NotesShala Class 11 · Physics

Chapter05

Work, Energy and Power

Work · Kinetic Energy & Work-Energy Theorem · Potential Energy & Conservation of Energy · Power & Collisions

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Work

In daily life, any physical or mental effort is called work. In physics, the word work has an exact meaning. Work is done only when a force moves a body. Energy is the ability of a body to do work. Power tells us how fast work is done. These three ideas are linked by the work-energy theorem, the conservation of energy and collisions.

1.1Work

Definition

Work is done by a force when it moves a body some distance along its own direction.

So a force does work only when two conditions are met:

iCondition

A force acts on the body.

iiCondition

The point where the force acts moves along the force.

Everyday examples of work

An engine pulls a train on the railsWork is done ✓

A man climbs a hillWork is done ✓

A horse pulls a cart along a roadWork is done ✓

1.2Work Done by a Constant Force

Definition

The work done by a constant force = (component of the force along the displacement) × (magnitude of the displacement).

Let a constant force $\vec{F}$ act on a body. The body moves through a displacement $\vec{s}$ in the same direction as the force. Then the work done is simply the force multiplied by the displacement.

Force Along the Displacement

W = \vec{F}\cdot\vec{s} = Fs

where, $F$ = magnitude of the constant force, $s$ = magnitude of the displacement

Force

F

moves the block through a displacement

s

along the force

Note

Work has magnitude only. It has no direction. So work is a scalar quantity. We get it from the dot (scalar) product of $\vec{F}$ and $\vec{s}$ .

1.3Work Done by an Inclined Force

Often the force acts at an angle $\theta$ to the displacement $\vec{s}$ . Then we resolve $\vec{F}$ into two rectangular components.

Along

\vec{s}

F_x = F\cos\theta

This part points along the displacement. So all the work is done by it.

Perpendicular to

\vec{s}

F_y = F\sin\theta

This part is at 90° to the displacement. So its work is zero.

Force at an angle

\theta

to the displacement

Only $F\cos\theta$ does work. So the work done is $F\cos\theta$ multiplied by $s$ :

Work Done by an Inclined Force

W = \vec{F}\cdot\vec{s} = Fs\cos\theta

$W$ is the dot (scalar) product of $\vec{F}$ and $\vec{s}$ , so work is a scalar.

\theta = 0^\circ

· Same Direction

$W = Fs\cos 0^\circ = Fs$

The work done is maximum.

\theta = 90^\circ

· Perpendicular

$W = Fs\cos 90^\circ = 0$

The force does zero work.

\theta = 180^\circ

· Opposite

$W = Fs\cos 180^\circ = -Fs$

The work done is negative.

Note

If the displacement is perpendicular to the force, the work done is zero. The type of motion does not matter. The body may move with constant velocity, constant acceleration, and so on. e.g. The centripetal force in circular motion does no work.

1.4Dimensions and Units of Work

Dimensional Formula of Work

W = \text{force}\times\text{distance} = [\text{MLT}^{-2}]\times[\text{L}] = [\text{ML}^2\text{T}^{-2}]

Work has two types of units: absolute units and gravitational units.

1Absolute Units

One absolute unit of work is done when a unit force moves a body through a unit distance along the force.

Joule (SI): $1\ \text{J} = 1\ \text{N} \times 1\ \text{m} = 1\ \text{N-m}$
Erg (CGS): $1\ \text{erg} = 1\ \text{dyne}\times 1\ \text{cm} = 1\ \text{dyne-cm}$

2Gravitational Units

One gravitational unit of work is done when a gravitational unit of force moves a body through a unit distance along the force.

kg-m (SI): $1\ \text{kg-m} = 9.8\ \text{J}$
g-cm (CGS): $1\ \text{g-cm} = 981\ \text{erg}$

1\ \text{J} = 10^{7}\ \text{erg}

1\ \text{kg-m} = 9.8\ \text{J}

1\ \text{g-cm} = 981\ \text{erg}

1.5Positive, Negative and Zero Work

We know that $W = Fs\cos\theta$ . So the sign of the work depends on the angle $\theta$ between the force and the displacement. The work done may be positive, negative or zero.

1Positive Work ·

0 \le \theta < 90^\circ

The force has a part along the displacement. So $\cos\theta$ is positive, and the work is positive.

Horse pulling a cart: the force and the motion are in the same direction.
Lifting a load: the lifting force and the displacement both point up.
Stretching a spring: the stretching force acts along the displacement.

2Negative Work ·

90^\circ < \theta \le 180^\circ

The force has a part opposite to the displacement. So $\cos\theta$ is negative, and the work is negative.

Brakes on a moving vehicle: the braking force acts against the motion.
Friction when a body is pulled along a rough floor.
Gravity when a body is raised upward.

3Zero Work ·

\theta = 90^\circ

The body moves at right angles to the force. Since $\cos 90^\circ = 0$ , the work is $W = Fs\cos 90^\circ = 0$ .

A coolie walks on level ground with a load on his head.
The centripetal force when a body moves in a circle at constant speed.
The weight of a body sliding on a smooth (frictionless) level floor.

Nature of Work	Angle $\theta$	Sign of $\cos\theta$	Work Done
Positive	$0 \le \theta < 90^\circ$	Positive (+)	$W = Fs\cos\theta$ is positive
Negative	$90^\circ < \theta \le 180^\circ$	Negative (−)	$W = Fs\cos\theta$ is negative
Zero	$\theta = 90^\circ$	Zero (0)	$W = 0$

1.6Work Done by a Variable Force

Sometimes the force is not constant. It changes with position, say along the $x$ -axis. Then $W = Fs$ cannot be used. So we break the path into very small displacements $\Delta x$ . Over each small step, the force is almost constant. Then we add the work of all the steps.

