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 [ML^{2}T^{-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 kx^{2}

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 (x^{2}x_{2} – x^{2}x_{1})

(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 [ML^{2}T^{-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 [ML^{3}T^{-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 mv^{2} = p^{2} / 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 kx^{2}

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 * 10^{10} N-m^{2} / C^{2} 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 = Δmc^{2}

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

v_{1} = (m_{1} – m_{2}) u_{1} + 2m_{2}u_{2} / (m_{1} + m_{2})

and v_{2} = (m_{2} – m_{1}) u_{2} + 2m_{1}u_{1} / (m_{1} + m_{2})

When masses of two colliding bodies are equal, then after the collision, the bodies exchange their velocities.

v1 = u_{2} and v_{2} = u_{1}

If second body of same mass (m_{1} = m_{2}) is at rest, then after collision first body comes to rest and second body starts moving with the initial velocity of first body.

v_{1} = 0 and v_{2} = u_{1}

If a light body of mass m_{1} collides with a very heavy body of mass m_{2} at rest, then after collision.

v_{1} = – u_{1} and v_{2} = 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 m_{1} collides with a light body of mass m_{2}(m_{1} > > m_{21}) at rest, then after collision

v_{1} = u_{1} and v_{2} = 2u_{1}

In Inelastic One Dimensional Collision

Loss of kinetic energy

ΔE = m_{1}m_{2} / 2(m_{1} + m_{2}) (u_{1} – u_{2})^{2} (1 – e2)

**In Perfectly Inelastic One Dimensional Collision**Velocity of separation after collision = 0.

Loss of kinetic energy = m

_{1}m

_{2}(u

_{1}– u

_{2})

^{2}/ 2(m

_{1}+ m

_{2})

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 v

_{1}and rises to a height h

_{1}, 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,

m_{1}u_{1} cos α_{1} + m_{2}u_{2} cos α_{2}= m_{1}v_{1} cos β_{1} + m_{2}v_{2} cos β_{2}

In vertical direction.

m_{1}u_{1} sin α_{1} – m_{2}u_{2} sin α_{2} = m_{1}u_{1} sin β_{1} – m_{2}u_{2} sin β_{2}

If m_{1} = m_{2} and α_{1} + α_{2} = 90°

then β_{1} + β_{2} = 90°

If a particle A of mass m_{1} moving along z-axis with a speed u makes an elastic collision with another stationary body B of mass m_{2}

**From conservation law of momentum**

m_{1}u = m_{1}v_{1} cos α + m_{2}v_{2} cos β

O = m_{1}v_{1} sin α – m_{2}v_{2} 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 s_{1} to s_{2} 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 s_{1} to s_{2} will be given by the area enclosed under the graph from s_{1} to s_{2}.

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.

E**NERGY**

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, v^{2} = u^{2} + 2as

0 = v0^{2} + 2as

or, 2as = – v0^{2}

or, a = -vo^{2}/2s

Hence force acting on the body,

F = ma

Fon body = – mvo^{2}/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

=+mvo^{2}/2s

Therefore work done by body,

W = ^{→}F. ^{→}s

or, W = mv0

2.s.Cos 0^{2}s

or, W = 1/2 mvo^{2}

Thus K.E. stored in the body is,

K.E.= 1/2 mvo^{2}

**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 mv

^{2}x m/m

= 1/2 m

^{2}v

^{2}/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 = v_{1}

v = v_{2}

s = s

Applying, v^{2} = u^{2} + 2as

v_{2}^{2} = v1^{2} + 2as

or, 2as = v_{2}^{2} – v_{1}^{2}

or, a = v_{2}^{2} – v_{1}^{2}/2s

Hence external force acting on the body is

F = ma

F = m v_{2}^{2} – v_{1}^{2}/2s

Therefore work done on body by external force

W = ^{→}F . ^{→}s

or, W = m v_{2}^{2} – v_{1}^{2}/2s . s .Cos 0

(since force and displacement are in same direction)

or, W = 1/2 – mv_{2}^{2} -1/2mv_{1}^{2}

or, W = K_{2} – K_{1}

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.

