WORK, ENERGY AND POWER

1. Dot or Scalar Product – Properties of Scalar Product
2. Work – Examples and Definition; Work done by Parallel Force and Oblique Force
3. When is Work Zero? Some Points about Work
4. Positive, Zero and Negative Work – Examples
5. Work done by a Variable Force
6. Energy – Different Forms of Energy
7. Kinetic Energy – Definition and Formula – Note on Kinetic Energy
8. Work–Energy Theorem for Constant Force and Variable Force
9. Potential Energy – Definition, Examples and Formula – Note on Potential Energy
10. Law of Conservation of Energy – Examples
11. Conservative Force and Non-conservative Force
12. Conservation of Mechanical Energy – Example using Gravitational Force & Spring
13. Collisions – Elastic, Completely Inelastic and Inelastic Collisions
14. Elastic Collision in One Dimension and Inelastic Collision in One Dimension
15. Elastic Collision in Two Dimensions
16. Power and Note on Power

Created by C. Mani, Assistant Commissioner, KVS RO Silchar

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Dot Product or Scalar Product

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A

P

O

B

θ

Q

In dot product, the symbol of multiplication between the vectors is

represented by ‘ . ‘ The result of the product is a scalar value.

Consider two vectors A and B making an angle θ

with each other.

The dot or scalar product is given by

A . B

| A | | B |

A . B = | A | | B | cos θ ^{and cos θ =}

Geometrically, B cos θ is the projection of B onto A and A cos θ is the

projection of A onto B.

or

P

A

B

θ

A . B = A (B cos θ)

Q

^OBcosθ

P

A

θ

A . B = B (A cos θ)

Q

B

Acosθ

O

Properties of Dot Product

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1. Dot product results in a scalar value.
2. ^{Dot product or Scalar product is commutative.} A . B = B . A

6. Component of A along B = ^{| A | cos θ =}

A . B

| B |

= A . B

A . B

7. Component of B along A = ^{| B | cos θ =}

| A |

= B . A

3. Dot product is distributive.

4. i ^. i ₌ j ^. j _{= k}^. k _{= 1} 5. i ^. j ₌ j ^.k _{= k}^. i _{= 0}

A . (B + C) = A . B + ^{A . C}

8. If

^{a = a}_xi ^{+ a}_y j ^{+ a}_z k

^{b = b}_xi ^{+ b}_y j ^{+ b}_z k

then

a² = (a ² + a ² + a ²)

x y z

a . b = (a_x b_x + a_y b_y + a_z b_z) a . a = (a_x a_x + a_y a_y + a_z a_z) b . b = (b_x b_x + b_y b_y + b_z b_z)

b² = (b ² + b ² + b ²)

x y z

WORK

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A horse pulling a cart does work.

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A man climbing the stairs does work.

An engine of a moving vehicle does work.

A body dropped from a height falls towards the earth. It indicates that the earth exerts a force of attraction (called gravity) on the body. The gravitational force does work in pulling the body towards it.

Work is said to be done scientifically when a

force applied on a body produces motion in it.

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in the direction of the displacement and the magnitude of this displacement.

Work done by the force is defined to be the product of component of the force

Work done = Component of Force x Displacement

W = (F cos θ) s = F s cos θ

F

s

s

θ

W = (F cos θ) s

= F s cos θ

Work done by a parallel force:

W = F s cos θ

= F s cos 0°

= F s

Work done by an oblique force:

W = F . s

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Force ‘F’ is resolved into two rectangular components.

Work done = Force component x Displacement

W = (F cosθ) s = Fs cosθ

F cosθ

F sinθ

F sinθ acting in the vertical direction does not displace the body in that direction.

Therefore, work done by the component

F sinθ is zero.

Fcosθ acting in the horizontal direction

displaces the body in that direction.

F

θ

s

Therefore,

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When is work zero?