Derivation

Over a tiny step $\Delta x$ , the force $F(x)$ hardly changes. The work in this step equals the area of the thin strip $abcd$ .
$\Delta W = F(x)\,\Delta x = \text{area of strip}$
Add the areas of all the strips from $x_i$ to $x_f$ . This gives the total work approximately.
$W \approx \sum\limits_{x_i}^{x_f} F(x)\,\Delta x$
Now let $\Delta x \to 0$ . The sum becomes an integral. It gives the exact area $PQRS$ under the curve.
$W = \lim\limits_{\Delta x\to 0}\sum F(x)\,\Delta x = \displaystyle\int_{x_i}^{x_f} F(x)\,dx$

(a) Work ≈ total area of the thin strips · (b) Exact work = shaded area

PQRS

Work Done by a Variable Force

W_{x_i \to x_f} = \int_{x_i}^{x_f} F\,dx = \int_{x_i}^{x_f} (F\cos\theta)\,dx = \textbf{Area under the } F\text{–}x \textbf{ curve}

Force in Three Dimensions

When both the force and the displacement change in 3-D, split each one into rectangular components:

\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}, \qquad d\vec{s} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}

Then the work done is:

W = \displaystyle\int_{x_i}^{x_f} F_x\,dx + \int_{y_i}^{y_f} F_y\,dy + \int_{z_i}^{z_f} F_z\,dz

Note

A body in equilibrium (static or dynamic) has zero resultant force. So no net work is done on it. The total work is zero.

1.7Conservative Forces

Definition

A conservative force is a force whose work is fixed by the initial and final positions alone. Its work does not depend on the path taken between them.

Examples

Gravitational force: its work depends only on the change in heightConservative ✓

Spring (elastic) force: its work depends only on the stretch or compressionConservative ✓

Properties

iEnd Points Only

The work done by or against the force is decided by the starting and ending positions alone.

iiPath Independent

The work does not depend on the path that the body follows between the two positions.

iiiClosed Path

For a closed path (round trip), the total work done is zero.

Take gravity as an example. Gravity does work $W_{PQ}$ from $P$ to $Q$ , and $W_{QP}$ from $Q$ to $P$ . These two are equal in size but opposite in sign. So for the round trip $P \to Q \to P$ :

Net Work over a Closed Path

W_{PQ} = -W_{QP} \quad\Rightarrow\quad W_{PQ} + W_{QP} = 0 \quad\Rightarrow\quad \oint \vec{F}\cdot d\vec{s} = 0

1.8Non-conservative Forces

Definition

A non-conservative force is a force whose work depends on the path followed between two positions.

Move a body from $A$ to $B$ along three different paths. Let the work done be $W_1$ , $W_2$ and $W_3$ . For a non-conservative force, these values are not equal:

W_1 \ne W_2 \ne W_3

So for a closed path $A \to B \to A$ , the total work is not zero. The lost work turns into heat.

Different paths give different work (path dependent)

Some examples

1Viscous force 2Tension 3Force of friction

Work Done by Friction

Friction always opposes the relative motion between surfaces. Its work can be negative, zero or even positive.

F < f_s

The push is less than the static friction. So the block stays at rest ( $s = 0$ ).

W = 0

F > f_s

The block moves. Kinetic friction acts against the displacement. So it does negative work.

W = -\mu_k Mg\,s

IIIBlock on Block

Block $A$ rests on block $B$ , and $B$ is pulled. Friction does positive work on $A$ and negative work on $B$ .

W_A = +fs,\ \ W_B = -fs

Remember

Conservative vs Non-conservative

Round a closed path, a conservative force does zero net work ( $\oint \vec{F}\cdot d\vec{s} = 0$ ). Then mechanical energy is conserved. A non-conservative force does non-zero work round the same path. Then mechanical energy is not conserved. Most of the lost energy becomes heat.

Part 2

Kinetic Energy & Work-Energy Theorem

A moving body can do work, so it has energy. The energy of motion is called kinetic energy. The work-energy theorem connects the work done on a body with the change in its kinetic energy.

2.1Energy

Definition

Energy is the capacity (ability) of a body to do work.

Nature

Scalar quantity, like work. It has magnitude only, no direction.

Units

joule (SI) and erg (CGS).

Dimensions

$[\text{ML}^2\text{T}^{-2}]$ , the same as work.

Practical Units of Energy

Practical Unit of Energy	Symbol	Value in SI
Erg	erg	$10^{-7}\ \text{J}$
Electron volt	eV	$1.6 \times 10^{-19}\ \text{J}$
Calorie	cal	$4.186\ \text{J}$
Kilowatt hour	kWh	$3.6 \times 10^{6}\ \text{J}$

2.2Kinetic Energy

Definition

Kinetic energy is the energy that a body has because of its motion. It is equal to the work that the moving body can do before it comes to rest.

Kinetic Energy

\text{KE} = \frac{1}{2}mv^2

where, $m$ = mass of the body, $v$ = velocity of the body

Everyday examples

A bullet from a gun goes through a target.
A fast stream of water turns a water mill.
Moving air (wind) turns a windmill.
A hammer in motion pushes a nail into wood.

Never Negative

A moving body always has positive KE. KE can never be negative.

Depends on the Frame

The value of KE changes with the frame of reference. e.g. A person sitting in a moving train has KE when seen from the ground. Seen from the train, the person's KE is zero.

2.3Measurement of Kinetic Energy

Take a body of mass $m$ that is at rest. A force $F$ acts on it and moves it along the direction of the force. The body speeds up and reaches a velocity $v$ .

All the work done by the force gets stored in the body. This stored energy is its kinetic energy.

