Work Energy and Power Class 11 Physics Notes And Questions

Please refer to Work Energy and Power Class 11 Physics notes and questions with solutions below. These revision notes and important examination questions have been prepared based on the latest Physics books for Class 11. You can go through the questions and solutions below which will help you to get better marks in your examinations.

Class 11 Physics Work Energy and Power Notes and Questions

Work

When a force acts on an object and the object actually moves in the direction of force, then the work is said to be done by the force.

Work done by the force is equal to the product of the force and the displacement of the object in the direction of force.

If under a constant force F the object displaced through a distance s, then work done by the force
W = F * s = F s cos θ
where a is the smaller angle between F and s.
Work is a scalar quantity, Its S1 unit is joule and CGS unit is erg.
∴ 1 joule = 107 erg
Its dimensional formula is [ML²T^-2].
Work done by a force is zero, if
(a) body is not displaced actually, i.e., s = 0
(b) body is displaced perpendicular to the direction of force, i.e.,
θ = 90°
Work done by a force is positive if angle between F and s is acute angle.
Work done by a force is negative if angle between F and s is obtuse angle.
Work done by a constant force depends only on the initial and final Positions and not on the actual path followed between initial and final positions.

Work done in different conditions

(i) Work done by a variable force is given by
W = ∫ F * ds
It is equal to the area under the force-displacement graph along with proper sign.

Work done = Area ABCDA
(ii) Work done in displacing any body under the action of a number of forces is equal to the work done by the resultant force.
(iii) In equilibrium (static or dynamic), the resultant force is zero therefore resultant work done is zero.
(iv) If work done by a force during a rough trip of a system is zero, then the force is conservative, otherwise it is called non-conservative force.

♦ Gravitational force, electrostatic force, magnetic force, etc are conservative forces. All the central forces are conservative forces.
♦ Frictional force, viscous force, etc are non-conservative forces.
(v) Work done by the force of gravity on a particle of mass m is given by W = mgh where g is acceleration due to gravity and h is height through particle one displaced.
(vi) Work done in compressing or stretching a spring is given by W = 1 / 2 kx²
where k is spring constant and x is displacement from mean position.
(vii) When on end of a spring is attached to a fixed vertical support and a block attached to the free end moves on a horizontal
table from x = x1 to x = x2 then W = 1 / 2 k (x²x₂ – x²x₁)
(viii) Work done by the couple for an angular displacement θ is given by W = i * θ where i is the torque of the couple.

power

The time rate of work done by a body is called its power.

Power = Rate of doing work = Work done / Time taken
If under a constant force F a body is displaced through a distance s in time t, the power
p = W / t = F * s / t
But s / t = v ; uniform velocity with which body is displaced.
∴ P = F * v = F v cos θ
where θ is the smaller angle between F and v.
power is a scalar quantity. Its S1 unit is watt and its dimensional formula is [ML²T^-3].
Its other units are kilowatt and horse power,
1 kilowatt = 1000 watt
1 horse power = 746 watt

Energy

Energy of a body is its capacity of doing work.
It is a scalar quantity.
Its S1 unit is joule and CGS unit is erg. Its dimensional formula is [ML³T^-3].
There are several types of energies, such as mechanical energy (kinetic energy and potential energy), chemical energy, light energy, heat energy, sound energy, nuclear energy, electric energy etc.

Mechanical Energy

The sum of kinetic and potential energies at any point remains constant throughout the motion.
It does not depend upon time. This is known as law of conservation of mechanical energy.

Mechanical energy is of two types:

1. Kinetic Energy

The energy possessed by any object by virtue of its motion is called its kinetic energy.
Kinetic energy of an object is given by k = 1 / 2 mv² = p² / 2m

where m = mass of the object, U = velocity of the object and p = mv = momentum of the object.