Examples:

1. A man holding a book in his hands in a stationary position does not do any work since the weight of the book acting on him does not produce any displacement.
2. A man trying to push a rigid and stationary wall does not do work since

the force applied by him does not move the wall.

For work to be done, both force (or component of force) and

displacement must be present in a body.

(i) If a force applied on a body does not produce displacement in the body, scientifically work is not done.

W

F

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(Here, friction is absent and the body is under uniform motion.)

(ii) If a force applied on a body is zero, scientifically work is not done.

9. 1 kilowatt hour (kWh) = 3.6 x 10⁶ J

Some points about work:

1. Work is a scalar quantity; work can be positive, zero or negative.
2. Negative work does not mean it a vector quantity. Work done by friction is

negative as the displacement is opposite to the friction.

3. Dimension of work is [ML²T^-2].
4. SI unit of work is ‘joule’ or ‘J’. 1 joule = 1 newton x 1 metre or 1 J = 1 Nm

1 joule is the amount of work done when 1 newton of force acting on a

body displaces it through 1 metre.

5. CGS unit of work is ‘erg’. 1 erg = 10^-7 joule
6. 1 joule = 10⁷ erg
7. 1 electron volt (eV) = 1.6 x 10^-19 J
8. 1 calorie (cal) = 4.186 J

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Positive, Zero and Negative Work done

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1. The work done is positive if the force (or component of force) acts on a

body in the direction of its motion. (0º ≤ θ < 90º)

W = Fs cos θ

W = Fs cos 0º

W = Fs (cos 0º = 1)

W = Fs cos 90º

W = 0 (cos 90º = 0)

3. The work done is negative if the force acts on a body in the direction

opposite to its motion. (90º < θ ≤ 180º)

W = Fs cos θ

W = Fs cos 180º

W = - Fs (cos 180º = -1)

F s

2. The work done is zero if the force acts on a body in the direction perpendicular to its motion. (θ = 90º)

F

W = Fs cos θ

s

90º

f

s

180º

Examples:

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Earth

Satellite’s Orbit

1. Work done by gravitational force on a satellite is zero because the force

acts perpendicular to the direction of motion of the satellite.

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F₁

F₂

Sun

Elliptical Orbit

2. Similarly, work done by the sun on the earth is zero.

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3. When we throw a ball vertically upwards, the work done by our force is positive whereas the work done by the gravitational force is negative since it acts in the direction opposite to the motion of the ball.

​

• When the ball falls back, the work done by the gravitational force is positive since it acts along the direction of motion of the ball.

Wo

rk

Δx

O

F(x)

Area = ΔA = F(x) Δx = ΔW

Work done by a variable force More commonly, force is variable in practice.

The displacement covered by the body from x_i to x_f can be considered to be made up of infinite number of infinitesimally small Δx.

Force F(x) may be considered to be

constant over this Δx.

Then, the work done is

ΔW = F(x) Δx

xx_i

xx_f

The total work done is obtained by adding the rectangular areas of the strips shown under the curve.

x_f

^{W ≈ ∑} F(x) Δx

x_i

If the displacements are allowed to approach zero, then the summation

approaches a definite value equal to the area under the curve.

Then, the work done is

^∑ F(x) Δx

i

x_f

W =

lim

Δx→0 _x

^F(x)const

x

i

x_f

W = ∫ F(x) Δx

x

or

Thus, for a varying force, the work done can be expressed as a definite integral of force over displacement.

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ENERGY

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Energy is the ability to do work.

or

The amount of energy possessed by a body is equal to the amount of

work it can do when its energy is released.

1. Energy is a scalar quantity.
2. SI unit of energy is ‘joule’ or ‘J’.
3. CGS unit of energy is ‘erg’.

1 joule = 10⁷ erg or 1 erg = 10^-7 joule

​

Different Forms of Energy

1. Mechanical energy ---> Potential energy and Kinetic energy

KINETIC ENERGY

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Kinetic energy is defined as the energy of a body by virtue of its motion.

It is the measure of work a body can do by virtue of its motion.

i.e. Every moving body possesses kinetic energy.

