Spotlight A+ Physics Form 4.5

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Extra Features of This Book







CHAPTER
7 Quantum Physics
SMART SCOPE • Explain the initiation of the quantum theory. Page 472 CONCEPT MAP
Important Learning Standard
7.1 Quantum Theory Of • Describe quantization of energy. 476 477
• Explain wave-particle duality.
Light
• Explain concept of photon. 478
Contains learning objectives • Solve problems using photon energy, E = hf and power, P = nhf. 478 479 Contents of the whole topic
• Explain photoelectric effect.
7.2 Photoelectric Effect • Identify four characteristics of photoelectric effect that cannot be 482
explained using classsical theory.
that need to be achieved in • State minimum energy required for a photoelectron to be emitted are summarised in the form
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from a metal surface using Einstein’s equation, hf = W + 1 —mv 2 max 483
2
each topic. • Explain threshold frequency, f 0 and work function, W. 485 of a concept map.
7.3 Einstein's • Determine work function of metal, W = hf 0 = hc —– λ 0 485
Photoelectric Theory
• Solve problems using Einstein’s equation for photoelectric effect. 486
Form • Explain production of photoelectric current in a photocell circuit. 488
4 Physics Chapter 2 Force and Motion I • Describe applications of photoelectric effect in daily life. 489
1. It is more difficult to move or to stop a heavier (c) The inertia of the passengers in the bus
object such as a bowling ball compared to a keep them in their initial state of rest or
lighter object such as a football. motion.
2. In Experiment 2.2 the horizontal oscillations of • Continuous energy / Tenaga selanjar
• Black body radiation / Sinaran jasad hitam
the load an inertial balance are not influenced
by gravity. • Ideal absorber / Penyerap unggul • Photoelectric effect / Kesan fotoelektrik 266
• Radiation intensity / Keamatan sinaran
3. The period of the horizontal oscillations of the • Threshold frequency / Frekuensi ambang
• Work function / Fungsi kerja
• Wave-particle duality / Kedualan gelombang-zarah
load depends on the mass of plasticine only. • The passengers fall backwards when a Pressure
• Discrete energy / Tenaga diskrit
4. The inertia of an object has a direct relationship stationary bus suddenly moves forward.
with its mass, whereby the inertia of the object
CHAP increases when its mass increases. Important Formula and Tips
2 Pressure in Atmospheric Gas Pressure Pascal’s Principle Archimedes’ Principle Bernoulli’s Principle
• Einstein’s photoelectric equation, • Photon power, P = nhf Liquids Pressure
hf = W + 1 — 2 mv 2 max in which n is the number of photons emitted per
second
Video of inertial balance • The passengers are thrown forward when a Formula, Venturi
–—
http://bit.ly/36NykFm • Work done, W = hf 0 = hc moving the bus suddenly stops. • de Broglie's wavelength, λ = h — P Formula, Applications: Manometer F 1 tube

λ 0
whereby, f 0 is threshold frequency of metal and λ 0 • Momentum of particle, p = ABBBB P = hρg • Water tank —— = F 2 —— A 2
Figure 2.43
2mK
(d) An oil tanker lorry has large inertia. Thus,
is the threshold wavelength of metal whereby, K = 1 — 2 mv 2 • Intravenous liquid Water Buoyant Formula,
A 1
The Effects of Inertia in Daily Life • Photon energy, E = hf = hc the storage tank is divided into three • Dam manometer force, F F = ρVg Aerofoil
–—
sep
λarate compartments to reduce the
1. The astronauts in the International Space impact of the inertia of petrol on the walls • Siphon Applications:
Mercury
Station (ISS) are in zero gravity contition, hence of the tank when the lorry stops abruptly. Factors manometer • Piston’s Pascal
470
they need the inertial balance to measure mass. • Depth in • Hydraulic system Weight of object in Lift force
(Hydraulic jack,
floating state, W
2. Examples of inertia situation in daily life and
liquid
SPOTLIGHT PORTAL its effects: • Density of Angle of
hydraulic brake)
(a) The rain drops are in motion as when the
liquid
umbrella rotates. When it stops rotating,
the inertia of the rain drops causes the
Figure 2.44
rain drops to continue in motion and leave (e) When a roller coaster changes direction Measuring instruments: Units of pressure W = F W . F W , F attack
the surface of the umbrella. suddenly, the inertia of the passengers • Mercury barometer • Pascal, Pa Floating at Moving Moving Applications:
keeps them to maintain their original • Fortin barometer • mm Hg a stationary downwards with upwards with • Bunsen burner
• Aneroid barometer
• m H 2 O
• Racing car
Scan QR code to visit state of motion. The passengers must wear • millibar position an acceleration an acceleration • Sports
safety belts to remain in their seats and not
• Aeronautics
thrown out of the carriage when it moves
at sudden changes of speed and direction. Effects of atmospheric pressure • Ship and Plimsoll line
Applications
websites or videos related Photograph 2.2 • at high altitude • Submarine
• at extreme depth
• Hot air balloon
(b) The chilli or tomato sauce can be poured
out from the glass bottle by jerking the
to subtopics learnt. There bottle downwards and stopping it abruptly.
When the bottle is stopped, the inertia of
the sauce will cause the sauce to continue
moving downwards and out of the bottle. Chapter 6 Light and Optics Physics Form 4
are videos for certain Photograph 2.4
(f) When a car brakes abruptly, the driver and
B Optical fibre
passengers in the car are thrown forward Doctors use endoscope to see and examine organs
Outer cladding
due to inertia. Thus, seat belts are designed
(Low refractive index)
activities or experiments. Photograph 2.3 to prevent them from being thrown Normal Light signal inside the human body. Engineers use fibre optics
forward and injuring themselves.
to monitor performance of complex machinery.
Communication experts use fibre optics to send INFORMATION GALLERY
high speed data.
inner cladding
44 (High refractive index)
2.4.3
Figure 6.23 C Cat’s eye reflector
• Widely used in the telecommunication and • Can be used as safety devices for drivers at
medical fields. night.
• Made up of pure plastic or glass fibres. • Light rays from the headlight of a car will be
• The inner core of high refractive index is reflected by the reflector inside through total Additional information
surrounded by an outer cladding of lower internal reflection.
refractive index.
• When laser signal carrying information
such as telephone signals is introduced into related to the topic.
the inner core at one end, it will propagate
along the fibre undergoing a series of total
internal reflections until reaching the other
Form
end of the fibre. Hence the signal will be sent
5
Chapter 3 Electricity Physics with high speed and free from electrical noise
disturbances. Photograph 6.1 Cat's eye reflector on the road
3.4
Aim: To investigate how the resistivity of a wire, ρ affects its resistance.
Solve Problems Involving Total Internal Reflection
Problem statement: How the resistivity of a wire, ρ affects the resistance of the wire?
Example 3
Hypothesis: Materials with high resistivity gives higher resistance.
Figure 6.24 shows a light ray is traversing from air to a prism of refractive index 1.49.
Variables: (a) What is the critical angle of the prism?
(a) Manipulated: Resistivity of wire, ρ (b) Draw the light paths in the prism until it is emerging again into the air.
(b) Responding: Resistance of wire, R 45°
(c) Constant: Length of wire, temperature, area of cross-section of wire
ACTIVITY / EXPERIMENT (a) sin c = 1 (b) Figure 6.24 Chapter 7 Quantum Physics Physics Form 5
Solution
Apparatus and Materials:
50 cm constantan wire (s.w.g. 24), 50 cm copper wire (s.w.g. 24), 50 cm tungsten wire (s.w.g. 24), connecting
wires, three dry cells, switch, ammeter (0 – 1 A), voltmeter (0 – 5 V), rheostat and battery holder
Operational definition: n 1 C 6. The relationship between the momentum of
The resistance of the conductor, R, is given by the ratio of the reading of voltmeter to the reading of the sin c = 1.49 Example 1 45° particle, p and its wavelength, λ is
CHAP
ammeter. c = sin −1  1.49  Calculate the photon energy for light with λ = h
45°
3
sin 60°
45°45°are both
wavelengths 450 nm and 700 nm. Comp
Complete activity or Procedure: c = 42.2° photons. A 45° B —– p
1. An electrical circuit is set up as shown in
Figure 3.28.
Solution
Figure 6.25
2. A constantan wire is connected across Ammeter A The critical angle of the prisim is 42.2° Planck’s constant, h = 6.63 × 10 –34 J s which h is Planck’s constant (6.63 × 10 –34 J s)
Angle of incidence (i = 45°) larger than critical

experiment including results, Speed of light in vacuum, c = 3.00 × 10 8 m s –1
angle (c = 42.2°) at boundaries AB and AC.
terminals P and Q. The length of the wire
Wavelength, λ 1 = 450 × 10 –9 m
across P and Q is adjusted to be 30 cm long.
Total internal reflection occurred and the light
ray emerging from BC again into the air along
3. The switch is connected and the rheostat is
——— = p 2
E = hf = hc
the normal direction.
adjusted so that the ammeter gives a reading Rhesotat P Q Constantan wire Wavelength, λ 2 = 700 × 10 –9 m —— K = 1 —mv 2 × m —– ➞ K = (mv) 2 ——
2
data analysis, discussion Voltmeter Photon energy of 450 nm: λ which p = mv, p = ABBBB 2m 2m
of 0.5 A. The reading of voltmeter is taken
m
V
2mK
and recorded in Table 3.6.
CHAP
6
—————– 2
4. The constantan wire is removed and Steps
450 × 10 –9
2 and 3 are repeated for copper wire and Figure 3.28 E 1 = 6.63 × 10 –34 1 3.00 × 10 8
and conclusion to increase 6.2.3 6.2.4 = 4.42 × 10 –19 J 7. The larger the momentum of particle, the shorter BRILLIANT TIPS
tungsten wire.
the wavelength produced. The momentum of
195
Photon energy of 700 nm:
Table 3.6
Results:
E 2 = 6.63 × 10 –34 1 3.00 × 10 8 particle is p = mv, then p 2 —— = K
—————– 2
700 × 10 –9
2m
students' scientific skill. Wire Reading of ammeter, Reading of voltmeter, R = V — I / Ω = 2.84 × 10 –19 J λ = h —–– = ———— h
Resistance,
V / V
I / A
The shorter the wavelength of light, the higher
ABBBB
mv
2mK
the photon energy.
Useful tips for students
Constantan 0.5 6.3 12.6 where m is the mass of particle, v is the velocity
Form of particle and K is the kinetic energy of particle.
Copper 0.5 0.2 0.4 5
Wave-Particle Duality
to solve problems in the
Chapter 1 Force and Motion II Physics 8. Since the value of h is too small, the particle
1.2
wavelength to be detected. Thus, the wave
Tungsten 0.5 0.6 Solving Problems Involving Resultant Force and Resolution of Forces CHAP with large mass will have too short of de Broglie
1 characteristics cannot be observed.
Discussions: Example 4 Example 5 Wavelength simulation 9. In 1927, the presence of wave properties of
1. From Table 3.6, different conductor gives different resistance. http://bit.ly/39xNYqu related subtopic.
A wooden block is pulled by force, T that inclines Figure 1.17 shows the free body diagram of a electrons was confirmed through the electron
at an angle of 30° above the horizontal surface as
2. Constantan gives the highest resistance, followed by tungsten and copper gives the lowest resistance. block sliding down a smooth inclined plane. diffraction experiments.
shown in Figure 1.16.
3. Different material gives difference resistance because different material has different resistivity, ρ.
Resistance of a material increases when the resistivity of the material increases. R = 5 N T = 40 N 1. Light has wave properties because it shows the 10. Photograph 7.1 shows the diffraction pattern of
Normal reaction = 10 N
Wooden phenomena of diffraction and interference. electrons through a thin layer of graphite. The
block pattern in Photograph 7.2 resembles the light
Block
30° 2. Object has particle properties because it has diffraction pattern through an aperture.
3.2.3 F R = 18 N 335 momentum, kinetic energy and also collide with
each other.
3. Louis de Broglie introduced a hypothesis states
W = 25 N Weight,
60°
W = 20 N that all particles can exhibit wave characteristics
Figure 1.16 in year 1924.
Figure 1.17
(a) What are the magnitudes of the horizontal 4. However, it is experimentally difficult to show
component T x and vertical component T y of (a) Sketch and label the component of the weight
the wave characteristics of particles with large
the pulling force, T? of the block parallel to the inclined plane and
masses.
(b) Calculate the magnitude and direction of the the component of the weight of the block
resultant force acting on the block. perpendicular to the inclined plane. Photograph 7.1 Photograph 7.2
5. Thus, Louis de Broglie predicted that the wave
(c) Calculate the acceleration of the block if its (b) Calculate the resultant force acting on the 11. This observation proved de Broglie’s hypothesis. CHAP 7
characteristics can be shown by light particles
block.
EXAMPLE (a) Magnitude of the pulling force, (c) What is the acceleration of the block if its 477
mass is 1.8 kg.
such as electron, proton and neutron.
mass is 2.0 kg?
Solution
7.1.3
T = 40 N

