Work, Energy and Power: Class 11 Physics · Chapter 5

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
28 practice questions
26 MCQs

What you will learn in Work, Energy and Power

Calculate the scalar product of two vectors from their magnitudes and angle or from components, and use it to find angles and projections
Calculate the work done by constant and variable forces, with its sign, and read it as an area under a force–displacement graph
State, derive and apply the work-energy theorem for constant and variable forces
Distinguish conservative from non-conservative forces and use potential energy for gravity near the Earth and for a spring
Apply conservation of mechanical energy, and its modified form with friction, to falling bodies, vertical circles and springs
Define power, use P = F·v, and convert between watts, horsepower and kilowatt hours
Analyse elastic and inelastic collisions in one and two dimensions using conservation of momentum and kinetic energy

Original companion text aligned to NCERT Physics Class 11, Chapter 5, pp. 71–91. Not affiliated with or endorsed by NCERT.

5.1 Introduction

Ask three people what work means and you may hear about a dabbawala pedalling lunch boxes across Mumbai, a student revising for an entrance test, and a porter at a railway station balancing a trunk on his head. In everyday speech all three are working hard. Physics is fussier. It counts work only when a force moves the thing it acts on. The student at her desk does almost no work on her books, however tired she feels, and the porter does none on the trunk while he stands waiting for his customer.

Energy and power have everyday meanings too. We say a child who never sits still is full of energy, and that a fast bowler who sends the ball down at $145 k m / h$ bowls with power. In this chapter the three words become measurable quantities with units. Work gets a precise definition in terms of force and displacement, energy becomes the capacity to do work, and power measures how quickly work is done. Expect a few surprises along the way: holding a heavy bag perfectly still feels like hard work, yet it involves no work at all.

One piece of vector algebra comes first. Work combines two vectors, force and displacement, yet the result has no direction. The operation that does this is the scalar product.

5.1.1 The scalar product

Adding forces is familiar from Chapter 3. Work needs a different combination: a force and a displacement multiplied to give a single number with no direction. Vector algebra provides exactly this, the scalar product, also called the dot product. (The vector product, which gives a vector, comes in Chapter 6.) Place two vectors $𝐀$ and $𝐁$ tail to tail, with angle $θ$ between them ( $0 \leq θ \leq 180^{\circ}$ ). Their scalar product is

Scalar (dot) product

𝐀 \cdot 𝐁 = A B \cos θ

(5.1a)

Read it as 'A dot B'. Every factor on the right is a plain number, so the product has a size and a sign but no direction. Grouping the factors in two ways gives it a picture (Fig. 5.1). As $A (B \cos θ)$ , it is the length of $𝐀$ times the projection of $𝐁$ on $𝐀$ : the shadow that $𝐁$ casts on the line of $𝐀$ when light falls at right angles to that line. As $B (A \cos θ)$ , it is the length of $𝐁$ times the shadow of $𝐀$ on $𝐁$ . Both groupings give the same number.

2/4The perpendicular from the tip of $𝐁$ marks its projection $B \cos θ$ on $𝐀$ , so $𝐀 \cdot 𝐁 = A (B \cos θ)$ .

Fig. 5.1NCERT Fig. 5.1 · p. 72

Vector A points right from O and vector B points up and right at angle theta. Step 1: just the vectors and the angle. Step 2: a perpendicular from B's tip marks a green bar, B cos theta, along A. Step 3: a perpendicular from A's tip marks a green bar, A cos theta, along B. Step 4: B swings past 90 degrees and its shadow, marked in vermillion as negative, falls behind O.

The dot product obeys rules much like ordinary multiplication. For any vectors $𝐏$ , $𝐐$ , $𝐑$ and real number $λ$ :

$𝐏 \cdot 𝐐 = 𝐐 \cdot 𝐏$ : the order does not matter (commutative law).
$𝐏 \cdot (𝐐 + 𝐑) = 𝐏 \cdot 𝐐 + 𝐏 \cdot 𝐑$ : dotting with a sum gives the sum of the dots (distributive law).
$𝐏 \cdot (λ 𝐐) = λ (𝐏 \cdot 𝐐)$ : a number can be pulled outside.

DerivationThe dot product from components

The unit vectors $\hat{𝐢}$ , $\hat{𝐣}$ , $\hat{𝐤}$ have length 1 and are mutually perpendicular. So a unit vector dotted with itself gives 1 ( $\hat{𝐢} \cdot \hat{𝐢} = 1$ , and the same for $\hat{𝐣}$ and $\hat{𝐤}$ ), while two different ones give 0 ( $\hat{𝐢} \cdot \hat{𝐣} = \hat{𝐣} \cdot \hat{𝐤} = \hat{𝐤} \cdot \hat{𝐢} = 0$ ).

Write $𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} + A_{z} \hat{𝐤}$ , write $𝐁$ the same way, and multiply out with the distributive law. Of the nine products, the six that pair different unit vectors vanish, leaving

\begin{aligned} 𝐀 \cdot 𝐁 & = A_{x} B_{x} \\ + A_{y} B_{y} \\ + A_{z} B_{z} \end{aligned}

(5.1b)

With $𝐁 = 𝐀$ (so $θ = 0$ ) this gives the square of a length, $A^{2} = 𝐀 \cdot 𝐀$ :

\begin{aligned} A^{2} & = A_{x}^{2} + A_{y}^{2} \\ + A_{z}^{2} \end{aligned}

(5.1c)

And two non-zero vectors are perpendicular exactly when their dot product is zero: a test for right angles that needs no protractor.

Example 5.1NCERT Example 5.1 · p. 72

The bond angle of methane

In a methane molecule ( $C H_{4}$ ) the carbon atom sits at the centre of a cube and the four hydrogen atoms at alternate corners. Two of the C–H bonds point along $𝐚 = \hat{𝐢} + \hat{𝐣} + \hat{𝐤}$ and $𝐛 = \hat{𝐢} - \hat{𝐣} - \hat{𝐤}$ (in units of half the cube's edge). Find (a) $𝐚 \cdot 𝐛$ , (b) the angle between the bonds, and (c) the projection of $𝐚$ on $𝐛$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

What is

\hat{𝐣} \cdot (\hat{𝐢} + \hat{𝐣})

1

0

\sqrt{2}

\hat{𝐤}

Show answer

Answer (A) $1$

\hat{𝐣} \cdot \hat{𝐢} = 0

and

\hat{𝐣} \cdot \hat{𝐣} = 1

, so the result is 1. A dot product is always a number, never a vector such as

\hat{𝐤}

5.2 Notions of work and kinetic energy: the work-energy theorem

Work and kinetic energy can be discovered rather than simply defined. Start from a result of Chapter 2 for motion along a line with constant acceleration $a$ , linking the initial speed $u$ , the final speed $v$ and the displacement $s$ :

v^{2} - u^{2} = 2 a s

(5.2)

Multiply both sides by $m / 2$ and use Newton's second law, $F = m a$ , for the net force on the body:

\begin{aligned} \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} \\ = m a s = F s \end{aligned}

(5.2a)

No new physics has entered: this is the second law rearranged. In three dimensions the kinematics reads $v^{2} - u^{2} = 2 𝐚 \cdot 𝐝$ , with $𝐝$ the displacement, and the same step gives

\begin{aligned} \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} \\ = m 𝐚 \cdot 𝐝 = 𝐅 \cdot 𝐝 \end{aligned}

(5.2b)

Equation (5.2b) invites two definitions. Call $\frac{1}{2} m v^{2}$ the kinetic energy $K$ of the body, and call $𝐅 \cdot 𝐝$ the work $W$ done by the force. With these names the equation reads

Work-energy theorem

K_{f} - K_{i} = W

(5.3)

where $K_{i}$ and $K_{f}$ are the initial and final kinetic energies (Fig. 5.2).