Example:

A moving cricket ball possesses kinetic energy.

Formula for Kinetic Energy

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Suppose a body of mass ‘m’ moving with a velocity ‘u’ is acted upon by a force ‘F’ for time ‘t’. Let ‘v’ be the final velocity and ‘s’ be the displacement of the body at the end of the time.

The work done by the force in displacing the body is given by

Work done = Force x Displacement

W = F x s

= ma x s

= m (as) ……………..(1)

We know that or

v² = u² + 2as as = ½(v² – u²)

Substituting for as in (1)

W = m x ½(v² – u²)

W = ½m(v² – u²)

This work done is possessed by the body in the form of kinetic energy.

KE = ½mv²

Therefore, KE = ½m(v² – u²)

If the body is initially at rest and its final velocity is ‘v’, then

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Note:

1. Kinetic energy is
1. directly proportional to the mass of the body, and
2. directly proportional to the square of velocity of the body.

If mass is doubled, KE is doubled and if the mass is halved, KE is also halved.

If velocity is doubled, KE increases four times and if the mass is halved, KE reduces to ¼ of its original value.

2. Kinetic energy is always positive. Why?

Mass is always positive. Even if the velocity is negative, square of

velocity will be positive. Therefore, kinetic energy is always positive.

3. Kinetic energy is a scalar quantity.
4. The term ‘speed’ can be used in place of ‘velocity’ in the formula for

kinetic energy.

WORK – ENERGY THEOREM

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Suppose a body of mass ‘m’ moving with a velocity ‘u’ is acted upon by a force ‘F’ for time ‘t’. Let ‘v’ be the final velocity and ‘s’ be the displacement of the body at the end of the time.

We know that v² – u² = 2as

Multiplying both the sides by m/2, we have

½mv² – ½mu² = mas

½mv² – ½mu² = F s

½mv² – ½mu² = W

K_f – K_i = W

The change in kinetic energy of a body is equal to the work done on it by the net force.

The above equation can be generalised for 3-dimensions by employing

vectors.

(since v . v = v² and u . u = u²)

v² – u² = 2as becomes v² – u² = 2 a . s

Multiplying both the sides by m/2, we have

½mv² – ½mu² = m a . s

½mv² – ½mu² = F . s

K_f – K_i = W

WORK–ENERGY THEOREM FOR A VARIABLE FORCE

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Kinetic energy is given by K = ½ mv²

Differentiating w.r.t. ‘t’,

dK

dt

₌ d

dt

½ mv²

_{= m v} dv

dt

_{= m} dv

dt

v

= F v

(from Newton’s Second law)

x_i

_{∫ dK =} ∫ F dx

K_i

dK _{= F} dx ^dt dt

dK = F dx

Integrating from initial position x_i to final position x_f, we have ^K_f x_f

where K_i and K_f are initial and final kinetic

energies corresponding to x_i and x_f.

or

x_f

K_f – K_i = ∫ F dx

x_i

K_f – K_i = W

or

which is work–energy theorem for variable force.

POTENTIAL ENERGY

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Work (driving the nail into the wooden block) is done by the hammer when

its energy is released.

Potential energy is defined as the energy of a body by virtue of its position

or configuration.

Examples:

The raised hammer possesses potential energy.

Wait…tank is getting filled…

The water stored in overhead tank

possesses potential energy.

Water, when released, is made to flow

(work is done) by this potential energy.

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F

Potential energy is stored in a compressed spring.

​

When the spring is released the potential energy stored in the spring does work on the ball and the ball starts moving.

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Potential energy is stored in a stretched catapult.

When released, it does the work of firing the stone.

Formula for Potential Energy (by virtue of position)

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m

h

g

Suppose a body of mass ‘m’ is raised to a height ‘h’ above the surface of the earth against the acceleration due to gravity ‘g’. (Here, g is taken constant as h << R_E.)

