Reading list

1. Murray R Spiegel. Theoretical mechanics. Kin Keong Printing CO. PTE.LTD.

2. David William. Element of mechanics
3. W.D Mc Comb. Dynamics and Relativity
4. Sear/Zemansky/young. College Physics.
5. Paul A Tipler (2008). Physics for Scientists and Engineers (6th). W.H. freeman and
Company, United Kingdom.
6. Bv Narayana Rao. First year BSc Physics (2nd edition)

1.0 Invariance and Conservation laws:

Newton`s Laws of motion:
Representation of common terms to be used:
i) Position of a particle as a function of time: � (t)
�
ii) Velocity of a particle as a function of time: � (t) = �� � (t). We refer to the

magnitude of the velocity, v = |�|, as the speed.

� �2
iii) Acceleration of a particle as a function of time: � (t) =�� � (t) =��2 � (t).

iv) Momentum of a particle: � (t) = m � (t)

Aristotle held that objects move because they are somehow impelled to seek out their natural
state. Thus, a rock falls because rocks belong on the earth, and flames rise because fire belongs
in the heavens. To paraphrase Wolfgang Pauli, such notions are so vague as to be “not even
wrong.” It was only with the publication of Newton’s Principle in 1687 that a theory of motion
which had detailed predictive power was developed.
Newton’s three Laws of Motion are stated as follows:
1. An isolated particle, independent of any external interactions remains indefinitely in
its state of rest or uniform motion in a straight line.
2. The rate of change of momentum is proportional to applied force and is parallel to the
direction in which the applied force acts.
d (m�)/dt = d�/dt = �
1|Page
 3. Any two bodies exert equal and opposite forces on each other and lie along the line
joining them.

Newton’s First Law states that a particle will move in a straight line at constant (possibly zero)
velocity if it is subjected to no forces. Now this cannot be true in general, for suppose we
encounter such a “free” particle and that indeed it is in uniform motion, so that r(t) = r0 + v0t.
Now r(t) is measured in some coordinate system, and if instead we choose to measure r(t) in a
different coordinate system whose origin R moves according to the function R(t), then in this
new “frame of reference” the position of our particle will be
r ′ (t) = r(t) − R(t)
= r0 + v0 t − R(t)

If the acceleration d2R/dt2 is non-zero, then merely by shifting our frame of reference we have
apparently falsified Newton’s First Law – a free particle does not move in uniform rectilinear
motion when viewed from an accelerating frame of reference. Thus, together with Newton’s
Laws comes an assumption about the existence of frames of reference – called inertial frames –
in which Newton’s Laws hold. A transformation from one frame K to another frame K′ which
moves at constant velocity V relative to K is called a Galilean transformation. The equations of
motion of classical mechanics are invariant (do not change) under Galilean transformations.

Newton’s “quantity of motion” is the momentum p, defined as the product p = mv of a particle’s

mass m (how much stuff there is) and its velocity (how fast it is moving). In order to convert the
Second Law into a meaningful equation, we must know how the force F depends on the
coordinates (or possibly velocities) themselves. This is known as a force law. Examples of force
laws include:

Constant force: F = −mg

Hooke’s Law: F = −kx
�
Gravitation: F = −GMm �2
�
Lorentz force: F = q E + q � × B

Fluid friction : F = −b �
2|Page
Note that for an object whose mass does not change we can write the Second Law in the familiar
��
form F = ma, where a = ��
= d2r/dt2 is the acceleration. Most of our initial efforts will lie in using

Newton’s Second Law to solve for the motion of a variety of systems.

The Third Law is valid for the extremely important case of central forces which we will discuss
in great detail later on. Newtonian gravity – the force which makes the planets orbit the sun – is a
central force. One consequence of the Third Law is that in free space, two isolated particles will
accelerate in such a way that F1 = −F2 and hence the accelerations are parallel to each other, with
�1 �
�2
=− �2
1

where the minus sign is used here to emphasize that the accelerations are in opposite directions.
We can also conclude that the total momentum P = p1 + p2 is a constant, a result known as the
conservation of momentum.

Inertial reference frames, Galilean invariance, and Galilean

transformation.
The cornerstone of the theory of special relativity is the Principle of Relativity: The Laws
of Physics are the same in all inertial frames of reference.
Briefly reviewing Newton’s mechanics in terms of frames of reference.