T y Solution
T x = T cos 30° T (a)
= 40 cos 30° 30° Normal reaction = 10 N
= 36.64 N (to the right) T x
Example and complete T y = T sin 30° W y 60° W x
= 40 sin 30°
= 20 N (upward)
(b) Resultant of horizontal components
solution to enhance students' 60°

= T x – F R

= 36.64 + (–18)
= 16.64 N (to the right) (b) W x = 20 sin 60° = 17.32 N
Resultant of horizontal components
W y = 20 cos 60° = 10 N

understanding. = T y + R + W Resultant of the forces perpendicular to the
= 20 + 5 + (–25)
inclined plane = 10 + (–10)
= 0 N = 0 N
Resultant force on the block, F is 16.64 N to Resultant force on the block = 17.32 N
the right. (c) F = 17.32 N; Mass of the block, m = 2.0 kg
(c) F = 16.64 N; Mass of block, m = 1.8 kg F = ma
F = ma Acceleration, a = F —– m Acceleration of the block, a = F —– m
= 16.64 1.8 = 17.32
——––
——––
2.0
a = 9.24 m s –2 a = 8.66 m s –2
Try question 2 in Formative Zone 1.2
1.2.2 235
ii

TAG OF 'TRY QUESTION...
IN FORMATIVE ZONE...' Form 5 Physics Chapter 3 Electricity
9. The resistance of a wire is directly proportional
to its resistivity, ρ (Ω m) and length, l (m) but Example 17
inversely proportional to the area of cross- A copper wire with the length of l and diameter d
section, A (m 2 ), that is has the resistance of R. What is the resistance of
R ∝ ρl and R ∝ 1 — A another copper wire with length 2l and diameter
Tag placed at the end of By combining the relationships shown, therefore 1 — 2 d in terms of R?
R = ρl
—–
Solution
examples to guide students Example 15 A Both wires are of the same material, therefore, the
resistivity, ρ, are the same.
For the first wire, area of cross section,
What is the resistance of a copper wire with a A = π × 1 — 2 2 d 2 Form
in answering related diameter of 0.5 mm and length of 2 m? = πd 2 —— 4 5 Physics Chapter 6 Nuclear Physics
Assume that the wire has the shape of a cylinder.
Resistivity of copper = 1.60 × 10 –8 Ω m.
∴ Resistance, R = ρl
6.1
Solution
CHAP
questions in Formative Zone. Length, l = 2 m = ρl —– A ——– 1. Fill in the blanks to complete the following radioactive decay equation. C1 228 Ra → 0
3
πd 2
Area of cross-section, A
(c)
+ –1 e
(a)
(b)
→ 1 p + –1 e

209 Po → 205 Pb +
0
88
84
82
1
= π × 1 —————– 2 2 = 4ρl —— 4 2. Calculate the number of α and β particles that are emitted from the radioactive decay series of
0.5 × 10 –3
2
= 1.96 × 10 –7 m 2 ——– 238 U → ... → 206 Pb C4
πd 2
Resistance of wire, R = ρl —– A For the second wire, area of cross section, Aʹ 3. Table 6.2 shows the record of the activity of a radioactive sample stored in the laboratory.
92
82
———————–
= 1.60 × 10 –8 × 2 1 —d 2 Table 6.2
1.96 × 10 –7 = π × 1 2
= 0.163 Ω ——–2 Date 15 March 2021 25 March 2021 4 April 2021
2
= πd 2 Activity / s –1 1 520 380 95
Try question 9 in Formative Zone 3.2 —— 16 (a) Calculate the half-life, T 1 — of the radioactive sample. C2
2
Example 16 ∴ Resistance = ρ × 2l A (b) Sketch a radioactive decay curve for the sample. C3
———–
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Wire X with the length of 120 cm and the cross- = ρ × 2l
©PAN ASIA PUBLICATIONS
section area of 0.80 mm 2 has the resistance of ———– FORMATIVE ZONE
πd 2
3.2 Ω. Calculate the resistivity of wire X. —— 16 6.2 Nuclear Energy
Solution = 8 × 4ρl ——– 1. Nuklear energy is an atomic energy, which is Nuclear fusion
Length, l = 120 cm πd 2 energy that tied the nucleus of an atom. http://bit.ly/2Kb1gBy
= 1.2 m = 8R 2. This energy is released during nuclear reactions
such as radioactive decay, nuclear fission and
SPM SIMULATION Area of cross-section, A Try question 11 in Formative Zone 3.2 nuclear fusion. 3. The loss of mass (mass defect) occurs during
these nuclear reactions. Questions to test
= 0.80 mm 2
= 0.80 × 10 –6 m 2
Resistance, R = 3.2 Ω
Nuclear Fission
students' understanding
HOTS QUESTIONS Resistance of wire, R = ρl —– A Ensure units are converted to S.I. while performing Nuclear fission 1. Nuclear fission is a nuclear reaction when a
http://bit.ly/3mqw0vm

= 3.2
heavy nucleus splits into two or more lighter
and stable nuclei while releasing a large amount
ρ × 1.2
0.80 × 10 –6
For example,
—————— = 3.2 the calculation. of energy. at the end of each
ρ = 2.13 × 10 –6 Ω m
1 cm 2 = 1 × 10 –4 m 2 Form A barium–141 nucleus, a kripton–92
Try question 10 in Formative Zone 3.2 1 mm 2 = 1 × 10 –6 m 2 Chapter 6 Light and Optics Physics 4 nucleus and three neutrons are produced.
A uranium–235 Energy is also released.
Provides complete solution 334 SPM Simulation HOTS Questions nucleus is bombarded An unstable uranium–236 + + + subtopic.
nucleus is produced.
by a neutron.
3.2.3
3.2.6
+
92 Kr
1. Figure 1 shows refraction of light when the + + + + + + + + + + + + + + + + + + + 36
CHAP
light travels from air to plastic.
with examiner's comment Incident ray 6 1 n Examiner's Comment: + + + + + + + + + + + + + + + + + + + Energy 3 1 n 0
+
+
+ +
+
+ + + +
+ + + +
+ + Light travel slower in medium with high
+
+
+ + + + +
+
+
+
+ +
+ +
+
+ + + + + +
0 optical density. When the ray of light
235 U
enters the plastic medium, the ray bends ++ + + + + + + + +
92
to help students to answer Plastic Air refraction is smaller than incident ray. If + + + + + + + + + + + 141 Ba 56
236 U
towards the normal. Therefore, angle of
92
the plastic is replaced with less dense
medium, the bending of ray of light will 36 0
235 U + 1 n → 236 U → 141 Ba + 92 Kr + 3 1 n + energy
0
92
92
56
be reduced.
HOTS questions. Figure 1 Answer: D
Figure 6.8 Nuclear fission involving uranium-35 bombarded by a single neutron
Which is the correct pathway of the light when 452 2. Figure 2 shows the position of an image 6.2.1
plastic is replaced with less dense medium? formed by a concave lens. Which position of
A C A, B, C or D, the viewer cannot see the image?
B
Form A
5 C
Chapter 4 Electromagnetism Physics Object Image