Force

F

moves the body from

A

(at rest) to

B

(velocity

v

)

Method 1 · Using Calculus

Derivation

Let the force move the body through a small displacement $ds$ along the force ( $\theta = 0^\circ$ ). The small work done is:
$dW = \vec{F}\cdot d\vec{s} = F\,ds\cos 0^\circ = F\,ds$
By Newton's second law, the force gives the body an acceleration $a$ . So $F = ma$ :
$dW = ma\,ds$
Write the acceleration as $a = \dfrac{dv}{dt}$ :
$dW = m\,\dfrac{dv}{dt}\,ds$
Rearrange the terms. Then use $\dfrac{ds}{dt} = v$ , the velocity of the body:
$dW = m\,ds\,\dfrac{dv}{dt} = m\,\dfrac{ds}{dt}\,dv = mv\,dv$
Integrate from rest ( $0$ ) to the final velocity $v$ . This gives the total work:
$W = \displaystyle\int_0^v mv\,dv = m\left[\dfrac{v^2}{2}\right]_0^v = \dfrac{1}{2}mv^2$

Result

This work becomes the kinetic energy of the body:

\text{KE} = \dfrac{1}{2}mv^2

So the kinetic energy of a body is half of its mass multiplied by the square of its velocity.

Method 2 · Using an Equation of Motion

For a constant force, we get the same result without calculus.

Derivation

A constant force moves the body through a distance $s$ . The work done is:
$W = Fs = ma\,s$
The body starts from rest ( $u = 0$ ) and reaches the speed $v$ . From $v^2 = u^2 + 2as$ :
$v^2 = 2as \;\Rightarrow\; as = \dfrac{v^2}{2}$
Put $as = \dfrac{v^2}{2}$ in $W = m(as)$ . This gives the kinetic energy:
$W = m(as) = m\cdot\dfrac{v^2}{2} = \dfrac{1}{2}mv^2$

Note

How KE Depends on m and v

From $\text{KE} = \tfrac{1}{2}mv^2$ , we get $\text{KE} \propto m$ and $\text{KE} \propto v^2$ . So a heavier body or a faster body has more kinetic energy. KE depends on the square of the speed:

Speed is doubledKE becomes 4 times

Speed is tripledKE becomes 9 times

2.4Kinetic Energy and Momentum

Kinetic energy and linear momentum both depend on mass and velocity. So we can link them by removing $v$ .

Derivation

Linear momentum is $p = mv$ , so $v = \dfrac{p}{m}$ . Put this in the formula for kinetic energy:
$\text{KE} = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dfrac{p}{m}\right)^2 = \dfrac{p^2}{2m}$
Rearrange this to get momentum in terms of kinetic energy:
$p^2 = 2m\,\text{KE} \;\Rightarrow\; p = \sqrt{2m\,\text{KE}}$

KE from Momentum

\text{KE} = \frac{p^2}{2m}

Momentum from KE

p = \sqrt{2m\,\text{KE}}

This relation gives three special cases. Each case has its own graph.

p

is constant

$\text{KE} \propto \dfrac{1}{m}$ . For the same momentum, a lighter body has more KE.

iiKE is constant

$p^2 \propto m$ , so $p \propto \sqrt{m}$ . The momentum grows as $\sqrt{m}$ .

iii

m

is constant

$p^2 \propto \text{KE}$ , so $p \propto \sqrt{\text{KE}}$ . The KE grows as $p^2$ .

2.5Work-Energy Theorem (Constant Force)

Law

Work-Energy Theorem

The work done by the net force on a body equals the change in its kinetic energy. This is called the work-energy theorem or the work-energy principle.

Let a constant force $F$ give a body an acceleration $a$ . Over a distance $s$ , its velocity changes from $u$ to $v$ .

Derivation

Use the equation of motion that links velocity and distance:
$v^2 - u^2 = 2as$
Multiply both sides by $\dfrac{1}{2}m$ :
$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = mas$
By Newton's second law, $F = ma$ . So $mas = Fs = W$ :
$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = Fs = W$

Work-Energy Theorem

W_{net} = K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \Delta\text{KE}

where, $K_f$ = final kinetic energy, $K_i$ = initial kinetic energy of the body

Positive net workKE increases · body speeds up

Negative net workKE decreases · body slows down

Zero net work (

W_{net} = 0

)Speed does not change

2.6A Block over Another Block

Block $A$ rests on block $B$ , and $B$ is accelerating. We must use the work-energy theorem carefully here. The example also proves a useful point. The theorem is true in every frame of reference, for every kind of force.

Block $B$ accelerates to the right with respect to the ground. Block $A$ sits on it. The weight ( $w$ ) and the normal reaction ( $N$ ) on $A$ are perpendicular to its displacement. So both do zero work.

Only friction can do work on $A$ . In the frame of $B$ , we must also count the work of the pseudo force.

Block

A

on block

B

, which accelerates to the right

iGround Frame (Inertial)

No pseudo force acts here. Only friction does work on $A$ :

W_{\text{friction}} = \Delta\text{KE}

$A$ speeds up along with $B$ . So the work done by friction on $A$ is positive.

iiFrame of B (Non-inertial)

Here the pseudo force also acts on $A$ . Apply the theorem to $A$ :

W_{\text{friction}} + W_{\text{pseudo}} = \Delta\text{KE}

$A$ is at rest relative to $B$ . So $\Delta\text{KE} = 0$ , which gives $W_{\text{pseudo}} = -\,W_{\text{friction}}$ .

Work-Energy Theorem in Any Frame

W_{\text{external}} + W_{\text{internal}} + W_{\text{pseudo}} + W_{\text{other}} = \Delta\text{KE}

The pseudo-force term is used only in a non-inertial frame. In an inertial frame, $W_{\text{pseudo}} = 0$ .

Note

Key Point

The theorem is true for every system of particles. The forces may be conservative or non-conservative. When KE rises, the net work is positive. When KE falls, the net work is negative.