The work done in lifting the body against the force of

gravity is given by

Work done = Force x Displacement

W = mg x h

= mgh

This work done is stored in the body in the form of potential energy.

Therefore, Potential energy V(h) = mgh

​

Potential energy is a scalar quantity.

But, it is to be taken with proper + or – sign depending on whether work is

done against the force or by the force.

Formula for Potential Energy (by virtue of configuration)

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F_s F^F_s F

m m m ^x_c x_e

​

x = 0

0 0

c

W_s = ∫ F_s dx = – ∫ kx dx = – ½ kx ²

W_s

c

= – ½ kx ²

The spring force is an example of a variable force which is conservative.

Press UP (↑) arrow key!

Consider a block of mass ‘m’ attached to a spring (say massless) of spring

_{constant ‘k’. Un}P_itla_oy_f U_spP_r(_in↑)_ga_cn_od_nD_stO_aW_ntN_is(↓_N)_/a_mrr_.ow keys alternately to

understand the oscillations of the spring!

The spring force F in an ideal spring is directly proportional to the

displacement of^Pt^rh^ee^ssbl^Do^Ock^Wf^Nro⁽m^↓)t^ah^re^roe^wqu^ki^eli^ybr^si ^um^ccp^eso^sii^vti^eo^ln^y.^{to stop the}

oscillations!

F_s = – kx

Suppose the block is pushed by an external force F and the spring is

compressed by x_c.

Then, the work done by the spring force F_s is

x_c x_c

^F_s ^{is n}p^eo^gs^ait^tiⁱv^ve^e

^ec

^{x is p}n^oeg^sa^ittⁱi^vv^ee

Note that the work done by the external pushing force F is positive since it

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overcomes the spring force F_s.

The work done by the external force F is ^W_s = + ½ kx_c

2

​

​

The same is true when the spring is extended through a displacement of x_e.

Then, the work done by the spring force F_s is ^W_s

= – ½ kx_e

2

W_s

= + ½ kx_e

2

The work done by the external force F is

F

O

x

x_e

B

A

​

​

The work done is given by the area of triangle OAB from the graph.

Note:

1. Potential energy is path independent.

i.e. it depends on the net vertical displacement (height) of the body but not on the path through which it is raised.

2. Potential energy in a round trip (i.e. over a closed path) is zero. PE gained by the body = + mgh

PE lost by the body Total PE in round trip

= - mgh

= + mgh – mgh = 0

Therefore, gravitational force is a conservative force.

h

g

m ^m

h

g

m

m

m

PE = mgh

PE = mgh – mgh = 0

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3. If h is taken as variable, then the gravitational force F equals the negative

of the derivative of V(h) w.r.t. h. Thus,

dh

F = – ^d V(h)

dh

F = – ^d (mgh)

F = – mg

​

The negative sign indicates that the gravitational force is downward.

​

4. If the body with V(h) = mgh is released, the PE is converted into K = ½ mv².

v² = 2gh

Multiplying both the sides by m/2, we have

½ mv² = mgh

5. The potential energy V(x) is defined if the force F(x) can be written as

F(x) = – ^dV

dx

This implies that

x_i V_i

x_f V_f

∫ F(x) dx _{= –} ∫ dV _{= Vi – Vf}

LAW OF CONSERVATION OF ENERGY

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Law of conservation of energy states that energy can neither be created nor

destroyed.

Whenever energy changes from one form into another, the total amount of

energy remains constant.

​

When a body is thrown upwards or dropped from height, the total mechanical energy (i.e. sum of potential and kinetic energy) at each and every point on its path remains constant.

Note that at the highest point of its motion, the energy is fully in the form of potential energy and at the lowest point, the energy is fully in the form of kinetic energy.

At all other points, the energy is partially

potential and partially kinetic.

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Potential energy is maximum and kinetic energy is zero at extreme positions.

​

Kinetic energy is maximum and potential energy is minimum at the mean position.

Conservation of Mechanical Energy

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Suppose that a body undergoes displacement Δx under the action of a

conservative force F.