TheoremWork-energy theorem

Over any displacement, the work done by the net force acting on a particle equals the increase in the particle's kinetic energy.

This work-energy theorem has been proved here only for a constant force; Section 5.6 removes that restriction.

3/3The work $𝐅 \cdot 𝐝$ equals the gain $\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ .

Fig. 5.2

A cart of mass m on a track is pulled right by a constant force. Step 1: a short arrow shows its initial speed u. Step 2: the cart has moved far right, a displacement arrow runs beneath, and a longer arrow shows speed v. Step 3: two equal bars appear, one for the work done and one for the gain in kinetic energy.

Example 5.2NCERT Example 5.2 · p. 73

Freewheeling down a ghat road

A cyclist (total mass with cycle $80 k g$ ) stops pedalling at the top of a hill road and freewheels $800 m$ along it, dropping $40 m$ in height. Her speed rises from $2.0 m / s$ to $12 m / s$ . Friction and air drag act on her in ways we do not know in detail. Taking $g = 9.8 m / s^{2}$ , find (a) the work done by gravity, (b) the total work done by friction and drag, and (c) the average resistive force along the road.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

2.0 k g

trolley speeds up from

3.0 m / s

5.0 m / s

. How much work did the net force do on it?

34 J

25 J

4.0 J

16 J

Show answer

Answer (D) $16 J$

Δ K = \frac{1}{2} (2.0) ({5.0}^{2} - {3.0}^{2}) = 16 J

. The tempting

4.0 J

comes from

(5.0 - 3.0)^{2}

: kinetic energy goes with the square of each speed, not the square of the change.

5.3 Work

Now look at work on its own. A constant force $𝐅$ acts on a body while the body moves through a displacement $𝐝$ , at an angle $θ$ to the force. Only the part of the force along the displacement, $F \cos θ$ , speeds up or slows down the motion; the part across it does neither. So the work is that component times the distance moved:

Work by a constant force

\begin{aligned} W & = (F \cos θ) d \\ = 𝐅 \cdot 𝐝 \end{aligned}

(5.4)

Definition

Work (by a constant force)

The displacement multiplied by the component of the force along it, $W = 𝐅 \cdot 𝐝$ . It is a scalar and can be positive, negative or zero.

ExamplePulling a suitcase $10 m$ with a $40 N$ force at $60^{\circ}$ to the floor does $200 J$ of work.

1/4For $θ < 90^{\circ}$ the component $F \cos θ$ points along the displacement, so $W > 0$ .

Fig. 5.3

A box slides right along a floor. Step 1: the force points forward at about 35 degrees, with a green component along the motion, and W greater than 0. Step 2: the force points straight up, at right angles, and W equals 0. Step 3: the force points backward at about 145 degrees, with a vermillion backward component, and W less than 0. Step 4: the force points straight back, like friction, and W equals minus F d.

This definition parts company with everyday speech. Picture a tug-of-war that ends in a stalemate: both teams strain for a full minute, the rope does not move, and neither team does any work on it, however exhausted everyone feels. Their muscles burn food energy just to stay tense, but that energy ends up as heat in their own bodies, not as work on the rope.

A force therefore does zero work on a body in three situations (Fig. 5.4):

No displacement. A porter standing still on the platform with a $40 k g$ trunk on his head does no work on the trunk, however long he waits.
No force along the motion. A puck gliding over smooth ice has no horizontal force acting on it, yet it travels a long way: zero force, zero work.
Force perpendicular to the displacement. When the porter walks along the level platform at steady speed, his upward push on the trunk is at right angles to its displacement, and $\cos 90^{\circ} = 0$ . Likewise the tension in a pendulum string does no work on the bob, which always moves at right angles to the string.

3/3Walking on level ground, the upward force is perpendicular to the displacement.

Fig. 5.4

Step 1: a porter stands still holding a box on his head with an upward force F; the text says d equals 0, so W equals 0. Step 2: a puck slides right over ice with no force along its motion. Step 3: the porter walks right with the box, his upward force at right angles to the displacement, so W equals 0.

The sign of the work tells you whether a force helps or hinders. A force with a component along the motion ( $θ < 90^{\circ}$ ) feeds energy in and does positive work; one with a component against the motion ( $θ > 90^{\circ}$ ) drains energy and does negative work. Kinetic friction on a sliding body points straight back, so its work on that body is always negative.

From Eq. (5.4), work has the dimensions of force times length, $[M L^{2} T^{- 2}]$ , the same as kinetic energy. Its SI unit is the joule (J): the work done when a force of $1 N$ moves its point of application $1 m$ along the force. The unit honours James Prescott Joule (1818–1889), whose experiments showed that a given amount of work always produces the same amount of heat. Older and specialist units of energy are still in use (Table 5.1).

Table 5.1Other units of work and energyNCERT Table 5.1 · p. 74

Unit	Value in joules	Where you meet it
erg	$10^{- 7} J$	CGS system, older books
electron volt (eV)	$1.6 \times 10^{- 19} J$	atoms, electrons, nuclei
calorie (cal)	$4.186 J$	heat; food labels use kcal
kilowatt hour (kWh)	$3.6 \times 10^{6} J$	the 'unit' on an electricity bill

Example 5.3NCERT Example 5.3 · p. 74

A parcel on a conveyor belt

At an airport a $6.0 k g$ parcel is dropped onto a conveyor belt that moves at a steady $2.0 m / s$ . Until the parcel moves with the belt, kinetic friction of $30 N$ acts between them. (a) How far does the parcel move while it slips, and how much work does the belt's friction do on it? (b) How far does the belt move in that time, and how much work does the parcel's friction do on the belt? (c) Add the two works and interpret the result.

Try it yourself first. Then reveal the solution one step at a time.

Notice what this example shows. Newton's third law pairs the forces, not the works: the belt's push on the parcel matches the parcel's push on the belt, but the two surfaces move through different distances, so the works differ. For the forces between two bodies $𝐅_{12} + 𝐅_{21} = 0$ always, while $W_{12} + W_{21}$ is zero only in special cases, such as when the two surfaces do not slip over each other.

5.4 Kinetic energy

From Section 5.2, a body of mass $m$ moving with velocity $𝐯$ has kinetic energy

Kinetic energy

\begin{aligned} K & = \frac{1}{2} m 𝐯 \cdot 𝐯 \\ = \frac{1}{2} m v^{2} \end{aligned}

(5.5)

Because $𝐯 \cdot 𝐯 = v^{2}$ , kinetic energy depends on speed alone, not on direction: it is a scalar, and never negative. Reading the work-energy theorem backwards shows what it measures. A body moving at speed $v$ does work $\frac{1}{2} m v^{2}$ on whatever brings it to rest. A hammer head does work on a nail in this way, driving it into the wood, and moving air does work on the blades of the wind turbines at Muppandal in Tamil Nadu.

