Spotlight A+1 Form 4.5 Physics

Extra Features of This Book

7CHAPTER Quantum Physics

SMART SCOPE 7.1 Quantum Theory Of Important Learning Standard Page CONCEPT MAP
Light • Explain the initiation of the quantum theory. 472
Contains learning objectives • Describe quantization of energy. 476 Contents of the whole topic
that need to be achieved in 7.2 Photoelectric Effect • Explain wave-particle duality. 477 are summarised in the form
each topic. • Explain concept of photon. 478 of a concept map.
7.3 Einstein's • Solve problems using photon energy, E = hf and power, P = nhf. 478
Photoelectric Theory 479
• Explain photoelectric effect. 482
• Identify four characteristics of photoelectric effect that cannot be
483
explained using classsical theory.
485
• State minimum energy required for a photoelectron to be emitted 485
from a metal surface using Einstein’s equation, hf = W + —21 mvm2ax 486

• Explain threshold frequency, f0 and work function, W.
• Determine work function of metal, W = hf0 = —hλc0–
• Solve problems using Einstein’s equation for photoelectric effect.

Fo4rm Physics Chapter 2 Force and Motion I • Explain production of photoelectric current in a photocell circuit. 488
• 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. tlTIbohnyhaeeEdglrxpodapeaevredpiirtoeiyamnd.ndeosinfnotetn2hrt.e2tiahhtlehobrmeaizhlaaoosnnsrcitozaefol anportlseaacslintloilocastict•••••nii iolenDWBRIlnadfollastiaeaunsidoacvocelikleynnrfa .e- astctbptibehoeooasde fndroet nyriibn cerlertaeegrd n dy/is au/Pit atieTylonie nt/yny eKa/ r/ geS aKa•aipnmeduTadiarhnsutakegaarnglnpiautajnsalsissngaeeadnlrogahmenirtsbaafmanllgb-zaackrawhards • Continuous energy / Tenaga selanjar 266
3. • Photoelectric effect / Kesan fotoelektrik
• Threshold frequency / Frekuensi ambang
4. The inertia of an object has a direct relationship • Work function / Fungsi kerja Pressure
when a

stationary bus suddenly moves forward.

with its mass, whereby the inertia of the object
2 ImportantCHAP increases when its mass increases.
Formula and Tips Pressure in Atmospheric Gas Pressure Pascal’s Principle Archimedes’ Principle Bernoulli’s Principle
Liquids Pressure
• Einstein’s photoelectric equation, • Photon power, P = nhf
hf = W +• —21Thme vpm2aasxsengers in which n is the number of photons emitted per
forwasredcwohnedn a
Video of inertial balance are thrown Formula, Venturi
http://bit.ly/36NykFm wiWPsh htoohertrekoe tnbdhy oer,n en fsee0h, ri osWg lytdh, = Erwe h=asfh(v 0dh eo=)fll ed =–nh λAts —fg–herc0htnλ—epemh cqao ourosivaetlfiot nntmeragcaneytghtck oeFaeoeiflmrgb mtuulaposrnearetskrraut2yldm .ai4hdsne3eandnsdtl ylλsiavs0r itgdotoepesdi ••.n reewdMirndethtoui oBaemc.rreeToetbhnhgytrltu,hieu esKeem,'s = wo —af21 vpemalrevtn2icglteh, ,p λ = = A —BPh2BmBKB Manometer —AF—11 = —AF—22 tube
• Formula, Applications: Water Formula, Aerofoil
The Effects of Inertia in Daily Life P = hρg • Water tank Applications: F = ρVg Lift force
• Intravenous liquid manometer • Piston’s Pascal Angle of
• • Dam Mercury • Hydraulic system Buoyant attack
• Siphon force, F Applications:
manometer (Hydraulic jack, • Bunsen burner
1. The astronauts in the International Space impact of the inertia of petrol on the walls hydraulic brake) • Racing car
Station (ISS) are in zero gravity contition, hence of the tank when the lorry stops abruptly. • Sports
they need the inertial balance to me4as7u0re mass. Factors Weight of object in • Aeronautics
• Depth in floating state, W
2. Examples of inertia situation in daily life and
SPOTLIGHT PORTAL its effects: liquid
(a) The rain drops are in motion as when the • Density of
Scan QR code to visit umbrella rotates. When it stops rotating,
the inertia of the rain drops causes the liquid
rain drops to continue in motion and leave
the surface of the umbrella. Figure 2.44 Measuring instruments: Units of pressure W=F W.F W,F
(e) When a roller coaster changes direction • Mercury barometer
• Fortin barometer • Pascal, Pa Floating at Moving Moving
suddenly, the inertia of the passengers • Aneroid barometer • mm Hg a stationary downwards with upwards with
keeps them to maintain their original • mm iHlli2bOar an acceleration an acceleration
state of motion. The passengers must wear • position
safety belts to remain in their seats and not
thrown out of the carriage when it moves Effects of atmospheric pressure Applications
at sudden changes of speed and direction. • at high altitude • Ship and Plimsoll line
• at extreme depth • Submarine
websites or videos related Photograph2.2 • 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. Chapter 6 Light and Optics Physics Fo4rm
are videos for certain When the bottle is stopped, the inertia of
the sauce will cause the sauce to continue
moving downwards and out of the bottle.

