NATURE OF LIGHT

Newton’s corpuscular theory of light
Newton’s corpuscular theory of light is based on the following points 1. Light consists of very tiny particles known as “corpuscular”. 2. These corpuscles on emission from the source of light travel in straight line with high velocity 3. When these particles enter the eyes, they produce image of the object or sensation of vision. 4. Corpuscles of different colours have different sizes.
Huygen’s wave theory of light
In 1679, Christian Huygens proposed the wave theory of light. According to huygen’s wave theory: 1. Each point in a source of light sends out waves in all directions in hypothetical medium called "ETHER". 2. Light is a form of energy 3. Light travels in the form of waves. 4. A medium is necessary for the propagation of waves & the whole space is filled with an imaginary medium called Ether 5. Light waves have very short wave length
Quantum theory of light
Quantum theory was put forward by MAX-PLANCK in 1905. According to quantum theory “Energy radiated or absorbed can not have any fractional value. This energy must be an integral multiple of a fixed quantity of energy. This quantity is called “QUANTUM” OR Energy released or absorbed is always in the form of packets of energy or bundles of energy. These packets of energy are known as QUANTA or PHOTONS

SIMPLE PENDULUM

SIMPLE PENDULUM
	simple pendulum consists of a heavy mass particle suspended by a light, flexible and in-extensible string.
	MOTION OF THE BOB OF SIMPLE PENDULUM
	The motion of the bob of simple pendulum simple harmonic motion if it is given small displacement. In order to prove this fact consider a simple pendulum having a bob of mass 'm' and the length of pendulum is 'l'. Assuming that the mass of the string os pendulum is negligible. When the pendulum is at rest at position 'A', the only force acting is its weight and tension in the string. When it is displaced from its mean position to another new position say 'B' and released, it vibrates to and fro around its mean position.
	Suppose that at this instant the bob is at point 'B' as shown below :
	FORCES ACTING ON THE BOB
	1. Weight of the bob (W) acting vertically downward. 2. Tension in the string (T) acting along the string.
	The weight of the bob can be resolved into two rectangular components:
	a. Wcosq along the string. b. Wsinq perpendicular to string.
	Since there is no motion along the string, therefore, the component Wcosq must balance tension (T) i.e.                                                                      Wcosq = T
	This shows that only Wsinq is the net force which is responsible for the acceleration in the bob of pendulum.
	According to Newton's second law of motion Wsinq will be equal to m x a
	i.e.                                                                      Wsinq = m a
	Since Wsinq is towards the mean position, therefore, it must have a negative sign. i.e.                                                                     m a = -  Wsinq
	But W = mg
	m a = -  mgsinq
	a = -  gsinq
	In our assumption q is very small because displacement is small, in this condition we can take sinq = q
	Hence                                                                     a = -  gq ----------- (1)
	If x be the linear displacement of the bob from its mean position, then from figure, the length of arc AB is nearly equal to x
	From elementary geometry we know that:
	Where s= x, r = l
	Putting the value of q in equation (1)
	As the acceleration of the bob of simple pendulum is directly proportional to displacement and is directed towards the mean position, therefore the motion of the bob is simple harmonic when it is given a small displacement

STATES OF EQUILIBRIUM

States of equilibrium
There are three states of equilibrium:
Stable equilibrium    Unstable equilibrium    Neutral equilibrium
Stable equilibrium
When the center of gravity of a body lies below point of suspension or support, the body is said to be in    STABLE EQUILIBRIUM. For example a book lying on a table is in stable equilibrium.
Explanation
A book lying on a horizontal surface is an example of stable equilibrium. If the book is lifted from one edge    and then allowed to fall, it will come back to its original position.    Other examples of stable equilibrium are bodies lying on the floor such as chair, table etc.
Reason of stability
When the book is lifted its center of gravity is raised . The line of action of weight passes through the    base of the book. A torque due to weight of the book brings it back to the original position.
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Unstable equilibrium
When the center of gravity of a body lies above the point of suspension or support, the body is said to    be in unstable equilibrium
Example
pencil standing on its point or a stick in vertically standing position.
Explanation:    If thin rod standing vertically is slightly disturbed from its position it will not come back to its original    position. This type of equilibrium is called unstable equilibrium, other example of unstable equilibrium are    vertically standing cylinder and funnel etc.
Reason of instability
when the rod is slightly disturbed its center of gravity is lowered . The line of action of its weight lies    outside the base of rod. The torque due to weight of the rod toppled it down.
Neutral equilibrium
When the center of gravity of a body lies at the point of suspension or support, the body is said to be in    neutral equilibrium. Example: rolling ball.
Explanation
If a ball is pushed slightly to roll, it will neither come back to its original nor it will roll forward rather it will    remain at rest. This type of equilibrium is called NEUTRAL EQUILIBRIUM.
Reason of neutral equilibrium
If the ball is rolled, its center of gravity is neither raised nor lowered. This means that its center of gravity    is at the same height as before.

