PhysicsICSE Class 10 · Chapter 6

Conservation of Energy ICSE Class 10: Principle, Free Fall and the Simple Pendulum

Updated 14 October 2026 · Fact-checked

The principle of conservation of energy says energy can neither be created nor destroyed, only changed from one form to another. For a freely falling body with no air resistance, U + K stays constant. With friction, mechanical energy decreases, but the total energy is still conserved as heat and sound appear.

What Conservation of Energy is about

Energy keeps changing form: a falling ball turns potential energy into kinetic energy; a swinging pendulum turns them back and forth; brakes turn kinetic energy into heat. Through all these changes the total energy stays the same.

The CISCE 2027 syllabus covers: the statement of the principle of conservation of energy; the theoretical verification that $U + K$ is constant for a freely falling body; the application to a simple pendulum (qualitative only); and simple numerical problems.

The CISCE examiners' comments for 2024 point to two big ideas that candidates mixed up: mechanical energy ( $U + K$ ) is conserved only when there is no friction, but total energy is always conserved. In a 2025 numerical, candidates also converted kgf wrongly and forgot to subtract the energy lost to friction.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Conservation of Energy is in the 1. Force, Work, Power and Energy (vi) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

The principle of conservation of energy

Energy can neither be created nor destroyed. It can only be changed from one form to another, and the total energy of an isolated system remains constant.

Whenever energy seems to disappear, look for where it has gone: heat (friction), sound, light, or the energy of other bodies.

Worked example

A moving car is brought to rest by its brakes. Where has its kinetic energy gone?

Try it first, then show the solutionSolution

Friction at the brakes and tyres opposes the motion.
The kinetic energy changes mainly into heat in the brakes, tyres and road, with some sound.

Answer: Into heat (and some sound); total energy is unchanged.

Mechanical energy and total energy

Mechanical energy $= U + K$ .

With no friction or air resistance, only gravity does work, and $U + K$ stays constant: mechanical energy is conserved.
With friction, part of the mechanical energy changes into heat, so $U + K$ decreases.
In both cases, the total energy (mechanical + heat + sound …) is conserved.

So 'mechanical energy is conserved' needs ideal conditions; 'energy is conserved' is always true.

Worked example

A block slides down a rough ramp. Is mechanical energy conserved? Is the law of conservation of energy obeyed?

Try it first, then show the solutionSolution

Friction turns some energy into heat, so $U + K$ at the bottom is less than at the top: mechanical energy is not conserved.
The lost mechanical energy equals the heat produced: total energy is conserved.

Answer: Mechanical energy: no. Conservation of energy: yes.

Freely falling body: U + K = constant

A body of mass $m$ is dropped from rest at height $h$ (point A). Ignore air resistance.

At A (height $h$ , $v = 0$ ): $U = m g h$ , $K = 0$ ; total $= m g h$ .
At B, after falling distance $x$ : height $h - x$ , so $U = m g (h - x)$ . From $v^{2} = 2 g x$ : $K = \frac{1}{2} m (2 g x) = m g x$ . Total $= m g h$ .
At C, just before the ground: $U = 0$ , $v^{2} = 2 g h$ , $K = m g h$ . Total $= m g h$ .

The total $U + K = m g h$ at every point: mechanical energy is conserved.

Worked example

A 2 kg ball is dropped from 20 m. Find $U$ and $K$ at the top, after falling 5 m, and just before it hits the ground. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

Top: $U = 2 \times 10 \times 20 = 400$ J, $K = 0$ .
After 5 m: $U = 2 \times 10 \times 15 = 300$ J; $v^{2} = 2 \times 10 \times 5 = 100$ , $K = \frac{1}{2} \times 2 \times 100 = 100$ J.
Ground: $U = 0$ , $K = 400$ J. Total is 400 J each time.

Answer: 400 J total at every point

Speed from height

With no friction, all the PE lost becomes KE:
$m g h = \frac{1}{2} m v^{2}$ , so $v = 2 g h$ .

The mass cancels: in the absence of air resistance, all bodies dropped from the same height reach the ground with the same speed. The same idea gives the speed at the bottom of a smooth slope or ramp, whatever its shape.

Worked example

A ball is dropped from 5 m. Find its speed just before it hits the ground. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

$v = 2 \times 10 \times 5 = 100$ .
$v = 10$ m/s.