Photograph 2.4
(f) When a car brBakesOabprtuicpatllyf,ibthreedriver and

passengers in the car are thOrouwtenr cflaodrwdinagrd
due to inertia. Thus, seat (bLeoltws raerferadcetisviegnineddex)
activities or experiments. to prevent them from being thrown Normal Doctors use endoscope to see and examine organs
forward and injuring themselves. inside the human body. Engineers use fibre optics
44 Photograph 2.3 Light signal to monitor performance of complex machinery. INFORMATION GALLERY
Communication experts use fibre optics to send
(Higihnnreefrracclat2idvd.e4ini.ng3dex) high speed data.

Figure 6.23 C Cat’s eye reflector

• Widely used in the telecommunication and • Can be used as safety devices for drivers at Additional information
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
internal reflection.
surrounded by an outer cladding of lower
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
ienFntod5errmonfatlhreeffilbecreti.oHnsenucnettihl eresaigcnhainl gwitlhl ebeotshenert
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.
(a) What is the critical angle of the prism?
Variables: (b) Draw the light paths in the prism until it is emerging again into the air.
(a) Manipulated: Resistivity of wire, ρ
(b) Responding: Resistance of wire, R 45°
(c) Constant: Length of wire, temperature, area of cross-section of wire

ACTIVITY / EXPERIMENTApparatusandMaterials: Solution Figure 6.24
5(b) Form
50 cm constantan wire (s.w.g. 24), 50 cm copper wvoirletm(se.wte.gr .(024–),55V0),cmrhetuonstgastteanndwbireat(tse.wry.gh.o2ld4e),rconne(cat)ingsin c = 1 Chapter 7 Quantum Physics Physics
wires, three dry cells, switch, ammeter (0 – 1 A), n
C
Operational definition: cCH=AP1.149 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 sin Example 1 particle, p and its wavelength, λ is
the λ = —hp–
wCaavlceulelantgethtsh4e50pnhmotoannde7n0e0rgnym44.f55oC°°4ro5°m4li5pg°ahrte with
ammeter. 3  c = sin−1 both
Complete activity sin 60° photons.
orProcedure: 1.49 45°
1. An electrical circuit is set up as shown in
Figure 3.28. c = 42.2° AB

The critical angle of the prisim is 42.2° Solution Figure 6.25 which h is Planck’s constant (6.63 × 10–34 J s)

2. A constantan wire is connected across Ammeter A Planck’s constanAtn, ghle= o6.f6i3nc×id1e0n–3c4eJ s(i = 45°) larger than critical
Speed of light inanvgalceu(ucm=, c42=.23°.0) 0at×b1o0u8 nmdas–r1ies AB and AC.
experiment including results,terminals P and Q. The length of the wire Wavelength, λλ21rtT==ahoye4t7a5e0nl00mEoin×r×e=tmer11ghar00nifl–n–9a=9dgmlmi—rrfeehrλo—cfcltmeiocntBi.oCnaogcacinurirnetdo and the light
across P and Q is adjusted to be 30 cm long. Rhesotat PQ Wavelength, the air along —2pm—2

3. The switch is connected and the rheostat is Constantan wire K= —21 mv2 × —mm– ➞ K = —(m2—mv—)2 =
adjusted so that the ammeter gives a reading V
Voltmeter Photon energy of 450 nm: 6which p =CHmAvP, p = AB2mBBKB
data analysis, discussionof 0.5 A. The reading of voltmeter is taken
and recorded in Table 3.6. Figure 3.28 1 2E1 = 6.63 × 10–34 —435.—000—××—110—0–89–

4. The constantan wire is removed and Steps = 4.42 × 10–19 J
2 and 3 are repeated for copper wire and Photon energy of 700 nm:

and conclusion to increasetungstenwire. 1 2E2 = 6.63 × 10–34 —730.—000—××—110—0–89– 7. The larger the momentum of particle, the shorter BRILLIANT TIPS
Results: 1tp9ha5ertiwclaevieslepn=gtmh vp, rthodenuc—2epdm—2. The momentum of
Table 3.6 6.2.3 6.2.4 =K

students' scientific skillW.ire Reading of ammeter, Reading of voltmeter, Resistance, = 2.84 × 10–19 J
Constantan —V The shorter the wavelength of light, the higher
I/A V/V R = I / Ω the photon energy. λ = —mh–v– = —AB—2Bmh—BK—B

0.5 6.3 12.6 Form Useful tips for studentswhere m is the mass of particle, v is the velocity