Definition

Kinetic energy

The energy a body has because of its motion, $K = \frac{1}{2} m v^{2}$ : the work it can do on whatever brings it to rest.

ExampleA $0.16 k g$ cricket ball at $40 m / s$ has $128 J$ .

The factor $v^{2}$ makes speed far more important than mass: doubling the mass doubles $K$ , but doubling the speed multiplies it by four. Fig. 5.5 shows how enormously kinetic energies range.

Fig. 5.5Rough kinetic energies of moving things, on a logarithmic scale (masses and speeds are typical values).NCERT Table 5.2 · p. 75

A bar chart on a logarithmic energy scale. A mosquito in flight has about 10 to the minus 6 joule, a fast-bowled cricket ball 1.3 times 10 squared, a cyclist 10 cubed, a loaded auto-rickshaw 4.3 times 10 to the 4, a loaded truck 3.2 times 10 to the 6, an express train 4.5 times 10 to the 8, and the Earth in orbit about 2.7 times 10 to the 33 joules, drawn with a broken bar.

Example 5.4NCERT Example 5.4 · p. 75

Why speed limits matter

A $1200 k g$ car travelling at $40 k m / h$ stops in $10 m$ under full braking, once the brakes are on. (a) What is the average braking force? (b) With the same braking force, how far would it take to stop from $80 k m / h$ ?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

A scooter speeds up from

20 k m / h

60 k m / h

. By what factor does its kinetic energy grow?

36927

Show answer

Answer (C) 9

The speed triples and

K \propto v^{2}

, so

K

grows

3^{2} = 9

times. Choosing 3 forgets the square.

5.5 Work done by a variable force

Most real forces change as things move. The pull of a spring grows as you stretch it, the thrust of a rocket changes as its fuel burns, and your push on a stuck almirah changes from moment to moment. How do we find the work then?

Take motion along the $x$ -axis under a force $F (x)$ that varies with position. Chop the path from $x_{i}$ to $x_{f}$ into short pieces of width $Δ x$ . Over one short piece the force hardly changes, so its work is nearly $F (x) Δ x$ , the area of a thin rectangle under the $F$ – $x$ graph (Fig. 5.6). Adding the pieces,

W \approx \sum_{x_{i}}^{x_{f}} F (x) Δ x

(5.6)

Make the strips ever thinner and their jagged tops hug the curve ever more closely. In the limit their total becomes exactly the area under the curve, which calculus writes as a definite integral:

Work by a variable force

\begin{aligned} W & = \lim_{Δ x \to 0} \sum F Δ x \\ = \int_{x_{i}}^{x_{f}} F (x) d x \end{aligned}

(5.7)

So the work done by a varying force is the area under its force–displacement graph between the start and the end, with area below the $x$ -axis counting as negative work. (Appendix 3.1 of the textbook recalls the integrals you need.)

2/3Thinner strips fit the curve better, and their total gets closer to the true work.

Fig. 5.6NCERT Fig. 5.3 · p. 76

A curve shows a force F varying with position between x i and x f. Step 1: four wide rectangles sit under the curve, one labelled delta W equals F of x delta x. Step 2: thirteen thinner rectangles fit the curve more closely. Step 3: the exact shaded region under the curve appears, labelled W equals area.

Example 5.5NCERT Example 5.5 · p. 76

A lift starting up

A lift and its passengers, total mass $400 k g$ , start from rest at the ground floor. As it rises, the cable tension falls steadily from $4320 N$ to $3920 N$ over the first $4.0 m$ and then stays at $3920 N$ up to $10 m$ . Take $g = 9.8 m / s^{2}$ and ignore friction. Find the work done over the $10 m$ (a) by the cable and (b) by gravity, and (c) the lift's speed at the $10 m$ mark.

Fig. 5.7Cable tension and weight against height for Example 5.5.NCERT Fig. 5.4 · p. 76

Force is plotted against height. The cable tension falls in a straight line from 4320 newtons to 3920 newtons over the first 4 metres, then stays at 3920 newtons up to 10 metres; its region is labelled plus 40 000 joules. The weight, minus 3920 newtons throughout, is a dashed rectangle below the axis labelled minus 39 200 joules.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

A force on a particle grows in a straight line from

0

x = 0

10 N

x = 4.0 m

. How much work does it do over these

4.0 m

2.5 J

10 J

40 J

20 J

Show answer

Answer (D) $20 J$

The area is a triangle,

\frac{1}{2} \times 4.0 \times 10 = 20 J

. Using the final force for the whole distance gives the wrong

40 J

5.6 The work-energy theorem for a variable force

We can now prove the theorem without assuming a constant force. Stay in one dimension and follow how quickly the kinetic energy changes.

DerivationWork-energy theorem with a varying force

Differentiate $K = \frac{1}{2} m v^{2}$ with respect to time (chain rule), then use Newton's second law, $m d v / d t = F$ :

\begin{aligned} \frac{d K}{d t} & = m v \frac{d v}{d t} \\ = F v = F \frac{d x}{d t} \end{aligned}

So $d K = F d x$ : over a tiny step the kinetic energy grows by exactly the work done in that step. Add up the steps from position $x_{i}$ , where the kinetic energy is $K_{i}$ , to $x_{f}$ , where it is $K_{f}$ :

\begin{aligned} K_{f} - K_{i} \\ = \int_{x_{i}}^{x_{f}} F d x \end{aligned}

(5.8a)

The right side is the work of Eq. (5.7), so

K_{f} - K_{i} = W

(5.8b)

This is the work-energy theorem again, now for any force that depends on position.

Think of Newton's second law as ball-by-ball commentary and the work-energy theorem as the final scorecard. The scorecard gives the result, the final speed, quickly, because the effect of the force has been summed over the whole stretch of path. But you cannot read from it when anything happened, and since it is a single number rather than a vector, it cannot tell you which way the body ends up moving. For those questions you go back to $𝐅 = m 𝐚$ .

Example 5.6NCERT Example 5.6 · p. 77

A carpet strip that gets rougher

A toy car of mass $0.40 k g$ rolls freely at $5.0 m / s$ onto a strip of carpet whose pile thickens along its length. The drag on the car grows with the distance $x$ it has travelled on the strip as $F = - c x^{2}$ , with $c = 3.0 N / m^{2}$ . (a) Find its speed after $1.0 m$ of carpet. (b) If the strip were long enough, how far would the car get?

Fig. 5.8The drag

F = - 3.0 x^{2}

and its work over the first metre.

The force on the toy car is plotted against the distance x along the carpet. It starts at zero and curves downwards as minus 3.0 x squared, reaching minus 3.0 newtons at x equals 1.0 metre. The region between the curve and the axis from 0 to 1.0 metre is shaded and labelled W equals minus 1.0 joule.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

The work-energy theorem can tell you how long a body takes to slow down from one speed to another.

TrueFalse

Show answer

Answer False

It links speeds to positions through work; time has been summed away. To find durations you need Newton's second law or kinematics.

5.7 The concept of potential energy

Kinetic energy belongs to motion, but energy can also be held in waiting. Pull back the rubber of a catapult and let go, and the stone flies off fast. Wind up a toy car and set it down, and it races across the floor. Water held behind a dam such as Bhakra rushes down through the turbines when the gates open. In each case energy was banked in the way the parts were placed or arranged, their configuration, and paid back later as kinetic energy. Energy banked like this is called potential energy.