From work – energy theorem, we have

ΔK = F(x) Δx

If the force is conservative, the potential energy function V(x) can be defined

such that

– ΔV = F(x) Δx

The above equations imply that

ΔK + ΔV = 0

Δ(K + V) = 0

which means that K + V, the sum of the kinetic and potential energies of the

body is a constant.

Over the whole path, x_i to x_f, this means that

K_i + V(x_i) = K_f + V(x_f)

The quantity K + V(x), is called the total mechanical energy of the system. Individually the kinetic energy K and the potential energy V(x) may vary from point to point, but the sum is a constant.

The total mechanical energy of a system is conserved if the forces, doing

work on it, are conservative.

Conservative Force

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• A force F(x) is conservative if it can be derived from a scalar quantity V(x)

by the relation ΔV = – F(x) Δx.

(The 3-dimensional generalisation requires the use of a vector derivative.)

• The work done by the conservative force depends only on the end points. This can be seen from the relation, W = K_f – K_i = V(x_i) – V(x_f), which depends on the end points.
• The work done by this force in a closed path is zero. This is once again

apparent from W = K_f – K_i = V(x_i) – V(x_f) since x_i = x_f.

​

​

Non - Conservative Force

• The work done by the non-conservative force depends on the path in which it is done.
• The work done by this force in a closed path is non - zero.
• Friction is an example for non-conservative force.

The heat energy produced by friction gets dissipated and hence can not be converted back into original form of energy.

Conservation of Mechanical Energy by Gravitational Force

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Suppose a ball of mass ‘m’ dropped from a height H.

Let the velocity of the ball at the height H (position A) be v(H) = 0, v(h) at the height h (position B) and v(0) at the ground level (position C).

h

H-h

B

v(h)

Mechanical Energy at A:

The ball at A is at rest and at the height H from the ground

level (or reference level).

Therefore, the ME possessed by the ball is wholly by

virtue of its position and hence only potential energy.

PE_A = mgH KE_A = 0

ME_A = PE_A + KE_A

ME_A = mgH

H

A

v(H) = 0

v(0) _C

m

m v(H) = 0

Mechanical Energy at B:

The ball at B has fallen through the height (H-h) and

gained velocity v(h).

Therefore, the ME possessed by the ball is partially

potential and partially kinetic.

PE_B = mgh

KE_B = ½ m [ v(h)² ]

But, v(h)² = 2g (H – h)

KE_B = mgH – mgh

ME_B = mgh + mgH – mgh

or ME_B = mgH

H

H-h

A

B

v(h)

h

C

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m v(H) = 0

Mechanical Energy at C:

The ball at C has fallen through the height H and gained

velocity v(0) (maximum velocity).

Therefore, the ME possessed by the ball is wholly by

virtue of its motion.

But,

PE_C = 0

KE_C = ½ m [ v(0)² ]

v(0)² = 2gH

KE_C = mgH

ME_C = PE_C + KE_c ME_C = mgH

The Mechanical Energy of the ball at all the positions A, B and C is the same and equal to mgH which indicates that the Mechanical Energy is conserved.

H

A

v(0)

_C

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Graphical representation of variation of PE and KE w.r.t. ‘h’ and

conservation of total ME of a body under vertical motion:

Energy Energy

O

H/2

h

E = K + V E = K + V

^H H

^{Height h} H/2 ^Height

When an object is dropped from a height, the PE decreases from its maximum value to zero and its KE increases from zero to its maximum value simultaneously.

i.e. one type of ME is converted into another type of ME.

At the height H/2, the mechanical energy is half potential and half kinetic. E = K + V remains constant at any point of its motion, i.e. at any height.