Answer: 10 m/s

The simple pendulum (qualitative)

A bob swings between two extreme positions through the mean (lowest) position.

At the extremes: the bob is momentarily at rest and highest: energy is all potential.
At the mean position: lowest and fastest: energy is all kinetic (taking the mean position as reference).
In between: partly potential, partly kinetic; the sum is constant if there is no air resistance.

In practice air resistance and friction at the support slowly convert mechanical energy into heat, so the amplitude decreases and the pendulum eventually stops. Total energy is still conserved.

Worked example

Why does a swinging pendulum gradually come to rest?

Try it first, then show the solutionSolution

At each swing, work is done against air resistance and friction at the point of support.
This changes some mechanical energy into heat, so the height reached at each extreme falls until the bob stops.

Answer: Mechanical energy is changed into heat by air resistance and friction.

Including friction in numericals

Write an energy balance:
$energy at start = energy at end + energy lost to friction$ .

For example: $m g h = \frac{1}{2} m v^{2} + W_{f}$ . Convert units first: a weight of $W$ kgf means a mass of $W$ kg (and a weight of $W \times g$ newtons).

Worked example

A child of mass 50 kg slides from rest down a smooth slide 7.2 m high, then crosses a rough patch where 900 J is lost. Find her speed at the foot of the slide and after the rough patch. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

At the foot: $K = m g h = 50 \times 10 \times 7.2 = 3600$ J; $v = 2 \times 10 \times 7.2 = 12$ m/s.
After the rough patch: $K = 3600 - 900 = 2700$ J.
$v = \frac{2 \times 2700}{50} = 108 \approx 10.39$ m/s.

Answer: 12 m/s; about 10.39 m/s

Physics in real life: Conservation of Energy

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Energy lost on a water slide

Suppose a 30 kg child slides from rest down a 5 m high water slide and reaches the bottom at 8 m/s.

How much energy was turned into heat and sound? ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

PE at the top $= 30 \times 10 \times 5 = 1500$ J.
KE at the bottom $= \frac{1}{2} \times 30 \times 64 = 960$ J.
Energy lost $= 1500 - 960 = 540$ J (36%).

Answer: 540 J

Memory hook: Start energy − end energy = losses.

A bouncing ball

Imagine a 0.2 kg ball dropped from 5 m rises only to 4 m after its first bounce.

How much mechanical energy is lost in the bounce, and where does it go? ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

PE at 5 m $= 10$ J; PE at 4 m $= 8$ J.
Lost $= 2$ J, turned into heat (in the ball and floor) and sound.

Answer: 2 J, as heat and sound

Memory hook: Lower bounce, warmer ball.

Pendulum clock

Suppose the bob of a pendulum clock rises 5 cm above its lowest point at each extreme.

Find the speed of the bob at its lowest point, and say why the clock needs a spring or battery.

Try it first, then show the solutionSolution

$v = 2 \times 10 \times 0.05 = 1$ m/s.
Air resistance and friction slowly take away energy; the spring or battery replaces it to keep the swing steady.

Answer: 1 m/s; to make up energy lost as heat

Memory hook: Clocks need topping up because friction never sleeps.

Topics in this chapter

Conservation of Energy formula sheet

Mechanical energy: $E = U + K = m g h + \frac{1}{2} m v^{2}$; Constant if there is no friction or air resistance.
Speed after falling height h: $v = 2 g h$; Independent of mass.
Height reached with speed v: $h = \frac{v ^{2}}{2 g}$; For a body thrown straight up.
Energy balance with friction: $U_{1} + K_{1} = U_{2} + K_{2} + W_{f}$; W_f = energy turned into heat.
Weight in kgf: $W kgf \Rightarrow m = W kg$; Weight in newtons = m g.

Conservation of Energy: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Freely falling body: energy at three points

To show theoretically that U + K is constant (3 marks).

Draw

A vertical line from point A at height h down to point C on the ground.
Point B on the line, a distance x below A.
The ball at A (at rest), B and C, with a downward velocity arrow at B and C.
A small table or labels beside each point with U, K and total.

Label

A: U = mgh, K = 0
B: U = mg(h − x), K = mgx
C: U = 0, K = mgh
Height h, distance fallen x
Total = mgh at each point

Watch out: Use v² = 2gx at B; the total at each point must come to mgh.

Simple pendulum and its energy

Qualitative questions on where energy is PE or KE.