D
B D Figure 2
SPM MODEL PAPER
Paper 1 Examiner's Comment:
Virtual image is the image that cannot
be formed on the screen.Therefore, the
1. Which diagram shows the correct catapult field? What needs to be done so that the conductor virtual image is not visible on the same
C2 moves to the right with a larger angle of side as the object. Paper 1
A C S deflection? C2 Answer: A [40 marks]
U U U U S S S
A Reverse the poles of the magnet and increase Instruction: Answer all questions.
the value of the current
B Maintain the poles of the magnet and
decrease the value of the current 3. Figure 3 shows the situation at the junction where Which of the following is a base unit A The gradient of the graph is a — b
1.
U C Reverse the direction of the current and
S S S S U U U car P is turning left while car Q is moving toward A kJ B p increases linearly with q
B D decrease the strength of the magnet the junction. The driver in car P cannot see car B µm House Car P C If q = 1 then p = a — b + c
U U U U S S S S D Maintain the direction of the current and Q because the view is obstructed by the house C cd
increase the strength of the magnet wall on the roadside. To overcome the problem D The equation of the graph is p = a — b q + b — a
2.
4. Figure 2 shows the force, F acting in the direction faced by driver P, an appropriate mirror need to be Which of the following is a derived quantity?
installed in the correct place.
I
Mass of a satellite
S on a current-carrying conductor in a magnetic (a) In Figure 3, mark Y, where the mirror need to be II Proton charge 5. Assume that the length, L, time, T and density,
field. placed. Give one reason for the answer. Wall ρ are chosen as the new base quantities of a
S S S S U U U U P (b) State the suitable mirror to be used. Give one III Electrical power physics, what is the relationship of the derived
A I and II only
reason for the answer.
quantity of mass in terms of ρ, L and T?
(c) State one feature of the mirror suggested in B I and III only
Car Q
Q
SUMMATIVE ZONE 2. A current-carrying conductor is placed in a C3 F Current 3(b) so that driver P can see the clearer image. C II and III only Figure 3 CHAP A ρL 3 3 (ρLT)
magneticfield produced by a pair of Magnadur
B LT 3
magnets.
CHAP
6 C ρ 3 L
3. The following statement provides information
4
A magnetic force acts on the conductor if
D ABBBB
A the direction of the magnetic field is the
same as the direction of the current. S about a particular system. 213 6. Figure 2 shows five steel balls P, Q, R, S and T
B the direction of the magnetic field is opposite The coordination of systems of measurement with the same densities and size. SPM MODEL PAPER
to the direction of the current. R units worldwide facilitates the field of
C the direction of the magnetic field is Figure 2 science, sports, trade, medicine and so on.
Questions of various levels What is the direction of the magnetic field?
perpendicular to the direction of the
current.
C3
3. Figure 1 shows the arrangement of apparatus to A P Which is the system mentioned above? P
A Metric System
B R
study the effect of the force acting on a current-
of thinking skill to evaluate C Q B Imperial System Q R S T SPM MODEL PAPER
carrying conductor in a magnetic field. The
C Universal Measurement System
D S
conductor moves to the left with a small angle D International System of Units
of deflection when current flows in the direction 5. Figure 3 shows a current-carrying conductor
students' understanding Current between the poles of a permanent magnet. 4. Figure 1 shows the relationship between the When ball P is pulled to the left, released and hit
Figure 2
that shown.
physical quantities of p and q.
Force – + p ball Q with a displacement, the ball P becomes
stationary while the ball T moves forward with
of each topic. S N a situation. SPM-oriented practice
the same displacement. Give a reason for this
A Impulsive force before collision = Impulsive
based on latest SPM
N Conductor (0, c) b force after collision
Magnet B Velocity of ball P before collision = Velocity
Figure 3 q of ball T after collision
S Which action will make the conductor 0 C Kinetic energy before collision = Kinetic
energy after collision
Figure 1 experience a force downwards? C1 Figure 1 D Momentum before collision = Momentum
Which of the following statements is incorrect? format to assess students
Form after collision
4 Physics Chapter 6 Light and Optics 397
on all the topics learnt in
Reinforcement & Assessment of Science Process Skills 503
1. In this experiment, you will determine the value of refractive index for water.
Carry out the following instructions, refering to Figure 1. Observer Form 4 & 5 textbooks.
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ruler
Retort stand Beaker
Water
REINFORCEMENT & ASSESSMENT Cork 9 8 7 6 5 4 3 2 1 Image pin P o Complete answers
http://bit.ly/3bHkxpp
Pin P o
Pin P 1 0 cm ANSWERS
Figure 1
SCIENCE PROCESS SKILL (a) Carry out the experiment according to the following instructions Chapter 1 Measurement (b) (i) Duration, T = 24 = 1.2 s —— 20
FORM 4
(i) Fix a pin P o at the bottom of the beaker with cellophane tape.
(ii) Fill the beaker with water until P o is at a depth of 6.0 cm from the water surface.
10
1. D 2. A 3. A
(iii) Adjust pin P i so as it is in line with the image of P o (no parallax between pin P i and image of P o ). Paper 1 7. A 8. C 4. A 9. B 10. C 5. D (ii) l = ————––— × (1.2) 2
6. B
4 × (3.142) 2
(iv) Measure the distance y between pin P i and the bottom of the beaker. 1.1 11. B 12. D 13. C 14. B 15. D = 0.36 m
(v) Calculate the apparent depth, h = H – y. 1. (a) Length 16. D 17. C 18. A 19. B 20. C
(vi) Repeat the experiment by varying the real depth, H = 8.0 cm, 10.0 cm, 12.0 cm and 14.0 cm. (b) Base unit: Meter; 21. A 22. B 23. A 24. B 25. D Section B
4. (a)
Record all the readings in the Table 1.
Symbol of unit: m

Questions for students to master the Table 1 Apparent depth, h / cm 2. (a) Scalar quantity has magnitude Paper 2 Base quantity S.I. unit Car A Speed of car / m s –1

Magnitude: 1.55;
Section A

Symbol of physical quantity: l
30.00
1. (a)
Real depth, H/ cm
y / cm
6.0
only and vector quantity has both
meter
science process skill (SPS) presented in (b) Scalar quantities: Distance, speed, Length kilogram B C D 37.49 25.20 33.30
magnitude and direction.
Time
8.0
second
10.0
Mass
time
12.0

Vector quantities: Acceleration
————
Science Practical Test. Scan the QR code 3. Base quantities: Time, mass, length (b) P = Force × Length —– s (b) Both of them use walkie talkie
E
24.20
Time
14.0
= kg m
———— × m
Derived quantities: Volume, area
s × s
or smart phone. The time taken
= kg m 2 s –3
(b) Plot a graph of H (y-axis) against h (x-axis). Start your axes from the origin (0,0). Draw a best-fit line.
below to get Tip dan Teknik Prihatin Murid FORM 4 ANSWERS 1. (a) Graph functions as a visual tool (c) S.I. unit is not same as metric to pass through point O and P is ANSWERS
recorded and the time interval, t
(c) (i) Based on the graph drawn in 1(b), determine the value of the refractive index for water.
unit because the time unit cannot
is determined or other reasonable
1.2
be expressed in terms of 10.
(ii) State two precautions that you took in this experiment in order to obtain reliable readings.

For example, 1 hour = 60 minutes
method.
to represent the relationship
in answering Science Practical Test. (b) Steps to analyse graph: = 60 × 60 seconds (c) Malaysia speed limit

= 110 km h –1
= 3 600 seconds.
between two physical quantities.

CHAP
Metric unit is based upon power
= 30.56 m s –1
6

Thus, car B exceeds the speed
(i) State the relationship
units of time cannot be expressed
between two variables of tens or decimal-based but S.I. limit on Malaysia highway.
in terms of 10.
(d) Use the ultrasound detector,
(ii) Determine the gradient of 2. (a) (i) Base quantity: Mass Complete answers are
time tracker application or other
224 graph reasonable method. Record
(iii) Determine the area under (ii) Derived quantity: Force several readings on speed within
(iii) Vector quantity: Force
the graph (b) S.I. unit G provided. Scan QR code to
5 km and calculate the average
Tip dan Teknik (iv) Determine the value of = Fr 2 ——— 5. (a) In old system: F = mlt –2
values.
a physical quantity from
interpolation
m = Ft 2 l –1
Mm
(v) Make a prediction through = ———————– so, In new FAω system: l = A 2 , 1 —
kg m × m 2
s × s × kg × kg
extrapolation
Prihatin Murid 2. (a) V increases = kg –1 m 3 s –2get explanations for objective
t = ω –1
– 1 — 2

so, m = Ft 2 l –1 = Fω –2 A
(b) V= 0 cm 3 , θ = –250 °C 3. (a) (i) Time taken, T = 26 – 2 (b) There is no suitable measuring
= 24 s
θ = 300 °C, V =22 cm 3 (ii) To get the value of one device to measure force, area and
https://bit.ly/2JfnKAo V / cm 3 complete oscillation questions.
frequency accurately.
– There is no standard tool and
accurately.
standard objects to determine
30 (iii) Repeat the experiment to get force, area and frequency.
22 cm 3 20 the average value of two sets – The unit for derived quantity
of readings for 20 complete
becomes very complex and
10
–250°C oscillations. Use electronic or hinders the communications
θ / °C digital stopwatch to measure between physicist.
–300 –200 –100 0 100 200 300 the time more accurately. – or other reasonable answers.
520
iii

CONTENTS





FORM 4 Revision Revision
Date
Date
Theme 1 Elementary Physics 4.2 Specific Heat Capacity 95
Chapter 1 Measurement 1 4.3 Specific Latent Heat 103
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1.1 Physical Quantities 3 4.4 Gas Laws 110
1.2 Scientific Investigation 6 Summative Zone 121
Summative Zone 11



Theme 2 Newtonian Mechanics Theme 4 Waves, Light and Optics
Chapter 2 Force and Motion I 20 Chapter 5 Waves 132

2.1 Linear Motion 23 5.1 Fundamentals of Waves 135

2.2 Linear Motion Graphs 31
5.2 Damping and Resonance 141
2.3 Free Fall Motion 38
5.3 Reflection of Waves 144
2.4 Inertia 42
5.4 Refraction of Waves 148
2.5 Momentum 45
5.5 Diffraction of Waves 153
2.6 Force 49
2.7 Impulse and Impulsive Force 52 5.6 Interference of Waves 159
2.8 Weight 53 5.7 Electromagnetic Waves 166
Summative Zone 55 Summative Zone 170



Chapter 3 Gravitation 64
Chapter 6 Light and Optics 181
3.1 Newton's Universal Law of 66
Gravitation 6.1 Refraction of Light 184

3.2 Kepler's Laws 74 6.2 Total Internal Reflection 191
3.3 Man-made Satellites 78 6.3 Image Formation by Lenses 197
Summative Zone 84 6.4 Thin Lens Formula 203

6.5 Optical Instruments 207
Theme 3 Heat 6.6 Image Formation by

Chapter 4 Heat 90 Spherical Mirrors 209

4.1 Thermal Equilibrium 92 Summative Zone 214




iv

FORM 5 Revision Revision
Date
Date
Theme 1 Newtonian Mechanics Chapter 4 Electromagnetism 366
Chapter 1 Force and Motion II 226 4.1 Force on a Current-carrying
Conductor in a Magnetic 368
1.1 Resultant Force 228 Field
1.2 Resolution of Forces 234 4.2 Electromagnetic Induction 378
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1.3 Forces in Equilibrium 236 4.3 Transformer 390
1.4 Elasticity 240 Summative Zone 397

Summative Zone 251
Theme 3 Applied Physics
Chapter 5 Electronics 409

Chapter 2 Pressure 264
5.1 Electron 411
2.1 Pressure in Liquids 267 5.2 Semiconductor Diode 417

2.2 Atmospheric Pressure 274 5.3 Transistor 422

2.3 Gas Pressure 278 Summative Zone 430

2.4 Pascal's Principle 281
Theme 4 Modern Physics
2.5 Archimedes' Principle 286
Chapter 6 Nuclear Physics 445
2.6 Bernoulli's Principle 293
6.1 Radioactive Decay 447
Summative Zone 298 6.2 Nuclear Energy 452
Summative Zone 459

Theme 2 Electricity and Electromagnetism
Chapter 7 Quantum Physics 470
Chapter 3 Electricity 313
7.1 Quantum Theory of Light 472
3.1 Current and Potential 316
Difference 7.2 Photoelectric Effect 479

3.2 Resistance 323 7.3 Einstein's Photoelectric 483
Theory
3.3 Electromotive Force (e.m.f.) 339 490
and Internal Resistance Summative Zone

3.4 Electrical Energy and Power 347
SPM Model Paper 503
Summative Zone 353
Answers 520 – 562




v

CHAPTER
3 Gravitation








Important Learning Standard Page
• Explain Newton’s Law of Gravitation. 66
• Ellipse / Elips ©PAN ASIA PUBLICATIONS
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3.1 Newton's • Solve problems involving Newton’s Law of Gravitation for two
Universal Law static objects on the Earth; objects on the Earth’s surface; Earth 67
of Gravitation and satellites; Earth and Sun.
• Relate the gravitational acceleration, g on the surface of the Earth 68
with the universal gravitational constant, G.
• Justify the importance of knowing the values gravitational 69
acceleration of the planets in the Solar System.