2.7Work-Energy Theorem (Variable Force)

The theorem also holds when the force changes. We use very small quantities and then integrate along the path.

Derivation

A variable force $F$ acts on a body of mass $m$ . It moves the body through a small displacement $ds$ along the force ( $\theta = 0^\circ$ ). The small work done is:
$dW = \vec{F}\cdot d\vec{s} = F\,ds\cos 0^\circ = F\,ds$
The kinetic energy is $K = \dfrac{1}{2}mv^2$ . Differentiate it with respect to time, keeping $m$ constant. Use the chain rule on $v^2$ :
$\dfrac{dK}{dt} = \dfrac{d}{dt}\!\left(\dfrac{1}{2}mv^2\right) = \dfrac{1}{2}m\cdot 2v\cdot\dfrac{dv}{dt} = mv\,\dfrac{dv}{dt}$
By Newton's second law, $m\dfrac{dv}{dt} = ma = F$ . So:
$\dfrac{dK}{dt} = \left(m\dfrac{dv}{dt}\right)v = (ma)\,v = Fv$
Put $v = \dfrac{ds}{dt}$ . Now KE changes at the same rate as the work is done. So for a small step, $dK = dW$ :
$\dfrac{dK}{dt} = F\dfrac{ds}{dt} \;\Rightarrow\; dK = F\,ds = dW$
Integrate from the initial state (position $x_i$ , KE $K_i$ ) to the final state (position $x_f$ , KE $K_f$ ):
$\displaystyle\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F\,ds \;\Rightarrow\; K_f - K_i = \int_{x_i}^{x_f} F\,ds = W$
This proves the work-energy theorem for a variable force.

Work-Energy Theorem (Variable Force)

K_f - K_i = \int_{x_i}^{x_f} \vec{F}\cdot d\vec{s} = W

The work done by the net (variable) force equals the change in KE

Note

The work-energy theorem is not a separate law. We get it from Newton's second law.
Newton's second law in 2-D or 3-D is a vector equation. The work-energy theorem is a scalar equation.
It can be used for a system of many particles too. It works whether the forces are conservative or not.

Part 3

Potential Energy & Conservation of Energy

A body can also store energy because of its position or its shape. This stored energy is called potential energy. When only conservative forces act, the sum of kinetic and potential energy stays constant.

3.1Potential Energy

Definition

The energy that a body has because of its position or configuration inside a field is called its potential energy. A change in configuration changes the potential energy too.

iDue to Position

An object on the roof of a building has potential energy. If it falls, it can do work.

iiDue to Configuration

A stretched or compressed spring has potential energy because its shape is changed.

Units and dimensions (same as kinetic energy)joule (J) ·

[\text{ML}^2\text{T}^{-2}]

Note

Potential Energy Changes into Kinetic Energy

Potential energy is defined only for forces where the work done against the force is stored as energy. The energy is stored due to the body's position or configuration. When the body is set free, the stored energy changes back into kinetic energy. Its position or configuration changes at the same time.

3.2Gravitational Potential Energy

Definition

Gravitational potential energy is the energy that a body has because of its height above the surface of the Earth.

Derivation

Lift a body of mass $m$ at constant velocity. There is no acceleration, so the lifting force balances gravity:
$F = w = mg$
Lift it through a height $h$ . Take $h \ll R_E$ , so $g$ stays constant. The work done is stored as potential energy:
$W = Fh = mgh = U$

Lifted through height

h

Gravitational Potential Energy

U = mgh

where, $m$ = mass, $g$ = acceleration due to gravity, $h$ = height measured from the reference level we choose

Force from PE

The force of gravity is the negative gradient of $U$ :

F = -\dfrac{dU}{dh} = -mg

The minus sign shows that the force acts downward.

PE Changes into KE on Falling

Drop the body from rest at height $h$ . It reaches the ground with speed $v = \sqrt{2gh}$ :

\dfrac{1}{2}mv^2 = \dfrac{1}{2}m(2gh) = mgh

All the potential energy changes into kinetic energy.

3.3Potential Energy of a Spring

Definition

The energy stored in a compressed or stretched elastic spring is called its potential energy. Such a spring is usually helical or spiral. Take the spring as massless. Then all the work done to deform it is kept as elastic potential energy.

Take a massless elastic spring. One end is fixed to a rigid support. The other end is attached to a block of mass $m$ on a smooth horizontal surface. Take the equilibrium position as $x = 0$ . The spring can be in three states:

aEquilibrium

$x = 0$ . The spring is relaxed, so $F_s = 0$ .

bStretched

$x > 0$ . The restoring force pulls the block back. $F_s = -kx$ is negative.

cCompressed

$x < 0$ . The restoring force pushes the block out. $F_s = -kx$ is positive.

(a) Equilibrium, (b) stretched and (c) compressed spring

Hooke's Law · Spring (Restoring) Force

F_s = -kx

where, $k$ = spring constant (SI unit $\text{N m}^{-1}$ ). The minus sign shows that the force acts opposite to the displacement $x$ .

Derivation

In a tiny displacement $dx$ , the spring force does work:
$dW_s = F_s\,dx = -kx\,dx$
Integrate as the block moves from $x_i$ to $x_f$ . The work is the area under the $F_s$ – $x$ line:
$W_s = -\displaystyle\int_{x_i}^{x_f} kx\,dx = -k\left[\dfrac{x^2}{2}\right]_{x_i}^{x_f} = \dfrac{1}{2}kx_i^2 - \dfrac{1}{2}kx_f^2$

From the Mean Position (

x_i = 0

)

Stretch or compress the spring from rest to $x$ :

W_s = -\dfrac{1}{2}kx^2

PE stored $= -W_s = \dfrac{1}{2}kx^2$ .