Draw

A rigid support with a thread and a bob.
The bob at the mean (lowest) position and at the two extreme positions.
A dotted arc joining the positions.
A horizontal line at the mean position as the reference level, and the height of the extreme above it.

Label

Extreme positions: all PE, speed zero
Mean position: all KE, maximum speed
Height h of the extreme above the mean
Point of suspension

Watch out: Say what happens to the energy over many swings: it turns into heat.

How the board asks Conservation of Energy

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ: pendulum, falling body or statement	Section A, Question 1: where KE is maximum, or what is conserved.	Extremes: all PE; mean: all KE.
Mechanical vs total energy	Smooth and rough paths; which conserves mechanical energy, which obeys conservation of energy.	Mechanical only without friction; total always.
Theoretical verification (3 marks)	Show that U + K is constant for a freely falling body.	Three points A, B, C with v² = 2gx.
Numericals	Speed after a fall or slide, height reached, energy at a point, loss to friction.	Energy balance; v = √(2gh); convert kgf.
Reasoning	Why a pendulum stops; where the energy goes in braking or bouncing.	Name the work against friction or air resistance and the heat produced.

Tricks to remember Conservation of Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Top = PE, bottom = KE
Highest point: all potential. Lowest point: all kinetic (no friction).
Use it when: Falling bodies and pendulums.
Mechanical needs smooth; total never fails
U + K is conserved only on smooth paths; total energy is always conserved.
Use it when: Conservation MCQs.
Mass cancels
v = √(2gh): mass is not in it.
Use it when: Speed after a drop.
Missing energy = heat
If energy at the end is less, the difference became heat (and sound).
Use it when: Friction numericals.

Conservation of Energy: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Pendulum at an extreme · 1 mark · MCQBoard level

At an extreme position of a swinging simple pendulum, the bob has:
(a) maximum kinetic energy
(b) maximum potential energy and zero kinetic energy
(c) equal potential and kinetic energy
(d) zero total energy

Try it first, then show the solutionSolution

It is momentarily at rest and at its highest point.

Answer: (b)

Example 2: Rough surfaces · 1 mark · Assertion-reasonBoard level

Assertion (A): The mechanical energy of a body sliding on a rough surface decreases.
Reason (R): Work done against friction changes part of the mechanical energy into heat.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

The lost mechanical energy appears as heat.

Answer: (c)

Example 3: Dropped ball · 2 marks · Short answerBoard level

A ball of mass 0.5 kg is dropped from a height of 10 m. Find its kinetic energy and speed just before it hits the ground. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

$K = m g h = 0.5 \times 10 \times 10 = 50$ J.
$v = 2 \times 10 \times 10 = 200 \approx 14.14$ m/s.

Answer: 50 J; about 14.14 m/s

Example 4: Verify U + K = constant · 3 marks · Short answerHarder board level

Show that the sum of potential and kinetic energy of a freely falling body is the same at the point of release, at an intermediate point and just before it reaches the ground.

Try it first, then show the solutionSolution

At A (height $h$ , rest): $U + K = m g h + 0 = m g h$ .
At B (fallen $x$ ): $U = m g (h - x)$ ; $v^{2} = 2 g x$ so $K = m g x$ ; sum $= m g h$ .
At C (ground): $U = 0$ ; $v^{2} = 2 g h$ so $K = m g h$ ; sum $= m g h$ . The sum is constant.

Answer: U + K = mgh at all three points

Example 5: Smooth track, then a rough slope · 3 marks · Short answerHarder board level

A ball of mass 0.5 kg is released from rest at a height of 3.2 m on a smooth curved track. At the bottom it runs up a rough slope, losing 4 J of energy to friction before it stops. Find (a) its speed at the bottom of the track, (b) the greatest height it reaches on the slope. ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

(a) Smooth track: $v = 2 g h = 2 \times 10 \times 3.2 = 8$ m/s; $K = \frac{1}{2} \times 0.5 \times 64 = 16$ J.
(b) Energy balance on the slope: $16 = m g h^{'} + 4$ , so $m g h^{'} = 12$ J.
$h^{'} = \frac{12}{0.5 \times 10} = 2.4$ m (less than 3.2 m because of the friction loss).

Answer: (a) 8 m/s (b) 2.4 m

Example 6: Simple pendulum · 2 marks · Long answerBoard level

For a simple pendulum, state where its energy is (a) entirely potential, (b) entirely kinetic. (c) Why does its amplitude decrease with time?