When an object is projected vertically upwards, the KE decreases from its

maximum value to zero and its PE increases from zero to its maximum value

simultaneously.

m

_Om

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Conservation of Mechanical Energy by Spring Force

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x

i

W_s = – ∫ kx dx

= ½ kx ² – ½ kx ²

i f

Thus the work done by the spring force depends only on the end points. If the block is pulled from x_i and allowed to return to x_i, then

^W_s = ½ kx_i – ½ kx_i = 0

2 2

​

Since the work done in round trip is zero, the spring force is conservative.

F_s F^F_s F

m m m ^x_c x_e

​

x = 0

​

If the block is moved from initial displacement x_i to final displacement x_f,

the work done by the spring force F_s is

x_f

We define the potential energy V(x) = 0 for a spring when the system of

spring and block is at equilibrium position.

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For extension x_e or compression x_c

V(x) = ½ kx²

The above equation easily verifies that – dV/dx = – kx which is the

spring force.

If the block of mass m is extended to x_e and released from rest, then its total mechanical energy at any arbitrary point x is the sum of its potential energy V(x) and kinetic energy K.

​

½ kx_e = ½ kx + ½ mv

2 2 2

​

The above equation suggests that the speed and the kinetic energy will

be maximum at the equilibrium position x = 0.

i.e.,

e

½ kx ² = ½ mv

m

2

e

where x is the maximum displacement and v

m

is the

maximum speed as given below.

v_m =

k

m

x_e

Graphical representation of variation of PE and KE w.r.t. ‘x’ and conservation of total ME of an oscillating spring:

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Energy

E = K + V

O

– x_e x_e x

The plots of PE and KE are parabolic.

They are complementary. i.e. one increases at the expense of the other. The total ME of the system remains constant at any arbitrary point

between – x_e and x_e.

COLLISIONS

some of he initial KE is lost.

Elastic Collision

In elastic collision,

1. The total linear momentum is conserved,
2. The total kinetic energy of the system is also conserved and
3. After the collision, the bodies completely regain from their deformities.

Completely Inelastic Collision

In completely inelastic collision,

1. The total linear momentum is conserved,
2. The total kinetic energy of the system is not conserved and
3. After the collision, the bodies stick together and move with common velocity.

Inelastic Collision

In inelastic collision,

1. The total linear momentum is conserved,
2. The total kinetic energy of the system is not conserved and
3. After the collision, the bodies partly regain from their deformities and

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Note:

1. Total momentum of an isolated system is always conserved, whether

the collision is elastic or inelastic.

2. But, the total KE is conserved only in elastic and not in inelastic

collision.

In inelastic collision, the loss of KE appears in any other form of energy (usually heat energy).

3. No collision is perfectly elastic nor completely inelastic. Most of the collisions are inelastic.

Elastic Collision in One Dimension

Consider two bodies A and B of mass m₁ and m₂ moving with velocities u₁ and u₂ respectively. Let the bodies collide, get apart and move with final velocities v₁ and v₂ respectively along the same direction as that before collision.

Let us consider the scalar treatment of the quantities for simplicity and because the motion is in one dimension (in the same direction) and energy is a scalar concept.

Since momentum is conserved,

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Click to see the collision…

u₁

u₂

v₁

v₂

^F21 ^F12

m₁

m₂

m₁

m₂

m

1

m

2

m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂ m₂(v₂ – u₂) = m₁(v₁ – u₁)

……….(1)

……….(2)

or

In elastic collision, kinetic energy is also conserved.

1 1 2 2 1 1

2

2

1

1

1

½ m v ² + ½ m v ² = ½ m u ² + ½ m u ²

m₂(v ² – u ²) = m (v ² – u ²)

2 2

……….(3)

or

If the initial and final velocities of both the bodies are along the same straight

line, then the collision is called ‘one-dimensional’ or ‘head-on’ collision.