Copper 0.5 0.2 0.4 5 of particle and K is the kinetic energy of particle.
Tungsten 0.5 Chapter 1 FWoracveea-nPdarMtoictiloenDIIualPithyysics CHAP to solve problems in the8. Since the value of h is too small, the particle
with large mass will have too short of de Broglie
0.6 Solving Prob1l.e2ms Involving Resultant Force and Resolution of Forces 1 wavelength to be detected. Thus, the wave
characteristics cannot be observed.
Example 4 ExampleW5avelength simulation
Discussions: Figure 1.17 hsthtopw://sbitth.lye/3f9rexeNYbqoudy diagram of a related subtopic.9. In 1927, the presence of wave properties of
block sliding down a smooth inclined plane. electrons was confirmed through the electron
1. From Table 3.6, different conductor gives different resistance. A wooden block is pulled by force, T that inclines diffraction experiments.
1. LigNhotrmhaalsrewaacvtieonp=ro1p0eNrties because it shows the 10. Photograph 7.1 shows the diffraction pattern of
2. Constantan gives the highest resistance, followed by tungsten and copperatgaivneasnthgleelofw3e0s°t arebsoivsteatnhcee.horizontal surface as phenomena of diffraction and interference.
3. Different material gives difference resistance because different materiaslhhoawsndiinffeFrieguntrere1.s1is6t.ivity, ρ. electrons through a thin layer of graphite. The
2. Object haBslopcakrticle properties because it has pattern in Photograph 7.2 resembles the light
Resistance of a material increases when the resistivity of the material increases. R=5N T = 40 N momentum, kinetic energy and also collide with diffraction pattern through an aperture.
Wooden each other.
block

30°

3.2.3 FR = 18 N 335

W = 25 N WW3.e=igtL2hho0ta,utNiaslld6pe0a°Brtriocglelisecianntreoxdhuibceitdwaavheypchoathraecsitsersitsattiecss

Figure 1.16 in year 1924.
(a) What are the magnitudes of the horizontal Figure 1.17
toShkfeethtccehobm4alon.pcdoktHmnlhapoeaebanwserswtleaetavlsohlv.eefeerl,tcthcoiothemtaihsrwpeaoeecixinntgepcehrelniitrsntitoomeicfdfsettpnhholteaefawnlplbeyealiaorgdntchiidkcftfliecsulwt ittoh show
tchoemppuolnlienngtfTorxcaen, dT?vertical component Ty of (a) large
(b) Calculate the magnitude and direction of the
Cpearlpcuenla5dt.eicutcThlhhaeaurrsrta,oecsLttueholruteiasiintsnitcdcslfeioncBraecrndeopgballeciaetnisneph.groewdoninctbethydeltihgahtt the wave Photograph 7.1 Photograph 7.2
resultant force acting on the block. particles
(c) Calculate the acceleration of the block if its (b) block. such as electron, proton and neutron. 711. This observation proved de Broglie’s hypothesis. CHAP

mass is 1.8 kg. (c) What is the acceleration of the block if its

EXAMPLE Solution mass is 2.0 kg?

Example and (a) Magnitude of the pulling force, Solution 7.1.3 477
T = 40 N
Ty (a)
Tx = T cos 30° T
= 40 cos 30° Tx Normal reaction = 10 N
30°
= 36.64 N (to the right)
Wy
complete Ty = T sin 30° 60°
= 40 sin 30°
= 20 N (upward) Wx

solution to enhance students'(b) Resultantofhorizontalcomponents = 3T6x .–64F+R (–18) 60°
=
= 16.64 N (to the right) (b) WWReyxs=u=l2t2a00ncstoionsf66t00h°°e==fo11r70c.e3Ns2 N
Resultant of horizontal components perpendicular
understanding. = T20y ++ + to the
= R + W inclined plane = 10 + (–10)
=0N 5 (–25) =0N

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
Acceleration, a = —mF– Acceleration of the block, a = —mF–
F = ma = —127—..03–2–
a = 8.66 m s–2
= —116—..86–4–
a = 9.24 m s–2

Try question 2 in Formative Zone 1.2

1.2.2 235

ii

TAG OF 'TRY QUESTION...

IN FORMATIVE ZONE...' Fo5rm Physics Chapter 3 Electricity

9. The resistance of a wire is directly proportional Example 17
to its resistivity, ρ (Ω m) and length, l (m) but A copper wire with the length of l and diameter d
inversely proportional to the area of cross- has the resistance of R. What is the resistance of
section, A (m2), that is another copper wire with length 2l and diameter
R ∝ ρl and R∝ —1 —12 d in terms of R?
A
Tag placed at the end of By combining the relationships shown, therefore

R = —ρA–l Solution

examples to guide students Example15 Both wires are of the same material, therefore, the
resistivity, ρ, are the same.
For the first wire, area of cross section,
A = π × 1—d2 22
What is the resistance of a copper wire with a Fo5rm
in answering related diameter of 0.5 mm and length of 2 m? = —π4d—2
Assume that the wire has the shape of a cylinder. Physics Chapter 6 Nuclear Physics
Resistivity of copper = 1.60 × 10–8 Ω m.
∴ Resistance, R = —ρA–l 6.1
questions in Formative Zone.3CHAP Solution = ——πρ—4d—l2–
Length, l = 2 m 1. Fill in the blanks to complete the following radioactive decay equation. C1
Area of cross-section, A
1 2= π × —0—.5—×2—10—–3– 2 (a) 20894Po → 20852Pb + (b) → 11p + –01e (c) 22888Ra → + –01e
= 1.96 × 10–7 m2
Resistance of wire, R = —ρA–l = —π4—dρl2– 2. Calculate the number of α and β particles that are emitted from the radioactive decay series of C4
= —1.—610.—9×6—×1—01–—08 –×—7 2–
= 0.163 Ω U238 → ... → 28026Pb
92