Not every force lets you bank energy. Push a box across a rough floor and your work is gone as heat; nothing pushes the box back. Lift the box onto a shelf, though, and the work can be recovered: let the box drop and gravity hands it back as kinetic energy. Forces of this recoverable kind are called conservative. The work such a force does between two points depends only on the two points, not on the route between them. That lets us attach to every position $x$ a number $V (x)$ , the potential energy, such that the force's work from $x_{i}$ to $x_{f}$ is $V (x_{i}) - V (x_{f})$ . In one dimension this is the same as saying

Force from potential energy

F (x) = - \frac{d V}{d x}

since then $\int_{x_{i}}^{x_{f}} F d x = - \int d V = V_{i} - V_{f}$ .

Gravity near the ground is the standard example. For heights much smaller than the Earth's radius $R_{E}$ , $g$ hardly changes (Chapter 7 deals with the change), so a mass $m$ feels a constant pull $m g$ downwards. Measure the height $h$ upwards from the ground. As the body rises from the ground to $h$ , gravity does work $- m g h$ , so

Gravitational potential energy near the Earth

V (h) = m g h

and indeed $- d V / d h = - m g$ , a downward force. Drop the body from height $h$ and kinematics gives its landing speed from $v^{2} = 2 g h$ , that is, $\frac{1}{2} m v^{2} = m g h$ : every joule banked on the way up is returned on the way down.

Because gravity's work depends only on the change of height, a block released on any frictionless slide of height $h$ arrives at the bottom with speed $\sqrt{2 g h}$ , whatever the slope or shape of the slide (Fig. 5.9). Friction spoils this: its work grows with the length of the path and depends on the motion, so it has no potential energy and is called non-conservative.

3/4A curved frictionless slide still gives the same speed: gravity's work depends only on the drop $h$ .

Fig. 5.9

Three slides lead from the top of a tower of height h to the ground: steep, gentle and curved. In steps 1 to 3 each frictionless slide in turn is highlighted, and the ball at its foot has the same speed, root 2 g h. In step 4 the gentle slide is rough, and the ball arrives with a much shorter arrow, a speed less than root 2 g h.

Potential energy has the dimensions $[M L^{2} T^{- 2}]$ and the unit joule, like work and kinetic energy. For a conservative force, a small displacement changes the potential energy by minus the work done:

Δ V = - F (x) Δ x

(5.9)

Definition

Potential energy

Energy stored by the position or configuration of a body acted on by a conservative force; its change is minus the work done by that force, $Δ V = - F Δ x$ .

ExampleLifting a $2.0 k g$ bag onto a shelf $1.5 m$ higher stores about $29 J$ .

Definition

Conservative force

A force whose work on a body depends only on the start and end points, so that its work round any closed path is zero and it has a potential energy.

ExampleGravity and the force of an ideal spring are conservative; friction is not.

Quick check

1 of 2

3.0 k g

box is carried from the ground to a landing

4.0 m

higher, once up the stairs and once up a long ramp. Compare the work done by gravity on the two trips.

Greater on the ramp, which is longerGreater on the stairs, which are steeperEqual, about

- 118 J

each timeZero both times, as the box ends at rest

Show answer

Answer (C) Equal, about $- 118 J$ each time

Gravity is conservative, so its work depends only on the rise:

- m g h = - (3.0) (9.8) (4.0) \approx - 118 J

for any route. The box ending at rest only tells us the net work is zero; gravity's share is still negative.

5.8 The conservation of mechanical energy

Toss a ball upwards and watch it slow, stop and speed up again on the way down. Kinetic energy disappears and returns. Potential energy tells us where it went in the meantime.

DerivationWhy K + V stays fixed

Let a body move through a small displacement $Δ x$ under a conservative force $F (x)$ alone. The work-energy theorem gives $Δ K = F (x) Δ x$ , and Eq. (5.9) gives $Δ V = - F (x) Δ x$ . Adding,

\begin{aligned} Δ K + Δ V & = 0 \\ Δ (K + V) & = 0 \end{aligned}

(5.10)

The sum does not change over any small step, so it does not change over the whole path from $x_{i}$ to $x_{f}$ :

\begin{aligned} K_{i} + V (x_{i}) \\ = K_{f} + V (x_{f}) \end{aligned}

(5.11)

The sum $E = K + V (x)$ is the total mechanical energy of the body. Think of $K$ and $V$ as money in two pockets: the amount in each pocket changes as the body moves, but the total in both does not. That is what the name 'conservative' records. We now have three equivalent ways to recognise a conservative force; any one implies the other two:

It is minus the slope of a potential-energy graph, $F (x) = - d V / d x$ (Eq. (5.9)). Motion in three dimensions needs a vector version of this slope, met in higher classes.
Its work between two points is fixed by the end points alone, $W = V (x_{i}) - V (x_{f})$ , whatever the route.
Round any closed loop it does no net work: start and finish coincide, so $V (x_{i}) - V (x_{f}) = 0$ .

PrincipleConservation of mechanical energy

If only conservative forces do work on a system, its total mechanical energy $K + V$ stays constant.

Apply the principle to a coconut that breaks off a palm at height $H$ and falls freely (Fig. 5.10). Take $V = 0$ at the ground and ignore air resistance. At the top, at a height $h$ on the way down, and just before landing, its mechanical energy is

E_{H} = m g H

(5.11a)

E_{h} = m g h + \frac{1}{2} m v_{h}^{2}

(5.11b)

E_{0} = \frac{1}{2} m v_{f}^{2}

(5.11c)

Gravity here is conservative, so the three are equal. Setting $E_{H} = E_{0}$ gives the landing speed $v_{f} = \sqrt{2 g H}$ , as in Section 5.7; setting $E_{H} = E_{h}$ gives the speed at any height:

v_{h}^{2} = 2 g (H - h)

(5.11d)

This kinematic result has come out without any mention of time. For a $1.5 k g$ coconut falling from $15 m$ it gives a landing speed of about $17 m / s$ and $220 J$ of kinetic energy, which is why nobody sensible naps under a coconut palm.

2/3At $h = H / 2$ the energy is shared equally between $V$ and $K$ .

Fig. 5.10NCERT Fig. 5.5 · p. 78

A coconut falls from a palm of height H; bars show its potential energy V and kinetic energy K under a fixed total-energy line. Step 1: at the top, V is full and K empty. Step 2: halfway down, at height h, both bars are half full. Step 3: just above the ground, V is empty and K full.

3/4At C the normal force has just fallen to zero: the weight alone bends the path, and the speed is $\sqrt{g R}$ .

Fig. 5.11NCERT Fig. 5.6 · p. 79

A vertical loop track of radius R has centre O, bottom A, side point B and top C. Step 1: the block at A moves right, with the normal force pointing up towards O. Step 2: at B it moves straight up, the normal force points to O and its weight points down. Step 3: at C it moves left with only its weight acting, and N equals 0. Step 4: the upper left quarter of track is removed, and the block flies from C along a parabola, landing level with A, 2R away.

Example 5.7NCERT Example 5.7 · p. 79

Looping the loop

A small block of mass $50 g$ slides without friction down a ramp that leads into a vertical circular loop of radius $R = 0.20 m$ (Fig. 5.11). It is released from rest at a height $H$ above the bottom A of the loop. B is level with the centre O and C is the top. (a) Find the least $H$ for which the block stays in contact with the track at C. For this $H$ , find (b) the speeds at A, B and C and the ratio $K_{B} / K_{C}$ , and (c) the normal force on the block at A. (d) What would happen at C if the track there had a gap?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Three identical stones are thrown from a cliff top at

10 m / s

: one straight up, one horizontally and one straight down. Ignoring air resistance, which reaches the sea below with the greatest speed?