2. Potential Energy

The energy possessed by any object by virtue of its position or configuration is called its potential energy.
There are three important types of potential energies:
(i) Gravitational Potential Energy If a body of mass m is raised through a height h against gravity, then its gravitational potential energy = mgh,

(ii) Elastic Potential Energy If a spring of spring constant k is stretched through a distance x.
then elastic potential energy of the spring = 1 . 2 kx²
The variation of potential energy with distance is shown in figure.
Potential energy is defined only for conservative forces. It does not exist for non-conservative forces.
Potential energy depends upon frame of reference.
(iii) Electric Potential Energy The electric potential energy of two point charges ql and q’l.
separated by a distance r in vacuum is given by
U = 1 / 4πΣ0 * q1q2 / r
Here 1 / 4πΣ0 = 9.0 * 10¹⁰ N-m² / C² constant.
Work-Energy Theorem
Work done by a force in displacing a body is equal to change in its kinetic energy.
where, Ki = initial kinetic energy and Kf = final kinetic energy.
Regarding the work-energy theorem it is worth noting that
(i) If Wnet is positive, then Kf – Ki = positive, i.e., Kf > Ki or kinetic energy will increase and vice-versa.
(ii) This theorem can be applied to non-inertial frames also. In a non-inertial frame it can be written as:
Work done by all the forces (including the Pseudo force) = change in kinetic energy in noninertial frame.

Mass-Energy Equivalence

According to Einstein, the mass can be transformed into energy and vice – versa.
When Δm. mass disappears, then produced energy
E = Δmc²
where c is the speed of light in vacuum.

Principle of Conservation of Energy

The sum of all kinds of energies in an isolated system remains constant at all times.
Principle of Conservation of Mechanical Energy For conservative forces the sum of kinetic and potential energies of any object remains constant throughout the motion.
According to the quantum physics, mass and energy are not conserved separately but are conserved as a single entity called ‘mass-energy’.

Collisions

Collision between two or more particles is the interaction for a short interval of time in which
they apply relatively strong forces on each other.
In a collision physical contact of two bodies is not necessary. rrhere are two types of collisions:

1. Elastic collision

The collision in which both the momentum and the kinetic energy of the system remains conserved are called elastic collisions.

In an elastic collision all the involved forces are conservative forces.
Total energy remains conserved.

2. Inelastic collision

The collision in which only the momentum remains conserved but kinetic energy does not
remain conserved are called inelastic collisions.
In an inelastic collision some or all the involved forces are non-conservative forces.
Total energy of the system remains conserved.
If after the collision two bodies stick to each other, then the collision is said to be perfectly inelastic.

Coefficient of Restitution or Resilience
The ratio of relative velocity of separation after collision to the velocity of approach before
collision is called coefficient of restitution resilience.
It is represented by e and it depends upon the material of the collidingI bodies.
For a perfectly elastic collision, e = 1
For a perfectly inelastic collision, e = 0
For all other collisions, 0 < e < 1

One Dimensional or Head-on Collision
If the initial and final velocities of colliding bodies lie along the same line, then the collision is called one dimensional or head-on collision.

Inelastic One Dimensional Collision
Applying Newton’s experimental law, we have

Velocities after collision
v₁ = (m₁ – m₂) u₁ + 2m₂u₂ / (m₁ + m₂)
and v₂ = (m₂ – m₁) u₂ + 2m₁u₁ / (m₁ + m₂)
When masses of two colliding bodies are equal, then after the collision, the bodies exchange their velocities.
v1 = u₂ and v₂ = u₁
If second body of same mass (m₁ = m₂) is at rest, then after collision first body comes to rest and second body starts moving with the initial velocity of first body.
v₁ = 0 and v₂ = u₁
If a light body of mass m₁ collides with a very heavy body of mass m₂ at rest, then after collision.
v₁ = – u₁ and v₂ = 0
It means light body will rebound with its own velocity and heavy body will continue to be at rest.
If a very heavy body of mass m₁ collides with a light body of mass m₂(m₁ > > m₂₁) at rest, then after collision
v₁ = u₁ and v₂ = 2u₁
In Inelastic One Dimensional Collision
Loss of kinetic energy
ΔE = m₁m₂ / 2(m₁ + m₂) (u₁ – u₂)² (1 – e2)

In Perfectly Inelastic One Dimensional Collision
Velocity of separation after collision = 0.
Loss of kinetic energy = m₁m₂ (u₁ – u₂)² / 2(m₁ + m₂)
If a body is dropped from a height ho and it strikes the ground with velocity vo and after inelastic collision it rebounds with velocity v₁ and rises to a height h₁, then

If after n collisions with the ground, the body rebounds with a velocity vn and rises to a height hn then
en = vn / v^o = √hn / ho