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.
Table 6.2
1 2= π × ——21—2d– 2
Date 15 March 2021 25 March 2021 4 April 2021
= —π1d—62 Activity / s–1 1 520 380 95
∴ Resistance = —ρ—×A—2l–
Try question 9 in Formative Zone 3.2 (a) Calculate the half-life, T—1 of the radioactive sample. C2
= —ρ—π—1×d—6—22 l– 2
Example 16 = 8 × —π4—dρl2–
Wire X with the length of 120 cm and the cross- = 8R (b) Sketch a radioactive decay curve for the sample. C3
section area of 0.80 mm2 has the resistance of
3.2 Ω. Calculate the resistivity of wire X. Try question 11 in Formative Zone 3.2 6.2 Nuclear Energy Nuclear fusion FORMATIVE ZONE
http://bit.ly/2Kb1gBy
Solution 1. Nuklear energy is an atomic energy, which is
SPM SIMULATION Length, l = 120 cm energy that tied the nucleus of an atom. Questions to test3. The loss of mass (mass defect) occurs during
HOTS QUESTIONS these nuclear reactions.
= 1.2 m 2. This energy is released during nuclear reactions
such as radioactive decay, nuclear fission and
Area of cross-section, A nuclear fusion.
= 0.80 mm2
= 0.80 × 10–6 m2 Nuclear Fission
Resistance, R = 3.2 Ω students' understanding1. Nuclear fission is a nuclear reaction when a
Resistance of wire, R = —ρA–l Ensure units are converted to S.I. while performing Nuclear fission heavy nucleus splits into two or more lighter
the calculation. http://bit.ly/3mqw0vm and stable nuclei while releasing a large amount
= 3.2 For example, at the end of eachofenergy.
—0—.8ρ0—××—11.—20–—6 = 3.2 Chapter 6 Light and Optics Physics Fo4rm
1 cm2 = 1 × 10–4 m2
ρ = 2.13 × 10–6 Ω m 1 mm2 = 1 × 10–6 m2

Try question 10 in Formative Zone 3.2 A barium–141 nucleus, a kripton–92
nucleus and three neutrons are produced.

Provides complete solution334 A uranium–235 An unstable uranium–236 subtopic.Energy is also released.++ ++++++++++++++92
nucleus is produced. 36
SPM Simulation 3H.2O.3TS3.2Q.6uestionnbuyscalenuesuitsrobno.mbarded

61. Figure 1 shows refraction of light wChHeAPn the Kr
light travels from air to plastic.
with examiner's comment LetEo++o++i2n+px+g+93w++t+2a5+t++h+e+i+aUmc+tr++++r+a+sd++it+ln+++rst++aehdvrteeh'esnelpsClsaintlosoyomt.wircmmeWmraehlie.nnedtTn:imhuemtehrd,ee+i+ft++uo+h+2++m+93r+r+e+++62+ae+++++U+y+,+r+++wa+++a++yiotnhfgblelheingigdohhstf 1
Incident ray 1 n refraction is smaller than incident ray. If Energy 3 0 n
0

+ ++
to help students to answer +++ +++++++ +++
Air +++ + 141 Ba
+ 56

Plastic the plastic is replaced with less dense 3926Kr + 301n + energy
medium, the ben239d25Uing+ 01onf →ray29362oUf →ligh145t16Bwail+l
be reduced.
HOTS questions.
AnFsiwguerre: 6D.8 Nuclear fission involving uranium-35 bombarded by a single neutron

Figure 1 2. Figure 2 shows the position of an image 6.2.1
formed by a concave lens. Which position of
Which is the correct pathway of the light whe4n52 A, B, C or D, the viewer cannot see the image?
plastic is replaced with less dense medium?
AC

B

Chapter 4 Electromagnetism Physics Fo5rm A C
Object Image

D

SPM MODEL PAPERB D
Paper 1 Figure 2
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

AU U U CU SS S dASefleRcetivoenr?se C2 Answer: A [40 marks]
the Instruction: Answer all questions.
poles of the magnet and increase
the value of the current
B Maintain the poles of the magnet and
RiMddneeecaccvrirreneeerataasassseeieentttthhhhteheeeesvstdtardrieleriunerneecggcttotithhofiontoohnffeottohchffeuetrtmhmhreeeaanggctcnnuueerrttrreennt3t .aanntQFwcddhaiagerlbulPrejuoecinnsa3cutttssiuhoehrenno.tiwnrhoTgseahtdlevhesiefeidtdwsrewiitv.huiesiTarleotoiiobnconsavtcrareautQrrcctothPeimsedcemjaubnontyhnvcieotntihtogpensrteoowhebwohluaeec1srramdee.r Which of the following is a base unit A The gradient of the graph is —ab
A kJ B p increases linearly with q
S S SS U U U CU B µm C If q = 1 then p = —ab + c
U BU U S DS S S D C cd House D The equation of the graph is p = —ab q + —ab
U Car P