Round Trip (

x_i \to x_f \to x_i

)

Pull the block to $x_f$ and bring it back to $x_i$ :

W_s = \tfrac{1}{2}kx_i^2 - \tfrac{1}{2}kx_i^2 = 0

The net work is zero. So the spring force is conservative.

F_s = -kx

: work done = shaded area

Elastic Potential Energy of a Spring

U = \frac{1}{2}kx^2

where, $x$ = stretch or compression of the spring from its natural length

Note

Conclusion

The spring force depends on position: $F_s = -kx$ (Hooke's law).
Its work depends only on the initial and final positions.
Its work in a complete cycle is zero.
Its work equals the initial value minus the final value of a potential energy function. So the spring force is a conservative force.

Remember

Zero of Potential Energy

For a conservative force, the potential energy has no fixed absolute value. The reason is that the point of zero potential energy is our own choice. For the spring energy $\tfrac{1}{2}kx^2$ , we take PE = 0 at the equilibrium position of the mass that oscillates.

3.4Conservation of Mechanical Energy

Principle

If only conservative forces act on a system, its total mechanical energy (kinetic + potential) stays constant. Energy only changes from KE to PE, or from PE to KE.

Derivation

For a conservative force, the work done equals the negative of the change in PE:
$W = -\Delta U = U_i - U_f$
By the work-energy theorem, the same work equals the change in KE:
$W = \Delta\text{KE} = K_f - K_i$
Equate the two and rearrange. The total energy does not change:
$K_f - K_i = U_i - U_f \;\Rightarrow\; K_f + U_f = K_i + U_i$

Free fall: PE changes into KE

Total Mechanical Energy is Conserved

E = \text{KE} + \text{PE} = \frac{1}{2}mv^2 + U = \text{constant}

Remember

Condition

Mechanical energy is conserved only when non-conservative forces are absent. Friction and air resistance are such forces. If they act, a part of the mechanical energy is lost as heat. Then KE + PE does not stay constant.

Take a body that falls freely from a height $H$ . Its total energy is $mgH$ at every point. As it falls, PE keeps changing into KE.

Position	KE	PE	Total
At the top (released, $h = H$ )	$0$	$mgH$	$mgH$
On the way (after falling $x$ )	$mgx$	$mg(H-x)$	$mgH$
Just before the ground ( $h = 0$ )	$mgH$	$0$	$mgH$

PE falls, KE rises, and the total stays the same

3.5Motion in a Vertical Circle

Tie a particle of mass $m$ to a light string that cannot stretch (inextensible). Its length $L$ is the radius $R$ of the circle. The particle moves round a vertical circle with its centre at a fixed point $O$ . At the lowest point $A$ , it is given a horizontal speed $u$ .

Take any point $B$ at a height $h = R(1-\cos\theta)$ , where the speed is $v$ . Resolve the weight $mg$ into two parts. The radial part $mg\cos\theta$ acts along the string. The tangential part $mg\sin\theta$ acts along the circle. Two equations decide the motion.

Derivation

Conservation of energy between $A$ and $B$ (the particle rises by $h$ ):
$\dfrac{1}{2}m(u^2 - v^2) = mgh,\quad h = R(1-\cos\theta)$
At $B$ , the centripetal force comes from the tension $T$ and the radial part $mg\cos\theta$ . The tangential part $mg\sin\theta$ only changes the speed.
$T - mg\cos\theta = \dfrac{mv^2}{R} \;\Rightarrow\; T = \dfrac{mv^2}{R} + mg\cos\theta$

Particle in a vertical circle ·

h = R(1-\cos\theta)

For the particle to just complete the circle, the minimum speeds and the tensions at three key points are:

Point on the Circle	Minimum Speed	Tension in the String
Highest point (C)	$v_{top} = \sqrt{gR}$	$T = 0$
String horizontal	$v = \sqrt{3gR}$	$T = 3mg$
Lowest point (A)	$u_{min} = \sqrt{5gR}$	$T = 6mg$

Note

The speed and the tension are greatest at the lowest point. They are least at the highest point. To complete the circle, the string must stay tight ( $T \ge 0$ ) all the way round. The most critical point is the top.

The speed $u$ at the lowest point $A$ decides the type of motion.

1Completes the Circle ·

u \ge \sqrt{5gR}

The string stays tight all the way. The particle goes over the top with a speed of at least $v_{top} = \sqrt{gR}$ .

2Leaves the Circle ·

\sqrt{2gR} < u < \sqrt{5gR}

The tension becomes zero before the top, where $90^\circ < \theta < 180^\circ$ and $h > R$ . The particle leaves the circle at $P$ . After that, it moves on a projectile (parabolic) path.

Leaves at

P

(

T = 0,\ v \ne 0,\ h > R

)

3Oscillates ·

0 < u \le \sqrt{2gR}

The speed becomes zero in the lower half ( $0^\circ < \theta \le 90^\circ$ ), while $T \ne 0$ . It moves to and fro like a pendulum.

Swings in the lower half (

T \ne 0,\ v = 0,\ h \le R

)

3.6Various Forms of Energy

Energy exists in many forms. It can change from one form to another. The total amount of energy stays the same.

Form of Energy	What It Is
Mechanical energy	It is kinetic energy + potential energy. KE comes from motion. PE comes from position or configuration.
Internal energy	It is the sum of the KE and PE of all the molecules in a body. It comes from molecular motion and the forces between molecules.
Heat energy	It comes from the random (disordered) motion of molecules. It is related to internal energy. e.g. Rubbing our hands makes them warm.
Chemical energy	It is the difference in energy between a stable compound and the atoms that form it. Exothermic reactions give it out. Endothermic reactions take it in.
Electrical energy	It is carried by an electric current, which is a flow of charges in a field. We use it to light bulbs, run motors and work heaters.
Nuclear energy	Heavy nuclei can split (fission). Light nuclei can join (fusion). Both give out a huge amount of energy. Nuclear reactors and bombs use it.