Two Dimensional or Oblique Collision
If the initial and final velocities of colliding bodies do not lie along the same line, then the collision is called two dimensional or oblique Collision.
In horizontal direction,
m₁u₁ cos α₁ + m₂u₂ cos α₂= m₁v₁ cos β₁ + m₂v₂ cos β₂

In vertical direction.
m₁u₁ sin α₁ – m₂u₂ sin α₂ = m₁u₁ sin β₁ – m₂u₂ sin β₂
If m₁ = m₂ and α₁ + α₂ = 90°
then β₁ + β₂ = 90°
If a particle A of mass m₁ moving along z-axis with a speed u makes an elastic collision with another stationary body B of mass m₂

From conservation law of momentum
m₁u = m₁v₁ cos α + m₂v₂ cos β
O = m₁v₁ sin α – m₂v₂ sin β

WORK

PHYSICAL DEFINITION
When the point of application of force moves in the direction of the applied force under its effect then work is said to be done.

MATHEMATICAL DEFINITION OF WORK

Work is defined as the product of force and displacement in the direction of force

If force and displacement are not parallel to each other rather they are inclined at an angle, then in the evaluation of work component of force (F) in the direction of displacement (s) will be considered.

VECTOR DEFINITION OF WORK

Force and displacement both are vector quantities but their product, work is a scalar quantity, hence work must be scalar product or dot product of force anddisplacement vector. 

WORK DONE BY VARIABLE FORCE
Force varying with displacement In this condition we consider the force to be constant for any elementary displacement and work done in that elementary displacement is evaluated. Total work is obtained by integrating the elementary work from initial to final limits.

Force varying with time
In this condition we consider the force to be constant for any elementary displacement and work done in that elementary displacement is evaluated.

Total work is obtained by integrating the elementary work from initial to final limits.

WORK DONE BY VARIABLE FORCE FROM GRAPH
Let force be the function of displacement & its graph be as shown.

To find work done from s₁ to s₂ we consider two points M & N very close on the graph such that magnitude of force (F) is almost same at both the points. If elementary displacement from M to N is ds, then elementary work done
from M to N is.
dW = F.ds
dW = (length x breadth)of strip MNds
dW = Area of strip MNds

Thus work done in any part of the graph is equal to area under that part. Hence total work done from s₁ to s₂ will be given by the area enclosed under the graph from s₁ to s₂.
W = Area (ABS2S1A)

DIFFERENT CASES OF WORK DONE BY CONSTANT FORCE

Case i) Force and displacement are in same direction
θ = 0
Since, W = Fs Cos θ
Therefore W = Fs Cos 0
or, W = Fs
Ex – Coolie pushing a load horizontally

Case ii) Force and displacement are mutually perpendicular to each other
θ = 90
Since, W = Fs Cos θ
Therefore W = Fs Cos 90
or, W = 0
Ex – coolie carrying a load on his head & moving horizontally with constant velocity.
Then he applies force vertically to balance weight of body & its displacement is horizontal.

(3) Force & displacement are in opposite direction 
θ = 180
Since, W = Fs Cos θ
Therefore W = Fs Cos 180
or, W = – Fs

Ex – Coolie carrying a load on his head & moving vertically down with constant velocity. Then he applies force in vertically upward direction to balance the weight of body & its displacement is in vertically downward direction.

ENERGY
Capacity of doing work by a body is known as energy.
Note – Energy possessed by the body by virtue of any cause is equal to the total work done by the
body when the cause responsible for energy becomes completely extinct.

TYPES OF ENERGIES
There are many types of energies like mechanical energy, electrical, magnetic, nuclear, solar, chemical etc.

MECHANICAL ENERGY
Energy possessed by the body by virtue of which it performs some mechanical work is known as mechanical energy.
It is of basically two types-
(i) Kinetic energy
(ii) Potential energy

KINETIC ENERGY
Energy possessed by body due to virtue of its motion is known as the kinetic energy of the body. Kinetic energy possessed by moving body is equal to total work done by the body just before coming out to rest.