Dividing eqn. (3) by eqn. (2),

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or

v₂ + u₂ = v₁ + u₁ ……….(4)

v₂ = v₁ + u₁ – u₂ ……….(5)

(to eliminate v₂ to get v₁)

Substituting for v₂ from eqn. (5) in eqn. (1),

m₁v₁ + m₂ (v₁ + u₁ – u₂) = m₁u₁ + m₂u₂

v₁ (m₁ + m₂) = u₁ (m₁ – m₂) + 2 m₂u₂

​

u₁ (m₁ – m₂) + 2 m₂u₂

v₁ =

(m₁ + m₂)

……….(6)

1

v =

(m₁ + m₂)

(m₁ – m₂)

1

u +

2 m₂

u

2

(m₁ + m₂)

or

v₂ + u₂ = v₁ + u₁ ……….(4)

v₁ = v₂ + u₂ – u₁ ……….(7)

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Substituting for v₁ from eqn. (7) in eqn. (1),

m₁ (v₂ + u₂ – u₁) + m₂ v₂ = m₁u₁ + m₂u₂

​

v₂ (m₁ + m₂) = u₂ (m₂ – m₁) + 2 m₁u₁

​

2 m₁u₁ + u₂ (m₂ – m₁)

v₂ =

(m₁ + m₂)

Again,

or

(to eliminate v₁ to get v₂)

Eqns. (6) and (8) give the final velocities after collision in terms of their

masses and initial velocities.

2

or v =

(m₁ + m₂)

(m₂ – m₁)

u

2

2 m₁

1

u +

(m₁ + m₂)

……….(8)

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2

v =

1 2

(m₂ – m₁)

u

2

2 m₁

1

u +

^(m₁ ^{+ m}₂⁾ (m + m )

……….(8)

……….(6)

1

v =

(m₁ + m₂)

(m₁ – m₂)

1

u +

2 m₂

u

2

(m₁ + m₂)

Special Cases:

Case I:

If the two masses are equal, i.e. m₁ = m₂ = m (say), then

v₁ = u₂ and v₂ = u₁

It is clear that the bodies interchange their velocities after the collision.

Sub - Case:

If the second body was at rest before collision i.e. u₂ = 0, then v₁ = 0 and v₂ = u₁

The first body comes to rest and pushes the second body with its (first)

initial speed.

Next time when you play carom….. just enjoy Physics!

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2

v =

(m₁ + m₂)

(m₂ – m₁)

u

2

2 m₁

1

u +

(m₁ + m₂)

……….(8)

……….(6)

1

v =

(m₁ + m₂)

(m₁ – m₂)

1

u +

2 m₂

u

2

(m₁ + m₂)

Case II:

If m₁ >> m₂, i.e. m₂ ≈ 0, then

v₁ ≈ u₁

and v₂ ≈ 2u₁ – u₂

The first body continues almost undisturbed and the second body flies

off with greater velocity.

Sub - Case:

If the second body was at rest before collision i.e. u₂ = 0, then

v₁ ≈ u₁

and v₂ ≈ 2u₁

Try to play with heavier striker and observe!

Or Imagine a heavy truck hitting a cyclist!

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2

v =

(m₁ + m₂)

(m₂ – m₁)

u

2

2 m₁

1

u +

(m₁ + m₂)

……….(8)

……….(6)

1

v =

(m₁ + m₂)

(m₁ – m₂)

1

u +

2 m₂

u

2

(m₁ + m₂)

Case III:

If m₂ >> m₁, i.e. m₁ ≈ 0, then

v₁ ≈ – u₁ + 2u₂

and v₂ ≈ u₂

The second body continues almost undisturbed.

Sub - Case:

If the second body was at rest before collision i.e. u₂ = 0, then

v₁ ≈ – u₁

and v₂ ≈ 0

The second body continues to be at rest and the first body reverses its direction with its initial speed.

Imagine a cyclist bumping onto a road-roller!

Completely Inelastic Collision in One Dimension

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Click to see the collision…

Consider two bodies A and B of mass m₁ and m₂ moving with velocities u₁ and u₂ respectively. Let the bodies collide, stick together and move with common final velocity v along the same direction as that before collision.

Let us consider the scalar treatment of the quantities for simplicity and because the motion is in one dimension (in the same direction) and energy is a scalar concept.