Try it first, then show the solutionSolution

(a) At the extreme positions. (b) At the mean position.
(c) Air resistance and friction at the support convert mechanical energy into heat.

Answer: (a) extremes (b) mean position (c) energy lost as heat

Example 7: Thrown upwards · 3 marks · Long answerHarder board level

A ball of mass 0.2 kg is thrown vertically upwards from the ground at 20 m/s. Find (a) the maximum height, (b) its speed, KE and PE at a height of 15 m, (c) the total energy. ( $g = 10$ m s⁻², no air resistance)

Try it first, then show the solutionSolution

(a) $h = \frac{2 0 ^{2}}{2 \times 10} = 20$ m.
(b) $v^{2} = 400 - 2 \times 10 \times 15 = 100$ , $v = 10$ m/s; $K = \frac{1}{2} \times 0.2 \times 100 = 10$ J; $U = 0.2 \times 10 \times 15 = 30$ J.
(c) Total $= 40$ J ( $= \frac{1}{2} \times 0.2 \times 400$ ).

Answer: (a) 20 m (b) 10 m/s, 10 J, 30 J (c) 40 J

Example 8: Roller coaster · 4 marks · Long answerHOTS / competency

A roller coaster car of mass 400 kg starts from rest at A, 30 m above the ground, on a smooth track. Find its speed at B (10 m high) and at C (25 m high). Can it get over a hill 32 m high? If friction actually removes 20 kJ between A and B, what is its speed at B? ( $g = 10$ m s⁻²)

Try it first, then show the solutionSolution

At B: drop of 20 m, $v = 2 \times 10 \times 20 = 20$ m/s.
At C: drop of 5 m, $v = 100 = 10$ m/s.
A 32 m hill is higher than the starting point; its total energy ( $400 \times 10 \times 30$ J) is not enough to reach 32 m, so it cannot.
With friction: $K_{B} = 400 \times 10 \times 20 - 20000 = 60000$ J, $v = 300 \approx 17.32$ m/s.

Answer: 20 m/s; 10 m/s; no; about 17.32 m/s

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

A water slide

Suppose a child of mass 30 kg starts from rest at the top of a water slide 5 m high and reaches the bottom at 8 m/s. ( $g = 10$ m s⁻²)

(i) · 3 marks
What speed would the child have at the bottom if the slide were perfectly smooth?
Answer
Answer: 10 m/s
$v = 2 g h = 2 \times 10 \times 5 = 10$ m/s.
(ii) · 3 marks
Find the energy lost on the actual slide and the percentage lost.
Answer
Answer: 540 J; 36%
Start: $U = 1500$ J. End: $K = \frac{1}{2} \times 30 \times 64 = 960$ J. Lost $= 540$ J $= 36%$ .
(iii) · 4 marks
Is mechanical energy conserved on the slide? Is the principle of conservation of energy obeyed? Explain.
Answer
Answer: No; yes
Mechanical energy falls from 1500 J to 960 J, so it is not conserved, because of friction and water resistance. The 540 J appears as heat and sound, so total energy is conserved: the principle holds.

A pendulum clock

Imagine the bob of a pendulum clock, of mass 0.1 kg, rises 5 cm above its lowest position at each extreme of its swing. ( $g = 10$ m s⁻²)

(i) · 3 marks
Find the potential energy of the bob at an extreme position (lowest position as reference).
Answer
Answer: 0.05 J
$U = 0.1 \times 10 \times 0.05 = 0.05$ J.
(ii) · 3 marks
Find the speed of the bob at the lowest point, assuming no losses.
Answer
Answer: 1 m/s
$\frac{1}{2} \times 0.1 \times v^{2} = 0.05$ , so $v^{2} = 1$ , $v = 1$ m/s.
(iii) · 4 marks
Explain why the clock must be wound or have a battery, using the principle of conservation of energy.
Answer
Answer: To replace energy lost as heat
Each swing loses a little mechanical energy to air resistance and friction at the pivot, which becomes heat. The swing would die away unless energy is supplied; the wound spring (elastic PE) or battery (chemical energy) provides it. Energy is converted, not created.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A stone is dropped from a height of 45 m. Find its speed just before it reaches the ground. ( $g = 10$ m s⁻²) (2 marks)