ADVANTAGES AND DISADVANTAGES OF FRICTION - METHODS OF REDUCING FRICTION

ADVANTAGES AND DISADVANTAGES OF FRICTION - METHODS OF REDUCING FRICTION

ADVANTAGES OF FRICTIONFriction plays a vital role in our daily life. Without friction we are handicap. 1. It is becomes difficult to walk on a slippery road due to low friction. When we move on ice, it     becomes difficult to walk due to low friction of ice. 2. We can not fix nail in the wood or wall if there is no friction. It is friction which holds the nail. 3. A horse can not pull a cart unless friction furnishes him a secure Foothold. DISADVANTAGES OF FRICTION Despite the fact that the friction is very important in our daily life, it also has some disadvantages like: 1. The main disadvantage of friction is that it produces heat in various parts of machines. In this way     some useful energy is wasted as heat energy. 2. Due to friction we have to exert more power in machines. 3. It opposes the motion. 4. Due to friction, noise is also produced in machines. 5. Due to friction, engines of automobiles consume more fuel which is a money loss. METHODS OF REDUCING FRICTIONThere are a number of methods to reduce friction in which some are discussed here. USE OF LUBRICANTS:The parts of machines which are moving over one another must be properly lubricated by using oils and lubricants of suitable viscosity. USE OF GREASE:Proper greasing between the sliding parts of machine reduces the friction. USE OF BALL BEARING:In machines where possible, sliding friction can be replaced by rolling friction by using ball bearings. DESIGN MODIFICATION:Friction can be reduced by changing the design of fast moving objects. The front of vehicles and airplanes made oblong to minimize friction.

Momentum- Law of conservation of Momentum

MOMENTUM
Quantity of motion of a body is referred to as "MOMENTUM".
Definition
Momentum of a moving body defined as :
"the product of mass and velocity of a body is called MOMENTUM."
Mathematically
Momentum = mass x velocity
It is a vector quantity. Momentum is always directed in the direction of velocity.    The unit of momentum is in S.I system kg .m/s or NS.    Momentum depends upon mass and velocity of body.
LAW OF CONSERVATION OF MOMENTUM.
The law of conservation of momentum states that:
"when some bodies constituting an isolated system act upon one another, the total momentum of the system remains constant." OR "the total momentum of an isolated system of interacting bodies remains constant." OR "Total momentum of an isolated system before collision is always equal to total momentum after collision."
Consider an isolated system of two bodies 'A' and 'B' as shown. The masses of bodies are ma and mb and
MATHEMATICAL REPRESENTATION
Consider two bodies of mass m1 and m2 moving initially with velocities u1 and u2.
Total momentum before collision = m1u1 + m2u2
Let after collision their velocities become v1 and v2.
Total momentum after collision = m1v1 + m2v2
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2

EQUATIONS OF MOTION EQUATIONS OF MOTION

FIRST EQUATION OF MOTION Vf = Vi + at
Consider a body initial moving with velocity "Vi". After certain interval of time "t", its velocity becomes "Vf". Now
Change in velocity = Vf - Vi  OR DV =Vf – Vi
	Due to change in velocity, an acceleration "a" is produced in the body. Acceleration is given by
	a = DV/t
	Putting the value of "DV"
	a = (Vf – Vi)/t at = Vf – Vi at + Vi =VfOR
	SECOND EQUATION OF MOTION OR S = Vit + 1/2at2
	Consider a car moving on a straight road with an initial velocity equal to ‘Vi’. After an interval of time ‘t’ its velocity becomes ‘Vf’. Now first we will determine the average velocity of body.
	Average velocity = (Initial velocity + final velocity)/2 OR Vav = (Vi + Vf)/2
	but Vf = Vi + at
	Putting the value of Vf
	Vav = (Vi + Vi + at)/2 Vav = (2Vi + at)/2    Vav = 2Vi/2 + at/2   Vav = Vi + at/2                            Vav = Vi + 1/2at.......................................(i)
	we know that
	S = Vav x t
	Putting the value of ‘Vav’
	S = [Vi + 1/2at] t
	THIRD EQUATION OF MOTION OR 2aS = Vf2 – Vi2
	Initial velocity, final velocity, acceleration, and distance are related in third equation of motion.
	Consider a body moving initially with velocity ‘Vi’. After certain interval of time its velocity becomes ‘Vf’. Due to change in velocity, acceleration ‘a’ is produced in the body. Let the body travels a distance of ‘s’ meters.  According to first equation of motion:
	Vf = Vi + at      OR Vf – Vi = atOR                          (Vf – Vi)/a = t....................(i)
	Average velocity of body is given by:
	Vav = (Initial velocity + Final velocity)/2                               Vav = (Vi + Vf)/2.................. (ii)
	we know that :
	S = Vav x t.................. (ii)
	Putting the value of Vav and t from equation (i) and (ii) in equation (iii)
	S = { (Vf + Vi)/2} { (Vf – Vi)/a} 2aS = (Vf + Vi)(Vf – Vi)
	According to [ (a+b)(a-b)=a2-b2]