3|Page
A “frame of reference” is just a set of coordinates: something you use to measure the things that
matter in Newtonian problems, that is to say, positions and velocities, so we also need a clock. A
point in space is specified by its three coordinates (x, y, z) and an “event” like, say, a little
explosion, by a place and time: (x, y, z, t). An inertial frame is defined as one in which Newton’s
law of inertia holds—that is, anybody which isn’t being acted on by an outside force stays at rest
if it is initially at rest, or continues to move at a constant velocity if that’s what it was doing to
begin with. An example of a non-inertial frame is a rotating frame, such as a carousel. Non-
inertial reference is defined as a frame of reference that is undergoing acceleration with respect
to an inertial frame.

The “laws of physics” we shall consider first are those of Newtonian mechanics, as expressed by
Newton’s Laws of Motion, with gravitational forces and also contact forces from objects pushing
against each other. For example, knowing the universal gravitational constant from experiment
(and the masses involved), it is possible from Newton’s Second Law, force = mass × acceleration,
to predict future planetary motions with great accuracy. Suppose we know from experiment that
these laws of mechanics are true in one frame of reference. How do they look in another frame,
moving with respect to the first frame? To find out, we have to figure out how to get from
position, velocity and acceleration in one frame to the corresponding quantities in the second
frame. Obviously, the two frames must have a constant relative velocity, otherwise the law of
inertia won’t hold in both of them. Let’s choose the coordinates so that this velocity is along the
x-axis of both of them.

Notice we also throw in a clock with each frame. Suppose S′ is proceeding relative to S at speed
v along the x-axis. For convenience, let us label the moment when O′ passes O as the zero point
of timekeeping. Now what are the coordinates of the event (x, y, z, t) in S′? It’s easy to see t′ = t;
4|Page
we synchronized the clocks when O′ passed O. Also, evidently, y′ = y and z′ = z, from the figure.
We can also see that x = x′ +vt. Thus (x, y, z, t) in S corresponds to (x′, y′, z′, t′ ) in S′, where

x’ =x - vt
y’ = y
z’ = z
t′=t
That’s how positions transform; these are known as the Galilean transformations. What about
velocities ? The velocity in S′ in the x′ direction
��' ��' � ��
�'� = ��'
= ��
= �� � − �� = ��
− � = �� − �

This is obvious anyway: it’s just the addition of velocities formula

�� = �'� + �

How does acceleration transform?

��'� ��'� � ���
��'
= ��
= �� �� − � = ��

since v is constant. That is to say,

�'� = ��

the acceleration is the same in both frames. This again is obvious—the acceleration is the rate of
change of velocity, and the velocities of the same particle measured in the two frames differ by a
constant factor-the relative velocity of the two frames. If we now look at the motion under
gravitational forces, for example,
��1 �2
�1 � = �
�2

we get the same law on going to another inertial frame because every term in the above equation
stays the same. Note that � � is the rate of change of momentum—this is the same in both
frames. So, in a collision, say, if total momentum is conserved in one frame (the sum of
individual rates of change of momentum is zero) the same is true in all inertial frames.
5|Page
Since the laws of mechanics are invariant under Galilean transformation, they are often referred
to as principle of Newtonian relativity or Galilean invariance

Exercise:
1. Describe the fundamental interactions of nature
2. Explain the assumptions of classical mechanics
3. Particle of mass, m starts from rest in gravitational force field given that the resistance
force of cmv2 acts on the particle. Show that the distance, S that the particle falls
accelerating from vo to v1 is given by
1 �−��2�
�= ��
2� �−��21

Conservation of linear momentum.

a) Conservation of Linear Momentum.
Momentum is the product of mass, m of a body and velocity, v at which the body is moving.
Momentum can be mathematically written as;
� = �� where � is the momentum

In proving the conservation of linear momentum, we shall consider two particles of masses, �1
and �2 with velocities �1 and �2 respectively. We shall also consider force, �1 is exerted by the
body of mass, �1 onto the body of mass, �2 ; force, �2 is exerted by the body of mass, �2 onto
the body of mass, �1 .