Consider a body of man (m) moving with velocity (vo).After travelling through distance (s) it comes to rest.
u = vo
v = 0
s = s
Applying, v² = u² + 2as
0 = v0² + 2as
or, 2as = – v0²
or, a = -vo²/2s
Hence force acting on the body,
F = ma
Fon body = – mvo²/2s

But from Newton’s third law of action and reaction, force applied by body is equal and opposite to the force applied on body
Fby body = -Fon body
=+mvo²/2s
Therefore work done by body,
W = ^→F. ^→s
or, W = mv0
2.s.Cos 0²s
or, W = 1/2 mvo²
Thus K.E. stored in the body is,
K.E.= 1/2 mvo²

KINETIC ENERGY IN TERMS OF MOMENTUM
K.E. of body moving with velocity v is
K.E. = 1/2 mvo2
Multiplying and dividing by m  K = 1/2 mv² x m/m
= 1/2 m²v²/m
But, mv = p (linear momentum)
Therefore, K = p2/2m

POTENTIAL ENERGY

Energy possessed by the body by virtue of its position or state is known as potential energy. Example:- gravitational potential energy, elastic potential energy, electrostatic potential energy etc.

GRAVITATIONAL POTENTIAL ENERGY

Energy possessed by a body by virtue of its height above surface of earth is known as gravitational potential energy. It is equal to the work done by the body situated at some height in returning back slowly to the surface of earth.
Consider a body of mass m situated at height h above the surface of earth. Force applied by the body in vertically downward direction is
F = mg
Displacement of the body in coming back slowly to the surface of earth is
s = h
Hence work done by the body is
W = FsCosθ
or, W = FsCos0
or, W = mgh
This work was stored in the body in the form of gravitational potential energy due to its position. Therefore
G.P.E = mgh

ELASTIC POTENTIAL ENERGY
Energy possessed by the spring by virtue of compression or expansion against elastic force in the spring is known as elastic potential energy.

Spring
It is a coiled structure made up of elastic material & is capable of applying restoring force & restoring torque when disturbed from its original state. When force (F) is applied at one end of the string, parallel to its length, keeping the other end fixed, then the spring expands (or contracts) & develops a restoring force (FR) which balances the applied force in equilibrium.

On increasing applied force spring further expands in order to increase restoring force for balancing the applied force. Thus restoring force developed within the spring is directed proportional to the extension produced in the spring.

F_R ∝ x
or, F_R = kx (k is known as spring constant or force constant)
If x = 1, F R = k
Hence force constant of string may be defined as the restoring force developed within spring when its length is changed by unity.
But in equilibrium, restoring force balances applied force.
F = FR = k x
If x = 1, F = 1
Hence force constant of string may also be defined as the force required to change its length by unity in equilibrium.

Mathematical Expression for Elastic Potential Energy

Consider a spring of natural length ‘L’ & spring constant ‘k’ its length is increased by xo. Elastic potential energy of stretched spring will be equal to total work done by the spring in egaining its original length.
If in the process of regaining its natural length, at any instant extension in the spring was x then force applied by spring is
F = kx
If spring normalizes its length by elementary distance dx opposite to x under this force then work done by spring is
dW = F. (-dx) . Cos0
(force applied by spring F and displacement –dx taken opposite to extension x are in same direction)
dW = -kxdx
Total work done by the spring in regaining its original length is obtained in integrating dW from x0 to 0 

This work was stored in the body in the form of elastic potential energy.
E.P.E = 1/2 kxo2

WORK ENERGY THEOREM

It states that total work done on the body is equal to the change in kinetic energy.(Provided body is confined to move horizontally and no dissipating forces are operating).

Consider a body of man m moving with initial velocity v1. After travelling through displacement s its final velocity becomes v2 under the effect of force F.
u = v₁
v = v₂
s = s
Applying, v² = u² + 2as
v₂² = v1² + 2as
or, 2as = v₂² – v₁²
or, a = v₂² – v₁²/2s
Hence external force acting on the body is
F = ma
F = m v₂² – v₁²/2s
Therefore work done on body by external force
W = ^→F . ^→s
or, W = m v₂² – v₁²/2s . s .Cos 0
(since force and displacement are in same direction)
or, W = 1/2 – mv₂² -1/2mv₁²
or, W = K₂ – K₁
or, W = ΔK