S 4. Figure 2 shows the force, F acting in the mdiaregcnteiot(ifiannac)scteaIdnllebFdiygidunrreitvhe3er, P, an appropriate mirror need to b2e. Which of the following is a derived quantity?
S on a current-carrying conductor in a correct place. mirror need to be I Mass of a satellite
SUMMATIVE ZONE mark Y, where the II PEWrleoactltolrniccahl aprogweer 5. Assume that the length, L, time, T and density,
field. placed. Give one reason for the answer. III ρ are chosen as the new base quantities of a
SS SU UU U physics, what is the relationship of the derived
P (b) State the suitable mirror to be used. Give one A I and II only
reason for the answer. B I and III oCnarlyQ quantity of mass in terms of ρ, L and T?
2. A current-carrying conductor is placed in a (c) State one feature of the mirror suggested in C II and III only
magneticfield produced by a pair of Magnadur Q 3(b) so that driver P can see the clearer image. CHAP A ρL3
LT 3
magnets. CHAP 3. The following statemFeingtuprero3vides information 6 B ρ3L
A magnetic force acts on the conductor if C3 4 about a particular system. C
A the direction of the magnetic field is the F Current SPM MODEL PAPER D 3AB(ρBLBTB)

same as the direction of the current. S 3 213 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.
R units worldwide facilitates the field of
to the direction of the current. Figure 2 science, sports, trade, medicine and so on.
What is the direction of the magnetic field?
Questions of various levelsC the direction of the magnetic field is C3 Which is the system mentioned above? P
perpendicular to the direction of the AP A Metric System
current. BR B Imperial System QR S T SPM MODEL PAPER
CQ C Universal Measurement System Figure 2
of thinking skill to evaluate3. Figure 1 shows the arrangement of apparatus to DS D International System of Units
study the effect of the force acting on a current- 4. Figure 1 shows the relationship between the When ball P is pulled to the left, released and hit
carrying conductor in a magnetic field. The 5. Figure 3 shows a current-carrying conductor physical quantities of p and q. ball Q with a displacement, the ball P becomes
conductor moves to the left with a small angle between the poles of a permanent magnet.
p SPM-orientedstationary while the ball T moves forward with
students' understandingof deflection when current flows in the direction Force –+ practice
that shown. S a the same displacement. Give a reason for this
Current N situation.
A Impulsive force before collision = Impulsive
of each topic. b based on latest SPMforceaftercollision

N Conductor (0, c) B Velocity of ball P before collision = Velocity
Magnet of ball T after collision
Figure 3 0q
S Which action will make the conductor C Kinetic energy before collision = Kinetic
experience a force downwards? C1
Figure 1 Figure 1 format to assess studentsenergyaftercollision
Which of the following statements is incorrect?
Fo4rm D Momentum before collision = Momentum
Physics Chapter 6 Light and Optics 397 after collision

Reinforcement & Assessment of Science Process Skills on all the503 topics learnt in

1. In this experiment, you will determine the value of refractive index for water.4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Form 4 & 5 textbooks.
Carry out the following instructions, refering to Figure 1.

Observer

Ruler

Retort stand

Beaker

Water

REINFORCEMENT & ASSESSMENT Cork 012 Image pin Po ANSWERS Complete answers
Pin P1 cm Pin Po http://bit.ly/3bHkxpp

SCIENCE PROCESS SKILL Figure 1

Questions for students to (a) Carry out the experiment according to the following instructions FORM 4 (b) (i) Duration, T = —22—40
(i) Fix a pin Po at the bottom of the beaker with cellophane tape. = 1.2 s
(ii) Fill the beaker with water until Po is at a depth of 6.0 cm from the water surface. Chapter 1 Measurement Paper 1
(iii) Adjust pin Pi so as it is in line with the image of Po (no parallax between pin and image of Po). (ii) l = —4 —× —(13—0.14––2—)2 × (1.2)2
(iv) Measure the distance y between pin and the bottom of the beaker. Pi 1.1 1. D 2. A 3. A 4. A 5. D = 0.36 m
(v) Calculate the apparent depth, h =H y. 6. B 7. A 8. C 9. B 10. C
P–i 1. (a) Length 11. B 12. D 13. C 14. B 15. D Section B
(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; 16. D 17. C 18. A 19. B 20. C
master theRecord all the readings in the Table 1. Symbol of unit: m 21. A 22. B 23. A 24. B 25. D 4. (a)
Real depth, H/ cm Magnitude: 1.55;
Table 1 Symbol of physical quantity: l Paper 2 Car Speed of car / m s–1
y / cm Section A 30.00
Apparent depth, h / cm 2. (a) Scalar quantity has magnitude A 37.49
only and vector quantity has both 1. (a) Base quantity
science process skill (SPS) presented6.0 in magnitude and direction. S.I. unit B
8.0 Length meter
10.0 (b) Scalar quantities: Distance, speed,
time Time second C 25.20
Vector quantities: Acceleration
Mass kilogram D 33.30
3. Base quantities: Time, mass, length E 24.20
Science Practical Test. Scan the QR code12.0 Derived quantities: Volume, area (b) P = Force × —LTe—inm—gte—h (b) Both of them use walkie talkie
14.0 = —ks—g×—ms— × —ms–