……….(1)

Since momentum is conserved,

(m₁ + m₂)v = m₁u₁ + m₂u₂

u

1

m₁

u₂

m

2

m₁

m

2

v

m₁u₁ + m₂u₂

or v =

(m₁ + m₂)

In inelastic collision, kinetic energy is not conserved.

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Total kinetic energy before collision = ½ m₁u₁ + ½ m₂u₂

2 2

​

Total kinetic energy after collision = ½ (m₁ + m₂)v²

Loss in kinetic energy on collision is

ΔK = ½ m₁u ² + ½ m u ² – ½ (m + m )v²

1 2 2 1 2

= ½ m₁u ² + ½ m u ² – ½ (m + m )

1 2 2 1 2

m₁u₁ + m₂u₂ (m₁ + m₂)

2

or

ΔK = ½ m₁u ² + ½ m u ^{2 – ½}

1 2 2

m ²u ² + m ²u ² + 2 m m u u

1 1 2 2 1 2 1 2

(m₁ + m₂)

If u₂ = 0, then ΔK = ½ m₁u ^{2 – ½}

1

1 1

_m 2_u 2

(m₁ + m₂)

On simplification,

which is a positive quantity as

expected.

ΔK = ½

m m

1 2

(m₁ + m₂)

1

_u 2

Elastic Collision in Two Dimensions

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Click to see the collision…

u₁

m₁

m₂

m₂

v₁

m₁

θ₁

θ₂

x

m₁v₁ cos θ₁

m₂v₂ cos θ₂

y

m₁v₁ sin θ₁

m₂v₂ sin θ₂

v₂

​

Consider a body A of mass m₁ moving with velocity u₁ collides elastically with another body B of mass m₂ at rest. Let the bodies, after collision, move with velocities v₁ and v₂ respectively along the directions as shown in the figure.

The collision takes place in two – dimensions, say in x-y plane.

​

Conservation laws have to be applied along x and y axes separately.

Along x – axis:

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m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂

Along y – axis:

0 = m₁v₁ sin θ₁ – m₂v₂ sin θ₂

Kinetic energy is a scalar and hence, the law of conservation of energy is

given by

​

½ m₁u ² = ½ m v ² + ½ m v ² 1 1 1 2 2

​

We have four unknown quantities v₁, v₂, θ₁ and θ₂ and only 3 equations. Therefore, atleast one more quantity must be known to solve the mathematical problems.

POWER

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Average Power = =

Time taken Time taken

Power is defined as the time rate of doing work or consuming energy. or

Power is defined as the rate of conversion of one form of energy into

another form of energy.

Work done Energy consumed

t

av

P =

W _E

=

t

The instantaneous power is defined as the limiting value of the

average power as time interval approaches zero.

_{P =} dW

dW = F . dr

dt

The work done by a force F for a displacement dr is

The instantaneous power can also be expressed as

P = F.

dr

dt

or

P = F . v

1. Power is a scalar quantity.

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Note:

2. SI unit of power is ‘watt’.
3. 1 watt = 1 joule per second
4. 1 watt is the power when 1 joule of work is done in 1 second or 1 watt is

the power when 1 joule of energy is consumed in 1 second.

5. 1 kilowatt = 1000 watt or 1 kW = 1000 W
6. 1 megawatt = 1,000,000 watt or 1 MW = 10⁶ W
7. Another unit of power is called ‘horse power’ or ‘hp’ 8. 1 hp = 746 W

9. The power of engines of cars and other vehicles is measured by unit

called ‘brake horse power’ which is equal to 1 horse power.

COMMERCIAL UNIT OF ENERGY

The commercial unit or trade unit of energy is kilowatt-hour (kWh).

1 kWh is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour.

1 kilowatt-hour = 1000 watt x 3600 seconds = 3,600,000 Ws = 3.6 x 10⁶ Joule

Acknowledgement

1. Physics Part I for Class XI by NCERT

]

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