3.7Equivalence of Mass and Energy

Concept

Einstein, 1905

Mass can be changed into energy, and energy can be changed into mass. So mass and energy are equivalent. They are not conserved separately. They are conserved together as one quantity, called mass-energy.

Mass-Energy Equivalence

E = mc^2

where, $m$ = mass that disappears, $E$ = energy that appears, $c$ = speed of light in vacuum ( $\approx 3\times 10^8\ \text{m s}^{-1}$ )

Example

Suppose just 1 kg of matter changes fully into energy:

E = 1\times(3\times 10^8)^2 = 9\times 10^{16}\ \text{J}

This is about the yearly output of a large power station of 3000 MW.

3.8Principle of Conservation of Energy

Principle

Energy can neither be created nor destroyed. One form of energy can only change into another form. If some energy is lost in one form, the same amount is gained in another form. So the total energy of an isolated system stays constant.

Note

The whole universe is an isolated system. So the total energy of the universe never changes. When some part loses energy, some other part gains exactly that much. This principle cannot be proved. But it has never been found to fail.

Part 4

Power & Collisions

Power tells us how fast work is done. In a collision, two bodies push each other with large forces for a very short time. Linear momentum is conserved in every collision. Kinetic energy is conserved only in elastic collisions.

4.1Power

Definition

Power is the rate of doing work (or of transferring energy) by a person or a machine.

Average Power

P_{av} = \frac{\text{Work done}}{\text{Time taken}} = \frac{W}{t}

Instantaneous Power

P = \frac{dW}{dt} = \vec{F}\cdot\vec{v}

Make the time interval smaller and smaller ( $\Delta t \to 0$ ). The average power then becomes the instantaneous power. We also use $dW = \vec{F}\cdot d\vec{x}$ :

Derivation

Take the limit of the average power as $\Delta t \to 0$ :
$P = \lim\limits_{\Delta t\to 0}\dfrac{\Delta W}{\Delta t} = \dfrac{dW}{dt}$
Put $dW = \vec{F}\cdot d\vec{x}$ and use $\dfrac{d\vec{x}}{dt} = \vec{v}$ :
$P = \vec{F}\cdot\dfrac{d\vec{x}}{dt} = \vec{F}\cdot\vec{v}$

Units of Power

Unit	Symbol	Equivalent
Watt (SI unit of power)	W	$1\ \text{J s}^{-1}$
Kilowatt	kW	$10^{3}\ \text{W}$
Horsepower	hp	$746\ \text{W}$
Kilowatt-hour (a unit of energy)	kWh	$3.6\times 10^{6}\ \text{J}$

Nature

Scalar quantity. It is the ratio of two scalars, $W$ and $t$ .

Dimensions

$[\text{ML}^2\text{T}^{-3}]$

At an Instant

Power is the dot product of force and velocity, $\vec{F}\cdot\vec{v}$ .

Remember

Another name for the kilowatt-hour (kWh) is the Board of Trade (BOT) unit. It is the commercial unit of electrical energy. It is not a unit of power.

4.2Collisions

Definition

A collision is an isolated event. In it, two or more bodies exert large forces on each other for a short time. The bodies need not touch each other.

Example

No Contact Needed

In Rutherford's scattering experiment, an α-particle does not touch the nucleus. It is turned away by electrostatic repulsion. Still, we say that the α-particle collides with the nucleus.

Two Types of Collision

Feature	Elastic Collision	Inelastic Collision
Kinetic energy	No KE is lost	Some KE is lost
Forces involved	Conservative	May be non-conservative
Energy change	KE does not change into heat, light or sound	A part of the KE changes into heat, light or sound
Examples	Sub-atomic particles, glass balls	Two vehicles; a ball and the floor

Conservation of Linear Momentum in a Collision

Derivation

During the collision time $\Delta t$ , the two bodies exert impulsive forces on each other. So their momenta change:
$\Delta p_1 = F_{12}\,\Delta t,\qquad \Delta p_2 = F_{21}\,\Delta t$
By Newton's third law, $F_{12} = -F_{21}$ . So the two changes cancel:
$\Delta p_1 + \Delta p_2 = 0$
So the total momentum does not change:
$\Delta(p_1 + p_2) = 0 \;\Rightarrow\; p_1 + p_2 = \text{constant}$

Momentum Conserved in Every Collision

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

The total linear momentum is conserved at every instant of the collision. This holds in elastic as well as inelastic collisions.

4.3Elastic Collision in One Dimension

Definition

In an elastic collision in one dimension, two bodies first move along the same straight line. They collide with no loss of kinetic energy. After the collision, they keep moving along the same line.

Two bodies $A$ and $B$ have masses $m_1$ and $m_2$ . They move along the same line with velocities $u_1$ and $u_2$ , where $u_1 > u_2$ . After they collide, both still move the same way. Their velocities are now $v_1$ and $v_2$ , with $v_2 > v_1$ . In this collision, momentum and kinetic energy are both conserved.