The one thrown up, as it rises highestThe one thrown down, as it starts downwardsThe one thrown horizontally, as it goes furthestAll three arrive with the same speed

Show answer

Answer (D) All three arrive with the same speed

Each starts with the same

K

and the same

V

, so each arrives with the same total energy, hence the same speed. Directions do not enter the energy equation; only the flight times differ.

5.9 The potential energy of a spring

Squeeze the spring of a ballpoint pen and it pushes back gently; compress a scooter's suspension spring and it pushes back hard. For modest deformations, an ideal spring pulls or pushes in proportion to how far its free end $x$ has been moved from the relaxed position, always back towards that position:

Hooke's law

F_{s} = - k x

This is Hooke's law. The spring constant $k$ , in $N / m$ , measures stiffness: the pen spring has a small $k$ , the suspension spring a large one. In Fig. 5.12 a block on a smooth floor is tied to such a spring: stretched ( $x > 0$ ), the spring force is negative; compressed ( $x < 0$ ), it is positive. Unlike gravity near the ground, this force changes with position, yet it too turns out to be conservative.

4/4The shaded triangle below the axis is the spring's work, $- \frac{1}{2} k x_{m}^{2}$ , for a stretch $x_{m}$ .

Fig. 5.12NCERT Fig. 5.7 · p. 80

A block on a smooth floor is tied to a wall by a spring; below, a graph of spring force against x is a straight line sloping down through the origin, with a dot for the current state. Step 1: relaxed, dot at the origin. Step 2: stretched right, the spring pulls back, dot at positive x and negative force. Step 3: compressed left, the spring pushes out, dot at negative x and positive force. Step 4: stretched again, with the triangle below the axis shaded as the spring's work.

DerivationWork done by a spring

Let the block move slowly from $x_{i}$ to $x_{f}$ . The spring force varies, so integrate it (Eq. (5.7)):

\begin{aligned} W_{s} & = - \int_{x_{i}}^{x_{f}} k x d x \\ = \frac{1}{2} k x_{i}^{2} \\ - \frac{1}{2} k x_{f}^{2} \end{aligned}

(5.17)

Two special cases matter. Stretching from the relaxed length ( $x_{i} = 0$ ) to $x_{f} = x_{m}$ gives

W_{s} = - \frac{1}{2} k x_{m}^{2}

(5.15)

which is the area of the shaded triangle in Fig. 5.12 (base $x_{m}$ , height $k x_{m}$ ), negative because it lies below the axis: the spring pulls back while the block moves out. The hand that does the stretching pulls with $+ k x$ at each point, so it does the opposite amount:

W = + \frac{1}{2} k x_{m}^{2}

(5.16)

A compression to $x_{c} < 0$ gives the same results, since only $x_{c}^{2}$ enters. And a round trip, out from $x_{i}$ and back to $x_{i}$ , gives

\begin{aligned} W_{s} & = \frac{1}{2} k x_{i}^{2} - \frac{1}{2} k x_{i}^{2} \\ = 0 \end{aligned}

(5.18)

So the spring force depends only on position, its work depends only on the end points, and it does no net work round a closed cycle: it is conservative. Choosing $V = 0$ for the relaxed spring, Eq. (5.17) says its potential energy at extension or compression $x$ is

Potential energy of a spring

V (x) = \frac{1}{2} k x^{2}

(5.19)

Differentiating confirms it: $- d V / d x = - k x$ . Now release the block from rest at $x = x_{m}$ . With no friction its kinetic plus potential energy keeps the value $\frac{1}{2} k x_{m}^{2}$ it started with:

\begin{aligned} \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} \\ = \frac{1}{2} k x_{m}^{2} \end{aligned}

So the block moves fastest as it passes the relaxed position, where the spring stores nothing and $\frac{1}{2} m v_{m}^{2} = \frac{1}{2} k x_{m}^{2}$ :

Greatest speed

v_{m} = x_{m} \sqrt{\frac{k}{m}}

A $0.50 k g$ block on a $200 N / m$ spring, pulled out $10 c m$ , reaches $v_{m} = 0.10 \times \sqrt{400} = 2.0 m / s$ . Fig. 5.13 shows the energy shuttling between spring and block while the total stays put, and the lab in Fig. 5.14 lets you run it. You will meet this motion again in Chapter 13 as simple harmonic motion.

3/3At every position the two parts stack to the same height $E$ : energy changes form, the total does not.

Fig. 5.13NCERT Fig. 5.8 · p. 81

Energy against position from minus x m to plus x m, with a dashed line for the total E. Step 1: a purple upward parabola shows V, zero at the centre and E at the ends. Step 2: a dashed blue downward parabola shows K, largest at the centre. Step 3: a stacked bar at one position shows V and K adding up exactly to E.

Fig. 5.14Set

k

m

and the stretch, press Play and watch

K

and

V

trade places. Then make the floor rough and see where the energy goes.

Try this

Measure a spring constant at home

You’ll need

a steel spring or a thick rubber band
a cloth or polythene bag with handles
four full 500 mL water bottles
a ruler and a hook or nail

Predict first

One full bottle stretches the spring by

4.0 c m

. If the spring obeys Hooke's law, how much will three bottles stretch it?

4.0 c m

8.0 c m

12 c m

36 c m

Hang the spring from the hook and tie the bag to its lower end. Note where the bottom of the bag is against the ruler.
Put in one full $500 m L$ bottle (weight about $4.9 N$ ) and measure the extension $x$ .
Add the bottles one at a time, measuring $x$ each time.
Work out weight ÷ $x$ for each row. Then find the energy stored with all the bottles, $\frac{1}{2} k x^{2}$ .

Table 5.2

Bottles	Weight (N)	Extension x (m)	Weight ÷ x (N/m)
1	4.9
2	9.8
3	14.7
4	19.6

Your entries are saved automatically.

Example 5.8NCERT Example 5.8 · p. 81

A crate dropped onto a spring

A $2.0 k g$ crate is dropped from rest $0.30 m$ above the top of a vertical spring with $k = 2000 N / m$ , which stands on the floor. Ignoring air resistance and the spring's mass, find the greatest compression of the spring and the largest force it exerts. Take $g = 9.8 m / s^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Three remarks complete the picture of conservative forces.

No clock. Energy conservation gave the crate's greatest compression but not how long the squeeze took. Timing needs Newton's second law, solved as a function of time.
Not every force is conservative. Friction is not, and the energy principle then needs modifying, as Example 5.9 shows.
The zero of potential energy is yours to choose. We took $V = 0$ for the relaxed spring and at the ground. In Chapter 7 the zero for gravity will be placed infinitely far away. Any choice works, but once made it must be kept for the whole problem: switching zeros halfway through is like a scorer changing scoreboards in the middle of an innings.