• Describe the centripetal force in the motion of satellites and 70
planets system.
• Determine the mass of the Earth and the Sun using Newton’s Law 73
of Gravitation and centripetal force.
• Explain Kepler’s Laws. 74
3.2 Kepler's Laws • Express Kepler’s Third Law. 75

• Solve problems using Kepler’s Third Law. 76

• Describe how the orbit of a satellite is maintained at a specific 78
height by setting the necessary satellite velocity.
• Communicate on geostationary and non-geostationary satellites. 79
3.3 Man-made
Satellites • Conceptualise escape velocity. 80
• Solve problems involving the escape velocity, v for a rocket from
the Earth’s surface, the Moon’s surface, Mars' surface and the 81
Sun’s surface.


Important Formula and Tips

• Gravitational force / Daya graviti • Gravitational force, F = Gm m 2
1
• Universal gravitational constant r 2
Pemalar kegravitian semesta • Gravitational acceleration, g = GM
• Gravitational acceleration / Pecutan graviti mv 2 r 2
• Satellite / Satelit • Centripetal force, F = r
• Centripetal force / Daya memusat T 2 r 3
• Kepler's Third Law: 1 = 1
T 2 2 r 2 3
• Focus / Fokus 
GM
• Geostationary satellite / Satelit geopegun • Linear satellite speed, v =  r GMm
• Escape velocity / Halaju lepas • Gravitational potential energy, U = − r
• Gravitational potential energy 
2GM
Tenaga keupayaan graviti • Escape velocity, v =  r

64

Man-made satellites Linear satellite speed GM v =   r Non-geostationary satellite Escape velocity 2GM v =   r



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consist of
Geostationary satellite Earth-satellite system










used for

Gravitation Kepler's laws First Law Second Law Third Law T 2 ∝ r 3 Sun-planet system







can derive

consist of






Newton's Universal Law of Gravitation Gravitational force F = GMm r 2 Gravitational acceleration GM g = r 2 Centripetal force F = mv 2 r Mass of Earth M = 4π 2 r 3 GT 2



















65

Form
4 Physics Chapter 3 Gravitation


3.1 Newton's Universal Law of Object 1 Object 2
Gravitation m 1 F F m 2


r

In the 17 century, Isaac Figure 3.3 Gravitational force between two objects
th
Newton observed that an apple
fell vertically to the Earth and 5. Newton’s Universal Law of Gravitation
the motion of the Moon was states that the gravitational force between two
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around the Earth. He deduced objects is directly proportional to the product
that an attractive force exists of the masses of both objects and inversely
not only between the apple and proportional to the square of the distance
the Earth but also between the
Earth and the Moon. between them.
Gm m
1. Gravitational force is an attractive force that F = 1 2 2
acts between any pair of objects in the universe. r
Thus, it is known as a universal force. F = Gravitational force between two objects
2. Figure 3.1 shows the gravitational force between m = Mass of first object
1
the Sun, Earth and Moon. m = Mass of second object
CHAP r = Distance between centres of the two objects
2
3 Earth Gravitational force G = Universsal gravitational constant
between the Sun (G = 6.67 × 10 N m kg )
–2
–11
2
and Earth
Try question 1 in Formative Zone 3.1
Gravitational
force between
the Earth
and Moon Gravitational force
Moon
between the Sun The value of the gravitational constant, G has been
and Moon determined experimentally.
Sun
Figure 3.1 Gravitational force as universal force
6. The magnitude of the gravitational force
3. The gravitational force exists in pairs and both between two objects can be calculated if the
objects experience the gravitational force of the mass of the two objects and the distance
same magnitude. between them are known.
4. In 1687, Isaac Newton presented two Example 1
relationships involving the gravitational force
between two objects. The mass of a student is 54 kg. What is the
gravitational force between the student and
the Earth?
24
(a) The gravitational force is [Mass of the Earth = 5.97 × 10 kg,
6
directly proportional to the radius of Earth, r = 6.37 × 10 m]
product of the masses of Solution
the two objects, F ∝ m m
24
1 2 Mass of the Earth, m = 5.97 × 10 kg
1
m m Mass of the student, m = 54 kg
F ∝ 1 2 2
(b) The gravitational force is r 2 Distance between the student and centre
6
inversely proportional to of Earth, r = 6.37 × 10 m
the square of the distance Gm m 2
1
between the two objects, Gravitational force, F = r 2
1
24
−11
F ∝ 6.67 × 10 × 5.97 × 10 × 54
r 2 F =
6 2
(6.37 × 10 )
Figure 3.2 Formulation of Newton’s Universal Law of = 530 N
Gravitation Try question 3 in Formative Zone 3.1
66 3.1.1

Form
4 Physics Chapter 3 Gravitation


5. Saturn, Uranus and Neptune have gravitational 8. The human body was created to live in a
acceleration almost equal to Earth’s gravitational acceleration of 9.81 m s . Thus,
–2
gravitational acceleration, g = 9.81 m s even the knowledge of the value gravitational
–2
though they have larger masses. However, their acceleration is important in space exploration
densities are lower than the Earth. and sustainability of life.
6. Mercury and Mars have lower gravitational 9. During space exploration far away from the
acceleration than the Earth due to their smaller Earth or near other planets, the body of an
masses. astronaut may be exposed to situations of low
or high gravity.
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Gravitational acceleration
https://bit.ly/2Vmdbyp Effect of gravity
https://go.nasa.gov/2ZbuzXF
7. The magnitude of the gravitational force that
acts on an object at the surface of the planet can 10. The effect of low or high gravity on human
be calculated when the value of gravitational growth are shown in the Table 3.2.
acceleration is known.
Table 3.2 Effect of change in gravity on human growth
CHAP Factor Effect of low gravity Effect of high gravity
3
(a) Change in density • Density of the body decreases. • Density of the body increases.
(b) Fragility of the • Bones become more fragile due to loss of • No significant change in fragility.
bones calcium.
(c) Size of the lungs • Increased or expanded. • Reduced or contracted.

(d) Blood circulation • Blood collects in the upper parts of the • Blood collects in the lower parts of the
system body. body due to difficulty to flow upwards.
(e) Blood pressure • Lower blood pressure and heart rate. • Higher blood pressure and heart rate.




Centripetal Force in the Motion of Satellites and Planets
1. Figure 3.7 shows three positions of a satellite Direction of
velocity
that orbits the Earth with a uniform speed.
2. An object in circular motion is continuously
experiencing change in its direction of motion
even though the speed is constant. Earth
3. When an object does circular motion, the
resultant force acting is always directed towards
the centre of the circle and is known as the Direction of
centripetal force. velocity





Centripetal force in the motion of
satellites and planets
https://bit.ly/2Vd3Uc1
Direction of
velocity
Figure 3.7 Satellite in circular motion

70 3.1.4 3.1.5

Form
4 Physics Chapter 3 Gravitation


Example 10 Example 11

The Moon moves in a circular orbit around the Based on the following data about the Earth
Earth with an orbital radius of 3.83 × 10 m and orbiting around the Sun, calculate the mass of
8
6
period of orbit 2.36 × 10 s. What is the mass of the Sun.
the Earth? Radius of orbit, r = 1.50 × 10 m
11
Solution Period, T = 1 year
Mass of the Earth, Solution
4π r T = 1 year
2 3
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M =
GT 2 T = (365 × 24 × 60 × 60) s
2 3
4π r
8 3
2
4π × (3.83 × 10 ) Mass of the Sun, M =
= GT 2
(6.67 × 10 ) × (2.36 × 10 )
6 2
−11
4π × (1.50 × 10 )
11 3
2
M =
24
= 5.97 × 10 kg (6.67 × 10 ) × (365 × 24 × 60 × 60) 2
−11
= 2.01 × 10 kg
30
3.1
CHAP
3 1. State Newton’s Universal Law of Gravitation. C2
2. What are the factors that affect the magnitude of the gravitational force between two bodies? C3
3. A spacecraft of mass 150 kg is at a distance of 7.20 × 10 m from the centre of the Earth.
6
Find the gravitational force between the spacecraft and the Earth? C3
[G = 6.67 × 10 N m2 kg , mass of the Earth, M = 5.97 × 10 kg]
24
–2
–11
12
4. The planet Uranus orbits around the Sun with radius of orbit 2.87 × 10 m and period 2.64 × 10 s.
9
Calculate the mass of the Sun. C3
3.2 Kepler's Laws I, II and III
1. Kepler was a German astronomer and
mathematician who modified the heliocentric Johannes Kepler worked as an assistant to the
model of the planets moving around the Sun. astronomer Tycho Brahe. His great determination
2. Kepler succeeded in formulating three laws that drove him to study Brahe’s astronomical data for
more than ten years.
described the motion of planets around the Sun.


Activity 3.2


Aim: To sketch an elliptical shape based on the dual focus concept of an ellipse
Materials: Pencil, thread, two pieces of thumb tacks, white A4 paper, softboard and cellophane tape
Instructions:
1. The thread is tied to form a loop with a circumference of about 32 cm.
2. The two thumb tacks are stuck at points F and F on the A4 paper. The distance between F and F is
1
1
2
2
12 cm.
3. The thread loop is placed on the white paper and the pencil is held as shown in Figure 3.15.
74 3.1.6 3.2.1

Form
4
Chapter 3 Gravitation Physics

Thread loop with circumference 32 cm 4. The ellipse is drawn by moving the pencil
slowly around F and F .
1 2
5. The thumb tacks and thread are removed.
16 cm
6. A small circle is drawn at F to represent
Minor axis 1
the Sun and another small circle on the
circumference of the ellipse to represent the
Thread Earth.
Thumb Discussion:
tack 1. The shape of the ellipse shows that when
F F
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1 2 the Earth moves in its orbit along the
Major 12 cm circumference of the ellipse, the distance
axis between the Sun and the Earth changes.
Figure 3.15 2. The Earth is nearest to the Sun when it is on
the major axis to the left of F . The Earth is
1
furthest away from the Sun when it is on the
major axis to the right of F .
2
Video: Drawing an ellipse
https://bit.ly/2UhHkhH