From Newton’s second law,

�� ��
� = �� = � ��
= ��
where � is the force acting on the body, m is the mass of the body,

� is the acceleration of the body, � is the velocity, t is the time, and � is the momentum.

From Newton’s third law, �1 ��� �2 are equal in magnitude and opposite in direction, therefore;
�1 =− �2
��1 ��2
�1 =− �2
�� ��
6|Page
��1 ��2
��
=− ��

��1 ��2
��
+ ��
=0
�
(�1 + �2 ) = 0
��

This therefore shows that the total momentum is constant; which is the principle of conservation
of linear momentum.

The general formula for more than two bodies can then be written as;
�
(�1 + �2 + … + �� ) = 0
��

where �� is the momentum due to mass, �� and velocity, �� .

b) Kinetic energy of a system.

Consider a system of particles. The ith particle has mass mi and velocity �� with respect to a
reference frame O. The kinetic energy of the system of particles is given by
1 1
�= � �2
�2 � �
=2 �
���� . ��

1
�=2 �
�� ���,� + ��� . ���,� + ���

Where the equation has been used to express �� in terms of ���,� ��� ��� . Expanding the last dot
product in the equation,
1
�= �
�� ���,� ∙ ���,� + ��� ∙ ��� + 2���,� ∙ ���
2

1 1
�=2 �
�� ���,� ∙ ���,� + 2 �
�� ��� ∙ ��� + �
�� ���,� ∙ ���

1 1
�= � �2
� 2 � ��,�
+2 �
�� �2�� + �
�� ���,� ���

The last term in the third equation vanishes and reduces to;
1 1
�= � �2
� 2 � ��,�
+ �
�� �2�� --- (*)
2

7|Page
We interpret the first term as the sum of the individual kinetic energies of the particles of the
system in the center of mass reference frame ��� and the second term as the kinetic energy of
the center of mass motion in reference frame O.

At this point, it’s important to note that no assumption was made regarding the mass elements
being constituents of a rigid body. The Equation (*) is valid for a rigid body, a gas, a firecracker
(but K is certainly not the same before and after detonation), and the sixteen pool balls after the
break, or any collection of objects for which the center of mass can be determined.

c) Collisions: Impulse and average force.

Impulse:
From Newton’s second law, an unbalanced force always accelerates or decelerates an object.
Impulse is a force applied to an object for a certain length of time causing the object’s
momentum to change.
Impulse, I = � ∙ ∆� where F is the force and � ∆t is the time taken.

Exercise:
1. Prove that impulse of the forces is equal to change in momentum of an object.

�� −�0 ��� −��0

� =� ∆�
= ∆�

� ∆� = ��� − ��0 = �� − �0 where ��� = �� ��� ��0 = �0 are final and initial
momentum respectively.

But I = � ∙ ∆� ; Therefore, � = �� − �0
2. A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time
of 0.10 seconds. Determine the impulse experienced by the hockey puck.
Impulse, I = F*t = 80 N * 0.1 s
Impulse = 8 N s

3. If a ball of mass, 10 kg is thrown against a wall for 0.1 s, while moving horizontally to
the left at 30 m s-1 and it bounces to the right at 20 m s-1, find the average force on the

8|Page
 ball during the impact.

F = ma
�−�
�=� �
20−(−30)
� = 10
0.1

� = 5000 �

d) Collisions in Centre of mass Reference Frame.

The laboratory frame: this is the frame where measurements are actually made.
The centre of mass frame: this is the frame where the centre of mass of the system is at rest and
where the total momentum of the system is zero.
Centre of mass (CM) is the point where the mass weighted position vectors relative to the point
sum to zero. Taking a system of n-particles, each with mass, �� located at positions �� , the
position vector of the CM is defined by:

�
�
�=1 �
�� − ��� = 0
1 �
��� = � ��
�=1 � �

�
Where � = �
�=1 �

Under the influence of the total internal force, the centre of mass move as a single particle of
mass, M;
� �
�= �
�=1 �
= �=1
��� where � = ��
Hence, the total momentum of a system in the Lab frame is equivalent to that of a single particle
having a mass, M and moving at a velocity, ��� = ��

9|Page
As for momentum in the centre of mass (CM) frame;
Assignment 1.1: Show that the total momentum of a system of particles in the centre of
frame is equal to zero.

