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CC033943
ISBN: 978-629-7557-49-6
Fulfil UASA
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PELANGI
Format: 190mm X 260mm TPTV Focus 2023 Sc BI version F2_pgi CRC
Science FORM
Dual Language Programme 3 KSSM
• Sabariah Hakim
• Jariah Khalib
(Textbook Author)
• Maznah Omar
(Textbook Author)
• Chien Hui Siong
(Guru Cemerlang)
© Penerbitan Pelangi Sdn. Bhd. 2023
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ISBN: 978-629-7557-49-6
eISBN: 978-629-7557-50-2 (eBook)
First Published 2023
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CONTENTS
Maintenance and Continuity of
Theme 1 Theme 3 Energy and Sustainability of Life
Life
Chapter Chapter
1 Stimuli and Responses 1 6 Electricity and Magnetism 88
Concept Map 1 Concept Map 88
1.1 Human Nervous System 2 6.1 Generation of Electricity 89
1.2 Stimuli and Responses in Humans 4 6.2 Transformer 94
1.3 Stimuli and Responses in Plant 14 6.3 Transmission and Distribution of Electricity 98
1.4 Importance of Responses to Stimuli 6.4 Calculate the Cost of Electricity Consumption 106
in Animals 19 Summative Practice 6 110
Summative Practice 1 21
Chapter
Chapter 7 Energy and Power 112
2 Respiration 24
Concept Map 112
Concept Map 24
2.1 Human Respiratory System 25 7.1 Work, Energy and Power 113
2.2 Movement and Exchange of Gases in the 7.2 Potential Energy and Kinetic Energy 116
Human Body 30 7.3 Principle of Conservation of Energy 120
2.3 Health of Human Respiratory System 32 Summative Practice 7 122
2.4 Adaptations in Respiratory Systems 35 Chapter
2.5 Gaseous Exchange in Plants 38 8 Radioactivity 124
Summative Practice 2 41
Chapter Concept Map 124
3 Transportation 43 8.1 Discovery of Radioactivity 125
8.2 Atom and Nucleus 129
Concept Map 43 8.3 Ionising Radiation and Non-ionising Radiation 130
3.1 Transport System in Organisms 44 8.4 Uses of Radioactive Radiation 136
3.2 Blood Circulatory System 44 Summative Practice 8 139
3.3 Human Blood 50
3.4 Transport System in Plants 53 Theme 4 Earth and Space Exploration
3.5 Blood Circulatory System in Animals and Transport
System in Plants 61 Chapter
Summative Practice 3 62 9 Space Weather 141
Exploration of Elements in Concept Map 141
Theme 2
Nature 9.1 Activities of the Sun that Affect Earth 142
9.2 Space Weather 147
Chapter
4 Reactivity of Metals 64 Summative Practice 9 149
Chapter
Concept Map 64 10 Space Exploration 151
4.1 Variety of Minerals 65
4.2 Reactivity Series of Metals 69 Concept Map 151
4.3 Extraction of Metals from Its Ore 74 10.1 Development in Astronomy 152
Summative Practice 4 77 10.2 Development of Technology and its
Application in Space Exploration 153
Chapter
5 Thermochemistry 79 Summative Practice 10 159
ANSWERS 161
Concept Map 79
5.1 Endothermic and Exothermic Reactions 80 UPSA Form 3 https://qr.pelangibooks.
Summative Practice 5 86 com/?u=1h4e9m7v
https://qr.pelangibooks.
UASA Form 3 com/?u=fQUfIUei
ii
00 Contents Focus Science Tg3.indd 2 22/03/2023 4:20 PM
Theme 1: Maintenance and Continuity of Life
Chapter
2 Respiration
CONCEPT MAP
Respiration
Human respiratory Adaptations in Respiration in
system respiratory system plants
Animal Function of the
respiratory stoma
structure
Structure of Breathing
respiratory mechanism
system
Adaptations
Inhalation of human
Adaptation of respiratory
alveolus system in
different
condition
Health of human
respiratory system Exhalation
Access to
INFOGRAPHIC
24
02 Focus KSSM SC F3.indd 24 22/03/2023 4:22 PM
Science Form 3 Chapter 2 Respiration
5. The trachea branches into two bronchi
2.1 Human Respiratory System (singular: bronchus). One bronchus enters
the right lung, the other enters the left lung.
1. The human respiratory system provides 6. Each bronchus branches into smaller tubes
an area for gaseous exchange between the called bronchioles.
blood and the environment. It allows oxygen
acquisition and carbon dioxide elimination. 7. The bronchioles end with millions of tiny
air sacs called alveoli (singular: alveolus).
2. The gaseous exchange in lungs is called CHAPTER
breathing (external respiration). This 8. The lungs are made up of alveoli, bronchioles
exchange involves the taking in and letting and blood capillaries.
out of air. 9. The lungs are located in a space called the 2
thoracic cavity.
Structure of the Human Respiratory
System 10. The lungs are protected by the rib cage
which consists of ribs and intercostal
1. Air enters the respiratory system through muscles.
the nasal cavity.
11. A sheet of muscle called the diaphragm
2. As air passes through the nasal cavity, it is separates the thoracic cavity from the
warmed and moistened. Dust is trapped by abdomen.
the hairs.
12. Figure 2.1 shows the structure of the human
3. Air then leaves the nasal cavity and enters
the trachea. respiratory system.
4. The wall of the trachea contains C-shaped
rings of cartilage which give it support and
prevent it from collapsing during inhalation.
Bronchus Nasal cavity
Bronchiole Oral cavity
Alveolus Epiglottis
Blood capillary Larynx
Trachea
Bronchus Rib
Bronchiole
Intercostal muscle
Human Right lung Left lung
respiratory
system
3D MODEL Diaphragm
Figure 2.1 Human respiratory system
Mechanism of Human Breathing
1. The breathing mechanism involves two processes, which are inhalation and exhalation.
2. During breathing, the lungs expand to inhale air into it and return to their normal size to pump
air out of it.
3. The movements of the ribs and the diaphragm bring about breathing.
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Science Form 3 Chapter 2 Respiration
4. Figure 2.2 shows the comparison of the mechanisms of inhalation and exhalation.
Inhalation Exhalation
❶ Air flows in ❹ Air flows out
❶ Rib cage moves
downwards and
❷ Rib cage moves ❹ inwards ❸
upwards and
outwards Volume of the thoracic Volume of the
cavity increases, air thoracic cavity
❸ The diaphragm pressure decreases ❷ The diaphragm decreases, air
pressure
CHAPTER
relaxes and
contracts arches increases
and flattens upwards
2
out
Similarity
Both processes are physical process which causes air to flow
through the respiratory system.
Differences
• Outer intercostal muscles contract Intercostal muscles • Outer intercostal muscles relax
• Inner intercostal muscles relax • Inner intercostal muscles contract
Moves upwards and outwards Rib cage Moves downwards and inwards
Contracts and flattens out Diaphragm Relaxes and arches upwards
Volume of thoracic
Increases Decreases
cavity
Air pressure inside
Decreases Increases
thoracic cavity
Air is forced into the lungs Movements of air Air is forced out of the lungs
Figure 2.2 Comparison of the mechanisms of inhalation with exhalation
Activity 2.1
Aim: To build a model of the human respiratory system to investigate the relationship between air
pressure in the thoracic cavity with the process of inhalation and exhalation.
Materials and apparatus: Balloons, thin rubber sheets, glass jar, Y-shaped glass tube and cork.
Procedure:
Rubber stopper
1. A model is set up as shown in Figure 2.3.
2. The rubber sheet is pulled down. Are there any changes to the size of the Y-tube
balloons? Balloon
Bell jar
3. The rubber sheet is pushed up. Are there any changes to the size of the
balloons? Rubber sheet
4. All observations are recorded in a table.
Figure 2.3 A model of the
human respiratory system
26
02 Focus KSSM SC F3.indd 26 22/03/2023 4:22 PM
Science Form 3 Chapter 2 Respiration
Observation:
Rubber sheet Changes on the balloon
Pulled down Inflated
Pushed up Deflated
Discussion:
1. The table below shows the part of the human respiratory system represented by the material/ apparatus
in the model. CHAPTER
Material/Apparatus Respiratory system
Y-shaped glass tube Trachea 2
Glass jar Thoracic cavity
Balloon Lungs
Rubber sheet Diaphragm
2. When the rubber sheet is pulled down, the volume inside the glass jar increases.
Therefore, the air pressure inside the glass jar decreases. Air from outside enters the balloons.
3. When the rubber sheet is pushed up, the volume inside the glass jar decreases. Therefore, the air pressure
inside the glass jar increases. Air is forced out of the balloons.
Conclusion:
1. Inhalation occurs when the air pressure inside the thoracic cavity decreases.
2. Exhalation occurs when the air pressure inside the thoracic cavity increases.
Inhaled Air and Exhaled Air
1. The air taken into the lungs is called inhaled air whereas the air forced out of the lungs is called
exhaled air.
2. The content of gases in inhaled and exhaled air are different.
Experiment 2.1
Aim: To study the difference in oxygen content in inhaled air and exhaled air.
Problem statement: Is the oxygen content in inhaled air equal to exhaled air?
Hypothesis: The oxygen content in inhaled air is higher than exhaled air.
Manipulated variables: Type of air in gas jar
Responding variable: Final water level in gas jar
Constant variable: Air temperature and air pressure
Materials and apparatus: Gas jars with cover, rubber tubes, glass basin, gas jar stand, water, candles, plasticine,
ruler, inhaled air and exhaled air.
Procedure:
1. A gas jar is filled with water and is inverted in a water-filled glass basin.
2. The rubber tube is put in the glass jar and blown till it is filled with exhaled air.
27
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Science Form 3 Chapter 2 Respiration
Rubber tube
Gas jar
Water
Basin
Figure 2.4 Collecting exhaled air
CHAPTER
2
3. The gas jar filled with the exhaled air is covered with a gas jar cover and is removed from the glass basin
carefully.