Step	Marks
$m g h = \frac{1}{2} m v^{2}$ , so $v = 2 g h$	1
$v = 900 = 30$ m/s	1

A 2 kg body falls freely from 20 m. Find its PE, KE and total energy 15 m above the ground. ( $g = 10$ m s⁻²) (3 marks)

Step	Marks
$U = 2 \times 10 \times 15 = 300$ J	1
Fallen 5 m: $K = m g x = 2 \times 10 \times 5 = 100$ J	1
Total $= 400$ J (= $m g h$ at the top)	1

A roller coaster car of mass 400 kg starts from rest 30 m up a track. Friction removes 20 kJ before it reaches a point 10 m above the ground. Find its speed there. ( $g = 10$ m s⁻²) (4 marks)

Step	Marks
PE lost $= 400 \times 10 \times 20 = 80000$ J	1
KE gained $= 80000 - 20000 = 60000$ J	1
$\frac{1}{2} \times 400 \times v^{2} = 60000$	1
$v = 300 \approx 17.32$ m/s	1

Where students lose marks in Conservation of Energy

The slips examiners see most often, and what to write instead.

Mistake: Mechanical energy vs total energy
CISCE examiners' comments, 2024: candidates could not see that mechanical energy is conserved only without friction, and could not separate conservation of mechanical energy from conservation of energy.
Do this: Smooth path: U + K constant. Rough path: U + K decreases, but total energy (including heat) is conserved.
Usually costs: 1 to 2 marks.
Mistake: Vague answers to 'which path'
CISCE examiners' comments, 2024: candidates answered 'right/left' or 'A/B' instead of naming the path as asked, and some guessed.
Do this: Answer in the exact terms the question uses (path 1, path 2, …).
Usually costs: 1 mark.
Mistake: kgf converted wrongly
CISCE examiners' comments, 2025: in a skateboard problem with a weight of 40 kgf, candidates converted kgf to N or to g wrongly, and errors carried into later parts.
Do this: A weight of 40 kgf means a mass of 40 kg; in K = ½mv² use m = 40 kg.
Usually costs: 1 to 2 marks.
Mistake: Ignoring the friction loss
CISCE examiners' comments, 2025: candidates subtracted only the potential energy at the next point and forgot the energy lost against friction, or wrongly set U equal to K there.
Do this: Write the full energy balance: start energy = end energy + energy lost.
Usually costs: 1 mark.