Centre of mass of a continuous volume:

If the mass distribution is continuous with density, �(�) inside a volume V, then;

�
�
�=1 �
�� − ��� = 0 becomes

�
�(�)(� − ��� ) �� = 0

Where �� = �(�)dV
Solving for ���
1
��� = � �
�(�)� �� where M is the total mass in volume.

Assignment 1.2:
What is the height of the centre of mass of a 4,400 m height mountain that takes up the shape of
a cone; assuming that the density is uniform?

Exercise:
1. A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant, the 2 kg
body is 1 m from the origin and has a velocity of 3 m s-1; and the 3 kg body is 2 m from
the origin and has a velocity of -1 m s-1. Find the position and velocity of the center of
mass and also the total momentum.

Solution:
Let x be x-coordinate of the center of mass, then from;
�1 �1+�2 �2 +…
�= �1 +�2

10 | P a g e
 2∗1 +(3∗2)
�= = 1.6 �
2+3

Therefore, the x-component of velocity is;

�1 �1� +�2 �2� +…
�� = �1 +�2
2∗3 +(3∗−1)
�� = 2+3
= 0.6 � �−1

The x-component of momentum;

�� = 2 ∗ 3 + 3 ∗− 1 = 3 ��/ (� �−1 )

Total mass, M will be;

� = ��
3
�=
0.6

� = 5 ��

e) Continuously Varying Mass and Rocket Propulsion.

We shall be dealing with objects where the mass is changing. We analyze the motion of a
rocket, which changes its velocity (and hence its momentum) by ejecting burned fuel gases, thus
causing it to accelerate in the opposite direction of the velocity of the ejected fuel (see (Figure)).
 As the rocket engines operate, they are continuously ejecting burned fuel gases,
which have both mass and velocity, and therefore momentum. By conservation of
momentum, the rocket’s momentum changes by the same amount (with the
opposite sign). We will assume the burned fuel is being ejected at a constant rate,
which means the rate of change of the rocket’s momentum is also constant.
By (Figure), this represents a constant force on the rocket.
 However, as time goes on, the mass of the rocket (which includes the mass of the
remaining fuel) continuously decreases. Thus, even though the force on the rocket
is constant, the resulting acceleration is not; it is continuously increasing.
 So, the total change of the rocket’s velocity will depend on the amount of mass of
fuel that is burned, and that dependence is not linear.

Question: A rocket at time t = 0 starts moving with speed, v in the positive x-direction. The
rocket burns fuel with ∆� as the quantity of fuel burnt in an interval of time, ∆� and ejected
backward with velocity, u relative to the rocket. Assuming there is no gravitational field and no
air resistance, determine the rocket equation
11 | P a g e
 Solution:
 The speed of the rocket will increase by ∆�
 The velocity of the rocket at time (� + ∆�) will be (� + ∆�)
 The mass of the rocket at time (� + ∆�) will be � + ∆� < �
 The momentum of the rocket at time (� + ∆�) will be (� + ∆�)(� + ∆�)
 The velocity of the burnt fuel relative to inertial frame in which � is measured is
(� − �)
 The mass of burnt fuel ( − ∆�) is a negative quantity and its momentum gives
−∆�(� − �)

Using the principle of conservation of momentum

�� = � + ∆� � + ∆� − ∆�(� − �) --- (1)

For small change in � (∆�), ∆�∆� is too small leading to

�� =− ∆�(� + ∆�)

Dividing by ∆�
�∆� ∆� �+∆�
∆�
=− ∆�
--- (2)

As ∆� → 0, equation (2) becomes

��� ��
��
=− � ��
--- (3)
��
� = �� =− �
��
where a is the acceleration of the rocket, and F is the

resulting force/ thrust on the rocket

Thrust is proportional to both velocity, u of the ejected fuel and the mass of the fuel
ejected per unit time.

12 | P a g e
 � ��
� =− � ��

Let �0 and �� be the initial and final velocity respectively; �� and �� be the initial and final
masses of the rocket plus fuel respectively.
Integrating Equation (3),
�� �� �� ��
�0
� �� = ��
−� ��
�� �� �� �
�0 ��
�� =− � �� �
��
��
�� − �0 =− � �� � ��

�� ��
�� − �0 =− � �� ��
= � �� ��
--- (4)

Equation (4) is the rocket equation which gives the change of velocity that the rocket obtains
from burning a mass of fuel.