4. Inhaled air (normal air) is collected by covering another gas jar with a gas jar cover.
5. Both gas jars are covered over a burning candle, as shown in Figure 2.5.
6. The water level on the gas jar is marked when the flame in the gas jar has gone out.
7. The water level in the two gas jars is compared, then the percentage of oxygen in the air in the gas jar is
estimated.
Gas jar Candle
Plasticine
Glass basin
Gas jar stand
Water
Inhaled air Exhaled air
Figure 2.5
Result:
Type of air in gas jar Final water level in gas jar (cm) Percentage of oxygen in the air (%)
Inhaled air 2 20
Exhaled air 0.8 Less than 20
Discussion:
1. The oxygen in the air in the gas jar is used during the burning of the candle. So, water enters the gas jar
to fill the space originally filled by oxygen.
2. The gas jar containing inhaled air shows a high rise in water level because the oxygen composition of
the inhaled air is higher.
Conclusion:
The hypothesis is accepted. The oxygen content in inhaled air is higher than in exhaled air.
28
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Science Form 3 Chapter 2 Respiration
Experiment 2.2
Aim: To study the difference in carbon dioxide content in inhaled and exhaled air.
Problem statement: What is the difference in carbon dioxide content in inhaled and exhaled air?
Hypothesis: The carbon dioxide content in inhaled air is lower than exhaled air.
Manipulated variables: Type of air passed through limewater
Responding variable: Change of limewater colour CHAPTER
Constant variable: Concentration of limewater
Materials and apparatus: Limewater, inhaled air, exhaled air, boiling tube, connecting tube, rubber tubing, 2
glass tube and rubber stopper
Procedure:
1. The set up of apparatus is prepared as shown in Figure 2.6.
2. Clip A is closed. Air is inhaled, and breath is held. Then, clip B closes and clip A opens. After that, the
air is blown out.
3. The condition of the limewater in the boiling tube (either clear or cloudy) where inhaled and exhaled
air passes through is observed and recorded.
Clip A Clip B
A B
Boiling tube
Limewater
P Q
Figure 2.6 Set up of apparatus
Observation:
Type of air that passes through limewater Change in limewater colour
Inhaled air Clear
Exhaled air Cloudy
Discussion:
1. Limewater is used to test the presence of carbon dioxide gas. The limewater turns cloudy with the presence
of carbon dioxide.
2. Limewater in a boiling tube containing exhaled air with a high concentration of carbon dioxide will
become cloudy.
Conclusion:
The hypothesis is accepted. The carbon dioxide content in inhaled air is lower than exhaled air.
29
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Science Form 3 Chapter 2 Respiration
3. Table 2.1 shows the composition of inhaled air and exhaled air.
Table 2.1 Composition of inhaled air and exhaled air
Content Inhaled air Exhaled air
Oxygen 21% 16%
Carbon dioxide 0.04% 4%
Nitrogen 79% 79%
Water vapour Variable Saturated
Temperature Variable 34°C
CHAPTER
2
4. Based on Table 2.1, it can be concluded that:
• Inhaled air contains more oxygen as it is needed for the respiration process. The composition
of inhaled air is the same as the composition of atmospheric air.
• Exhaled air contains more carbon dioxide and water vapour. This is because respiration
produces carbon dioxide and water vapour.
Formative Practice 2.1
1. Name two mechanisms involved in respiration process.
2. Name the gases that are involved in the respiration process.
3. For each statement below, name the structure in the lungs accordingly.
(a) Area of exchange of oxygen and carbon dioxide gas
(b) Enables atmospheric air to flow into the lungs
(c) A muscle sheet that layers the bottom of the thoracic cavity
4. The oxygen content in inhaled air is higher than in exhaled air. Justify.
2.2 Movement and Exchange of Gases in the Human Body
1. During inhalation, air is taken into the lungs through the nasal cavity, passing through the trachea,
bronchi, bronchioles and finally alveoli.
2. The air that reaches the alveoli is rich in oxygen and poor in carbon dioxide.
3. Gaseous exchange takes place in the lungs by diffusion.
Diffusion is the movement of substances from an area of high concentration area to an
area of low concentration.
4. Oxygen diffuses from the alveoli into the red blood cells in the capillary (Figure 2.7).
5. Red blood cells contain the oxygen-carrying pigment called haemoglobin.
6. Oxygen combines with haemoglobin to form oxyhaemoglobin.
7. Blood cells are transported throughout the body.
8. Capillaries have a higher oxygen concentration compared to body cells (Figure 2.8).
30
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Science Form 3 Chapter 2 Respiration
3. Table 2.1 shows the composition of inhaled air and exhaled air. 9. Body cells have a higher carbon dioxide concentration than in the blood capillaries.
Table 2.1 Composition of inhaled air and exhaled air 10. Oxyhaemoglobin in the red blood cells releases oxygen which then diffuses into the body cells.
Content Inhaled air Exhaled air 11. At the same time, carbon dioxide from body cells diffuses into red blood cells to be taken to the
Oxygen 21% 16% lungs and then removed.
Carbon dioxide 0.04% 4% Blood flowing to the alveolus
(high concentration of CO 2 ,
low concentration of O 2 )
Nitrogen 79% 79%
Movement of air Air flow
Water vapour Variable Saturated Body cell CHAPTER
Temperature Variable 34°C CO 2 Inhaled air Carbon dioxide Oxygen
Alveolus
O 2 Exhaled air 2
4. Based on Table 2.1, it can be concluded that: Haemoglobin O 2 O 2 O 2 are carried as
in the red
• Inhaled air contains more oxygen as it is needed for the respiration process. The composition blood cell oxyhaemoglobin CO 2 CO 2 Blood
CO 2 Alveolus capillary
of inhaled air is the same as the composition of atmospheric air.
CO 2
• Exhaled air contains more carbon dioxide and water vapour. This is because respiration Capillary
produces carbon dioxide and water vapour. Vein Red blood Carbon Oxygen Body cell
dioxide
Key: Blood Blood leaving the cell
Oxyhaemoglobin
capillary alveolus (high
release oxygen
O 2 : Oxygen
Formative Practice 2.1 CO 2 : Carbon dioxide CO 2 concentration of O 2 ,
O 2
low concentration of CO 2 )
1. Name two mechanisms involved in respiration process. Figure 2.7 Exchange of gases across the surface of the Figure 2.8 Exchange of gases between blood and
Body cells
2. Name the gases that are involved in the respiration process. alveolus and blood capillary in the lung body cell
3. For each statement below, name the structure in the lungs accordingly.
Importance of the Adaptations of the Alveolar Structure
(a) Area of exchange of oxygen and carbon dioxide gas
• Large surface area Capillary that transports Capillary that transports • Very thin walls
(b) Enables atmospheric air to flow into the lungs The lungs have millions deoxygenated blood oxygenated blood The wall of each alveolus
of alveoli to provide a wall is one-cell thick to ease
(c) A muscle sheet that layers the bottom of the thoracic cavity large surface area for Bronchiole diffusion.
gaseous exchange.
4. The oxygen content in inhaled air is higher than in exhaled air. Justify.
Alveolus
The network of blood
2.2 Movement and Exchange of Gases in the Human Body capillaries that surround
the alveoli (where gas
exchange takes place)
• Moist inner surface
1. During inhalation, air is taken into the lungs through the nasal cavity, passing through the trachea, The inner surface of each alveolus is
bronchi, bronchioles and finally alveoli. moist so that oxygen can dissolve in • Surrounded by numerous blood
capillaries
2. The air that reaches the alveoli is rich in oxygen and poor in carbon dioxide. the moisture before diffusing across The numerous blood capillaries help
the alveolar wall. ease gaseous exchange.
3. Gaseous exchange takes place in the lungs by diffusion.
Diffusion is the movement of substances from an area of high concentration area to an
area of low concentration. Figure 2.9 Adaptation of the alveolus
4. Oxygen diffuses from the alveoli into the red blood cells in the capillary (Figure 2.7). Formative Practice 2.2
5. Red blood cells contain the oxygen-carrying pigment called haemoglobin.
1. By what physical process does oxygen enter the blood capillaries from the alveoli?
6. Oxygen combines with haemoglobin to form oxyhaemoglobin.
2. How is oxygen supplied to body tissues?
7. Blood cells are transported throughout the body. 3. State four alveolus structure adaptations for efficient gas exchange.
8. Capillaries have a higher oxygen concentration compared to body cells (Figure 2.8).
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Science Form 3 Chapter 2 Respiration
2.3 Health of Human Respiratory System
1. The quality of air can have significant effects on our respiratory system.
2. As air moves into our lungs, harmful substances in the air may enter too. These harmful substances
can cause diseases or even death in the long term.
3. Table 2.2 shows harmful substances in the air with their sources and effects.
Table 2.2 Harmful substances in the air with their sources and effects
CHAPTER
Substance Sources Effects
Tar • Irritates the respiratory tract, causing ‘smoker’s cough’
2
• Causes lung cancer, bronchitis and emphysema
Nicotine • Causes addiction to smoker
• Causes rapid heartbeat, increases blood pressure and
rapid breathing
• Causes emphysema and stroke
Cigarette smoke
Sulphur • Irritates the respiratory tract
dioxide • Causes temporary and permanent damage to the lungs
Carbon • Carbon monoxide combines with haemoglobin faster than
monoxide oxygen can. Hence, the oxygen content in red blood cells
decreases.
• Lack of oxygen can cause dizziness, headaches and
fatigue.
Vehicle smoke
Nitrogen • Irritates the lung tissue
dioxide
Factory smoke
Haze • Causes cough, asthma and bronchitis
Dust • Causes cough, nasal congestion, watery nose, sneezing
and shortness of breath
Open burning
Pollen • Causes allergies
• Causes shortness of breath and sneezing
Plants that spread pollen
through the wind
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Science Form 3 Chapter 2 Respiration
HOTS Challenge 1
Formatif 1
Pr
The haze outbreak has struck city X which is where Helmi lives. However, Helmi had to cycle to buy necessities
at a shop located about a kilometer from his house. In your opinion, what security measures should he take?
Justify.