The two bodies before, during and after an elastic collision

Derivation

Momentum is conserved. Bring the terms of each mass to one side:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \qquad …(i)$

$m_1 u_1 - m_1 v_1 = m_2 v_2 - m_2 u_2$

$m_1(u_1 - v_1) = m_2(v_2 - u_2) \qquad …(ii)$
Kinetic energy is conserved (elastic collision). Cancel the $\tfrac{1}{2}$ and group the terms of each mass. Then use $a^2 - b^2 = (a+b)(a-b)$ :

$\tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2$

$m_1 u_1^2 - m_1 v_1^2 = m_2 v_2^2 - m_2 u_2^2$

$m_1(u_1 + v_1)(u_1 - v_1) = m_2(v_2 - u_2)(v_2 + u_2) \qquad …(iii)$
Divide (iii) by (ii). The factors $m_1(u_1 - v_1)$ and $m_2(v_2 - u_2)$ cancel out:

$\dfrac{m_1(u_1 + v_1)(u_1 - v_1)}{m_1(u_1 - v_1)} = \dfrac{m_2(v_2 - u_2)(v_2 + u_2)}{m_2(v_2 - u_2)} \;\Rightarrow\; u_1 + v_1 = v_2 + u_2$
Rearrange to get an important result:

$u_1 - u_2 = v_2 - v_1 \qquad …(iv)$

Approach Velocity = Separation Velocity

u_1 - u_2 = v_2 - v_1

The speed with which the bodies move apart after the collision equals the speed with which they came together before it.

Velocities after the Collision

Now solve the relative velocity relation (iv) together with the momentum equation (i). This gives $v_1$ and $v_2$ using only $u_1$ , $u_2$ and the masses.

Derivation

From $u_1 - u_2 = v_2 - v_1$ , write $v_2$ in terms of $v_1$ :
$v_2 = u_1 - u_2 + v_1 \qquad …(v)$
Put (v) in the momentum equation (i). Then expand the right-hand side:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2(u_1 - u_2 + v_1)$

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 u_1 - m_2 u_2 + m_2 v_1$
Collect the $v_1$ terms on one side and the $u$ terms on the other:
$(m_1 - m_2)u_1 + 2m_2 u_2 = (m_1 + m_2)v_1$
Solve for $v_1$ . This is Eq. (vi). Put it back in (v) to get $v_2$ , which is Eq. (vii):
$v_1 = \dfrac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \qquad …(vi)$

Velocity of Body A · (vi)

v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2

Velocity of Body B · (vii)

v_2 = \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2 + \left(\frac{2m_1}{m_1 + m_2}\right)u_1

So (vi) and (vii) tell us the velocities after the collision when the velocities before it are known. Three special cases are important.

IEqual Masses (

m_1 = m_2 = m

)

Put $m_1 = m_2 = m$ in Eqs. (vi) and (vii). The terms with $(m_1 - m_2)$ become zero.

The two bodies simply exchange their velocities.

v_1 = \dfrac{2mu_2}{2m} = u_2

v_2 = \dfrac{2mu_1}{2m} = u_1

IILight Body Hits a Heavy Body at Rest (

m_1 \ll m_2,\ u_2 = 0

)

Put $u_2 = 0$ in Eqs. (vi) and (vii). Neglect $m_1$ in comparison with $m_2$ .

The light body bounces back with the same speed. The heavy body stays almost at rest.

v_1 = \dfrac{-m_2 u_1}{m_2} = -u_1

v_2 \approx 0

IIIHeavy Body Hits a Light Body at Rest (

m_1 \gg m_2,\ u_2 = 0

)

Put $u_2 = 0$ in Eqs. (vi) and (vii). Neglect $m_2$ in comparison with $m_1$ .

The heavy body moves on almost unchanged. The light body flies off at about twice that speed, $2u_1$ .

v_1 \approx \dfrac{m_1 u_1}{m_1} = u_1

v_2 \approx \dfrac{2m_1 u_1}{m_1} = 2u_1

Definition

Head-on (1-D) Collision

A collision is called head-on or one-dimensional when both bodies move along one straight line, before and after the collision.

Remember

During and After the Collision

In an elastic collision, KE is conserved only after the collision is over. It is not conserved during the short time of the actual collision. At the moment of impact, the two bodies are deformed. They may even be at rest relative to each other for an instant. At that moment, the KE is held for a while as elastic potential energy of the squeezed bodies.

4.4Perfectly Inelastic Collision

Definition

A collision is perfectly inelastic if the two bodies stick together after colliding. They then move as one body with a common velocity.

A body of mass $m_1$ moving with velocity $u_1$ hits a body of mass $m_2$ moving with velocity $u_2$ along the same line. They join and move together with a common velocity $v$ . Only momentum is conserved:

m_1 u_1 + m_2 u_2 = (m_1 + m_2)\,v

The bodies stick and move together after the collision

Common Velocity after the Collision

v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}

If the second body is at rest at first ( $u_2 = 0$ ), then $v = \dfrac{m_1 u_1}{m_1 + m_2}$ .

Momentum is conserved, but kinetic energy is not. Take $u_2 = 0$ and find the loss of kinetic energy.

Derivation

Find the total KE before the collision and after it:

$E_1 = \dfrac{1}{2}m_1 u_1^2 \qquad …(i)$

$E_2 = \dfrac{1}{2}(m_1 + m_2)\,v^2 \qquad …(ii)$
Put $v = \dfrac{m_1 u_1}{m_1 + m_2}$ in Eq. (ii):
$E_2 = \dfrac{1}{2}(m_1 + m_2)\left[\dfrac{m_1 u_1}{m_1 + m_2}\right]^2 = \dfrac{1}{2}\dfrac{m_1^2 u_1^2}{m_1 + m_2}$
The loss of kinetic energy is $E_1 - E_2$ :
$\Delta\text{KE} = E_1 - E_2 = \dfrac{1}{2}m_1 u_1^2 - \dfrac{m_1^2 u_1^2}{2(m_1 + m_2)}$
Take $2(m_1 + m_2)$ as the common denominator and simplify the numerator:
$\Delta\text{KE} = \dfrac{m_1^2 u_1^2 + m_1 m_2 u_1^2 - m_1^2 u_1^2}{2(m_1 + m_2)} = \dfrac{m_1 m_2\, u_1^2}{2(m_1 + m_2)}$

Loss of Kinetic Energy

\Delta\text{KE} = \frac{m_1 m_2\, u_1^2}{2(m_1 + m_2)}

This is always positive. The lost KE appears mainly as light, sound and heat.