CHAP
3
A Kepler's First Law B Kepler's Second Law
• Kepler’s First Law states that the orbit of • Kepler’s Second Law states that a line that
each planet is an ellipse, with the Sun at one connects a planet and the Sun sweeps out
focus. equal areas in equal time intervals as the
• The planets in the Solar System have elliptical planet moves in its orbit.
shaped orbits and the Sun is always at one • If a planet takes the same time to move from
focus of the ellipse as shown in Figure 3.16. A to B and C to D in Figure 3.17, the area of
region AFB is equal to the area of region CFD.
• The distance AB is larger than the distance
Minor axis
CD as the planet moves with a higher linear
speed from A to B compared to from C to D.
Ellipse
Planet
B
C
F
Major axis
Sun Sun
D
A
Equal area in equal time
Planet
Figure 3.17 The motion of a planet in its orbit
Figure 3.16 Try question 2 in Formative Zone 3.2
• The orbits of most planets in the Solar System C Kepler's Third Law
have the major axes and minor axes of almost
the same length. • Kepler’s Third Law states that the square of
• Therefore, elliptical shape of the orbits the period of a planet is directly proportional
2
3
of planets in the Solar System are actually to the cube of the radius of its orbit, T ∝ r .
almost round. • Therefore, planets that are further away from
• The radius of the orbit is the average distance the Sun will take a longer time to complete
between the planet and the Sun. one orbit around the Sun.
Try question 1 in Formative Zone 3.2

3.2.1 3.2.2 75

Form
4
Chapter 4 Heat Physics

Relationship between Pressure and Volume of a Gas


4.4

Inference: The volume of a gas affects the pressure of the gas.
Hypothesis: The smaller the volume of a gas, the higher the pressure of the gas.
Aim: To determine the relationship between the volume and pressure of a fixed mass of gas at a constant
temperature.
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Variables:
(a) Manipulated : Volume, V
(b) Responding : Pressure, P
(c) Constant : Temperature and mass of air
Apparatus: 100 ml syringe, rubber tube, pressure gauge and retort stand with clamp
Procedure:
1. The apparatus is set up as shown in Figure 4.24.
2. The piston is adjusted so that the volume of air in Piston
the syringe is 100 ml. Then, the end of the syringe Pressure gauge
is connected to the pressure gauge. 100 ml syringe
3. The readings of the volume and initial pressure of Retort stand
the air in the syringe are recorded in Table 4.10.
Rubber tube
4. The piston is pushed slowly until the volume of air
in the syringe becomes 90 ml. The reading of the
pressure of the air is recorded. Figure 4.24
5. Step 4 is repeated with volumes 80 ml, 70 ml and 60 ml.
Results:
Table 4.10
P / kPa
Volume, Pressure, 1 –1
V / ml P / kPa V / ml 180 CHAP
100 100 0.010 4
90 110 0.011 160
80 125 0.013
70 140 0.014 140
60 165 0.017
120
Data analysis:
1
A graph of pressure, P against is drawn as shown in 100
Figure 4.25. V
P
80

60


40
1
—
0 V
Figure 4.26 20
1 1
The graph of P against shows that the pressure is 0 – / ml –1
V 0.005 0.010 0.015 0.020 V
inversely proportional to the volume. Figure 4.25

4.4.2 111

Form
4 Physics Chapter 4 Heat



Conclusion:
The smaller the volume of a gas, the higher the pressure of the gas. The hypothesis is accepted.
Discussion:
1. A syringe with a large volume is used to obtain a higher change of volume and pressure of the air
trapped in it.
2. The piston is pushed slowly into the syringe so that the temperature of the air in the syringe is kept
constant.

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4. The rate of collision between the molecules and
the wall of the container increases.

Robert Boyle (1627-1691) was a scientist who 5. The force per unit area on the surface of the
emphasized on the scientific method when wall increases. Thus, the gas pressure increases.
conducting investigations. From experimental data,
he made the conclusion that the volume of a gas is
inversely proportional to the pressure of the gas.
For Boyle's law, temperature is constant.
1. Boyle’s law states that pressure is inversely P × V Constant
proportional to the volume of a fixed mass of
gas at constant temperature.
T
1
P ∝
V
1
P = k 1 2 PV = constant
P V = P V
V
1 1
2 2
where k is a constant
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P = pressure of the gas (Pa)
3
V = volume of the gas (m )
CHAP Eduweb TV: Boyle's law
4 Therefore, PV = k http://bit.ly/2tb1y2I
Try question 1 in Formative Zone 4.4
2. Suppose a gas experiences a change of pressure
and volume from an initial state to a final state. Example 6
Initial state of the gas, P V = k
1 1
Final state of the gas, P V = k The air in a closed syringe has a volume of
2 2
60 cm and pressure 108 kPa. The piston of
3
Therefore, P V = P V the syringe was pushed to compress the air to
1 1 2 2
a volume of 48 cm . Find the pressure of the
3
compressed air.
Solution
Volume is P = 108 kPa, P = pressure of compressed air
1
2
decreased V = 60 cm V = 48 cm 3
3
1 2
The temperature of gas is constant,
Boyle’s law is used:
P V = P V
1 1 2 2
108 × 60 = P × 48
Figure 4.27 A fixed mass of gas compressed at 2
constant temperature P = 108 × 60
2 48
3. When the volume of the gas is decreased, the
number of molecules per unit volume increases. = 135 kPa
112 4.4.2

Form
5
Chapter 3 Electricity Physics

3.4

Aim: To investigate how the resistivity of a wire, ρ affects its resistance.

Problem statement: How the resistivity of a wire, ρ affects the resistance of the wire?
Hypothesis: Materials with high resistivity gives higher resistance.
Variables:
(a) Manipulated: Resistivity of wire, ρ
Discussions: ©PAN ASIA PUBLICATIONS
©PAN ASIA PUBLICATIONS
(b) Responding: Resistance of wire, R
(c) Constant: Length of wire, temperature, cross-sectional area of wire
Apparatus and Materials:
50 cm constantan wire (s.w.g. 24), 50 cm copper wire (s.w.g. 24), 50 cm tungsten wire (s.w.g. 24), connecting
wires, three dry cells, switch, ammeter (0 – 1 A), voltmeter (0 – 5 V), rheostat and battery holder
Operational definition:
The resistance of the conductor, R, is given by the ratio of the reading of voltmeter to the reading of the CHAP
ammeter.
3
Procedure:
1. An electrical circuit is set up as shown in
Figure 3.28.
2. A constantan wire is connected across
Ammeter A
terminals P and Q. The length of the wire
across P and Q is adjusted to be 30 cm long.
3. The switch is connected and the rheostat is Rheostat P Q
adjusted so that the ammeter gives a reading
of 0.5 A. The reading of voltmeter is taken V Constantan wire
and recorded in Table 3.6.
Voltmeter
4. The constantan wire is removed. Steps 2 and Figure 3.28
3 are repeated for copper wire and tungsten
wire.

Results:
Table 3.6

Reading of ammeter, Reading of voltmeter, Resistance,
Wire V
I / A V / V R = — / Ω
I
Constantan 0.5 6.3 12.6


Copper 0.5 0.2 0.4

Tungsten 0.5 0.6 1.2




1. From Table 3.6, different conductor gives different resistance.
2. Constantan gives the highest resistance, followed by tungsten and copper gives the lowest resistance.
3. Different material gives difference resistance because different material has different resistivity, ρ.
Resistance of a material increases when the resistivity of the material increases.



3.2.3 335

Form
5 Physics Chapter 3 Electricity



Conclusion:
Different material has different resistivity and gives different resistance. When the resistivity of a material
increases, its resistance also increases. The hypothesis is accepted.







Assume the conductors as a road and the current as a car you are driving in. Constantan wire is as if an old road
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with potholes and badly damaged, copper is as if a newly built road and tungsten is as if uneven road. The
different road (different wires) gives you different obstructions (resistance) to drive through them. Old road
with potholes and badly damaged (constantan) is most difficult to drive through due to a lot of obstructions
(high resistance). Newly built road (copper) is easiest to drive through because it gives very little obstruction
(resistance).



CHAP
3 Application of Resistivity of Conductor in Daily life

1 Heating element

(a) Heating element is used in water heater and electric kettle to heat
up water.
(b) Heating element is usually made from material with higher
resistivity.

(c) The higher resistance enables the heating element to convert
electrical energy to heat energy to heat up the water.
(d) Other than that, heating element must also be made of:
(i) material with high melting point so that it does not melt Figure 3.29 Heating element
when heating up water,
(ii) material that can withstand oxidation so that the water heated
up is not contaminated by its oxide which is probably toxic.



2 Electrical wiring in homes

(a) Copper wires are usually used in the electrical wiring in
homes because copper have low resistivity.
(b) The low resistance of copper wire enables electrical current
to flow more efficiently by reducing the energy loss to the
surroundings in the form of heat.
(c) Other than that, copper wire can resist oxidation and do not
oxidise easily by the action of oxygen in air. Figure 3.30 Copper wires in
electrical wiring in homes
(d) Nevertheless, in the National Grid Network, aluminium
cables are used instead of copper even though the resistivity
of aluminium is higher than copper. This is because the
density of aluminium is way lower than copper. This causes
the aluminium cables to be lighter than copper cables.



336 3.2.5

Form
4 Physics Chapter 5 Waves


Applications of Diffraction of Waves in Daily Life
1. Diffraction of water, light and sound waves have their applications in daily life.

Gap
Water waves
• Diffraction of water waves when moving through
a gap produces a region of calm water suitable for Region of
calm water
berthing of ships and water recreational activities.
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Light waves
• Holograms produced by the diffraction of light is
used as a security mark in bank cards such as the
debit card and credit card.


Sound waves
• Infrasonic sound waves with long wavelength
produced by elephants enables long distance
communication between elephants.




5.5

1. Figure 5.48 shows water waves approaching a gap in a Barrier Gap
ripple tank.
(a) Draw the wave pattern formed to show the wave Ripple
phenomenon that occurs. C3 tank
(b) What happens to the amplitude of water waves after
passing through the gap? C4



Figure 5.48

2. Figure 5.49 shows an audio generator and loudspeaker placed near the corner of a building. Three
students, A, B and C are standing around the corner. The audio generator can produce sounds with
different pitch.
C
B A
CHAP
5
Loudspeaker






Audio generator
Figure 5.49
(a) State one factor that affects the pitch of the sound. C2
(b) When a high pitch sound is generated, only student A can hear the sound clearly. When a low pitch
sound is generated, all three students can hear the sound clearly. Explain this situation. C3


158 5.5.4

Form
5 Physics Chapter 2 Pressure









Paper 1

1. Figure 1 shows an excavator. 3. Figure 3 shows a flask connected to a mercury
manometer. The pressure of the gas in the flask
CHAP is 83 cm Hg.
2
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Flask Gas



20 cm
12 cm
Figure 1
What is the physics principle or law that is applied Figure 3
to move the arm of the excavator? C3 What is the value of the atmospheric pressure?
A Pascal’s principle C3
B Archimedes’ principle
C Bernoulli’s principle A 71 cm Hg
D Newton’s Third Law of Motion B 75 cm Hg
C 76 cm Hg
SPM Clone
2. Figure 2 shows the supply of water from a tank. D 91 cm Hg
The reading of a pressure meter at X is 2.8 × 10 5
Pa and the density of water is 1 000 kg m . 4. Figure 4 shows a hydraulic system to lift a car.
–3
–2
[g = 9.81 m s ]

