1
= m v = m(u¿¿ 2+2 as)¿
2 2
1
¿ m ( 0+2 gh ) =mgh
2
potential energy (P.E) =mgh=mg ( 0 )=0.
Total Energy (T.E) = K.E + P.E = mgh + 0 = mgh
At position B:
kinetic energy (K.E) of the body
 1 2 1
= m v ¿ m(u¿¿ 2+2 as)¿
2 2
1
¿ m ( 0+2 gx )=mgx
2
potential energy (P.E) = mg(h−x)
Total Energy (T.E) = K.E + P.E
= mgx + mg(h−x) = mgh

The principle of conservation of energy may also be stated as “the total energy in any
system is always remains constant.” The variation in energy is shown in figure.

It is obvious from figure that the total energy remains constant (mg h) throughout.

Linear Momentum (P):

The linear momentum of a particle is defined as the product of its mass “m” and its velocity
v.
P = mv ----- (1)
Conservation of linear momentum:
The principle of conservation of linear momentum states that when the vector sum of the
external forces acting on a system of particles is zero (no resultant force on the system), the
total linear momentum of the system remains constant.
We know that linear momentum P=mV ----- (1)

Where “m” = mass of the particle,

“V” = Velocity of the particle.
Differentiating above equation, we get
dp dV
=m =m a ----- (2)
dt dt
where ‘a’ is the acceleration of the particle.
But ma = Fext ----- (3)
 dp
From eq. (2), m a =
dt
dp
=F ext ----- (4)
dt
Eq. (4) shows that the role of change of momentum of a particle is
equal to the vector sum of all the external forces acting on the
particle provided no mass leaves or enters the system.
When no external force acts on the system, F ext =0
dp
=0 or P = constant
dt
This process the law of conservation of linear momentum.

Angular Momentum
Consider a particle of mass ‘m’ having position vector r⃗ with respect to origin ‘O’ of an
inertial frame as shown in Fig.
Let p = mv be the linear momentum of the particle
The angular momentum (l) of the particle about ‘O’ is defined as the product of r⃗ and ⃗p.
i.e., ⃗ r × ⃗p
l=⃗ ----- (1)

the direction of l⃗ is perpendicular to the plane containing r⃗ and ⃗p.

Conservation of angular momentum:

The law of conservation of angular momentum is states that if no external torque acts on a
body rotating about a fixed point, the angular momentum of the body remains constant.

When τ ext is the sum of external torque acting on the system. Above equation indicates that
the rate of change of total angular momentum of the system of particles about origin ‘O’ of an
inertial reference frame is equal to the external torque on the system.
dL
If τ ext =0, then =0 or L = constant
dt
If no external torque is acting on the system, the angular momentum remains constant. This is
the law of conservation of angular momentum.
Collision:
Collision is the interaction between two or more bodies in which sudden changes of
momenta takes place.
Types of Collisions:
(i) Elastic collision: The collision in which both kinetic energy and momentum are
conserved is called an elastic collision.
1 2 1 2 1 2 1 2
i.e., conservation of kinetic energy, m 1 u1 + m 2 u 2 = m 1 v 1 + m 2 v 2
2 2 2 2
conservation of linear momentum, m1 u 1+ m2 u2=m1 v 1 +m2 v 2
Ex: collision between ivory balls, atomic particles
(ii) Inelastic collision: The collision in which only the momentum is conserved is
called as inelastic collision.
i.e., conservation of linear momentum, m1 u 1+ m2 u2=m1 v 1 +m2 v 2
Ex: collision between a bullet and a wooden block.

Elastic collision In-elastic collision

Definition A collision where A collision where

momentum and kinetic momentum is conserved, but
energy are conserved kinetic energy is not
conserved

Kinetic Energy It remains the same before It does not remain the same
and after the collision before and after the
collision. Some energy is
lost due to sound, heat, and
friction

Physical change Highly unlikely Most collisions result in

deformation or dent

Examples Newton’s cradle and Ballistic pendulum and

collision between atoms collision between cars
Applications of Newton’s laws – rocket motion
Newton’s first law:
• Before the rocket engines are ignited, the rocket and all its contents are
at rest relative to each other.
• Once the rocket engines are ignited, they expel mass (propellant) at high
speeds in one direction.
• According to Newton's first law, the rocket would remain at rest or continue drifting
in space if there were no external forces acting on it.

Newton’s Second law:

• The expulsion of propellant out of the rocket engines generates a force called thrust
(F).
• The mass of the rocket (m) includes both the rocket
itself and the remaining mass of propellant.
• By expelling propellant at high speeds, the rocket
generates a thrust force that accelerates it in the
opposite direction (i.e., forward).
Newton’s Third law:
• When the rocket engines expel propellant (action), an equal and opposite force
is applied to the rocket in the opposite direction (reaction).
• This reaction force, which is the thrust, propels the rocket forward according
to Newton's third law.

Applications of Newton’s laws – Car Air Bag

Newton’s first law:

• When a car is at rest, passengers inside it are also at rest relative to the car.
 • During a collision, the car rapidly decelerates, but passengers tend to continue moving
at their original speed due to inertia.
• The airbag system uses sensors to detect the sudden deceleration, which triggers the
deployment of the airbag.
Newton’s Second law:
• When a collision is detected, the airbag system rapidly inflates the airbag with a gas,
typically within milliseconds.
• The rapid inflation of the airbag generates a force to slow down and cushion the
impact of the passengers.
Newton’s Third law:
• The rapid expansion of the airbag material creates a force directed towards the
passenger.
• This force is designed to counteract the force of the passenger's forward motion
during the collision.
• The result is a controlled deceleration of the passenger, reducing the risk of injury.