Respiratory Diseases and Their Symptoms CHAPTER
1. Harmful substances in the air can cause diseases such as:
Asthma 2
• This disease is caused by the narrowing of the
fine airways (bronchi and bronchioles) due to
the contraction of muscles in their walls. This Emphysema
is perhaps because of an individual’s exposure • A condition in which the structure of the alveoli
to air pollutants. is broken down due to cough. This reduces the
• A patient will find it difficult to breathe, surface area for gas exchange.
experience wheezing and feel tightness in the • The patient cannot take in enough oxygen and
chest. has very short breath.
Lung cancer
• This disease is caused by a change in the Bronchitis
cells inside the lungs that divide out of control • This disease is caused by the inflammation of
and produce lumps of cells (tumours) which the lining of the bronchioles. This may be due
interfere with the normal functions of the lungs. to infection by bacteria or viruses, or chemicals
• This disease occurs due to chemicals in such as those found in cigarette smoke.
cigarette smoke such as tar which contain • Chronic bronchitis is a condition characterised
cancer-causing substances. by recurrent or chronic cough every day over
• Lung cancer may not have any symptoms. a period of several months. Smoker’s cough’
When it is detected, the cancer cells may have is an early sign of chronic bronchitis.
spread from the lungs to other organs. • In severe cases, the patient’s fingers and lips
may turn blue due to poor oxygenation of
blood.
Effects of Smoking on the Lungs
1. Smoking is dangerous to human
health. However, the number of Stop
smokers keeps increasing in Malaysia
especially among teenagers.
2. Cigarette smoke contains thousands Smoking!
of toxic chemicals that affect the
respiratory system of smokers and
passive smokers.
A passive smoker is a non-smoking person who
inhales cigarette smoke from a nearby smoker. The
risk of getting diseases is the same as smokers.
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Science Form 3 Chapter 2 Respiration
Experiment 2.3
Aim: To show the effect of cigarette smoking on the lungs.
Problem statement: What is the effect of cigarette smoke on the lungs?
Hypothesis: Cigarette smoke contains substances that can damage the lungs.
Manipulated variable: Presence of cigarette smoke
Responding variables: Change of the colour of the cotton wool, change of the colour of the hydrogen
carbonate indicator solution, temperature inside the U-tube
Constant variables: Rate of suction of air using the filter pump
Materials and apparatus: Cigarettes, cotton wool, hydrogen carbonate indicator solutions, U-tube, glass
CHAPTER
tubes, rubber stopper, thermometer, boiling tube and filter pump.
Procedure:
2
Rubber tubing
Thermometer
Safety
Precaution
To filter
pump Make sure the cigarette is
lighted by your teacher.
Cigarette
U-tube Beware!
Hydrogen carbonate
indicator solutions Conduct this experiment
in a fume chamber
Cotton wool because cigarette smoke
is dangerous to your
Figure 2.10 Set-up of apparatus health.
1. The apparatus is set up as shown in Figure 2.10.
2. Without lighting the cigarette, the filter pump is turned on for a few minutes.
3. The temperature and the colour of the cotton wool and the hydrogen carbonate indicator solution are
observed.
4. Steps 1 to 3 are repeated with the cigarette lighted.
5. All observations are recorded.
Observation:
Material / Apparatus Beginning of the experiment End of the experiment
Cotton wool White Black
Hydrogen carbonate indicator solution Red Yellow
Thermometer 25°C 40°C
Discussion:
1. Cotton wool inside the U-tube represents the human lungs.
2. Smoke cigarette contains sulphur dioxide which are acidic. This substance causes the color of the hydrogen
carbonate indicator solution to turn yellow.
3. The cotton wool turns black because cigarette smoke contains tar.
4. The thermometer reading increases because the cigarette smoke releases heat.
5. The effect of cigarette smoke on human lungs are:
(a) Heat will dry the wall of the alveoli and make it difficult for oxygen to dissolve.
(b) The tar thickens the wall of the alveoli and complicates the process of diffusion of gas and can cause
lung cancer.
(c) Acids cause irritation on the respiratory tract and make the channel narrow.
Conclusion:
The hypothesis is accepted. Cigarette smoke contains substances that can damage the lungs.
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Science Form 3 Chapter 2 Respiration
Formative Practice 2.3
1. State three substances in the air which are harmful to the respiratory system.
2. Name one disease associated with the respiratory system. State the causes and symptoms that can be
detected in the patient.
3. Who is a ‘passive smoker’?
4. A woman is suffering from lung cancer. According to the doctor, the disease is caused by the presence of
tar in the lungs, but she is not a smoker. Justify. CHAPTER
2.4 Adaptations in Respiratory Systems 2
Adaptations in the respiratory system are important to maximise the exchange rate of respiratory
gases.
Adaptations in Animal Respiratory System
Animals have a modified respiratory system for efficient oxygen absorption according to their body
size and habitat.
Amphibians Fish
O 2 CO 2 Glottis Gill
Lung Water
Trachea Nostril Water
Bottom Operculum
of mouth
Water
Figure 2.12 Respiratory structure of fish
Figure 2.11 Respiratory structure of frog 1. Gills are respiratory structures for gas
exchange in water.
1. Amphibians such as frog have two
respiratory organs which are lungs and 2. When water passes through the thin
moist skin. membrane of the gills, the dissolved oxygen
2. While on land, frogs respire through the in the water diffuse into the blood capillaries
lungs but while in the water, they use moist contained in the membrane and transports
skin. them to the fish’s body cells.
3. The moist skin surface facilitates the 3. Layered gill surfaces increase the surface
diffusion of oxygen. Under the frog skin area for oxygen diffusion.
layer there are numerous blood capillaries
to increase the gas exchange rate.
Formatif 1 2
HOTS Challenge
Pr
SC I I I I I E N C E INFO Observing the condition of the gills is a method
C
C
C
E
E
C
SCIENCE INFO
E
SC
SC
SC
SC
E
N
N
N
N
E
E
SCIENCE INFO
INFO
E INFO
INFO
INFO
INFO
INFO
INFO
E INFO
INFO
INFO
INFO
that can determine whether the fish you want to
The frog breathes with the gills during the buy is fresh or not. Justify.
tadpole stage.
35
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Science Form 3 Chapter 2 Respiration
Insects
Air sac 1. Insects such as grasshopper have a simpler
Tracheole (in some insects) Muscle respiratory system known as a tracheal
Air sac Tracheole
system.
Trachea
Trachea 2. Air enters the trachea through the respiratory
pore called spiracles.
CO 2
Spiracle 3. The ends of the branching trachea form a
Spiracles tracheole which are embedded in the body
tissues. Oxygen from the tracheole diffuses
CHAPTER
O 2
into the body cells.
Figure 2.13 Respiratory structure of grasshopper
2
Adaptation in Human Respiratory System
1. For humans, the rate of gaseous exchange decreases in the state of oxygen shortage, for example
at the top of the mountain or at the bottom of the sea.
2. At high altitudes, atmospheric pressure is low; thereby causing partial pressures of oxygen in the
atmospheric air to decrease.
3. We will have difficulty breathing thus causing the oxygen level in the blood to be low.
4. Respiratory rate will increase to allow more oxygen to enter the lungs.
5. Pressure in the deep sea is much higher than on the surface of the earth.
6. This causes the oxygen content to decrease.
7. Therefore, the rate of respiration will increase to allow more oxygen to enter the lungs.
Mountain climbers also need
to carry oxygen tanks to help
get enough oxygen.
Photograph 2.1
Divers need to carry oxygen tanks to breathe in
the water.
Photograph 2.2
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Science Form 3 Chapter 2 Respiration
8. Sports activities and an individual’s lifestyle
also affect the ability of the respiratory
system.
9. An athlete needs more energy during
vigorous activities such as running and
swimming.
10. Energy is required by the muscles for
movement. Therefore, the breathing and CHAPTER
respiratory rate should also be improved.
Photograph 2.3
11. Intake of oxygen is increased by panting. 2
12. Anaemia refers to the body whose number of
red blood cells or haemoglobin content is below
normal levels. This condition may also be caused
by abnormal shape of the red blood cells.
Normal cells 13. This condition reduces the ability of the blood
to transport oxygen to the body cells.
14. Cellular respiration in an individual with this
condition does not work efficiently.
15. An individual with anaemia are often fatigued
Sickle cells
because of low energy production due to lack of
oxygen.
Figure 2.14
16. There are some diseases that can be associated with anaemia such as sickle cell anaemia and
also thalassaemia.
17. The best way to keep our respiratory system healthy is to practices a healthy lifestyle such as
eating nutritious food, not smoke and exercise continuously. Proper exercise is important for
smooth breathing.
Formative Practice 2.4
1. Match the correct respiratory structure of the following animals.
(a) Frog Gill
(b) Grasshopper Trachea
(c) Fish Moist skin
2. State the adaptations of fish respiratory structures to increase the surface area.
3. Insects have fine pores on the sides of their body. Name the pores and state its function.
4. Explain why does a person who is doing vigorous activities breathe rapidly.
5. State the difference between the shape of a red blood cell of a normal person with a person suffering from
sickle cell anaemia. Explain the effects of the difference in shape on the patient.
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Science Form 3 Chapter 2 Respiration
2.5 Gaseous Exchange in Plants
Nucleus Chloroplast
1. Plants undergo both photosynthesis and
respiration during the day. The rate of Air sac
photosynthesis usually exceeds the rate of
respiration. At night, green plants only do Guard cell Stoma
respiration. Stem
2. During respiration, plants absorb oxygen
and get rid of carbon dioxide. During Leaf
CHAPTER
photosynthesis, chlorophyll in the leaves
absorbs sunlight and converts water and Key:
2
carbon dioxide into glucose and oxygen. Oxygen gas (O 2 )
3. Plants do not have specific respiratory Carbon dioxide gas (CO 2 )
structures for exchange of gases.
Figure 2.15 Gaseous exchange during photosynthesis
4. Gaseous exchange in plants occur through
the stoma found in the leaves, stems and
Activity 2.2
Aim: To observe the gas released from the surface of the leaf.
Materials and apparatus: Fresh leaf, 500 ml of water, beaker, Bunsen burner, tripod stand, wire gauze.