E_{1} = P_{ 1} + K_{1}

or, E_{1} = mgh + ½ mvo^{2}

If the body falls under gravity through distance y, then it acquires velocity v_{1} and its height becomes (h-y)

u = vo

s = y

a = g

v = v_{1}

From v^{2} = u^{2} +2as

v_{1}^{2} = vo^{2} + 2gy

Energy of body in second situation

E_{2} = P_{2} + K_{2}

or, E_{2} = mg (h-y) + ½ mv^{2}

or, E_{2} = mg (h-y) + ½ m (vo^{2} + 2gy)

or, E_{2} = mgh – mgy + ½ mvo^{2} + mgy

or, E_{2} = mgh + ½ mvo^{2}

Now we consider the situation when body reaches ground with velocity v^{2}

u = vo

s = h

a = g

v = v^{2}

From v^{2} = u^{2} +2as

2^{2} = vo^{2} + 2gh

Energy of body in third situation

E_{3} = P_{3} + K_{3}

or, E_{3} = mg0 + ½ mv_{2}^{2}

or, E_{3} = 0 + ½ m (vo^{2} + 2gh)

or, E_{3} = ½ mvo^{2} + mgh

From above it must be clear that E_{1} = E_{2} = E_{3}. 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

W_{2} = FsCos0

or, W_{2} = mgSinθ. h/Sinθ . 1

or, W_{2} = mgh

From above W_{1} = W_{2} 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, W_{2} = FsCos180(since force and displacement are in opposite direction)

or, W_{2} = – mgh

Hence total work done against conservative forces in a complete cycle is

W = W_{1} + W_{2}

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

W_{1} = mgh + μmgh/Tanθ_{1}

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 = W_{1} + W_{2} = 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 m_{1} and m_{2} with their center of masses moving along the same straight line in same direction with initial velocities u_{1} and u_{2} with m_{1} after m_{2}. Condition necessary for the collision is u_{1} > u_{2} due to which bodies start approaching towards each other with the velocity of approach u_{1} – u_{2}. Collision starts as soon as the bodies come in contact. Due to its greater velocity and inertia m_{1} continues to push m_{2} 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 m_{2} 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. v_{2}>v_{1} is attained and the bodies get separated with velocity of separation v_{2} – v_{1}. 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,

m_{1} >> m_{2}

m_{2} will be neglected in comparison to m_{1}

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,

m_{2} >> m_{1}

m_{1} will be neglected in comparison to m_{2}

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}

m_{1}u_{1} + m_{2}u_{2} = m_{1}v + m_{2}v

or, m_{1}u_{1} + m_{2}u_{2} = (m_{1}+m_{2})v

or, v = m_{1}u_{1} + m_{2}u_{2}/(m_{1}+m_{2})

**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.

(v_{2} – v_{1}) < (u_{1} – u_{2})

**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 m_{1} and m_{2} be u_{1} and u_{2} 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 + m_{2}v_{2} Cos Ø = m_{1}u_{1} + m_{2}u_{2}

Applying momentum conservation along y-axis

p_{f} = p_{i}

m_{1}v_{1} Sinθ – m_{2}v_{2} Sin Ø = m_{1}0 + m_{2}0

or, m_{1}v_{1} Sinθ – m_{2}v_{2} Sin Ø = 0

or, m_{1}v_{1} Sinθ = m_{2}v_{2} Sin Ø

In case of elastic collision applying energy conservation can also be applied

Kf = Ki

1/2 m_{1}u_{1}^{2} + 1/2m_{2}u_{2}^{2} = 1/2 m_{1}v_{1}^{2} + 1/2m_{2}v_{2}^{2}

**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 = (v_{2} – v_{1})/(u_{1} – u_{2})

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