Example 5.9NCERT Example 5.9 · p. 82

A spring launcher on a rough floor

A $1.0 k g$ block rests against a spring ( $k = 400 N / m$ ) that is compressed by $0.15 m$ on a rough floor ( $μ_{k} = 0.30$ ). The block is released and leaves the spring when the spring reaches its natural length. Find (a) the block's speed as it leaves the spring, and (b) the total distance it slides from the release point. Take $g = 9.8 m / s^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5.9 used the general rule. Split the forces on a body into conservative ones, $F_{c}$ , and non-conservative ones, $F_{n c}$ . For a small displacement the work-energy theorem gives $(F_{c} + F_{n c}) Δ x = Δ K$ , while $F_{c} Δ x = - Δ V$ . Hence $Δ (K + V) = F_{n c} Δ x$ , and over a whole path

With non-conservative forces

E_{f} - E_{i} = W_{n c}

The mechanical energy changes by exactly the work done by the non-conservative forces. Unlike the work of a conservative force, $W_{n c}$ depends on the path from $i$ to $f$ : slide the block a longer way round and friction takes more.

Quick check

1 of 2

The same force stretches two springs: P with

k = 200 N / m

and Q with

k = 400 N / m

. Which stores more energy?

Both store the same energyQ, because it is stifferNeither, since the forces are equalP, because it stretches more

Show answer

Answer (D) P, because it stretches more

For a force

F

x = F / k

and

V = \frac{1}{2} k x^{2} = F^{2} / 2 k

. The softer spring P stretches twice as far and stores twice the energy. 'Stiffer stores more' is true only for equal stretches.

5.10 Power

Two students carry identical bags up the same staircase. One takes 10 seconds, the other 20. Each does the same work against gravity, yet we would call the first fitter, because she did it faster (Fig. 5.15). How quickly work gets done, or energy gets passed on, is measured by power.

If a force does work $W$ in a time $t$ , its average power is

Average power

P_{a v} = \frac{W}{t}

Letting the time interval shrink to zero gives the instantaneous power:

P = \frac{d W}{d t}

(5.20)

Over a small displacement $d 𝐫$ a force does work $d W = 𝐅 \cdot d 𝐫$ . Dividing by $d t$ ,

P = 𝐅 \cdot \frac{d 𝐫}{d t} = 𝐅 \cdot 𝐯

(5.21)

with $𝐯$ the velocity at that moment. Power is a scalar, with dimensions $[M L^{2} T^{- 3}]$ . Its SI unit is the watt (W), with $1 W = 1 J / s$ , named after James Watt (1736–1819), whose improvements to the steam engine helped drive the Industrial Revolution. Vehicle brochures still often quote engine output in horsepower (hp), where $1 h p = 746 W$ .

2/3A reaches the top in $10 s$ while B is halfway: the same work for A, done at twice B's rate.

Fig. 5.15

A staircase with a clock. Climber A is a circle, climber B a square. Step 1: both at the bottom at 0 seconds. Step 2: at 10 seconds A is at the top and B halfway, and A's power is W over 10 seconds. Step 3: at 20 seconds B arrives too, and B's power is W over 20 seconds.

Electricity bills charge for energy, not power. The 'unit' on a bill is the kilowatt hour (kWh), the energy a $1 k W$ appliance uses in an hour:

\begin{aligned} 1 k W h & = (1000 J / s) \\ \times (3600 s) \\ = 3.6 \times 10^{6} J \end{aligned}

A $1.5 k W$ geyser used for 20 minutes a day for 30 days consumes $1.5 \times \frac{1}{3} \times 30 = 15 k W h$ , which costs ₹105 at ₹7 per unit. The kilowatt hour measures energy; the kilowatt measures power.

Definition

Power

Work done (or energy passed on) per unit time, $P = d W / d t = 𝐅 \cdot 𝐯$ ; its SI unit is the watt, $1 W = 1 J / s$ .

ExampleLifting a $50 k g$ sack through $2.0 m$ in $4.0 s$ needs an average of $245 W$ .

Try this

Measure your own power

You’ll need

a staircase of at least two floors
a stopwatch or phone timer
a measuring tape
a friend

Predict first

50 k g

student climbs a total height of

6.0 m

10 s

. Roughly what is her power output against gravity?

30 W

300 W

3000 W

30 000 W

Measure the height of one step and count the steps, to get the total rise $h$ in metres.
Have your friend time you as you walk up briskly (do not run). Repeat at an easy pace.
Work out $m g h$ with your own mass, then divide by each time to get your power in watts, and by 746 to get horsepower.

Table 5.3

Climb	Time (s)	Work mgh (J)	Power (W)
Brisk
Easy

Your entries are saved automatically.

Example 5.10NCERT Example 5.10 · p. 83

A car on a hill road

A $1200 k g$ car climbs a straight hill road that rises $1 m$ for every $20 m$ travelled along it ( $\sin θ = 0.050$ ), at a steady $54 k m / h$ . Air drag and rolling friction together oppose the motion with $600 N$ . Find (a) the power the engine must deliver to the wheels, in kW and in hp, and (b) the power needed at the same speed on a level road. Take $g = 9.8 m / s^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Which of these is a unit of energy, not of power?

kilowatthorsepowerkilowatt hourjoule per second

Show answer

Answer (C) kilowatt hour

A kilowatt hour is a power multiplied by a time, which is an amount of energy. The other three measure a rate.

5.11 Collisions

When two bodies collide, the forces between them are huge, brief and complicated: a cricket ball is visibly flattened against the bat during a contact lasting about a millisecond. We could never track those forces in detail. Fortunately we do not need to, because two conservation laws, of momentum and of energy, tell us much of the outcome without them. Carrom, marbles and billiards are all games of collisions.

Take an idealised collision between two bodies of masses $m_{1}$ and $m_{2}$ . Body 1 moves with velocity $𝐯_{1 i}$ (i for initial) and body 2 is at rest. This loses no generality, since we can always watch the collision from a frame moving with body 2. Afterwards the bodies move off with velocities $𝐯_{1 f}$ and $𝐯_{2 f}$ , at angles $θ_{1}$ and $θ_{2}$ to the original line.

5.11.1 Elastic and inelastic collisions

DerivationMomentum is conserved in every collision

During the contact, body 2 pushes body 1 with a force $𝐅_{12} (t)$ and body 1 pushes body 2 with $𝐅_{21} (t)$ , both changing rapidly. The change in each momentum is the impulse of the force on it over the contact time:

\begin{aligned} Δ 𝐩_{1} & = \int 𝐅_{12} d t \\ Δ 𝐩_{2} & = \int 𝐅_{21} d t \end{aligned}

Newton's third law makes $𝐅_{12} (t) = - 𝐅_{21} (t)$ at each instant, so the two integrals cancel and $Δ 𝐩_{1} + Δ 𝐩_{2} = 0$ , however complicated the forces.

Kinetic energy is another story: colliding bodies get squashed, and squashed material can warm up, ring, crack or stay dented. Trolleys with a spring plunger make this easy to picture (Fig. 5.16). If the squeezed spring rebounds fully, the total kinetic energy afterwards equals that before: the collision is elastic, although during the contact part of the energy sits in the spring. If the trolleys latch together, the squeeze is never undone and they move off as one: the collision is completely inelastic. Most real collisions lie in between, with some kinetic energy lost; these are inelastic collisions.

2/4At maximum compression both carts move at $u / 2$ , and half the kinetic energy is stored in the spring.