Procedure:
1. 500 ml of water is filled in a beaker and heated by using Boiling
a Bunsen burner until it boils. water
Leaf
2. The Bunsen burner is turned off when the water has
boiled. Wire gauze
3. A fresh leaf is immersed in the boiling water for 30
minutes as shown in Figure 2.16.
Bunsen
4. The surface of the leaf is observed. burner Tripod stand
Observation:
There are tiny bubbles seen from the surface of the leaf. Figure 2.16 Set-up of apparatus
Discussion:
1. The bubbles are carbon dioxide released from the surface of the leaf.
2. The exchange of gas between plant cells and the environment occurs through the stoma by diffusion.
3. Fresh leaves carry out respiration by absorbing oxygen and releasing carbon dioxide.
4. The leaf is immersed in boiling water because heat activates the gas particle trapped in the leaf and diffuse
out.
Conclusion:
Plants carry out respiration by absorbing oxygen gas and releasing carbon dioxide gas on the surface of the
leaf.
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Science Form 3 Chapter 2 Respiration
Activity 2.3
Aim: To observe the structure of the stoma on the surface of a leaf.
Materials and apparatus: Green leaf, glass slides, slide covers, knives, forceps, mounting needles, iodine
solution, filter paper and microscope.
Procedure:
Epidermal CHAPTER
layer
Leaf Epidermal
Slide layer
cover 2
Glass slide
Figure 2.17 Preparation of the slide
1. The bottom of the green leaf is folded and the epidermis layer is removed by using forceps.
2. The epidermis layer is placed on a slide.
3. A drop of iodine solution is put on the specimen.
4. The specimen is closed with the slide cover.
5. A filter paper is used to absorb excess iodine solution.
6. The slide is observed under a microscope.
7. The structure of the stoma structure observed is drawn and labeled.
Observation:
Guard cells
Inner wall
Outer wall
Chloroplast
Stoma
Figure 2.18 Structure of the stoma
1. Stomata are found more abundantly on the lower surface of the leaf than the upper surface.
2. A pair of nut-shaped guard cell surrounds each stoma.
Conclusion:
A stoma is a pore on the leaf that is surrounded by a pair of guard cells.
roots.
Opening and Closing of Stoma
1. On the lower surface of the leaves, there are tiny pores called stomata. Each stoma is controlled
by two guard cells on the left and right. Guard cells contain chloroplasts.
2. The main function of the stoma is to allow the gas to diffuse in and out of the leaves. During
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02 Focus KSSM SC F3.indd 39 22/03/2023 4:22 PM
Science Form 3 Chapter 2 Respiration
the process of respiration, oxygen from the atmosphere diffuses into the
leaves while carbon dioxide diffuses out of the leaves into the atmosphere.
3. Most stomata open during the day and close at night.
4. During the day, the guard cells carry out photosynthesis. The glucose
produced increases the osmotic concentration of the guard cells. Water
from the neighbouring epidermal cells diffuses into the guard cells by
osmosis. The walls of the guard cells which are thick and less elastic are
pulled out. The guard cells become turgid and curved. The stoma opens. Figure 2.19 The
stoma opens
Osmosis is the movement of water molecules from a region of high
CHAPTER
water concentration (dilute) to a region of low water concentration
2
(concentrated).
5. During the night, the guard cells do not carry out photosynthesis. The
content of glucose and water in the guard cells decreases and causes the
cells to shrink. The guard cells become straight and the stoma is closed.
Figure 2.20 The
6. Besides producing glucose, photosynthesis also produces oxygen. Some stoma closes
oxygen is used by the plant to perform respiration while the rest is released
to the atmosphere via stomata.
Importance of Unpolluted Environment for the Survival of Plants
1. Air pollution does not only harm the health of the human respiratory system but also affects
plants.
2. Industrial factories and motor vehicles release smoke containing soot and dust.
3. This substance will be suspended in the air and finally settled over the surface of the leaves
forming a layer that will reduce the absorption of sunlight by chlorophyll.
4. This condition leads to the decreased rate of photosynthesis. The growth of trees will be disrupted
and the harvest will decrease.
5. The situation can be controlled by the following steps:
(a) Enforcement of laws such as jail sentence or fine.
(b) Awareness campaigns through mass media.
(c) Cleaning campaign around housing areas.
(d) Formal education in educational institutions.
Formative Practice 2.5
1. During respiration in plants, diffuses into the plant cell and diffuses
out of the plant cell through by .
2. Farid carries out an experiment to observe the structure of the stoma. He takes a sample of a leaf at 3.00
pm. He observes the slide he prepared under the microscope. Draw the structure of the stoma that he
might be observing.
3. Open burning produces dust particles that float in the air. These particles will settle over the surface of
the leaves. State the effect.
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Science Form 3 Chapter 2 Respiration
Summative Practice 2
SECTION A
1. Which of the following is the correct term for the 4. Which of the following statements are true
process of movement of oxygen from the alveoli about the characteristics of alveoli to increase
into the blood capillaries? its efficiency? CHAPTER
A Absorption I Thickness of one cell
B Path II Dry
C Diffusion III Has a narrow surface area
D Flow IV Surrounded by a network of blood capillaries 2
A I and II
2. Figure 1 shows both candles lit simultaneously. B II and III
C II and III
D I and IV
Gas jar
Inhaled Exhaled
air air 5. The process of inhaling and exhaling occurs
Candle
without us realising. Which of the following help
Cardboard both mechanisms to continuously occur?
A The change in position of the lungs in the
T U
thoracic cavity.
Figure 1 B The difference of oxygen and carbon dioxide
Why is the candle in container U extinguished content in inhaled and exhaled air.
first? C The change in air pressure in the thoracic
A Exhaled air contains a higher oxygen content cavity as compared to atmospheric pressure.
and a lower carbon dioxide content than D The periodic pumping of the heart which
inhaled air. changes the air pressure in the lungs.
B Exhaled air contains a lower oxygen content
and a higher carbon dioxide content than 6. Why do dust that is released from industrial
inhaled air. factories and vehicles reduce the rate of
C The quantity of exhaled air in container U is photosynthesis but does not influence much of
lower than that in container T. the rate of respiration of plants?
D The air pressure in container U is less stable A Dust is only released during the day.
than the air pressure in container T. B The plant breathes only at night.
C Sunlight will be dim with the presence of
3. Why do patients with sickle cell anemia often dust.
experience fatigue? D Respiration in plants only occurs through
A The shape of the blood cell that looks like a the stomata that are located on the surface
sickle lowers the amount of haemoglobin in under the leaf.
the blood.
B The red blood cell that is sickle-shaped does 7. Which of the following is involved during gas
not have haemoglobin in it. exchange in insects?
C The sickle-cell anaemia patient has less A Trachea and tracheole
appetite which causes the patient to lose B Trachea and body cell
energy. C Tracheole and body cell
D The quantity of blood of the sickle-cell anaemia D Blood and tracheole
patient is far lower than a normal person.
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Science Form 3 Chapter 2 Respiration
SECTION B
1. (a) Figure 1 shows a respiratory organ in humans.
A
D B
CHAPTER
2
C
Figure 1
In Figure 1, tick (✓) the box that shows the muscles involved in inhalation. [2 marks]
(b) Tick (✓) the correct inhalation mechanism.
(i) The diaphragm relaxes and moves downward
(ii) The rib cage moves upwards and outwards
(iii) The air pressure around the lungs decreases
(iv) The volume of the thoracic cavity decreases
[2 marks]
SECTION C
1. Figure 2 shows the source of pollutants that can endanger the respiratory system.
X Y
Figure 2
(a) (i) Name the pollutants in X and Y. [2 marks]
(ii) What is the effect of the pollutants towards the human respiratory system? [2 marks]
(iii) State the steps that should be taken to reduce the effects in X and Y. [2 marks]
(b) David is an amateur hiker who plans to go on an expedition to climb Mount Everest. He made the
necessary preparations for the expedition. Other than food, medicine and suitable clothing, David also
prepared an oxygen tank. Justify it. [2 marks]
(c) Several steps were taken by the authorities to overcome air pollution such as law enforcement and giving
informal education to the society. In your opinion, which method is more effective? Justify.
[2 marks]
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Theme 3: Energy and Sustainability of Life
Chapter
Science Form 3 Chapter 7 Energy and Power
7 Energy and Power
CONCEPT MAP
Potential to do Rate of doing work
Work Energy Power
Force × displacement Work
Time
Unit: Joule (J)
Unit: Watt (W)
Different forms of energy
Examples
( 1 )
Potential energy Kinetic energy 2 mv 2
Gravitational Elastic potential
potential energy energy ( )
1
Fx
(mgh) 2
Principle of conservation of energy
Changes in gravitational potential energy = Changes in kinetic energy
1
mgh = mv 2
2
Access to
INFOGRAPHIC
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Science Form 3 Chapter 7 Energy and Power
3. Energy is required to run, push a trolley,
7.1 Work, Energy and Power climb stairs and etc.
4. The total work done is equivalent to the
Work amount of energy used to do the task.
1. Work is done when a force results in motion.
1 J = 1 N m
2. This means work is done when a force
causes an object to move in the direction
of the force.
Example 1
3. The distance moved by the object is called
displacement (s). A wooden block is put on the floor. Vincent pulls the
4. Work (W) is defined as the product of the wooden block horizontally using 2 N along 0.5 m.
Calculate the amount of work done.
force, F and the displacement, s in the
direction of the force.
Work, W (J) = Force, F (N) × Displacement, s (m)
5. The metric unit for work is joule (J), or
Newton meter (N m).
6. One joule equals to the amount of work
done when one Newton of force is applied
to move an object by 1 meter. Work = Force Displacement
= 2 N 0.5 m
7. The amount of work increases if: = 1 N m or 1 J
(a) the force applied is large.
(b) the displacement of object is more.
8. Work is not done if: Example 2
(a) force applied does not move the object.