Note

This collision causes the greatest loss of kinetic energy that is possible while momentum is conserved.

4.5Collision in Two Dimensions

Definition

A collision is oblique (or two-dimensional) if it is not head-on. Here the force between the bodies is not along the line of the initial velocity. So after the collision, the bodies move along different lines.

Body $A$ (mass $m_1$ ) moves with speed $u_1$ along the $x$ -axis. It hits body $B$ (mass $m_2$ ), which is at rest. After the collision, $A$ moves at an angle $\theta$ and $B$ at an angle $\phi$ to the $x$ -axis. Their speeds $v_1$ and $v_2$ are resolved into components along the axes.

Before:

A

moves towards

B

at rest · After:

A

moves at

\theta

and

B

\phi

to the

x

-axis

Governing Equations

Momentum along $x$ :

m_1 u_1 = m_1 v_1\cos\theta + m_2 v_2\cos\phi \quad …(i)

Momentum along $y$ :

0 = m_1 v_1\sin\theta - m_2 v_2\sin\phi \quad …(ii)

KE (if elastic):

m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2 \quad …(iii)

Remember

Four Unknowns, Three Equations

We have four unknowns: $v_1$ , $v_2$ , $\theta$ and $\phi$ . But there are only three equations. So one quantity, usually the angle $\theta$ , must be measured in an experiment. Then the others can be found.

The three equations lead to three special cases.

IGlancing Collision

Here the incident particle loses no kinetic energy. It goes on almost undeflected. Here $\theta = 0^\circ$ and $\phi = 90^\circ$ .

From Eqs. (i) and (ii):

u_1 = v_1, \qquad v_2 = 0

So the target particle gets no kinetic energy: $\tfrac{1}{2}m_2 v_2^2 = 0$ .

IIHead-on Collision

The incident particle keeps moving along its original line, so $\theta = 0^\circ$ . The momentum equations become:

m_1 u_1 = m_1 v_1 + m_2 v_2\cos\phi, \qquad 0 = m_2 v_2\sin\phi

So $\phi = 0^\circ$ , and $B$ also moves on that line. Equation (iii) stays the same. The problem becomes a one-dimensional (head-on) collision.

IIIElastic Collision of Two Identical Particles

Let $m_1 = m_2$ and the incident speed $u_1 = u$ . The target is at rest. The three equations become:

v_1^2 + v_2^2 = u^2 \;…(iv)

v_1\cos\theta + v_2\cos\phi = u \;…(v)

v_1\sin\theta - v_2\sin\phi = 0 \;…(vi)

Derivation

Square Eq. (v). Both (iv) and (v) give $u^2$ , so equate them:
$v_1^2 + v_2^2 = (v_1\cos\theta + v_2\cos\phi)^2 = v_1^2\cos^2\theta + v_2^2\cos^2\phi + 2v_1 v_2\cos\theta\cos\phi$
Move the cosine terms to the left side. Use $1 - \cos^2 = \sin^2$ :
$v_1^2\sin^2\theta + v_2^2\sin^2\phi = 2v_1 v_2\cos\theta\cos\phi \qquad …(vii)$
From Eq. (vi), $v_2\sin\phi = v_1\sin\theta$ . So $v_2^2\sin^2\phi = v_1^2\sin^2\theta$ . Put this in (vii):
$2v_1^2\sin^2\theta = 2v_1 v_2\cos\theta\cos\phi \;\Rightarrow\; \cos\theta\cos\phi = \dfrac{v_1\sin^2\theta}{v_2} = \sin\theta\sin\phi$
So $\cos\theta\cos\phi - \sin\theta\sin\phi = 0$ , that is, $\cos(\theta + \phi) = 0$ :
$\cos(\theta + \phi) = 0 \;\Rightarrow\; \theta + \phi = 90^\circ$

Identical Particles · Right Angle

\theta + \phi = 90^\circ

A moving particle hits an identical particle at rest in an elastic 2-D collision. Then, after the collision, their paths are perpendicular (at 90°) to each other.

Note

Inelastic Collision in 2-D

In an inelastic oblique collision, KE is not conserved. We can use only the momentum equations along $x$ and $y$ . So we get three equations for an elastic 2-D collision and only two for an inelastic one.

4.6Coefficient of Restitution

Definition

The coefficient of restitution $(e)$ is the relative velocity of separation divided by the relative velocity of approach. Separation is measured after the collision, and approach before it. Another name for $e$ is the coefficient of resilience.

Coefficient of Restitution

e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}

where, $u_1, u_2$ = velocities before the collision, $v_1, v_2$ = velocities after it. The value of $e$ is decided by the materials of the two bodies.

e = 1

Perfectly elastic. The velocity of separation is equal to the velocity of approach.

0 < e < 1

Inelastic (real) collision. e.g. Glass balls: $e \approx 0.95$ . Lead balls: $e \approx 0.20$ .

e = 0

Perfectly inelastic. The bodies stick together and do not separate.

Rebounding of a Ball

A ball is dropped from a height $h$ . After each bounce, it rises to a smaller height. These two results are useful:

Height after n Rebounds

h_n = e^{2n}\,h

Total Distance before Stopping

s = h\left(\frac{1 + e^2}{1 - e^2}\right)

The rebound heights become smaller and smaller

Different Types of Collisions

Collision	Kinetic Energy	Coefficient ( $e$ )	Main Domain
Elastic	Conserved	$e = 1$	Between atomic and sub-atomic particles
Inelastic	Not conserved	$0 < e < 1$	Between ordinary objects
Perfectly inelastic	Maximum loss of KE	$e = 0$	Bodies stick together
Super-elastic	KE increases	$e > 1$	In explosions

End of Chapter 5 · notesshala.in