Practice questions on Conservation of Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
'Energy can neither be created nor destroyed' is a statement of:
(a) Newton's first law
(b) the principle of moments
(c) the principle of conservation of energy
(d) the law of inertia
Hint
Energy principle.
Answer
(c)
Q2 · 1 mark · MCQBoard level
The kinetic energy of the bob of a simple pendulum is maximum at:
(a) an extreme position
(b) the mean position
(c) halfway to an extreme
(d) the point of suspension
Hint
Fastest at the bottom.
Answer
(b)
Q3 · 1 mark · MCQHarder board level
Two balls of masses 1 kg and 5 kg are dropped together from the same height, with no air resistance. Just before hitting the ground:
(a) the heavier one is faster
(b) the lighter one is faster
(c) both have the same speed
(d) both have the same kinetic energy
Hint
v = √(2gh).
Answer
(c)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): Total energy is conserved even when friction acts on a moving body.
Reason (R): The mechanical energy lost against friction is converted into heat.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Where does the energy go?
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
A boy weighing 40 kgf climbs to a platform 3 m high. Find his mass and the potential energy he gains. ( $g = 10$ m s⁻²)
Hint
40 kgf = weight of a 40 kg mass.
Answer
Mass $= 40$ kg (weight 400 N); $U = 40 \times 10 \times 3 = 1200$ J.
Q6 · 2 marks · Short answerHarder board level
A stone dropped from a bridge hits the water at 25 m/s. Using the conservation of energy, find the height of the bridge above the water, neglecting air resistance. ( $g = 10$ m s⁻²)
Hint
mgh = ½mv², so h = v²/2g.
Answer
$h = \frac{v ^{2}}{2 g} = \frac{625}{20} = 31.25$ m
Q7 · 3 marks · Long answerBoard level
(a) State the principle of conservation of energy. (b) Describe the energy changes in one swing of a simple pendulum from one extreme to the other.
Hint
PE → KE → PE.
Answer
(a) Energy can neither be created nor destroyed, only converted; total energy is constant. (b) At one extreme all PE; moving down, PE changes to KE, all KE at the mean position; moving up, KE changes back to PE, all PE at the other extreme.
Q8 · 3 marks · Long answerHarder board level
A ball of mass 0.5 kg is thrown vertically upwards at 12 m/s. Find the maximum height and its PE there. ( $g = 10$ m s⁻², no air resistance)
Hint
h = v²/2g.
Answer
$h = \frac{144}{20} = 7.2$ m; $U = 0.5 \times 10 \times 7.2 = 36$ J (equal to the initial KE).
Q9 · 4 marks · Long answerHOTS / competency
A 0.2 kg ball dropped from 5 m rebounds to 80% of the height it fell from, each time. (a) How high does it rise after the first bounce? (b) How much energy is lost in the first bounce, and where does it go? (c) How high does it rise after the third bounce? ( $g = 10$ m s⁻²)
Hint
Multiply by 0.8 each time.
Answer
(a) 4 m. (b) $10 - 8 = 2$ J, into heat and sound. (c) $5 \times 0. 8^{3} = 2.56$ m.
Q10 · 3 marks · Long answerHarder board level
A 3 kg body falls freely from a height of 25 m. Find its PE, KE and speed when it is 5 m above the ground, and check that the total energy is unchanged. ( $g = 10$ m s⁻²)
Hint
It has fallen 20 m.
Answer
$U = 3 \times 10 \times 5 = 150$ J; $K = m g x = 3 \times 10 \times 20 = 600$ J; $v = 2 \times 10 \times 20 = 20$ m/s. Total $= 750$ J $= 3 \times 10 \times 25$ ✓
Q11 · 1 mark · MCQHOTS / competency
A ball is thrown vertically upwards from the ground. Taking the ground as the reference level and neglecting air resistance, its kinetic energy equals its potential energy at:
(a) the maximum height
(b) half the maximum height
(c) one quarter of the maximum height
(d) the ground
Hint
Total = mgH; U = K = mgH/2.
Answer
(b)
Q12 · 4 marks · Long answerHOTS / competency
The bob of a pendulum passes its lowest point at 2 m/s. (a) How high above the lowest point does it rise? (b) If 10% of its mechanical energy is lost in each swing, how high does it rise on the next swing? (c) Where does the lost energy go? ( $g = 10$ m s⁻²)
Hint
h = v²/2g.
Answer
(a) $h = \frac{4}{20} = 0.2$ m. (b) $0.9 \times 0.2 = 0.18$ m. (c) Into heat, by work against air resistance and friction at the support.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the principle of conservation of energy.Show answer
Energy can neither be created nor destroyed; it only changes form, and the total stays constant.
2.When is mechanical energy conserved?Show answer
When there is no friction or air resistance.
3.Speed after falling height h?Show answer
v = √(2gh)
4.Where does a pendulum bob have maximum KE?Show answer
At the mean (lowest) position.
5.Where is a pendulum's energy all potential?Show answer
At the extreme positions.
6.Why does a pendulum stop?Show answer
Air resistance and friction turn its energy into heat.

Best order to study it

Statement of the principle.
Mechanical energy vs total energy.
Freely falling body: three-point verification.
Speed from height; height from speed.
Simple pendulum (qualitative).
Numericals with friction losses.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state the principle precisely?
Can you show U + K = mgh at three points?
Do you know when mechanical energy is not conserved?
Can you find speed from a height and height from a speed?
Can you explain the pendulum's energy changes?
Do you include friction losses in the balance?

Conservation of Energy: frequently asked questions

Is the simple pendulum numerical in the syllabus?

The syllabus treats the simple pendulum qualitatively; you may still use U = mgh and K = ½mv² for its bob as simple energy numericals.

Does a heavier body fall faster?

Without air resistance, no: v = √(2gh) does not depend on mass.

Where does the energy go when a ball stops bouncing?

Into heat in the ball and the floor, and into sound; the total energy is unchanged.

Is energy really conserved when friction acts?

Yes. Friction changes mechanical energy into heat; the total energy remains constant.

What is meant by an isolated system?

A system that does not exchange energy with its surroundings, so its total energy stays the same.

Based on: CISCE ICSE 2027 Physics (Science Paper 1) syllabus; CISCE ICSE 2027 Physics (Science Paper 1) specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2025; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2024