Fig. 5.16

Two equal carts on a track, cart 2 with a spring bumper, and a total kinetic energy bar below. Step 1: cart 1 moves at u towards cart 2 at rest; the bar is full. Step 2: the spring is squeezed, both move at u over 2, and half the bar is stored in the spring. Step 3: the carts have separated, cart 1 at rest and cart 2 at u; the bar is full again. Step 4: the carts are latched together at u over 2 and half the bar is lost.

5.11.2 Collisions in one dimension

Begin with a completely inelastic collision along a line ( $θ_{1} = θ_{2} = 0$ ). The bodies stick together and move off with a common velocity $v_{f}$ , so momentum conservation alone settles the outcome.

DerivationCompletely inelastic collision

Momentum conservation, $m_{1} v_{1 i} = (m_{1} + m_{2}) v_{f}$ , gives

v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1 i}

(5.22)

The kinetic energy lost is $Δ K = \frac{1}{2} m_{1} v_{1 i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}$ . By Eq. (5.22) the second term is $\frac{1}{2} m_{1} v_{1 i}^{2} \cdot m_{1} / (m_{1} + m_{2})$ , and $1 - m_{1} / (m_{1} + m_{2}) = m_{2} / (m_{1} + m_{2})$ , so

\begin{aligned} Δ K \\ = \frac{m_{1} m_{2} v_{1 i}^{2}}{2 (m_{1} + m_{2})} \end{aligned}

This is always positive: bodies that stick together always waste some kinetic energy.

Next, an elastic collision in one dimension. Now both momentum and kinetic energy are conserved, giving two equations for the two unknown final velocities.

DerivationVelocities after a one-dimensional elastic collision

\begin{aligned} m_{1} v_{1 i} & = m_{1} v_{1 f} \\ + m_{2} v_{2 f} \end{aligned}

(5.23)

\begin{aligned} m_{1} v_{1 i}^{2} & = m_{1} v_{1 f}^{2} \\ + m_{2} v_{2 f}^{2} \end{aligned}

(5.24)

(the factors of $\frac{1}{2}$ cancel in the second). Collect the terms for body 1 on the left: $m_{1} (v_{1 i} - v_{1 f}) = m_{2} v_{2 f}$ and $m_{1} (v_{1 i}^{2} - v_{1 f}^{2}) = m_{2} v_{2 f}^{2}$ .

Factorise $v_{1 i}^{2} - v_{1 f}^{2} = (v_{1 i} - v_{1 f}) (v_{1 i} + v_{1 f})$ and divide the second line by the first, which is allowed because a collision does change $v_{1}$ :

v_{2 f} = v_{1 i} + v_{1 f}

(5.25)

In words, the bodies separate as fast as they approached. Substituting back into Eq. (5.23) gives

v_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} v_{1 i}

(5.26)

v_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} v_{1 i}

(5.27)

The final velocities are now fixed entirely by $m_{1}$ , $m_{2}$ and $v_{1 i}$ . Three special cases are worth remembering (Fig. 5.17):

Equal masses, $m_{1} = m_{2}$ : $v_{1 f} = 0$ and $v_{2 f} = v_{1 i}$ . The moving body stops dead and the target takes over its velocity, as when one carrom coin hits an identical coin squarely.
Very heavy target, $m_{2} ≫ m_{1}$ : $v_{1 f} \approx - v_{1 i}$ and $v_{2 f} \approx 0$ . The light body bounces straight back at almost its full speed, like a ball thrown at a wall.
Very light target, $m_{1} ≫ m_{2}$ : $v_{1 f} \approx v_{1 i}$ and $v_{2 f} \approx 2 v_{1 i}$ . The heavy body hardly notices, and the light one shoots off at nearly twice its speed.

1/3Equal masses: ball 1 stops and ball 2 leaves with velocity $u$ .

Fig. 5.17

Each step shows a before row and an after row. Step 1, equal balls: ball 1 at u hits ball 2 at rest, then ball 1 stops and ball 2 moves at u. Step 2, a small ball hits a large one: it bounces back at minus u while the large ball barely moves. Step 3, a large ball hits a small one: the large ball keeps nearly u and the small one leaves at nearly 2u.

Fig. 5.18Set the masses and velocities, choose the type of collision and press Play. The table compares momentum and kinetic energy before and after.

Example 5.11NCERT Example 5.11 · p. 85

Which target takes the most energy?

A $0.16 k g$ cricket ball moving at $20 m / s$ makes a head-on elastic collision with a ball at rest. Find what fraction of its kinetic energy it hands over if the target is (a) a $0.057 k g$ tennis ball, (b) another $0.16 k g$ cricket ball, (c) a $5.0 k g$ shot-put. First show that, for a target of mass $m_{2}$ at rest, the fraction is $4 m_{1} m_{2} / (m_{1} + m_{2})^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Everything so far assumed the motion stays on one line: a head-on or one-dimensional collision. With two smooth balls that needs perfect aim, the moving ball heading straight for the centre of the other. Aim slightly off and the balls fly apart at angles, in a two-dimensional collision.

Quick check

1 of 2

2.0 k g

ball moving at

6.0 m / s

makes a head-on elastic collision with a

4.0 k g

ball at rest. What is the velocity of the

2.0 k g

ball afterwards?

- 2.0 m / s

+ 2.0 m / s

0

+ 4.0 m / s

Show answer

Answer (A) $- 2.0 m / s$

v_{1 f} = \frac{2.0 - 4.0}{6.0} (6.0) = - 2.0 m / s

: it bounces back. The

4.0 k g

ball leaves at

+ 4.0 m / s

; check that the momentum (

12 k g m / s

) and kinetic energy (

36 J

) are unchanged.

5.11.3 Collisions in two dimensions

Now let body 1 strike body 2 off-centre, so the two fly apart at angles $θ_{1}$ and $θ_{2}$ on either side of the original line (Fig. 5.19). Momentum is a vector, so its conservation gives an equation for each direction. Take the $x$ -axis along $𝐯_{1 i}$ and the $x$ – $y$ plane to contain the two final velocities; the $z$ -equation then confirms that nothing leaves this plane, and the other two read

\begin{aligned} m_{1} v_{1 i} \\ = m_{1} v_{1 f} \cos θ_{1} \\ + m_{2} v_{2 f} \cos θ_{2} \end{aligned}

(5.28)

\begin{aligned} 0 & = m_{1} v_{1 f} \sin θ_{1} \\ - m_{2} v_{2 f} \sin θ_{2} \end{aligned}

(5.29)

(With $θ_{1} = θ_{2} = 0$ , Eq. (5.28) is just Eq. (5.23).) If the collision is elastic, kinetic energy adds a third equation:

\begin{aligned} \frac{1}{2} m_{1} v_{1 i}^{2} & = \frac{1}{2} m_{1} v_{1 f}^{2} \\ + \frac{1}{2} m_{2} v_{2 f}^{2} \end{aligned}

(5.30)

Count the unknowns: with $m_{1}$ , $m_{2}$ and $v_{1 i}$ given, we still need $v_{1 f}$ , $v_{2 f}$ , $θ_{1}$ and $θ_{2}$ , four numbers from at most three equations. The conservation laws cannot settle the outcome alone; it also depends on exactly where the bodies touch, which is how a carrom player aims a coin. In particle experiments one angle (say $θ_{1}$ ) is measured with a detector placed at chosen angles round the target, and Eqs. (5.28) to (5.30) then give the rest.