PRINCIPLE OF CONSERVATION OF ENERGY

It states that energy can neither be creased neither be destroyed. It can only be converted from one form to another.
Consider a body of man m situated at height h & moving with velocity vo. Its energy will be.
E₁ = P ₁ + K₁
or, E₁ = mgh + ½ mvo²
If the body falls under gravity through distance y, then it acquires velocity v₁ and its height becomes (h-y)
u = vo
s = y
a = g
v = v₁
From v² = u² +2as
v₁² = vo² + 2gy
Energy of body in second situation
E₂ = P₂ + K₂
or, E₂ = mg (h-y) + ½ mv²
or, E₂ = mg (h-y) + ½ m (vo² + 2gy)
or, E₂ = mgh – mgy + ½ mvo² + mgy
or, E₂ = mgh + ½ mvo²
Now we consider the situation when body reaches ground with velocity v²
u = vo
s = h
a = g
v = v²
From v² = u² +2as
2² = vo² + 2gh
Energy of body in third situation
E₃ = P₃ + K₃
or, E₃ = mg0 + ½ mv₂²
or, E₃ = 0 + ½ m (vo² + 2gh)
or, E₃ = ½ mvo² + mgh
From above it must be clear that E₁ = E₂ = E₃. This proves the law of conservation of energy.

CONSERVATIVE FORCE

Forces are said to be conservative in nature if work done against the forces gets conversed in the body in form of potential energy. Example:-
gravitational forces, elastic forces & all the central forces.

PROPERTIES OF CONSERVATIVE FORCES

1. Work done against these forces is conserved & gets stored in the body in the form of P.E.
2. Work done against these forces is never dissipated by being converted into nonusable forms of energy like heat, light, sound etc.
3. Work done against conservative forces is a state function & not path function i.e. Work done against it, depends only upon initial & final states of body & is independent of the path through which process has been carried out.
4. Work done against conservative forces is zero in a complete cycle.

TO PROVE WORK DONE AGAINST CONSERVATIVE FORCES IS A STATE FUNCTION
Consider a body of man m which is required to be lifted up to height h. This can be done in 2 ways.
(i) By directly lifting the body against gravity
(ii) By pushing the body up a smooth inclined plane. mg
Min force required to lift the body of mass m vertically is

F = mg
And displacement of body in lifting is 
s = h
Hence work done in lifting is
W1 = FsCos0^o (since force and displacement are in same direction) mg
W1 = mgh
Now we consider the same body lifted through height h by pushing it up a smooth inclined plane

Min force required to push the body is
F = mgSinθ
And displacement of body in lifting is
s = h/Sinθ
Hence work done in pushing is
W₂ = FsCos0
or, W₂ = mgSinθ. h/Sinθ . 1
or, W₂ = mgh
From above W₁ = W₂ we can say that in both the cases work done in lifting the body through height ‘h’ is same.

To Prove That Work Done Against Conservative Forces Is Zero In A Complete Cycle

Consider a body of man m which is lifted slowly through height h & then allowed to come back to the ground slowly through height h.

For work done is slowly lifting the body up,
Minimum force required in vertically upward direction is
F = mg
Vertical up displacement of the body is
s = h
Hence work done is
W = FsCosθ
or, W_I = FsCos0 (since force and displacement are in same direction)
or, W_I = mgh (since force and displacement are in same direction)
For work done is slowly bringing the body down,
Minimum force required in vertically upward direction is
F = mg
Vertical down displacement of the body is
s = h
Hence work done is
or, W₂ = FsCos180(since force and displacement are in opposite direction)
or, W₂ = – mgh
Hence total work done against conservative forces in a complete cycle is
W = W₁ + W₂
or, W = (mgh) + (-mgh)
or, W = 0

NON-CONSERVATIVE FORCES
Non conservative forces are the forces, work done against which does not get conserved in the body in the form of potential energy.

PROPERTIES OF NON-CONSERVATIVE FORCES

1. Work done against these forces does not get conserved in the body in the form of P.E.
2. Work done against these forces is always dissipated by being converted into non usable forms of energy like heat, light, sound etc.
3. Work done against non-conservative force is a path function and not a state function.
4. Work done against non-conservative force in a complete cycle is not zero.

PROVE THAT WORK DONE AGAINST NON–CONSERVATIVE FORCES IS A PATH FUNCTION

Consider a body of mass (m) which is required to be lifted to height ‘h’ by pushing it up the rough incline of inclination.