Tank


Area, Pressure, P Area,
2
A Pressure, P A
1 1 2
h
Figure 4
The multiplying factor of the hydraulic system is
X
C3
A
A —––
1
Figure 2 A 2
A
What is the value of h in the situation above? B —––
2
C3 A 1
A 2.85 m P 1
B 28.0 m C —––
P
2
C 28.5 m P
2
D 285 m D —––
P
1


298

Form
5
Chapter 2 Pressure Physics
SPM Clone
5. Figure 5 shows a boy floating at the surface of The speed of flow of water will increase if the
water. container is tilted in the direction shown because
C3
A the surface area of the water increases
B the volume of air in the container increases
C the tap is nearer to the floor
D the tap is at a deeper level below the water
surface
CHAP
8. Figure 8 shows two syringes used to study
©PAN ASIA PUBLICATIONS
Figure 5 Pascal’s principle. A 5 N force applied at the 2
©PAN ASIA PUBLICATIONS
Which relationship is correct? C4 small syringe produces a 20 N force at the large
[F = buoyant force; W = weight of the boy; piston. The cross-sectional area of the large
2
–4
B = weight of water displaced] syringe is 7.2 × 10 m .
A F = W and W = B
B F . W and W , B
C F , W and W = B
D F = W and W , B Small
syringe Large
6. Figure 6 shows a Venturi tube. syringe
Air

S T U


Figure 8
What is the cross-sectional area of the small
syringe? C4
A 9.0 × 10 m C 3.6 × 10 m 2
2
–5
–4
Figure 6 B 1.8 × 10 m D 2.88 × 10 m 2
–3
2
–4
Which comparison about the heights of the
water column in tube S, T and U is correct? 9. Figure 9 shows two aluminium cans placed
[h = height of water column in tube S; C4 on a row of glass tubes. When the pupil blows
S
h = height of water column in tube T; in between the two cans, both the cans move
T
h = height of water column in tube U] towards each other.
U
A h . h . h U
S
T
B h . h . h S
U
T
C h . h . h U
T
S
D h . h . h
U T S
7. Figure 7 shows water flowing out of a container. Glass tube
Glass tube
Figure 9
This situation can be explained by C4
A Newton’s second law of motion
B Archimedes’ principle
C Bernoulli’s principle
Figure 7 D Pascal’s principle


299

Form
4
Chapter 3 Gravitation Physics

The shortest distance between the probe and A C
the planet is r. The speed v of the probe at the
GM

nearest position is 1.5  , where G is the
r
universal gravitational constant. Which
diagram shows the correct path of the probe? B D
C3
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©PAN ASIA PUBLICATIONS








Paper 2

Section A

1. Figure 1.1 shows a durian tree on the surface of the Earth of mass M and radius R and Figure 1.2 shows CHAP
the Moon in orbit of radius 60.1 times the radius of the Earth around the Earth. 3

Moon




r = 60.1 R
moon

Durian
tree



R


Earth’s centre Earth’s centre
Figure 1.1 Figure 1.2

(a) Write down an equation in terms of the mass, M and radius R of the Earth, the universal gravitational
constant, G for the gravitational force, F acting on a durian of mass m by the Earth. C2 [1 mark]
(b) By relating the weight of the durian to the gravitational force, derive an equation to find the
gravitational acceleration, g for the free fall of the durian. C3 [3 marks]
(c) The mass of the Earth is 5.97 × 10 kg and its radius is 6.37 × 10 m. Calculate
24
6
(i) gravitational acceleration at the surface of the Earth. C3 [2 marks]
(ii) gravitational acceleration of the Earth at the surface of the Moon. C3 [2 marks]
(d) Based on your answers in 1(c)(i) and 1(c)(ii), explain the large difference in value of the gravitational
acceleration on the of the Earth at the Moon compared to at the surface of the Earth. C4
[2 marks]




87

CHAPTER
7 Quantum Physics









Important Learning Standard Page
• Explain the initiation of the quantum theory. 472
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©PAN ASIA PUBLICATIONS
• Describe quantization of energy. 476
7.1 Quantum Theory Of • Explain wave-particle duality. 477
Light
• Explain concept of photon. 478
• Solve problems using photon energy, E = hf and power, P = nhf. 478

• Explain photoelectric effect. 479
7.2 Photoelectric Effect
• Identify four characteristics of photoelectric effect that cannot be
explained using classsical theory. 482


• State minimum energy required for a photoelectron to be emitted
1 483
from a metal surface using Einstein’s equation, hf = W + —mv 2
2 max
• Explain threshold frequency, f and work function, W. 485
0
hc
7.3 Einstein's • Determine work function of metal, W = hf = —– 485
0
Photoelectric Theory λ 0
• Solve problems using Einstein’s equation for photoelectric effect. 486
• Explain production of photoelectric current in a photocell circuit. 488

• Describe applications of photoelectric effect in daily life. 489





• Black body radiation / Sinaran jasad hitam • Continuous energy / Tenaga selanjar
• Ideal absorber / Penyerap unggul • Photoelectric effect / Kesan fotoelektrik
• Radiation intensity / Keamatan sinaran • Threshold frequency / Frekuensi ambang
• Wave-particle duality / Kedualan gelombang-zarah • Work function / Fungsi kerja
• Discrete energy / Tenaga diskrit


Important Formula and Tips

• Einstein’s photoelectric equation, • Photon power, P = nhf
1
hf = W + —mv 2 in which n is the number of photons emitted per
2 max second
h
hc
• Work done, W = hf = –— • de Broglie's wavelength, λ = —
P
0
λ
0
whereby, f is threshold frequency of metal and λ • Momentum of particle, p = ABBBB
2mK
0
0
is the threshold wavelength of metal whereby, K = —mv 2
1
hc
• Photon energy, E = hf = –— 2
λ
470

Form
5












Einstein's photoelectric theory Einstein's photoelectric equation, 1 hf = W + —mv 2 2 hc Work function, W = hf 0 = –— λ 0 Photoelectric current production Applications Light detector of automatic door Image sensor ISS solar panel

©PAN ASIA PUBLICATIONS
©PAN ASIA PUBLICATIONS




explained by • • Solar cells • • • •








Four characteristics of photoelectric The higher the light photon frequency, the higher the kinetic energy of the emitted photoelectrons. The minimum frequency to emit a photoelectron is the threshold frequency, f 0 The kinetic energy of photoelectrons is independent of the intensity of the light The emission of photoelectrons is instantaneous


Quantum Physics
Photoelectric
effect






effect
•
•
•
•
hc λ
Wave with particle properties Discrete energy packet, E = hf Photon energy, E = –— Power, P = nhf





Quantum theory of light Explanation of black body spectrum Max Planck h p • •







Particle with wave properties de Broglie's hypothesis, λ = — Wave-particle duality Application Electron microscope










471

Form
5 Physics Chapter 7 Quantum Physics


7.1 Quantum Theory of Light
1. The electromagnet spectrum is a continuous spectrum that consists of seven types of waves with different
frequencies and wavelengths.
2. Figure 7.1 shows seven types of waves with their frequencies and wavelengths.



Increasing wavelength
Wavelength / m
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10 –11 10 –10 10 –9 10 –8 10 –7 10 –6 10 –5 10 –4 10 –3 10 –2 10 –1 1 10 1 10 2 10 3 10 4
©PAN ASIA PUBLICATIONS
Gamma X-rays Ultraviolet Infrared Microwave Radio waves
rays radiation radiation

10 20 10 19 10 18 10 17 10 16 10 15 10 14 10 13 10 12 10 11 10 10 10 9 10 8 10 7 10 6 10 5 10 4 Frequency / Hz
Increasing frequency
Visible light




400 500 600 700 750
Wavelength / nm
Figure 7.1
3. Any object with a temperature above absolute
zero emits electromagnetic radiation at all
wavelengths. The usual meaning of 'black' is absorbs visible light.
4. Objects with low temperature or cold objects But in Physics, a black body is one of that absorbs
emit waves with low frequencies such as radio all the radiation that falls on it, at all wavelengths.
waves or microwaves. It is the perfect absorber and radiator. Our Sun is a
5. Meanwhile, objects with high temperature or back body.
hot objects emit waves with high frequencies
such as visible light and ultraviolet radiation. 8. The emitted radiaton forms a continuous
6. A black body is an idealised body that is able to spectrum and is unaffected by the nature of the
absorb all electromagnetic radiation that falls black body surface.
on it.
9. When the temperature of an object rises, the
object (black body radiator) emits thermal
radiation at all wavelengths.
Simulation of black body
spectrum
https://bit.ly/35FlCtc
The light rays that
7. A black body also emit thermal radiation enter the ear cavity
depending on its temperature and is known as will undergo repeated
black body radiator. reflections on the inner Ear
walls of the ear cavity. cavity
At each reflection,
parts of the rays are
absorbed by the inner
CHAP Thermal radiation is electromagnetic radiation walls of the ear until all the rays are absorbed. Thus,
7 consist of visible light and radiation that cannot be the ear cavity acts like a black body.
seen by the human eye such as infrared radiation.


472 7.1.1

Form
5 Physics Chapter 7 Quantum Physics


Ideas that Sparked the Quantum Physics 4. The light energy produced able to reach
Theory unlimited high value due to the frequency of the
electrons has no limit.
1. In a hot object, electrons vibrate rapidly and 5. When the temperature of the black body
randomly in any direction and produce light. increases, the intensity of radiation also increase
2. The hotter the object, the higher the energy and will not become zero value.
of electron vibration. Thus, more light will be 6. However, the experimental results of black-
emitted. body radiation are inconsistent with classical
3. As stated in classical theory, the electron that physics theory.
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vibrates at the same frequency should have the 7. Several physicists that involve in the quantum
©PAN ASIA PUBLICATIONS
same energy content. theory development are shown as the following.


Classical Theory

(a) Isaac Newton (1643-1727)
• Described light as a single stream of
particles or corpuscles in 1704. Isaac Newton
• Unsuccessful in explaining the https://bit.ly/37FU9KI
phenomenon of light refraction due
to failure in comparing the speed of
light in glass and air.

(b) Thomas Young (1773-1829)

• Conducted doubleslit experiment
on light in 1801 and showed that Thomas Young
light is a wave. https://bit.ly/3ozx6aH
• Unable to explain the radiation
spectrum produced by black bodies.




(c) John Dalton (1766-1844)
• Matter consists of basic particles
that cannot be further divided John Dalton
called atoms. https://bit.ly/35BN6zU
• Same elements have the same type
of atoms.
• Unable to explain the light
spectrum produced by atoms.


(d) J.J. Thomson (1856-1940)

• Discovered negatively charged
subatomic particles called electrons J.J. Thomson
in 1897. https://bit.ly/3kpyFW1
• Designed experiment to study the
CHAP behaviour of electrons.
7 • Unable to explain the line spectrum
of light produced by atoms.