3/3The final momenta add up to the initial one; for equal masses in an elastic collision the triangle has a right angle.

Fig. 5.19NCERT Fig. 5.10 · p. 83

Step 1: coin 1 moves right along a line passing just above the centre of coin 2 at rest. Step 2: after the collision coin 1 moves off at theta 1 above the original line and coin 2 at theta 2 below it. Step 3: a vector triangle shows the initial momentum as the sum of the two final momenta, which meet at a right angle.

Example 5.12NCERT Example 5.12 · p. 85

A proton glancing off a proton

In a particle detector, a proton moving at $3.0 \times 10^{6} m / s$ collides elastically with a proton at rest. The detector shows the incoming proton deflected by $25^{\circ}$ from its original direction. Find (a) the direction of the struck proton, and (b) the speeds of both protons after the collision.

Try it yourself first. Then reveal the solution one step at a time.

Bodies can also 'collide' without touching. Voyager 2 swung past Jupiter in 1979 and left with a new speed and direction without ever touching the planet. Such an encounter through a force acting at a distance is called scattering. Momentum is conserved as in any collision, and the outcome depends on the initial velocities and on the nature of the force. In Class 12 you will meet the experiment in which the scattering of alpha particles revealed the atomic nucleus.

Quick check

1 of 2

A billiard ball strikes an identical ball at rest, off-centre, in an elastic collision. Afterwards the first ball moves at

35^{\circ}

to its original line. At what angle to that line does the second ball move?

90^{\circ}

, on the other side

35^{\circ}

, on the same side

145^{\circ}

, on the other side

55^{\circ}

, on the other side

Show answer

Answer (D) $55^{\circ}$ , on the other side

Equal masses, one at rest, elastic and not head-on: the final velocities are perpendicular, so the two angles add up to

90^{\circ}

, on opposite sides of the line.

Recap

Work, Energy and Power: key points and summary

The scalar product $𝐀 \cdot 𝐁 = A B \cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ is a scalar; it is commutative and distributive, and zero for perpendicular vectors (Section 5.1.1).
Work-energy theorem: the net work done by all the forces on a body equals the increase in its kinetic energy, $K_{f} - K_{i} = W$ (Section 5.2, Section 5.6).
A constant force does work $W = 𝐅 \cdot 𝐝 = F d \cos θ$ , which can be positive, negative or zero; its SI unit is the joule (Section 5.3).
Kinetic energy $K = \frac{1}{2} m v^{2}$ is a scalar that is never negative (Section 5.4).
The work done by a variable force is the area under its force–displacement graph, $W = \int F d x$ (Section 5.5).
A force is conservative if its work depends only on the end points (equivalently, is zero round every closed path); then $F (x) = - d V / d x$ . Near the Earth $V = m g h$ (Section 5.7).
If only conservative forces do work, mechanical energy $K + V$ is conserved; non-conservative forces change it by $W_{n c}$ (Section 5.8).
An ideal spring obeys $F_{s} = - k x$ and stores potential energy $V = \frac{1}{2} k x^{2}$ (Section 5.9).
Power is the rate of doing work, $P = d W / d t = 𝐅 \cdot 𝐯$ , measured in watts; the kilowatt hour is a unit of energy (Section 5.10).
Momentum is conserved in every collision; kinetic energy is conserved only in elastic ones, and some is always lost when bodies stick together (Section 5.11.1).
In a 1D elastic collision with body 2 at rest, $v_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} v_{1 i}$ and $v_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} v_{1 i}$ (Section 5.11.2).
In two dimensions the conservation laws leave one quantity undetermined; equal masses in an elastic glancing collision move off at right angles (Section 5.11.3).

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NCERT Exercises 5.14	Practice Q14
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NCERT Exercises 5.19	Practice Q19
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NCERT Exercises 5.21	Practice Q21
NCERT Exercises 5.22	Practice Q22
NCERT Exercises 5.23	Practice Q23

Quick check: MCQs on Work, Energy and Power

5.1.1 The scalar product2 questions
5.2 Notions of work and kinetic energy: the work-energy theorem2 questions
5.3 Work3 questions
5.4 Kinetic energy2 questions
5.5 Work done by a variable force2 questions
5.6 The work-energy theorem for a variable force2 questions
5.7 The concept of potential energy2 questions
5.8 The conservation of mechanical energy3 questions
5.9 The potential energy of a spring2 questions
5.10 Power2 questions
5.11.2 Collisions in one dimension2 questions
5.11.3 Collisions in two dimensions2 questions

Glossary · 13

Key terms (English–Hindi)

conservative forceसंरक्षी बलsanrakshi bal: A force whose work depends only on the start and end points, and is zero round any closed path.First seen in the chapter →

elastic collisionप्रत्यास्थ संघट्टpratyasth sanghatt: A collision after which the total kinetic energy is the same as before; momentum is conserved too.First seen in the chapter →

inelastic collisionअप्रत्यास्थ संघट्टapratyasth sanghatt: A collision in which some kinetic energy is lost; if the bodies stick together it is completely inelastic.First seen in the chapter →

jouleजूलjool: The SI unit of work and energy: the work done when a force of 1 N moves its point of application 1 m along the force.First seen in the chapter →

kinetic energyगतिज ऊर्जाgatij oorja: The energy of motion, $K = \frac{1}{2} m v^{2}$ .First seen in the chapter →

mechanical energyयांत्रिक ऊर्जाyaantrik oorja: The sum $K + V$ of a body's kinetic and potential energies.First seen in the chapter →

potential energyस्थितिज ऊर्जाsthitij oorja: Energy stored by position or configuration under a conservative force; $Δ V = - F Δ x$ .First seen in the chapter →
powerशक्तिshakti: The rate of doing work, $P = d W / d t = 𝐅 \cdot 𝐯$ , measured in watts.First seen in the chapter →

scalar productअदिश गुणनफलadish gunanphal: The product $𝐀 \cdot 𝐁 = A B \cos θ$ of two vectors; the result is a scalar.First seen in the chapter →
spring constantस्प्रिंग नियतांकspring niyatank: The constant $k$ in Hooke's law $F_{s} = - k x$ ; it measures stiffness, in N/m.First seen in the chapter →

wattवाटvaat: The SI unit of power: one joule per second.First seen in the chapter →
workकार्यkaarya: Displacement times the component of the force along it, $W = 𝐅 \cdot 𝐝$ ; a scalar measured in joules.First seen in the chapter →
work-energy theoremकार्य-ऊर्जा प्रमेयkaarya-oorja pramey: The net work done by all the forces on a body equals the increase in its kinetic energy.First seen in the chapter →

Work, Energy and Power: Class 11 Physics · Chapter 5

What you will learn in Work, Energy and Power

5.1 Introduction

5.1.1 The scalar product

5.2 Notions of work and kinetic energy: the work-energy theorem

5.3 Work

5.4 Kinetic energy

5.5 Work done by a variable force

5.6 The work-energy theorem for a variable force

5.7 The concept of potential energy

5.8 The conservation of mechanical energy

5.9 The potential energy of a spring

5.10 Power

5.11 Collisions

5.11.1 Elastic and inelastic collisions

5.11.2 Collisions in one dimension

5.11.3 Collisions in two dimensions

Work, Energy and Power: key points and summary

Practice questions with answers (NCERT “Exercises”)

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Quick check: MCQs on Work, Energy and Power

Key terms (English–Hindi)

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