Assignment 1.3:
A rocket is moving vertically upwards relative to the surface of the earth. The motion takes place
close to the surface of the earth and it is assumed that g is the constant gravitational acceleration.
At time t, the mass of the rocket is M (1- kt), where M and k are positive constants, and the
rocket is moving upwards with speed v. The rocket expels fuel vertically downwards with speed
u relative to the rocket. Given further that when t = 0, v = 0, determine an expression for v in
time t, in terms of u, g and k.

Assignment 1.4:
A spacecraft is moving in deep space in a straight line with speed 2u. At time t = 0, the mass of
the spacecraft is M and at that instant the engines of the spacecraft are fired in a direction
opposite to that of the motion of the spacecraft. Fuel is ejected at a constant mass rate k with
speed u relative to the spacecraft. At time t, the mass of the spacecraft is m and its speed is v.
a) Use the impulse momentum principle to show that
�� ��
��
= �−��

b) Hence determine, in terms of u, the speed of the spacecraft when the mass of the
1
spacecraft is 3 of its initial mass.

13 | P a g e
 Conservative force fields:
A force field, F is said to be conservative if and only if there is a differentiable curl, V
such that � =− ∇� or ∇ ∧ F = 0

Example:
Determine if the following vector fields are conservative or not

Solution:
a)

Since the partial derivatives are not the same, the vector fields are not conservative.

b)

Since the partial derivatives are the same, the vector fields are conservative.

Exercise:
a) Show that the force field, F = −m�2 � acting on mass, m as shown in the figure, is conserved

b) Find the potential energy at points A and B

c) Find the work done by the force in moving particle from point A to B.
d) Find the total energy of the particle and show that it is conserved.

14 | P a g e
 1.8 Conservation of Angular momentum.
The law of conservation of angular momentum states that when no external torque acts on an
object, no change of angular momentum will occur.
The symbol for angular momentum is the letter L. Just as linear momentum is conserved when
there are no net external forces, angular momentum is constant or conserved when the net torque
is zero. We can see this by considering Newton’s 2nd law for rotational motion:

Consider a particle with position vector, � moving in a force field, F; and torque, � = � ∧ �. The
magnitude of � is the measure of a turning effect produced on the particle by the force.
�
� ∧ � = �� � � ∧ � where � � ∧ � = � is the angular momentum. Therefore,

the theorem states that the torque acting on the particle is equal to the rate of change of angular
�
momentum i.e. � = �� �

�=� �∧�
� = � ∧ ��
�=�∧� where � is the linear momentum

Newtons second law states that the rate of change of momentum is directly proportional to the
applied force.
If the net external force acting on the particle is zero (F = 0), the momentum of the particle is
�
conserved i.e. �� �� = 0; mv = constant.

If the net external torque acting on the body is zero � = 0 , the angular momentum remains
�
unchanged or conserved i.e. ��
� �∧� = 0; � � ∧ � = ��������.

Example:
Prove that the moment of a force about the origin of a coordinate system is equal to the rate of

15 | P a g e
change of the angular momentum.
From � = � ∧ �
�
� = � ∧ �� �� but � = � � ∧ �
�� �
��
= �� � � ∧ �
�� �� ���
��
= �� ∧ �� + � ∧ ��
��
��
= �� ∧ � + � ∧ �
��
��
=�∧�

Therefore,
��
�∧�= ��

1.9 Conservation of energy.

Conservation of energy theorem states that in the conservative force field, the total energy is
constant.

The work done �12 , for conserved energy can be written as;
2 �2 � ��
�12 = 1
� ∙ �� = �1
(� �) ∙ ��
�� ��
�2 �� 1 �2 �
�12 = � �1 ��
� �� = � �1 ��
( � ∙ �) ��
2
1
�12 = 2 �(�22 − �21 ) = ∆�

This therefore shows that the work done leads to change in kinetic energy. For a conservative
force, this can be related to change in potential energy
2 �2
�12 = 1
� ∙�= �1
( − �� ) ∙ �� = �1 − �2 =− ∆�

Hence,
∆� =− ∆�
∆(� + �) = 0
� = � + � is conserved
16 | P a g e