below to get Tip dan Teknik Prihatin Murid(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. FORM 4 ANSWERS = kg m2 s–3 or smart phone. The time taken
(c) (i) Based on the graph drawn in 1(b), determine the value of the refractive index for water.
(ii) State two precautions that you took in this experiment in order to obtain reliable readings. 1.2 (c) S.I. unit is not same as metric to pass through point O and P is ANSWERS
unit because the time unit cannot recorded and the time interval, t
1. (a) Graph functions as a visual tool be expressed in terms of 10. is determined or other reasonable
to represent the relationship For example, 1 hour = 60 minutes method.
between two physical quantities.
in answering Science PracticaC6HAPl Test. = 60 × 60 seconds (c) Malaysia speed limit
(b) Steps to analyse graph:
(i) State the relationship = 3 600 seconds. = 110 km h–1
between two variables
(ii) Determine the gradient of Metric unit is based upon power = 30.56 m s–1
graph
(iii) Determine the area under of tens or decimal-based but S.I. Thus, car B exceeds the speed
the graph
(iv) Determine the value of limit on Malaysia highway.
a physical quantity from (d) Use the ultrasound detector,
interpolation
(v) Make a prediction through time tracker application or other
extrapolation reasonable method. Record
several readings on speed within
2. (a) V increases 5 km and calculate the average
values.
(b) V= 0 cm3, θ = –250 °C
θ = 300 °C, V =22 cm3 5. (a) In old system: F = mlt–2
224 units of time cannot be expressed
V / cm3 in terms of 10.
Tip dan Teknik to
Prihatin Murid Complete answers are2. (a) (i) Basequantity:Mass
https://bit.ly/2JfnKAo (ii) Derived quantity: Force
(iii) Vector quantity: Force

(b) S.I. unit G

provided. Scan QR code=—F—r2—
get explanations for objectiveMm so, m = Ft2l–1 A—21 ,
In new FAω system: l =
= —s —×kgs—×m—k—×g—m×—2kg–
= kg–1 m3 s–2 t = ω–1
so, m = Ft2l–1 = Fω–2A– —21

3. (a) (i) Time taken, T = 26 – 2 (b) There is no suitable measuring

questions.=24s

(ii) To get the value of one
complete oscillation
device to measure force, area and
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 cm3 20
the average value of two sets – The unit for derived quantity

of readings for 20 complete becomes very complex and

– 250°C 10 oscillations. Use electronic or hinders the communications

θ/ °C digital stopwatch to measure between physicist.
100 200 300
–300 –200 –100 0 the time more accurately. – or other reasonable answers.

520

iii

CONTENTS

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

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 5.6 Interference of Waves 159
2.7 Impulse and Impulsive Force 52 5.7 Electromagnetic Waves 166
2.8 Weight 53 Summative Zone 170
Summative Zone 55

Chapter 3 Gravitation 64 Chapter 6 Light and Optics 181

3.1 Newton's Universal Law of 66 6.1 Refraction of Light 184
Gravitation 74
78 6.2 Total Internal Reflection 191
3.2 Kepler's Laws 84
3.3 Man-made Satellites 6.3 Image Formation by Lenses 197
Summative Zone
6.4 Thin Lens Formula 203

6.5 Optical Instruments 207

Theme 3 Heat 90 6.6 Image Formation by 209
Chapter 4 Heat 92 Spherical Mirrors

4.1 Thermal Equilibrium Summative Zone 214

iv

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

4.3 Transformer 390

Summative Zone 397

Summative Zone 251 Theme 3 Applied Physics 409
Chapter 5 Electronics 411
Chapter 2 Pressure 264 417
2.1 Pressure in Liquids 267 5.1 Electron 422
2.2 Atmospheric Pressure 274 5.2 Semiconductor Diode 430
2.3 Gas Pressure 278 5.3 Transistor
Summative Zone

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

Theme 2 Electricity and Electromagnetism Chapter 7 Quantum Physics 470
472
Chapter 3 Electricity 313 7.1 Quantum Theory of Light 479
7.2 Photoelectric Effect 483
3.1 Current and Potential 316 7.3 Einstein's Photoelectric
Difference 490
Theory
3.2 Resistance 323 Summative Zone

3.3 Electromotive Force (e.m.f.) 339
and Internal Resistance

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

v

CHAPTER Gravitation

3

3.1 Newton's Important Learning Standard Page
Universal Law 66
of Gravitation • Explain Newton’s Law of Gravitation.
67
3.2 Kepler's Laws • Solve problems involving Newton’s Law of Gravitation for two
static objects on the Earth; objects on the Earth’s surface; Earth 68
3.3 Man-made and satellites; Earth and Sun.
Satellites 69
• Relate the gravitational acceleration, g on the surface of the Earth
with the universal gravitational constant, G. 70

• Justify the importance of knowing the values gravitational 73
acceleration of the planets in the Solar System. 74
75
• Describe the centripetal force in the motion of satellites and 76
planets system. 78
79
• Determine the mass of the Earth and the Sun using Newton’s Law 80
of Gravitation and centripetal force.
81
• Explain Kepler’s Laws.

• Express Kepler’s Third Law.

• Solve problems using Kepler’s Third Law.

• Describe how the orbit of a satellite is maintained at a specific
height by setting the necessary satellite velocity.

• Communicate on geostationary and non-geostationary satellites.

• Conceptualise escape velocity.

• Solve problems involving the escape velocity, v for a rocket from
the Earth’s surface, the Moon’s surface, Mars' surface and the
Sun’s surface.