Minimum force required to slide the body up the rough inclined plane having coefficient of kinetic friction μ with the body is
F = mgSinθ + fk
or, F = mgSinθ + μN
or, F = mgSinθ + μmgCosθ
Displacement of the body over the incline in moving through height h is
s = h/Sinθ
Hence work done in moving the body up the incline is
W = F.s.Cos0(since force and displacement are in opposite direction)
or, W = (mgSinθ + μmgCosθ).h/Sinθ.1
or, W = mgh + μmgh/Tanθ
Similarly if we change the angle of inclination from θ to θ1, then work done will be
W₁ = mgh + μmgh/Tanθ₁
This clearly shows that work done in both the cases is different & hence work done against non-conservative force in a path function and not a state function i.e. it not only depends upon initial & final states of body but also depends upon the path through which process has been carried out.

To Prove That Work Done Against Non-conservative Forces In A Complete Cycle Is Not Zero

Consider a body displaced slowly on a rough horizontal plane through displacement s from A to B.

Minimum force required to move the body is
F = fk = μN = μmg
Work done by the body in displacement s is
W = F.s.Cos0(since force and displacement are in same direction)
or, W = μmgs
Now if the same body is returned back from B to A

Minimum force required to move the body is
F = f_k = μN = μmg
Work done by the body in displacement s is
W = F.s.Cos0(since force and displacement are in same direction)
or, W = μmgs
Hence total work done in the complete process
W = W₁ + W₂ = 2μmgs
Note – When body is returned from B to A friction reverse its direction.

POWER
Rate of doing work by a body with respect to time is known as power.

Average Power

It is defined as the ratio of total work done by the body to total time taken.
P_avg = Total work done/Total time taken = ΔW/ Δt

Instantaneous Power

Power developed within the body at any particular instant of time is known as instantaneous power.
Or
Average power evaluated for very short duration of time is known as instantaneous power.

EFFICIENCY

It is defined as the ratio of power output to power input.
Or
It is defined as the ratio of energy output to energy input.
Or
I It is defined as the ratio of work output to work input.

PERCENTAGE EFFICIENCY
Percentage Efficiency = Efficiency x 100

COLLISION
Collision between the two bodies is defined as mutual interaction of the bodies for a short interval of time due to which the energy and the momentum of the interacting bodies change.

Types of Collision
There are basically three types of collisionsi)

i) Elastic Collision – That is the collision between perfectly elastic bodies. In this type of collision, since only conservative forces are operating between the interacting bodies, both kinetic energy and momentum of the system remains constant.

ii) Inelastic Collision – That is the collision between perfectly inelastic or plastic bodies. After collision bodies stick together and move with some common velocity. In this type of collision only momentum is conserved. Kinetic energy is not conserved due to the presence of non-conservative forces between the interacting bodies.

iii) Partially Elastic or Partially Inelastic Collision – That is the collision between the partially elastic bodies. In this type of collision bodies do separate from each other after collision but due to the involvement of non-conservative inelastic forces kinetic energy of the system is not conserved and only momentum is conserved.

Collision In One Dimension – Analytical Treatment

Consider two bodies of masses m₁ and m₂ with their center of masses moving along the same straight line in same direction with initial velocities u₁ and u₂ with m₁ after m₂. Condition necessary for the collision is u₁ > u₂ due to which bodies start approaching towards each other with the velocity of approach u₁ – u₂. Collision starts as soon as the bodies come in contact. Due to its greater velocity and inertia m₁ continues to push m₂ in the forward direction whereas m2 due to its small velocity and inertia pushes m1 in the backward direction. Due to this pushing force involved between the two colliding bodies they get deformed at the point of contact and a part of their kinetic energy gets consumed in the deformation of the
bodies. Also m1 being pushed opposite to the direction of the motion goes on decreasing its velocity and m2 being pushed in the direction of motion continues increasing its velocity. This process continues until both the bodies acquire the same common velocity v. Up to this stage there is maximum deformation in the bodies maximum part of their kinetic energy gets consumed in their deformation.