For example, pushing a wall or standing Amran uses a pulley to pull a box weighing 15 kg
while holding a book. vertically from the floor. The vertical displacement is
(b) displacement to the direction of force is 4 m. Calculate the amount of work done. CHAPTER
zero, so no work is done on the weight Smooth pulley
of the object. For example, bringing
objects with constant velocity. 7
9. In conclusion, work is not done when: Load 15 kg
(a) the motion is perpendicular to the
acting force.
(b) the force is applied upon the object but 4 m
the object does not move.
Energy
(Consider 1 kg = 10 N)
1. Any object that has energy can do work. Work = Force Displacement
2. Energy is the ability to do work. = (15 10) N 4 m
= 600 N m or 600 J
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Science Form 3 Chapter 7 Energy and Power
Force Force
No movement Direction of the trolley
Activity showing no work is done Activity showing work is done
Power
Can you determine between the two cars as shown in the figure below, which car has a higher power?
Moves 2 km in 10 minutes Moves 2 km in 5 minutes
Figure 7.1 Two cars that have different powers
1. Power is defined as the rate of work done 6. Horse power is a common term to a vehicle’s
in a unit of time. efficiency.
2. The metric unit for power is Watt (W). 7. A higher horse power sums to a greater
3. One watt is equivalent to one joule of work powered engine.
done in one second. The formula for power
is as follows:
CHAPTER
7
Work (J) Example 1
Power (W) =
Time (s)
A pulley is used to lift a table weighing 1000 N. The
table is carried to a height of 2.5 m in 50 seconds.
4. One watt is equivalent to one joule per Calculate the power of the pulley.
second.
Solution:
1 W = 1 J s –1 Power = Work
Time
1 000 N × 2.5 m
5. Have you heard of horse power? Horse = 50
power means the amount of work done by 2 500
a horse in one second. = 50
= 50 W
1 horse power = 735 W
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Science Form 3 Chapter 7 Energy and Power
Example 2
An elevator in a shopping complex could fit 20 people when move upward of 10 m in 8 seconds. What is the
power of the elevator if the total weight of the people is 16 000 N?
Solution:
Work
Power =
Time
16 000 N × 10 m
=
8 s
160 000
=
8
= 20 000 W
Formative Practice 7.1
1. A student with a mass of 45 kg climbed the stairs with height of 4.8 m. Calculate the amount of work done
by the student.
2. A car with a mass of 540 kg is pulled by lorry along 1.0 km. What is the amount of work done by the lorry?
(1 kg = 10 N)
3. Figure 7.2 shows a man climbing stairs and the time taken to do so.
3 m CHAPTER
4 m 7
Figure 7.2
(a) Determine the amount of time taken by the man to climb the stairs.
(b) If the mass of the man is 70 kg, calculate the amount of work done by him. (1 kg = 10 N)
(c) Calculate the man’s power.
4. Do elevators utilise high-powered motor? Give your justification.
5. State True or False for the following statements about work done.
Statement True / False
(a) Work is not done when Fatin pushes a brick wall at her school.
(b) Work is calculated based on the speed of the force to displace an object.
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Science Form 3 Chapter 7 Energy and Power
7.2 Potential Energy and Kinetic 2. When a box is lifted to a certain height,
h from surface of Earth, work is done to
Energy
oppose gravitational forces.
Potential Energy
1. Potential energy is energy stored in the A 2
object because of its position, condition or
shape. F
2. For instance, Farid who sits at the higher h
end of a see-saw stores more potential
energy.
A 1
3. The two types of potential energy are:
(a) gravitational potential energy
(b) elastic potential energy
W
Gravitational Potential Energy
3. A force, F that is equivalent to the weight
Potential energy stored in the object of a box, W is required to lift the box from
because of its higher position from ground. position A to A .
2
1
4. Gravitational potential energy that is stored
1. Examples include: in the box at position A is similar to the
(a) Rocks on a mountain work done to lift the box. 2
(b) Fruits on a tree
(c) Birds on a telephone pole Gravitational potential energy = Work done
(d) Bats sleeping hanging upside down
5. Gravitational potential energy can be
calculated using the formula as shown
below:
Force = F
Mass = m
Weight = W
Gravitational acceleration = g
CHAPTER
Displacement = Distance = h
7
F = W = m × g
Thus,
F = mg
Gravitational potential energy
= Work done
= Force × Displacement
= m × g × h
Figure 7.3
Thus,
Relationship between Work and Gravitational Gravitational potential energy = mgh
Potential Energy
1. Work done that is against the gravitational
force is known as gravitational potential
energy.
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Science Form 3 Chapter 7 Energy and Power
Example 1 Example 3
A weight that has a mass of 40 kg is lifted to the Calculate the gravitational potential energy of an
height of 5.2 m. Calculate the gravitational potential object with a mass of 2 kg at a height of 0.5 m.
energy gained by the weight.
Solution:
Solution: Gravitational potential energy = mgh
Gravitational potential energy = mgh = 2 × 10 × 0.5
= 40 × 10 × 5.2 = 10 J
= 2 080 J
6. The quantity of gravitational potential
Example 2 energy depends on the:
(a) mass of objects
Iffat and Sam each has a mass of 40 kg and (b) distance of objects from surface of Earth
55 kg respectively. They climb stairs with a vertical (c) strength of the Earth’s gravitational field
height of 2 m. Who has more gravitational potential
energy stored? on the object.
7. Amount of gravitational potential energy
Solution:
Gravitational potential energy of Iffat = 40 × 10 × 2 increases when the:
= 800 J (a) mass of object increases.
(b) height of object from surface of Earth
Gravitational potential energy of Sam = 55 × 10 × 2 increases.
= 1 100 J
(c) gravitational strength increases.
∴ Sam has more gravitational potential energy than
Iffat. 8. The gravitational potential energy for an
object on the surface of the Earth is zero.
9. The benefits of gravitational potential energy are: CHAPTER
A swimmer can dive 7
easily into the water
from an elastic board.
Water in a dams can move faster to turn turbines
to generate electrical energy in hydroelectric
power stations.
Kids can slide down
easily.
Figure 7.4
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Science Form 3 Chapter 7 Energy and Power
Elastic Potential Energy Example 1
Elastic potential energy is the energy stored Figure 7.5 shows a spring with length of 20 cm
as a result of deformation of an elastic was compressed with a 2 kg weight till the length
object as it is compressed or stretched. decreased to 12 cm. What is the value of the elastic
potential energy of the spring?
Relationship between Work and Elastic
Potential Energy
Compress
1. Energy is needed to compress and stretch 2 kg F = 20 N
elastic objects like springs and elastic bands. 20 cm
12 cm
2. Elastic objects like springs gain energy when
work is applied on it by compressing or
stretching it. Figure 7.5
3. When the elastic object is stretched or Solution:
compressed, the energy is known as elastic
potential energy. Displacement, x = Original length – Length upon
compression
4. The amount of work done to stretch or = 20 cm – 12 cm
compress elastic object is equivalent to the = 0.08 m
quantity of elastic potential energy stored Elastic potential energy = 1 Fx
in the object. 2
1
Elastic potential energy = Amount of work done = 2 × 20 × 0.08
= 0.8 J
5. The force required to stretch a spring to a
certain distance (displacement from the
equilibrium position, x) increases uniformly Example 2
with distance. Figure 7.6 shows an arrow pulled on an elastic bow
string. The bow string was pulled with 25 N force
and it was pulled 0.5 m from its original length.
Calculate the elastic potential energy that is stored
in the bow string.
x Bow
CHAPTER
F 2x
7
0.5 m
2F
Force
6. Thus, to calculate work done to stretch 25 N
spring, we use the average force value, that
is: Arrow
0 + F 1
Average force = = F
2 2 Figure 7.6
7. Potential elastic energy is calculated with Solution:
this formula: 1
Elastic potential energy = Fx
Elastic potential = Work Done 2
energy = Force × displacement = 1 × 25 × 0.5
1 2
= F × x = 6.25 J
2
1
= Fx
2
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Science Form 3 Chapter 7 Energy and Power
Kinetic Energy Example 3
Kinetic energy is the energy possessed by A car with a mass of 800 kg moves at a speed of
–1
an object due to its motion. 110 km h in the highway. Calculate the kinetic
energy of the car in motion.
1. The amount of kinetic energy on a moving Solution:
object depends on the: Mass, m = 800 kg
(a) mass of the moving object. 40 × 1 000
–1
(b) speed. Speed, v = 40 km h = 3 600
2. If the mass of the object increases and its speed = 11.11 m s –1
increases, thus the kinetic energy increases. 1
Kinetic energy of car = mv 2
3. The amount of kinetic energy of an object 2
in motion is determined using the formula = 1 × 800 × 11.11 2
2
below: = 49 372.84 J
1 = 49.37 kJ
Kinetic energy = × mass × (speed) 2
2
1 4. From example 1, it is concluded that a car
= mv 2
2 with higher speed has more kinetic energy.
Factor Unit 5. Higher kinetic energy of the car can often
cause injuries and serious damage during
Work Joule (J)
collision.
Mass Kilogram (kg)
6. Thus, we have to drive safely with
Speed Meter per second (m/s) appropriate speeds.
Example 1 Kinetic Energy and Work Done
Compare the kinetic energy for two cars of the Newton’s First Law
same mass, that is 950 kg which travel at different An object will remain in its motion
speeds: unless a force acts upon it to change
(a) Car P = 7 m/s
(b) Car Q = 33 m/s the motion. CHAPTER
Solution: 1. According to Newton’s First Law of motion,
1
(a) Kinetic energy of car P = × 950 × 7 2 an object that moves at a constant speed
2 7
= 23 275 J will maintain its kinetic energy.
= 23.28 kJ 2. Work is done when the kinetic energy of a
1 moving object increases or decreases.
(b) Kinetic energy of car Q = × 950 × 33 2
2 3. The change in the kinetic energy of an
= 517 275 J object is similar to the work done on the
= 517.28 kJ
object.
∴ Kinetic energy of car Q is higher than that of car P.