474 7.1.1

Form
5
Chapter 7 Quantum Physics Physics
Example 1 6. The relationship between the momentum of
particle, p and its wavelength, λ is
Calculate the photon energy for light with
wavelengths 450 nm and 700 nm. Compare both λ = —–
h
photons. p
Solution which h is Planck’s constant (6.63 × 10 J s)
–34
Planck’s constant, h = 6.63 × 10 J s
–34
Speed of light in vacuum, c = 3.00 × 10 m s
–1
8
Wavelength, λ = 450 × 10 m
–9
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1
©PAN ASIA PUBLICATIONS
Wavelength, λ = 700 × 10 m
–9
2 2
p
hc
(mv)
1
m
2
E = hf = —— K = —mv × —– ➞ K = ——— = ——
2
λ 2 m 2m 2m
Photon energy of 450 nm: which p = mv, p = ABBBB
2mK
8
1
E = 6.63 × 10 –34 3.00 × 10 –9 2
—————–
1 450 × 10
= 4.42 × 10 J 7. The larger the momentum of particle, the shorter
–19
Photon energy of 700 nm: the wavelength produced. The momentum of
p
2
8
1
E = 6.63 × 10 –34 3.00 × 10 –9 2 particle is p = mv, then —— = K
—————–
2 700 × 10 2m
= 2.84 × 10 J
–19
h
h
The shorter the wavelength of light, the higher λ = —–– = ————
mv
ABBBB
the photon energy. 2mK
where m is the mass of particle, v is the velocity
of particle and K is the kinetic energy of particle.
Wave-Particle Duality 8. Since the value of h is too small, the particle
with large mass will have too short of de Broglie
wavelength to be detected. Thus, the wave
characteristics cannot be observed.
Wavelength simulation 9. In 1927, the presence of wave properties of
http://bit.ly/39xNYqu
electrons was confirmed through the electron
diffraction experiments.
1. Light has wave properties because it shows the 10. Photograph 7.1 shows the diffraction pattern of
phenomena of diffraction and interference. electrons through a thin layer of graphite. The
pattern in Photograph 7.2 resembles the light
2. Object has particle properties because it has diffraction pattern through an aperture.
momentum, kinetic energy and also collide with
each other.
3. Louis de Broglie introduced a hypothesis states
that all particles can exhibit wave characteristics
in year 1924.
4. However, it is experimentally difficult to show
the wave characteristics of particles with large
masses.
5. Thus, Louis de Broglie predicted that the wave Photograph 7.1 Photograph 7.2
characteristics can be shown by light particles 11. This observation proved de Broglie’s hypothesis. CHAP
such as electron, proton and neutron. 7
7.1.3 477

Form
5 Physics Chapter 7 Quantum Physics


2. Figure 7.5 shows a photocell circuit to show the Based on Figure 7.5,
photoelectric effect. (a) When a light sensitive metal surface
(cathode) is illuminated with a certain light
Photocell beam, electrons will be emitted from the
metal surface. These electrons are called
Source of
light photoelectrons.
(b) The emitted photoelectrons are attracted to
Light sensitive
e - the anode which has positive potential.
e - metal
e - e - (c) The movement of the photoelectrons
+ - _
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Anode e from the cathode to the anode produces
©PAN ASIA PUBLICATIONS
e - Cathode
a photocurrent inside the circuit. The
Vacuum
milliammeter shows the value of this
current.
2 3
1 4
0 5
mA
+ _
− +
Milliammeter Video of demonstration of
Battery
photoelectric effect
Figure 7.5 http://bit.ly/31HRcpe
Activity 7.1

Tujuan: To determine the value of Planck’s constant using the Planck’s constant kit
Radas: Planck’s constant kit (9 V battery, 1 kΩ potentiometer, LEDs of different colours, milliammeter and
voltmeter)
Instructions: 2 3
1 0 4 5
1. The Planck's constant kit is connected as shown V
Voltmeter
in Figure 7.6. 1 2 3 4
0 5
2. The knob on the potentiometer is adjusted to mA − +
obtain the voltage, V = 0.2 V. The milliammeter
− +
reading is recorded in Table 7.2.
3. Step 2 is repeated for V = 0.4 V, 0.6 V, 0.8 V, and Milliammeter LED
3.0 V.
4. A graph of current against voltage is drawn.
Then, the activation voltage, V of the red LED Potentiometer
a – +
is determined from the intercept value on the
voltage axis.
5. Steps 2 to 4 are repeated using orange, green Battery
and blue LEDs.
Figure 7.6
Results:
Current, I / A


LEDs emit light when a current flows. The LED will
not emit a photon until the electron has enough
energy to release a photon. The smallest possible
voltage across the LED just emits light, the
activation voltage was determined in this activity.
1.0 1.5 2.0 2.5 3.0
CHAP Voltage, V / V
7 Figure 7.7


480 7.2.1

Form
5
Chapter 7 Quantum Physics Physics
Data analysis:
1. Table 7.2 shows the activation voltage value that is obtained from the graph current, l against voltage,
V for each LED colour.
Table 7.2

LED Wavelength, Activation voltage, 1
—/10 m –1 • Video to determine
6
colour λ λ / nm V / V λ λ
a
Planck’s Constant
White 793 1.35 1.261
http://bit.ly/395cDTW
Red 623 1.78 1.605
• Classroom
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©PAN ASIA PUBLICATIONS
Orange 586 1.90 1.706 fundamentals:
measuring the
Green 567 2.00 1.764 Planck’s constant
Blue 467 2.45 2.141 http://bit.ly/2XafllD
1
2. Based on the Table 7.2, the graph of V against — is plotted.
a λ
V / V
a
2.6
Note:
For simplicity, we
2.4 can assume the
energy loses inside
the p-n junction of
2.2
LED to be equal for
all the LEDs.
2.0


1.8

1.6


1.4

1
6
1.2 – / 10 m –1
1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 λ
Figure 7.8
3. From the plotted graph, the gradient of the graph, m and Planck’s constant, h are determined.
me
Given h = —— ; e = charge of one electron (1.60 × 10 C) and c = speed of light in vacuum
–19
c
8
(3.00 × 10 m s )
–1
–6
–19
2.50 – 1.38 (1.19 × 10 )(1.60 × 10 )
m = ———————– and h = ———————————–
(2.20 – 1.26)10 6 3.00 × 10 8
= 1.19 × 10 V m = 6.35 × 10 J s
–34
–6
Discussion:
The activation voltage, V can be obtained through V-intercept from the graph of I against V as shown in
a
1
Figure 7.9. The activation voltage, V has a linear relationship with — as shown in Figure 7.10. Gradient of
a λ
1 hc
a graph of V against —, m is equal to the value of ——. Therefore, the value of Planck’s constant can be CHAP
e
a λ
me 7
determined from the expression ——.
c
7.2.1
7.2.1 481

Form
5 Physics Chapter 7 Quantum Physics


I V
a


hc
m = __
e




©PAN ASIA PUBLICATIONS
0 V
©PAN ASIA PUBLICATIONS
V
a
1
Figure 7.9 Graph of I against V Figure 7.10 Graph of V against —–
a
λ
The Characteristics of the Photoelectric Effect (d) The threshold frequency, f is the minimum
0
frequency required to produce photoelectric
Source
1. According to the classical theory, light in wave effect on a metal of light
form is a spectrum with continuous energy.
2. Photoelectric effect should be able to occur at Metal
any light wave frequency.
3. Bright light which has high energy should be Thermionic emission Metal
able to emit electrons quickly. • The emission of electrons from a metal
4. Dim light has low energy so the electrons need surface by thermionic emission may take
a longer time to absorb sufficient amount of some time.
energy to escape from the metal surface.
Source
5. However, the results of the photoelectric effect of light
experiments show that:
(a) The higher the frequency of the photon of
light, the higher the kinetic energy of the Metal
photoelectrons emitted from the metal
surface. Metal

(b) The minimum frequency of light needed for Thermionic emission
a metal to emit electrons is known as the • The emission of electrons from a
threshold frequency, f for that metal. metal surface by photoelectric effect is
0
(c) The kinetic energy of photoelectrons does instantaneous.
not depend on the intensity of light. An Try questions 3 and 4 in Formative Zone 7.2
increase in the light intensity does not 6. Threshold frequency is the minimum
produce photoelectrons with a higher frequency required to produce photoelectric
kinetic energy. effect on a metal.



7.2
1. What is photoelectric effect? C1
2. Compare the emitted number of photons from the metal surface by a bright light and a dim light of
the same frequency. C2
3. Explain four characteristics of photoelectric effect from the experiment. C2
4. Why are photoelectrons emitted instantaneously from a metal surface when it is illuminated by a
CHAP light of certain frequency? C3
7 5. Does an increasing of light intensity increase the kinetic energy of the photoelectrons? Why? C4



482 7.2.2

Form
5 Physics Chapter 7 Quantum Physics









Paper 1


1. What is the energy for one γ-photon with 6. Figure 1 shows four energy levels for an electron
frequency 10 Hz? C3 in hydrogen atom.
20
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©PAN ASIA PUBLICATIONS
A 10 J C 10 J E
–13
–15
B 10 J D 10 J 4
–12
–14
E 3
2. The energy carried by one blue light photon of
450 nm is C3
E
A 2.21 × 10 J C 6.63 × 10 J 2
–19
–19
B 4.42 × 10 J D 8.84 × 10 J
–19
–19
3. In the Einstein’s photoelectric equation Figure 1 E 1
1
—mv = hf – W, for a metal, the symbol W The possible number of spectral lines produced
2
2
represents the C4 from the transition of energy levels is C4
A minimum energy electron escaped from the A 3
metal. B 6
B minimum energy for ionization of one atom C 9
in the metal. D 12
C minimum photon energy needed to remove 7. The consecutive light photons of wavelength
one photoelectron from the metal surface.
D minimum potential energy for one 590 nm produced by a laser source were
separated at a distance of 0.20 m. What is the
photoelectron to be liberated from the metal power of the laser source in nW?
surface. C4
A 0.1
4. Photoelectron cannot be liberated from clean B 0.3
copper surface by visible light because the C 0.5
C4 D 0.7
A photon energy for visible light is less than 8. Photon is the name given to C2
the work function of copper A one unit of energy quanta.
B threshold frequency of copper is less than B one quantum of electromagnetic radiation.
the frequencies of visible light C one electron emitted from a clean metal
C light intensity of visible light is insufficient to surface.
remove photoelectron from copper surface D one electron liberated from a metal surface
D kinetic energy of the free electron inside by light radiation.
copper is less than photon energy of visible
light 9. Which of the following experiment phenomena
provide evidence for the existence of discrete
5. What is the de Broglie wavelength of a particle energy levels in atoms? C2
with mass m moving at velocity of v ? C1 A Photoelectric effect
h
hm
A —— C —— B Line spectrum from sodium lamps
mv v C Energy distribution of black-body radiation
2πh
h
B ——— D ——– D Visible light spectrum from tungsten
filament lamps
2πmv
mv
CHAP
7
490

Form
5
Chapter 7 Quantum Physics Physics
Table 12 shows nine electronic components and gadgets that may be connected to complete the
electronic eye system circuit.
Table 12






Capacitor Photodiode Thermistor
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©PAN ASIA PUBLICATIONS




9 V d.c. power supply Resistor Electromagnetic coil






240 V a.c. power supply Alarm Variable resistor

Using the knowledge of physics, choose any six suitable electronic components and gadgets in Table 12
to complete the circuit in Figure 12.2. Give reasons for your choice. C6 [12 marks]






Reinforcement & Assessment of Science Process Skills


1. You are required to carry out an experiment to determine the value of Planck’s constant.
(a) Carry out the experiment by using the steps below:
(i) Use five LEDs of different colors whereby their
wavelengths were taken from the manufacturer’s
catalogue. The threshold voltage, V is the potential +
difference across the LED when it begins to conduct
is measured by adjusting the potentiometer to set the 6 V d.c.
value of ammeter to zero. –
(ii) The values of the wavelengths for the five LEDs are V
λ = 467 nm, 567 nm, 586 nm, 623 nm and 793 nm as A
labelled on them.
(iii) Tabulate your readings in a table for all values of λ, V Figure 1.1
1
and –—.
λ
Table 1

λ λ


V
1
–— CHAP
λ λ
7


501

Form
5 Physics Chapter 7 Quantum Physics


(b) Based on the experiment conducted, state
(i) the manipulated variable
(ii) the responding variable
(iii) the constant variable.