Important Formula and Tips

• Gravitational force / Daya graviti • Gravitational force, F = Gm1m2
• Universal gravitational constant r2
GM
Pemalar kegravitian semesta • Gravitational acceleration, g = r2
• Gravitational acceleration / Pecutan graviti
• Satellite / Satelit • Centripetal force, F = mv 2
• Centripetal force / Daya memusat r
• Ellipse / Elips T 2 r 3
• Focus / Fokus Kepler's Third 1 1
• Geostationary satellite / Satelit geopegun • Law: 2 = 3
• Escape velocity / Halaju lepas T 2 r 2
• Gravitational potential energy
• Linear satellite speed, v = GM
Tenaga keupayaan graviti • r
Gravitational potential energy, U = − GMm
64
• Escape velocity, v = 2GM r
r

Gravitation

Newton's Universal Law Kepler's laws Man-made satellites
of Gravitation

Gravitational force consist First Law consist Linear satellite speed
of Second Law of
Third Law v = GrM
F = GMm can derive
r2 T 2 ∝ r 3

Gravitational acceleration used for

g= GM Geostationary Non-geostationary
r2 satellite satellite

Centripetal force Escape velocity
F = mv2
r v = 2GrM

Mass of Earth Sun-planet system Earth-satellite system

M = 4π2r3
GT 2

65

Fo4rm Physics   Chapter 3 Gravitation Object 1 Object 2
3.1 m1 F m2
Newton's Universal Law of
Gravitation F

r

In the 17th century, Isaac Figure 3.3  Gravitational force between two objects
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
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 between them.
Earth and the Moon.

1. Gravitational force is an attractive force that F = Gm1m2
acts between any pair of objects in the universe. r2
Thus, it is known as a universal force.
F = Gravitational force between two objects
2. Figure 3.1 shows the gravitational force between mmr 12 = Mass of first object
the Sun, Earth and Moon. = Mass of second object
CHAP = Distance between centres
3 Earth of the two objects
G = Universsal gravitational constant
Gravitational force   (G = 6.67 × 10–11 N m2 kg–2)
between the Sun
and Earth
Try question 1 in Formative Zone 3.1
Gravitational
force between Gravitational force The value of the gravitational constant, G has been
the Earth between the Sun determined experimentally.
and Moon Moon and Moon

Sun

Figure 3.1  Gravitational force as universal force 6. The magnitude of the gravitational force
between two objects can be calculated if the
3. The gravitational force exists in pairs and both mass of the two objects and the distance
objects experience the gravitational force of the between them are known.
same magnitude. Example 1

4. In 1687, Isaac Newton presented two The mass of a student is 54 kg. What is the
relationships involving the gravitational force gravitational force between the student and
between two objects. the Earth?
[Mass of the Earth = 5.97 × 1024 kg,
(a) The gravitational force is radius of Earth, r = 6.37 × 106 m]
directly proportional to the
product of the masses of Solution
the two objects, F ∝ m1m2
Mass of the Earth, m1 = 5.97 × 1024 kg
(b) The gravitational force is
inversely proportional to F ∝ m1m2 Mass of the student, m2 = 54 kg
the square of the distance r2
Distance between the student and centre
bFe∝twer12en the two objects, of Earth, r = 6.37 × 106 m
Gm1m2
Gravitational force, F = r2

F= 6.67 × 10−11 × 5.97 × 1024 × 54
(6.37 × 106)2

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

Fo4rm 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–2. Thus,
gravitational acceleration, g = 9.81 m s–2 even the knowledge of the value gravitational
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.

Gravitational acceleration Effect of gravity
https://bit.ly/2Vmdbyp https://go.nasa.gov/2ZbuzXF

7. The magnitude of the gravitational force that 10. The effect of low or high gravity on human
acts on an object at the surface of the planet can growth are shown in the Table 3.2.
be calculated when the value of gravitational
acceleration is known.

Table 3.2  Effect of change in gravity on human growth

3CHAP Factor Effect of low gravity Effect of high gravity
• Density of the body increases.
(a) Change in density • Density of the body decreases.

(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 Direction of
velocity
1. Figure 3.7 shows three positions of a satellite
that orbits the Earth with a uniform speed. Earth

2. An object in circular motion is continuously Direction of
experiencing change in its direction of motion velocity
even though the speed is constant.

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
centripetal force.

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

70 3.1.4 3.1.5

Fo4rm 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 × 108 m and orbiting around the Sun, calculate the mass of
period of orbit 2.36 × 106 s. What is the mass of the Sun.
the Earth? Radius of orbit, r = 1.50 × 1011 m
Period, T = 1 year
Solution
Mass of the Earth, Solution

M = 4π2r3 T = 1 year
GT 2
T = (365 × 24 × 60 × 60) s

= 4π2 × (3.83 × 108)3 Mass of the Sun, M = 4π2r3
(6.67 × 10−11) × (2.36 × 106)2 GT 2

= 5.97 × 1024 kg M = (6.67 × 4π2 × (1.50 × 1011)3 × 60)2
10−11) × (365 × 24 × 60

= 2.01 × 1030 kg

CHAP 3.1

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 × 106 m from the centre of the Earth.

Find the gravitational force between the spacecraft and the Earth?  C3
[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth, M = 5.97 × 1024 kg]
4. The planet Uranus orbits around the Sun with radius of orbit 2.87 × 1012 m and period 2.64 × 109 s.
Calculate the mass of the Sun.  C3

3.2 Kepler's Laws I, II and III Johannes Kepler worked as an assistant to the
astronomer Tycho Brahe. His great determination
1. Kepler was a German astronomer and drove him to study Brahe’s astronomical data for
mathematician who modified the heliocentric more than ten years.
model of the planets moving around the Sun.