4. If a trolley of mass m kg moves along a
horizontal even surface with speed of u m s
–1
until it reaches v m s speed, the work done
–1
Example 2
to increase the speed of the trolley is:
Calculate the kinetic energy of a ball with a mass
of 200 g and moves at 12.2 m s . Work = Change in kinetic energy
–1
1 1
Solution: = mv – mu 2
2
1 2 2
Kinetic energy of the ball = × 0.2 × 12.2 2
2 v = Final speed
= 2.44 J u = Initial speed
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Science Form 3 Chapter 7 Energy and Power
If trolley moves form a resting position, then Principle of Conservation of
u is zero (0). 7.3
Thus, Energy
1 1
Work = mv – m(0) 2 1. Principle of conservation of energy states
2
2 2
that:
1
2
= mv – 0
2 • Energy cannot be created or
destroyed.
1
= mv 2 • Energy can be transformed from
2
one form to another.
1 • The sum of all energy is always the
Work done = mv 2 same.
2
2. The pendulum system abides to the
Formative Practice 7.2 principle of conservation of energy.
Example: simple pendulums and oscillation
1. Identify the energy stored in objects below: of a spring with weights.
Types of 3. In the pendulum swing, although the
Condition of object energy in potential and kinetic energy changes from
object one form to another, the sum of both energy
(a) Water in dams is always the same for any position of the
pendulum bob during the swing.
(b) Spinning ceiling fan
(c) A person sitting at the Transformation between Potential
end of a water slide
energy and Kinetic energy
2. Calculate the gravitational potential energy 1. In everyday life, transformation of potential
gained by the object at the final position. energy to kinetic energy of an object
Smooth Pull happens when the object falls down or is
pulley tossed upwards.
2. When an object falls from a higher position,
the speed increases.
The rope is
1.2 m held so that it
CHAPTER
5 kg does not move 3. This means when an object falling,
gravitational potential energy is decreasing
7
Initial Final
but kinetic energy is increasing.
Figure 7.7 4. The decrease in potential energy equals to
3. The length of a spring increases by 2 cm after it the increase in kinetic energy.
is hung upon a load of mass of 500 g. Calculate 5. Changes in the gravitational potential
the elastic potential energy of the spring.
energy is the same as that in kinetic energy.
The formula is as follows:
1
mgh = mv 2
2
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Science Form 3 Chapter 7 Energy and Power
Kinetic energy = 0
Potential energy = Maximum Kinetic energy = Maximum
Potential energy = 0
Height
increases
Original position
Kinetic energy = 0
Potential energy = Maximum
Original Height
position increases
Figure 7.8 Changes of energy in simple pendulum system
Example 1 Example 2
A rock with a mass of 4 kg is dropped from a height A durian with a mass of 1.5 kg fell from a tree of 20
of 4 m. Calculate the kinetic energy of the rock when m height. Calculate the
it drops to the ground. (a) speed of durian before it hit the ground.
(b) quantity of kinetic energy before it hits the
Solution: ground.
Loss of gravitational = Increase in
potential energy kinetic energy Solution:
(a) Kinetic energy = Potential energy
Loss of gravitational = mgh
potential energy = 4 × 10 × 4 1 mv = mgh CHAPTER
2
= 160 J 2
1 × 1.5 × v = 1.5 × 10 × 20
2
∴ Increase in kinetic energy = 160 J 2
v = 1.5 × 10 × 20 × 2 7
2
1.5
v = 400 m s –1
2
v = 20 m s –1
1
(b) Kinetic energy = mv 2
2
1
= × 1.5 × 400
2
= 300 J
Formative Practice 7.3
1. Define Principle of Conservation of Energy.
2. An elastic ball with a mass of 0.12 kg which is at rest is dropped from a height of 1.8 m (from ground).
Calculate the
(a) speed of the ball before hitting the ground.
(b) quantity of kinetic energy before the ball hits the ground.
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07 Focus KSSM SC F3.indd 121 22/03/2023 4:25 PM
Science Form 3 Chapter 7 Energy and Power
Summative Practice 7
SECTION A
1. Four motorcycles P, Q, R and S move on the 4. Which of the following are units for work?
road at different speeds. Which motorcycle has I Joule (J)
the highest kinetic energy? II Watt (W)
III Joule per second (Js )
-1
Motorcycle Mass (kg) Speed (km h ) IV Newton meter (N m)
-1
P 160 110 A I and II
B III and IV
Q 130 115 C II and III
R 140 120 D I and IV
S 120 125
5. Which of the following factors influence the
A Motorcycle P C Motorcycle R quantity of gravitational potential energy of a
B Motorcycle Q D Motorcycle S
body?
I Mass of the body
2. Which of the following stores elastic potential II Size of the body
energy? III Speed of the body
A A compressed spring IV Distance of the body from the surface of the
B A fruit on a tree Earth
C A bird nest on the tree A I and II
D Water in a dam B III and IV
C II and III
3. Which of the following statements is true? D I and IV
A The amount of gravitational potential energy
that is stored in two bodies with different mass
is the same at the same height.
B Gravitational potential energy is equal to work
done
C Energy can be created but cannot be
destroyed.
D Gravitational potential energy is equal to the
CHAPTER
product of force and displacement.
7
SECTION B
1. (a) Mark (✓) the situation that shows kinetic energy.
(i) (ii) (iii) (iv)
[2 marks]
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07 Focus KSSM SC F3.indd 122 22/03/2023 4:25 PM
Science Form 3 Chapter 7 Energy and Power
(b) Determine whether the statements below are True or False.
Statement True / False
(i) The ceiling fan has gravitational potential energy.
(ii) The kinetic energy of an individual hiking a mountain increases.
[2 marks]
SECTION C
1. (a) Figure 1.1 shows a man participating in an archery contest. The bowstring is pulled with 25 N of force
at a distance of 0.5 m from the original length.
0.5 m
25 N
Figure 1.1
(i) Name the type of energy stored in the bowstring and calculate the amount of energy. [3 marks]
(ii) Is work done by the male participant? If yes, what is the quantity of the energy done? [3 marks]
(b) Figure 1.2 shows Marissa swinging on a swing.
J L CHAPTER
7
K
Figure 1.2
(i) At which position does Marissa have maximum potential energy and maximum kinetic energy?
[2 marks]
(ii) Is the energy stored in Marissa’s body at position J the same as the amount of energy at position
K? Give a reason for your answer. [2 marks]
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07 Focus KSSM SC F3.indd 123 22/03/2023 4:25 PM
ANSWERS Science Form 3 Answers
Chapter 2. Statement Monocular Stereoscopic (c) (i) Absorbing moisture in the air
1 Stimuli and Responses causes the environment in the
(a) beaker to be dry / without air
(b) (ii) J - Roots grow upwards
towards the area of moisture
HOTS Challenge (c) that is found in the area
Formatif 1 1
Pr
(d) outside the beaker.
Sneezing is an involuntary action because it (e) K - Roots grow downwards in
the direction of water.
cannot be controlled. (iii) Roots grow in the direction of
3. To move in the dark. It detects the water.
Formative Practice 1.1 reflection of the sound waves of objects
in the surroundings, allowing the bats to
fly in the dark.
1. (a) receptor Chapter
(b) impulse 4. To attract the attention of the male 2 Respiration
(c) nerve cells cockchafer.
(d) brain
(e) effector Summative Practice 1 Formative Practice 2.1
1. Inhalation and exhalation
SECTION A
HOTS Challenge 2. Oxygen and carbon dioxide
Formatif 1 2
Pr
1. D 2. C 3. D 4. A 3. (a) Alveolus
The front of the eye lens is curved; it bends the (b) Trachea
light, produces inverted image on the retina. SECTION B (c) Diaphragm
Formative Practice 1.2 1. (a) P: Semicircular tube 4. The oxygen content in the inhaled air
is higher than that of the exhaled air to
Q: Cochlea
(b) P - Helps to balance the body supply enough oxygen for the use of
1. Sensory Stimuli detected Q - Changes sound waves to nerve the body’s cells.
organs impulses
Eyes Light 2. (a) (i) True Formative Practice 2.2
Ears Sound (ii) False 1. Diffusion
Nose Chemical substances (b) (i) short-sightedness 2. Oxygen diffuses into the blood in the
in air (ii) thin blood vessels and combines with the
Tongue Chemical substances SECTION C haemoglobin to form oxyhaemoglobin.
in food When it reaches the body tissues,
1. (a) (i) Monocular oxyhaemoglobin releases oxygen.
Skin Touch, temperature, (ii) Has a wide range of vision. Oxygen then diffuses into the body
pressure This animal can detect tissues.
predators around it more 3. The alveolus has a large surface
2. Rod cells and cone cells effectively. area, very thin walls, moist surfaces
3. Outer ear, middle ear and inner ear (b) (i) Sees small objects such as and is covered with numerous blood
bacteria // Sees far objects capillaries.
4. Reduce the risk of injury / bleeding
such as planets.
when a person falls (ii) Use microscope // Use
Formatif 1 1
5. Sour, salty, sweet, bitter and umami telescope HOTS Challenge
Pr
Formative Practice 1.3 (c) (i) The cornea of the eyes is not Wear a face mask to prevent dust / smoke
flat. The light from the object
cannot be focused onto one from the haze from entering the nose and
1. Geotropism, hydrotropism, phototropism point on the retina.
and thigmotropism (ii) Possible. Use cylindrical mouth which irritates the respiratory tract.
2. (a) Root spectacles. Formative Practice 2.3
(b) Root (d) Watching television from a close
(c) Shoot/ Leaf distance // Seeing the screen of a 1. Nicotine, sulphur dioxide and carbon
(d) Root smartphone in the dark. monoxide
3. A tropism depends on the direction 2. (a) (i) Tropism response happens 2. Disease: Asthma
of the stimulus whereas a nastic very slowly but nastic response Causes: Cigarette smoke
movement does not depend on the happen very quickly. Symptoms: Difficulty breathing and
direction of the stimulus. A nastic (ii) Phototropism / Hydrotropism / chest tightness
movement occurs more quickly Geotropism / Thigmotropism / (Accept other suitable answers)
compared to a tropism. Photonastic / Thermonastic / 3. A passive smoker is a non-smoker who
Seismonastic is exposed to cigarette smoke from the
Formative Practice 1.4 (b) No. smokers around them.