1
(c) Draw a graph of V against –—.
λ
(d) Based on your graph in (c), state the relationship between V and λ.
©PAN ASIA PUBLICATIONS
©PAN ASIA PUBLICATIONS
1
(e) Calculate the gradient, k of the graph V against –— in S.I. unit.
λ
hc
–19
(f) If k = –— where e = charge of an electron (1.60 × 10 C)
e
c = speed of light in vacuum (3.00 × 10 m s )
8
–1
Calculate Planck's constant, h
If you are unable to carry out the experiment described above, you can answer this question by using the data
obtained from the readings of the millivoltmeter shown in Figure 1.2.




1 2 1 2 1 2
0 3 0 3 0 3
V V V








λ = 467 nm λ = 567 nm λ = 586 nm
V = V V = V V = V






1 2 1 2
0 3 0 3
V V








λ = 623 nm λ = 793 nm

V = V V = V
CHAP
7 Figure 1.2



502

SPM MODEL PAPER





Paper 1

[40 marks]
Instruction: Answer all questions.

a
©PAN ASIA PUBLICATIONS
1. Which of the following is a base unit? A The gradient of the graph is —
©PAN ASIA PUBLICATIONS
b
A kJ B p increases linearly with q
B µm a
C cd C If q = 1 then p = — + c
b
b
a
D The equation of the graph is p = —q + —
2. Which of the following are derived quantities? b a
I Mass of a satellite 5. Assume that the length, L, time, T and density,
II Proton charge ρ are chosen as the new base quantities of a
III Electrical power physics, what is the relationship of the derived
A I and II only quantity of mass in terms of ρ, L and T?
B I and III only 3
C II and III only A ρL
B LT 3
3
3. The following statement provides information C ρ L
(ρLT)
3
about a particular system. D ABBBB
6. Figure 2 shows five steel balls P, Q, R, S and T
The coordination of systems of with the same densities and sizes.
measurement units worldwide facilitates SPM MODEL PAPER
the field of science, sports, trade, medicine
and so on.

Which is the system mentioned above?
A Metric System P
B Imperial System Q R S T
C Universal Measurement System
D International System of Units

4. Figure 1 shows the relationship between the Figure 2
physical quantities of p and q. When ball P is pulled to the left, released and hit
ball Q with a displacement, the ball P becomes
p stationary while the ball T moves forward with
the same displacement. Give a reason for this
a situation.
A Impulsive force before collision = Impulsive
b force after collision
(0, c)
B Velocity of ball P before collision = Velocity
of ball T after collision
q
0
C Kinetic energy before collision = Kinetic
Figure 1 energy after collision
Which of the following statements is incorrect? D Momentum before collision = Momentum
after collision




503

Physics SPM Model Paper

Paper 2

Section A
[60 marks]
Instructions: Answer all questions in this section.
kQq
1. Coulomb’s law can be expressed in the form of the equation F = ——— where F is the force, Q and q are
r 2
the charges and r is the distance between the two charges.
©PAN ASIA PUBLICATIONS
(a) Based on the equation, state one example of
(i) base quantity,
(ii) vector quantity. [2 marks]
(b) Derive the unit of k in terms of the base S.I unit. [2 marks]

2. Figure 2 shows the velocity against time graph of the motion of an MRT along a straight line track.
–1
Velocity / m s


20
15

SPM MODEL PAPER ©PAN ASIA PUBLICATIONS
10
5

Time / s
0
40
30
50
60
20
90
10
70
80
Figure 1
(a) What is the average speed travelled by the MRT in the first 70 s?
(b) What is the physical quantity represented by the gradient of the graph from 0 s to 10 s? [2 marks]
[1 mark]
(c) Describe the movement of the MRT,
(i) from 10 s to 70 s
(ii) from 70 s to 90 s. [2 marks]
3. Figure 3 shows a cargo ship being towed by two tug boats using the same force of 1 200 N. The resultant force
from the two tug boats causes the cargo ship to move forward.
P

Tug boats

20°
20°
Cargo ship
Q


Tug boats
Figure 2




510

Physics SPM Model Paper

Paper 3
[15 marks]
Instruction: Answer all questions.


1. You are required to carry out an experiment to investigate how resistance of a bulb changes with the potential
difference across it.
d.c. power supply
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Slide wire
Mark 0.0 cm


Meter rule
Crocodile clip
A
Bulb




V
Figure 1.1
SPM MODEL PAPER (i) Turn on the switch.
(a) The circuit has been set up for you. Carry out the following steps below:

(ii) Adjust the position of the crocodile clip on the slide wire until the potential difference across the
bulb is 0.2 V.
(iii) Record the current reading I, indicated by the ammeter.
(iv) Adjust the crocodile clip and record the corresponding value of I for V = 0.8 V, 1.4 V, 2.0 V and
2.6 V.
(v) Turn off the switch.
(vi) Calculate and record the resistance of bulb, R for each value of V in Table 1. Use the equation
V
R = —–.
I
Table 1
V / V I / A R / Ω
0.2
0.8

1.4
2.0
2.6
[6 marks]

(b) Plot the graph of R (Ω) against V (V). [3 marks]
(c) State what the shape of the graph tells you about the relationship of resistance of the bulb filament with
the potential difference across it. [2 marks]
(d) In a similar type of experiment, changes in current and potential difference for the bulb can be achieve
by using a variable resistor instead of a slide wire. Draw the new electrical circuit. [2 marks]
(e) What will happen to the shape of the graph if the bulb is replaced by a semiconductor diode in forward
biased configuration? [2 marks]
518

SPM Model Paper Physics
If you are unable to carry out the experiment described, you can answer this question by using the data obtained
from Figure 1.2.





0.2 0.4 0.6 0.8 0.2 0.4 0.6 0.8

0 1.0 0 1.0
A A
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When V = 0.2 V, When V = 0.8 V,
I = A I = A







0.2 0.4 0.6 0.8 0.2 0.4 0.6 0.8
0 1.0 0 1.0

A A SPM MODEL PAPER







When V = 1.4 V, When V = 2.0 V,
I = A I = A







0.2 0.4 0.6 0.8

0 1.0
A







When V = 2.6 V,
I = A

Figure 1.2


519

Complete answers
ANSWERS http://bit.ly/3bHkxpp




24
FORM 4 (b) (i) Duration, T = ——
20
= 1.2 s
Chapter 1 Measurement Paper 1
10
1. D 2. A 3. A 4. A 5. D (ii) l = ————––— × (1.2) 2
1.1 6. B 7. A 8. C 9. B 10. C 4 × (3.142) 2
11. B 12. D 13. C 14. B 15. D = 0.36 m
1. (a) Length
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16. D 17. C 18. A 19. B 20. C
(b) Base unit: Meter; 21. A 22. B 23. A 24. B 25. D Section B
Symbol of unit: m 4. (a) Car Speed of car / m s –1
Magnitude: 1.55; Paper 2
Symbol of physical quantity: l Section A A 30.00
1. (a)
2. (a) Scalar quantity has magnitude Base quantity S.I. unit
only and vector quantity has both Length meter B 37.49
magnitude and direction. Time second C 25.20
(b) Scalar quantities: Distance, speed, Mass kilogram
time D 33.30
Length
Vector quantities: Acceleration (b) P = Force × ————
3. Base quantities: Time, mass, length kg m Time E 24.20
m
= ———— × —–
s
Derived quantities: Volume, area (c) S.I. unit is not same as metric (b) Both of them use walkie talkie
s × s
= kg m s
or smart phone. The time taken
2
–3
FORM 4 ANSWERS 1. (a) Graph functions as a visual tool unit because the time unit cannot (c) Malaysia speed limit
to pass through point O and P is
recorded and the time interval, t
1.2
is determined or other reasonable
be expressed in terms of 10.
method.
For example, 1 hour = 60 minutes
= 60 × 60 seconds
to represent the relationship

–1
= 110 km h
= 3 600 seconds.
between two physical quantities.
(b) Steps to analyse graph:
Thus, car B exceeds the speed
of tens or decimal-based but S.I.
(i) State the relationship Metric unit is based upon power = 30.56 m s –1
limit on Malaysia highway.
between two variables units of time cannot be expressed (d) Use the ultrasound detector,
(ii) Determine the gradient of in terms of 10. time tracker application or other
graph 2. (a) (i) Base quantity: Mass reasonable method. Record
(iii) Determine the area under (ii) Derived quantity: Force several readings on speed within
the graph (iii) Vector quantity: Force 5 km and calculate the average
(iv) Determine the value of (b) S.I. unit G values.
a physical quantity from Fr 2 5. (a) In old system: F = mlt –2
interpolation = ——— so, m = Ft l
Mm
2 –1
1
—
2
kg m × m
(v) Make a prediction through = ———————– In new FAω system: l = A ,
2

extrapolation s × s × kg × kg t = ω –1
–1
= kg m s –2 – 1 —
3
2 –1
2. (a) V increases so, m = Ft l = Fω A 2
–2
(b) V= 0 cm , θ = –250 °C 3. (a) (i) Time taken, T = 26 – 2 (b) There is no suitable measuring
3
θ = 300 °C, V =22 cm 3 = 24 s device to measure force, area and
(ii) To get the value of one frequency accurately.
V / cm 3 complete oscillation – There is no standard tool and
accurately. standard objects to determine
30 (iii) Repeat the experiment to get force, area and frequency.
22 cm 3 the average value of two sets – The unit for derived quantity
20
of readings for 20 complete becomes very complex and
10 oscillations. Use electronic or hinders the communications
–250°C
digital stopwatch to measure between physicist.
θ / °C
–300 –200 –100 0 100 200 300 the time more accurately. – or other reasonable answers.
520

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