Addition of vectors by Head to Tail method (Graphical Method)

Head to Tail method or graphical method is one of the easiest method used to find the resultant vector of two of more than two vectors.
	DETAILS OF METHOD
	Consider two vectors  and  acting in the directions as shown below:
	In order to get their resultant vector by head to tail method we must follow the following steps:
	STEP # 1
	Choose a suitable scale for the vectors so that they can be plotted on the paper.
	STEP # 2
	Draw representative line  of vector
	Draw representative line  of vector  such that the tail of  coincides with the head of vector .
	STEP # 3
	Join 'O' and 'B'.  represents resultant vector of given vectors  and  i.e.
	STEP # 4
	Measure the length of line segment  and multiply it with the scale choosen initially to get the magnitude of resultant vector.
	STEP # 5
	The direction of the resultant vector is directed from the tail of vector  to the head of vector .

ADDITION OF VECTORS

ADDITION OF VECTORS4

PARALLELOGRAM LAW OF VECTOR ADDITION
Acccording to the parallelogram law of vector addition:
"If two vector quantities are represented by two adjacent sides or a parallelogram then the diagonal of parallelogram will be equal to the resultant of these two vectors."
	EXPLANATION
	Consider two vectors . Let the vectors have the following orientation
	parallelogram of these vectors is :
	According to parallelogram law:
	MAGNITUDE OF RESULTANT VECTOR
	Magintude or resultant vector can be determined by using either sine law or cosine law.

LAW OF CONSERVATION OF ENERGY

LAW OF CONSERVATION OF ENERGY
According to the law of conservation of energy :                        "Energy can neither be created nor it is destroyed, however energy can be                                            converted from one form energy to any other form of energy"
SHOW THAT THE MOTION OF A SIMPLE PENDULUM IS ACCORDING TO THE LAW OF CONSERVATION ENERGY.                                                                       OR PROVE THE LAW OF CONSERVATION WITH THE HELP OF A SUITABLE EXAMPLE.
	We know that the motion of the bob of a simple pendulum is simple harmonic motion. Here we have to prove that the energy is conversed during the motion of pendulum.  Proof: Consider a simple pendulum as shown in the diagram.
	Energy Conservation At Point ‘A’
	At point ‘A’ velocity of the bob of simple pendulum is zero. Therefore, K.E. at point ‘A’ = 0. Since the bob is at a height (h), Therefore, P.E. of the bob will be maximum. i.e. P.E. = mgh. Energy total = K.E. + P.E Energy total = 0 + mgh Energy total = mgh This shows that at point A total energy is potential energy.
	Energy Conservation At Point ‘M’
	If we release the bob of pendulum from point ‘A’, velocity of bob gradually increases, but the height of bob will decreases from point to the point. At point ‘M’ velocity will become maximum and the height will be nearly equal to zero. Thus , K.E. = maximum = 1/2mV2 but P.E. = 0. Energy total = K.E. + P.E Energy total = 1/2mV2 + 0 Energy total = 1/2mV2 This shows that the P.E. at point is completely converted into K.E. at point ‘M’.
	Energy Conservation At Point ‘B’
	At point M the bob of Pendulum will not stop but due to inertia, the bob will moves toward the point ‘B’. As the bob moves from ‘M’ to ‘B’, its velocity gradually decreases but the height increases. At point ‘B’ velocity of the bob will become zero. Thus K.E. at point ‘B’ = 0 but P.E. = max. P.E. = mgh. Energy total = K.E. + P.E. Energy total = 0 + mgh Energy total = mgh This shows that at point B total energy is again potential energy.
	CONCLUSION
	Above analysis indicates that the total energy during the motion does not change. I.e. the motion of the bob of simple pendulum is according to the law of conservation of energy.

KINETIC ENERGY

KINETIC ENERGY

KINETIC ENERGY
"Energy posses by a body by virtue of its motion is referred to as ‘Kinetic Energy’".
FORMULA
K.E. = 1/2 mv2
	Kinetic energy depends upon the mass and velocity of body. If velocity is zero than K.E. of body will also be zero.
	Kinetic energy is a scalar quantity like other forms of energies.
	DERIVE: K.E = 1/2 mv2
	PROOF
	Consider a body of mass "m" starts moving from rest. After a time interval "t" its velocity becomes V. If initial velocity of the body is Vi = 0 ,final velocity Vf = V and the displacement of body is "d". Then
	First of all we will find the acceleration of body.
	Using equation of motion
	2aS = Vf2 – Vi2 Putting the above mentioned values 2ad = V2 – 0 a = V2/2d
	Now force is given by F = ma Putting the value of accleration F = m(V2/2d) As we know that Work done = Fd
	Putting the value of F Work done = (mv2/2d)(d) Work done = mV2/2 OR Work done = ½ mV2
	Since the work done is motion is called "Kinetic Energy" i.e. K.E. = Work done OR K.E. =1/2mV2.