It will not grow towards the sunlight 4. The woman is a passive smoker.
1. Monocular and stereoscopic vision / window / door because there Passive smokers also face the same
is not enough light to carry out risk as smokers.
photosynthesis.
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Science Form 3 Answers
(c) – Enforce the law: Harsh actions Formative Practice 3.5
HOTS Challenge cause people to be afraid of
Formatif 1 2
Pr
doing the wrong things 1.
Bright red gill blood contains higher levels of – Awareness campaign: Increase Statement Human Plants
civic
and
responsibility
oxygen. This indicates that the fish is still new awareness to obey the rules (a)
and fresh compared to the dark red gill blood.
without force
(b)
Formative Practice 2.4
Chapter (c)
1. (a) Frogs – Moist skin 3 Transportation
(b) Grasshopper – Trachea (d)
(c) Fish – Gills Formative Practice 3.1
2. The gill surface is layered and covered (e)
with numerous blood capillaries. 1. To transport oxygen and nutrients from
3. Spiracles – Openings that allow air to the environment to all body cells and (f)
enter the body remove the waste materials from body
4. To gain more oxygen. This will be cells to the environment. 2. The materials needed for cell survival
used to generate energy for muscle 2. (a) Simple cannot be delivered to the cells and the
waste materials cannot be removed.
movement. (b) small / fine
5. Normal blood cells are biconcave, while (c) diffusion
the blood cells of a person who suffers (d) amoeba Summative Practice 3
from sickle cell anaemia are crescent-
shaped. The crescent red blood cells Formatif 1 1 SECTION A
Pr
have abnormal haemoglobin (oxygen- HOTS Challenge
carrier pigment), thus affecting the rate 1. B 2. D 3. C 4. B 5. B
of oxygen transported in the body. Arterial blood vessels are located close to the
skin. Pulse is easier to detect. SECTION B
Formative Practice 2.5 Formative Practice 3.2 1. (a) (i) Mammal
1. oxygen, carbon dioxide, stoma, diffusion (ii) Fish
2. 1. Left ventricle. The ventricle pumps (b) X: Red blood cell
oxygenated blood into other parts of the
body (except the lungs). Y: Platelet
SECTION C
Stoma open 2. To allow blood to flow in one direction
only 1. (a) (i) Has one ventricle // Has three
3. This increases blood pressure and flow lung cavities
rate, thereby speeding up the delivery (ii) Frog / Salamander / Newt
3. The stoma on the leaves will be coated of nutrients and oxygen to the body (iii) • Oxygenated blood and
deoxygenated blood mixes
by dust and dust particles. Therefore, tissues. • Blood that is pumped to
oxygen cannot enter and carbon dioxide 4. Artery, vein and capillary the body tissues have less
cannot get out of the leaves. The rate oxygen
of respiration and photosynthesis of the Formative Practice 3.3
plants will be affected. (b) (i) Pulmonary circulation system
1. Red blood cell, white blood cell, platelet (ii) Carries blood from the heart to
and plasma the lungs and back to the heart.
Summative Practice 2 (iii) – The amount of oxygen in R
2. Blood group A, B, AB and O.
is higher than in Q
SECTION A 3. Blood group A and O. – Q carries oxygenated blood
1. C 2. B 3. A 4. D 4. Patients with AB blood group can from the heart to the whole
5. C 6. D 7. C receive blood from all blood groups. body whereas R carries
deoxygenated blood from
SECTION B Formative Practice 3.4 the body tissues
– The pressure in R is higher
1. (a) B and C 1. (a) Transpiration is the evaporation of than in Q
(b) (ii) and (iii) water from the aerial parts of the
plant.
SECTION C (b) • Low light intensity: The stoma Chapter
4
1. (a) (i) X - Carbon monoxide closes in dark conditions. This Reactivity of Metals
Y - Sulphur dioxide reduces the rate of transpiration.
(ii) X - Competes with oxygen to • Absence of air movement: Water Formative Practice 4.1
bond with haemoglobin vapour tends to form at the
Y – Irritate the respiratory tract. bottom of the leaves and around 1. Earth’s crust
(iii) Install a catalytic converter the stoma. This will reduce the 2. Natural element minerals are less
// Use lead-free fuel // Use rate of transpiration. reactive towards oxygen. These
renewable energy // Carpool 2. Helps plants absorb and transport water. elements do not combine chemically
(b) – The amount of oxygen in 3. Xylem – Transports water and mineral with other natural elements. Natural
highlands is low. salts compound minerals are composed
– Humans need to breathe with the Phloem – Transports products of of elements that are reactive towards
help of an oxygen tank. photosynthesis oxygen. These elements tend to react
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11 Answer Focus Science Tg3.indd 162 22/03/2023 4:27 PM
Science Form 3 Answers
with other elements such as oxygen, (d) (i) Magnesium + Oxygen → Chapter
carbon dioxide and sulphur to form Magnesium oxide 6 Electricity and Magnetism
compounds. (ii) Copper + Oxygen → Copper
3. Natural element minerals: Gold, silver, oxide
mercury (e) The rate of reaction of metals with Formative Practice 6.1
Natural compounds mineral: Calcium oxygen slows down.
carbonate, aluminium oxide (any 1. – Renewable energy can be used
suitable answer) continuously but non-renewable
Chapter energy cannot be used continuously.
Formative Practice 4.2 5 Thermochemistry – Renewable energy does not pollute
the environment but non-renewable
1. To determine the method of extracting Formative Practice 5.1 energy does pollute the environment.
of metal from its ore. 2. Chemical energy → Heat energy →
Potential energy → Kinetic energy →
2. Zinc, Ferum, Tin, Lead 1. Endothermic: Photosynthesis, frying an Electric energy
egg, evaporation of sweat, melting of
3. Aluminium metal is located above ice 3. Hydroelectric generator – Water
carbon in the Reactivity Series of Exothermic: Respiration, combustion Thermal generator – Steam
Metals. Carbon is the less reactive than of ethanol, decomposition of waste by Biomass generator – Hot gases
aluminium, thus carbon cannot remove bacteria in the soil, thermite reaction
aluminium from bauxite. 2. (a) Exothermic reaction 4. (a) Alternating current
(b) Endothermic reaction (b) Direct current
Formative Practice 4.3 (c) Direct current, Alternating current
1. – Electrolysis of molten ores by using Summative Practice 5 (d) Direct current
carbon electrodes SECTION A (e) Alternating current, Direct current
– Heating metal oxide with carbon in a
blast furnace. 1. D 2. A 3. A 4. C Formative Practice 6.2
2. (a) Carbon 1. (a) ✓ (b) ✓ (c) ✗
(b) Carbon can remove lead from tin SECTION B
oxide, so pure tin metal can be 1. (a) (i) True (d) ✓ (e) ✓ (f) ✗
obtained. (ii) False 2. Voltmeter S
(c) Greenhouse effect. Carbon dioxide (b) (i) released V N
gas traps heat. The increase in (ii) heat energy 3. V s = N s
percentage of carbon dioxide gas p p
in the atmosphere will cause an SECTION C = 50 × 50
increase in atmospheric temperature. 1. (a) (i) Exothermic reaction, heat is 100
3. – Develop a former mining site to a released to the environment = 25 V
recreation site (ii) Endothermic reaction, heat is 4. 40
– Phytoremediation method in recovery absorbed from the environment
of soil contaminated by heavy (b) (i) Apparatus X: Thermometer 5. (a) Step-down transformer
metals. Parameter: Temperature (b) Step-up transformer
(ii) The student's mistake is using 6. N s = V s
Summative Practice 4 a glass cup because glass will N p V p
absorb heat that is released 60 V 1
SECTION A during the thermochemical 240 V = 4
reaction. The student needs to
1. B 2. D 3. A 4. D replace the glass cup with a ∴ Ratio of turns of coil = 4 : 1
polystyrene cup.
SECTION B Formative Practice 6.3
(c) (i)
1. (a) (i) Aluminium Bamboo 1. 240 V
(ii) Iron
2. (a) Neutral wire
(b) Element: Gold, Diamond Ziplock plastic bag (b) Live wire
Compound: Magnetite, Marble
(c) Earth wire
SECTION C Vinegar 3. (a) 240 V
Limestone
1. (a) Magnesium: Burns very brightly (b) 415 V
Copper: Glow brightly 4. Fuse, earth wire and circuit breaker.
(b) (i) X: Potassium manganate(VII) (ii) • Crush the limestone into 5. The 2-pin plug does not have the earth
small pieces and put them
Supplies oxygen into the zipped plastic bag. wire but the 3-pin plug does have the
(ii) Avoids mixing of potassium • Mix some vinegar into the earth wire.
manganate(VII) powder with plastic bag until it is half full.
the metallic powder which • Put a bamboo stick and zip Formative Practice 6.4
might cause an explosion the plastic bag.
(c) Magnesium • The reaction between the 1. Energy efficiency is the percentage of
Zinc more reactive limestone and vinegar energy input converted to useful form of
Copper will release heat because energy output.
Argentum the neutralisation reaction
of acid and base is an
exothermic reaction.
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11 Answer Focus Science Tg3.indd 163 22/03/2023 4:27 PM
Science Form 3 Answers
2. Not equal. Energy efficiency can be (f) The voltage of electricity that is
improved via application of efficient generated at the power station will Summative Practice 7
technology such as motor, lamp and be increase before passing through
other energy-efficient technology. a cable network to be sent to SECTION A
It involves investment and expert consumers. 1.
advice to choose the right technology. C 2. A 3. B 4. D 5. D
Meanwhile, energy saving can be SECTION B
improved via practices of correct use of Chapter
energy such as turning off the light and 7 Energy and Power 1. (a) (i) and (iv)
electrical appliances when not in use, (b) (i) True
using sunlight and many more. It does Formative Practice 7.1 (ii) False
not involve costs but only a change of
user attitude towards energy. 1. Work = Force × Distance SECTION C
3. By looking at the number of stars = 45 kg × 10 × 4.8 m 1. (a) (i) Elastic potential energy
on the energy efficiency label of the = 2 160 J 1
refrigerator. One star refer to low energy 2. Work = Force × Distance = 2 Fx
efficiency while five stars refers to very = 540 kg × 10 × 1 000 m 1
high energy efficiency. = 5 400 000 J = 2 x 25 N x 0.5 m
4. (a) Voltage = 240 V = 5 400 kJ = 6.25 Nm (6.25 J)
Power = 1 500 W 3. (a) 12 s
P (ii) Yes, work is done. The quantity
(b) I = V (b) Work = Force × Distance of work done is equal to the
1 500 = 70 kg × 10 × 3 m elastic potential energy that is
= = 2 100 J stored in the arrow string which
240
= 6.25 A (c) Man’s power = Work is 6.25 J.