Elastic collision

In case of elastic collision bodies are perfectly elastic. Hence after their maximum deformation they have tendency to regain their original shapes, due to which they start pushing each other. Since m₂ is being pushed in the direction of motion its velocity goes on increasing and m1 being pushed opposite to the direction of motion its velocity goes on decreasing. Thus condition necessary for separation i.e. v₂>v₁ is attained and the bodies get separated with velocity of separation v₂ – v₁. In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is again returned back to the system when the bodies regain their original shapes. Hence in such collision energy conservation can also be applied along with the momentum conservation.
Applying energy conservation

Hence in perfectly elastic collision between two bodies of same mass, the velocities interchange.ie. If a moving body elastically collides with a similar body at rest. Then the moving body comes at rest and the body at rest starts moving with the velocity of the moving body.
Case 2- If a huge body elastically collides with the small body,
m₁ >> m₂
m₂ will be neglected in comparison to m₁

Hence if a huge body elastically collides with a small body then there is almost no change in the velocity of the huge body but if the small body is initially at rest it gets thrown away with twice the velocity of the huge moving body.eg. collision of truck with a drum.
Case 3- If a small body elastically collides with a huge body,
m₂ >> m₁
m₁ will be neglected in comparison to m₂

Hence if a small body elastically collides with a huge body at rest then there is almost no change in the velocity of the huge body but if the huge body is initially at rest small body rebounds back with the same speed.eg. collision of a ball with a wall.

Inelastic collision
In case of inelastic collision bodies are perfectly inelastic. Hence after their maximum deformation they have no tendency to regain their original shapes, due to which they continue moving with the same common velocity. In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is permanently consumed in the deformation of the bodies against non-conservative inelastic forces. Hence in such collision energy conservation can-not be applied and only momentum conservation is applied.
Applying momentum conservation
p_i = p_f
m₁u₁ + m₂u₂ = m₁v + m₂v
or, m₁u₁ + m₂u₂ = (m₁+m₂)v
or, v = m₁u₁ + m₂u₂/(m₁+m₂)

Partially Elastic or Partially Inelastic Collision
In this case bodies are partially elastic. Hence after their maximum deformation they have tendency to regain their original shapes but not as much as perfectly elastic bodies. Hence they do separate but their velocity of separation is not as much as in the case of perfectly elastic bodies i.e. their velocity of separation is less than the velocity of approach.
In such collision the part of kinetic energy of the bodies which has been consumed in the deformation of the bodies is only slightly returned back to the system. Hence in such collision energy conservation can-not be applied and only momentum conservation is applied.
(v₂ – v₁) < (u₁ – u₂)

Collision In Two Dimension – Oblique Collision

When the centers of mass of two bodies are not along the same straight line, the collision is said to be oblique. In such condition after collision bodies are deflected at some angle with the initial direction. In this type of collision momentum conservation is applied separately along x-axis and y-axis. If the collision is perfectly elastic energy conservation is also applied.
Let initial velocities of the masses m₁ and m₂ be u₁ and u₂ respectively along x-axis. After collision they are deflected at angles θ and Ø respectively from x-axis, on its either side of the x axis.

Applying momentum conservation along x-axis
p_f = p_i
m1v1 Cosθ + m₂v₂ Cos Ø = m₁u₁ + m₂u₂
Applying momentum conservation along y-axis
p_f = p_i
m₁v₁ Sinθ – m₂v₂ Sin Ø = m₁0 + m₂0
or, m₁v₁ Sinθ – m₂v₂ Sin Ø = 0
or, m₁v₁ Sinθ = m₂v₂ Sin Ø
In case of elastic collision applying energy conservation can also be applied
Kf = Ki
1/2 m₁u₁² + 1/2m₂u₂² = 1/2 m₁v₁² + 1/2m₂v₂²

Coefficient Of Restitution

It is defined as the ratio of velocity of separation to the velocity of approach.
e = Velocity of separation / Velocity of approach
or, e = (v₂ – v₁)/(u₁ – u₂)

Case-1 For perfectly elastic collision, velocity of separation is equal to velocity of approach, therefore
e = 1
Case-2 For perfectly inelastic collision, velocity of separation is zero, therefore
e = 0
Case-3 For partially elastic or partially inelastic collision, velocity of separation is less than velocity of approach, therefore
e < 1

MEMORY MAP

We hope the above Work Energy and Power Class 11 Physics are useful for you. If you have any questions then post them in the comments section below. Our teachers will provide you an answer. Also refer to MCQ Questions for Class 11 Physics