(c) Cost = 1.5 kW × 2 h × RM0.20 per Time (b) (i) Maximum potential energy:
unit = 2 100 J J and L
= RM0.60 12 s Maximum kinetic energy: K
= 175 W (ii) Yes, because according to the
Summative Practice 6 4. Yes, because it needs more energy to Principle of Conservation of
Energy, the amount of energy
do work to oppose gravitational force
when moving upwards before the changes is the same
SECTION A as the amount of energy after
5. (a) True (b) False the change.
1. D 2. B 3. A 4. C
5. B 6. C 7. A 8. B Formative Practice 7.2
Chapter
9. A 1. (a) Gravitational potential energy 8 Radioactivity
(b) Kinetic energy
SECTION B
(c) Gravitational potential energy Formative Practice 8.1
1. (a) (i) True
(ii) False 2. Gravitational potential energy = mgh 1. (a) Antoine Henri Becquerel
(b) P: Thermal electric generator = 5 kg × 10 × 1.2 m (b) Wilhelm Roentgen
= 60 J
Q: Hydroelectric generator 1
3. Elastic potential energy = Fx (c) Marie Curie
SECTION C 1 500 2 2 (d) Pierre Curie
= × kg × 10 × m
1. (a) A network of cables and 2 1 000 100 (e) Marie Curie
transformers that connect all = 0.05 J 2. Becquerel (Bq) and Curie (Ci)
housing, schools, offices and
factories to the main power station Formative Practice 7.3 3. (a) Radioactive materials are elements
with an unstable nucleus.
(b) P: Step-up transformer 1. The principle of Conservation of Energy
Q: Step-down transformer states that energy cannot be created (b) Radioactivity is a process
R: Step-down transformer nor destroyed but can be transformed whereby radioactive rays released
(c) Alternating current from one form to another. spontaneously and randomly from
an unstable radioactive substance
(d) Similarity: Both transformers have 2. (a) Gained kinetic energy change into a more stable element.
two coils. = Loss of gravitational potential
Difference: The number of turns energy (c) The half-life decay concept of a
of the primary coil in transformer 1 radioactive material is the time
2
P is less than the number of turns 2 mv = mgh taken by radioactive atoms to
of the secondary coil whereas the 1 decay to half of its original number.
2
number of turns of the primary coil 2 × 0.12 × v = 0.12 × 10 × 1.8 m 4. Unstable elements can become
in transformer Q is more than the v = 36 stable elements when decay occurs
2
number of turns of the secondary v = 6 m s –1 spontaneously.
coil.
(e) Different energy requirements for (b) Kinetic energy of the ball
1
industrial and residential areas. = mv 2
2
1
= × 0.12 × 36
2
= 2.16 J
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11 Answer Focus Science Tg3.indd 164 22/03/2023 4:27 PM
Science Form 3 Answers
Formative Practice 8.2 3. Radioactive materials and waste must Chapter
be stored in a thick-walled container
9
1. (a) made of concrete and lead. This Space Weather
container with waste is buried in a mine
that is not in use or is not inhabited by Formative Practice 9.1
Neutron humans.
Nucleus 4. Short-term effects – Nausea, vomitting, 1. (a) ✓
Electron Proton lethargy (b) ✗
Long-term effects – Cell mutation, (c) ✓
tumours, infertility (d) ✗
5. Alpha rays cannot penetrate paper like 2. (a) Sunspot
beta rays
(b) Solar flare
6. – Sterilise food and surgical equipment
(b) Yes, because the number of protons – Kill cancer cells and tumours (c) Solar prominence
(3) is the same as the number of 3. The surface of the Sun, the
electrons (3). Summative Practice 8 photosphere is constantly turbulent
(c) Atoms become positive-charged due to the release of gases from the
ions. SECTION A convection zone. The surface of the
2. Four photosphere is seen uneven due to the
3. Positive ions are formed when atoms 1. D 2. B 3. B 4. A 5. A presence of granules.
lose electrons, while negative ions are SECTION B
formed when atoms receive electrons. Formative Practice 9.2
4. (a) ✓ 1. (a) (i) Discovered radium and polonium 1. Space weather is the environmental
(b) ✗ (ii) Discovered radioactivity change around the space between the
(c) ✗ phenomena atmosphere of the Sun and the Earth.
(b) X-rays, Gamma rays 2. Activities that occur on the surface of
Formative Practice 8.3 SECTION C the Sun.
3. (a) Communication disruption
1. (a) Ionising radiation is electromagnetic 1. (a) Substance P is a radioactive (b) Aurora
rays that have sufficient energy to substance which is an element
ionise atoms or molecules. whose nucleus is unstable. (c) Cell mutation
(b) Non-ionising radiation is (b) Substance P is undergoing 4. An aurora occurs when the cosmic rays
electromagnetic rays that do not radioactive decay (process X) to collide with the atoms and molecules in
have enough energy to ionise change into a more stable element. the Earth’s atmosphere.
atoms or molecules
(c) Before process X, the nucleus 9
2. (a) Non-ionising radiation of substance P is unstable but Summative Practice
(b) Ionising radiation after process X, the nucleus of SECTION A
(c) Ionising radiation substance P becomes more stable. 1. B 2. B 3. D 4. A
(d) Non-ionising radiation (d) Carbon-14 // Iodine-131 //
Cobalt-60 5. C 6. B 7. A 8. A
(e) Ionising radiation
(e)
3. Alpha, beta and gamma rays SECTION B
4. Sievert 1. (a) (i) and (iii)
5. Soil, rocks and plants (b) (i) Magnetosphere
(ii) Space weather
Formative Practice 8.4 2. (a) (i) Convection zone
(ii) Corona
1. – To control the population of pests
by hindering their breeding or killing (b) (i) type G
them with gamma rays. (ii) denser
– Gamma rays are used to mutate This logo must be displayed on the SECTION C
plant cells to generate new genes outer part of the storage container
with high quality. of substance P to warn others to 1. (a) The Sun is categorised as a star
2. Employees who handle radioactive be careful of the dangers of the because it can generate its own
waste should wear badges known substance. energy.
as dosimeters to detect their level (f) Shows that the substance inside it (b) – Consists of two types of gas,
of exposure to radioactive rays. releases dangerous ionising rays. namely hydrogen and helium
Photograph film on dosimeters will – The surface temperature is
change in colour when exposed to about 6 000°C
radioactive rays. The black colour of – About 109 times the diameter
of the Earth
film shows the level of exposure to the
rays. (c) Core
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11 Answer Focus Science Tg3.indd 165 22/03/2023 4:27 PM
Science Form 3 Answers
(d) The Sun generates energy through Formative Practice 10.2
nuclear fusion that occurs in its Summative Practice 10
core. Nuclear fusion is a nuclear 1. The technology used in space
reaction process whereby four exploration allows humans to study SECTION A
hydrogen atoms combine to form the universe and its objects in more
one helium atom. During this detail. For example, the use of space 1. B 2. B 3. D 4. C 5. B
reaction, energy is released in the telescopes and space probes enables SECTION B
form of photons (light particles). astronomers to obtain direct information
(e) – Effects on technology about planets in outer space. In 1. (a) (i) ✗
addition, the creation of rockets and
and equipment on Earth space shuttles allows astronauts and (ii) ✓
for example high energy appliances to be taken into space to (b) (i) Kepler
particles from the Sun hit conduct space studies. (ii) Ptolemy
the communication satellites,
causing a disruption of 2. Satellites are divided into several types SECTION C
television broadcasts and according to their respective functions:
telephones. • Communication satellites: Used for 1. (a) X: Satellite
– Effects on human health. radio communications and television Y: Space shuttle
Radiation that is released broadcasts Z: Space station
from the Sun can cause cell • Weather satellites: Used to predict (b) Y can be used repeatedly. Y can
mutations on the human body. Earth’s weather transport astronauts from the
• Navigation satellites: Used by
airplanes and ships to determine space station to the Earth again.
their position accurately. (c) (i) The place for astronauts to do
Chapter
10 Space Exploration 3. • As a research and development research while being in outer
centre in remote sensing technology space.
Formative Practice 10.1 and other related technologies such (ii) International Space Station
as GIS and GPS.
(ISS), Salyut 1, Skylab
1. Astronomy is a scientific field that • Involved in supplying satellite imagery (d) We need to continue space
studies the universe as well as the of remote sensing to users. exploration because the
objects in it. 4. The space exploration project is information obtained will allow us
2. Johannes Kepler, Aristarchus, Nicolaus desirable because through this to understand the evolution of the
Copernicus exploration, we can unearth the secrets Solar System, natural phenomena
of Earth, planets and space objects /
3. Ptolemy’s theory states that Earth search for new mineral sources from and forces that affect the Earth.
is the centre of the universe, while the Moon, other planets and asteroids.
Copernicus’s theory states that the Sun (Other correct answers are acceptable)
is the centre of the universe. Answers for UPSA &
4. The heliocentric theory is true because UASA Model Papers
according to this theory, the Sun is https://qr.pelangibooks.com/?u=kDTIow11
the centre of the solar system and the
planets circulate around the Sun in their
orbits.
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11 Answer Focus Science Tg3.indd 166 22/03/2023 4:27 PM
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CC033943
ISBN: 978-629-7557-49-6
Fulfil UASA
Assessment Format
PELANGI