2. Kepler succeeded in formulating three laws that
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 F1 and F2 on the A4 paper. The distance between F1 and F2 is
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

Chapter 3 Gravitation  Physics Fo4rm

Thread loop with circumference 32 cm 4. The ellipse is drawn by moving the pencil
slowly around F1 and F2.
16 cm
Minor axis 5. The thumb tacks and thread are removed.

6. A small cainrcdleainsodthraewr nsmatalFl 1ctiroclreepornetsheent
the Sun
circumference of the ellipse to represent the
Earth.
Thread
Thumb F1 F2 Discussion:
tack 12 cm 1. The shape of the ellipse shows that when

Major Figure 3.15 the Earth moves in its orbit along the
axis circumference of the ellipse, the distance
between the Sun and the Earth changes.

2. The Earth is nearest to the Sun when it is on
the major axis to the left of wF1h.eTnhiet Earth is
furthest away from the Sun is on the
major axis to the right of F2.

Video: Drawing an ellipse
https://bit.ly/2UhHkhH

A Kepler's First Law B Kepler's Second Law CHAP
• 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 3
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.
Minor axis • The distance AB is larger than the distance
CD as the planet moves with a higher linear
Ellipse speed from A to B compared to from C to D.

Planet
B

Major axis CF
Sun
D Sun
Planet A
Figure 3.16
Equal area in equal time
• The orbits of most planets in the Solar System Figure 3.17  The motion of a planet in its orbit
have the major axes and minor axes of almost
the same length. Try question 2 in Formative Zone 3.2

• Therefore, elliptical shape of the orbits C Kepler's Third Law
of planets in the Solar System are actually • Kepler’s Third Law states that the square of
almost round. the period of a planet is directly proportional
to the cube of the radius of its orbit, T 2 ∝ r3.
• The radius of the orbit is the average distance • Therefore, planets that are further away from
between the planet and the Sun. the Sun will take a longer time to complete
one orbit around the Sun.
3.2.1 3.2.2 Try question 1 in Formative Zone 3.2

75

Chapter 4 Heat  Physics Fo4rm

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: T o determine the relationship between the volume and pressure of a fixed mass of gas at a constant

temperature.

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: Piston Pressure gauge
1. The apparatus is set up as shown in Figure 4.24.
2. The piston is adjusted so that the volume of air in 100 ml syringe

the syringe is 100 ml. Then, the end of the syringe Retort stand
is connected to the pressure gauge.
Rubber tube
3. The readings of the volume and initial pressure of
the air in the syringe are recorded in Table 4.10. Figure 4.24

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.

5. Step 4 is repeated with volumes 80 ml, 70 ml and 60 ml.

Results:

Table 4.10

Volume, Pressure, 1 / ml–1 P / kPa CHAP
V / ml P / kPa V 180
4
100 100 0.010

90 110 0.011 160

80 125 0.013

70 140 0.014 140

60 165 0.017

Data analysis: 120
100
A graph of pressure, P against 1 is drawn as shown in
Figure 4.25. V 80

P

60

0 —V1 40

Figure 4.26 20

The graph of P against 1 shows that the pressure is 0 0.005 0.010 0.015 0.020 V1– / ml–1
V
inversely proportional to the volume.
Figure 4.25

4.4.2 111

Fo4rm 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.

Robert Boyle (1627-1691) was a scientist who 4. The rate of collision between the molecules and
emphasized on the scientific method when the wall of the container increases.
conducting investigations. From experimental data,
he made the conclusion that the volume of a gas is 5. The force per unit area on the surface of the
inversely proportional to the pressure of the gas. wall increases. Thus, the gas pressure increases.

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

P∝ 1
V
1 PV = constant
P = k1 V 2 P1V1 = P2V2

where k is a constant Eduweb TV: Boyle's law
P = pressure of the gas (Pa) http://bit.ly/2tb1y2I
V = volume of the gas (m3)

4CHAP Therefore, PV = k
Try question 1 in Formative Zone 4.4

2. Suppose a gas experiences a change of pressure Example 6
and volume from an initial state to a final state. The air in a closed syringe has a volume of
FIninitaial lstsatateteoof fththeeggaas,s,PP2V1V2 1==kk 60 cm3 and pressure 108 kPa. The piston of
Therefore, P1V1 = P2V2 the syringe was pushed to compress the air to
a volume of 48 cm3. Find the pressure of the
compressed air.

Solution

Volume is P1 = 108 kPa, P2 = pressure of compressed air
decreased V1 = 60 cm3 V2 = 48 cm3
The temperature of gas is constant,
Boyle’s law is used:

P1V1 = P2V2

Figure 4.27  A fixed mass of gas compressed at 108 × 60 = P2 × 48
constant temperature
P2 = 108 × 60
3. When the volume of the gas is decreased, the 48
number of molecules per unit volume increases.
= 135 kPa
112
4.4.2

Chapter 3 Electricity  Physics Fo5rm

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
(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: CHAP
The resistance of the conductor, R, is given by the ratio of the reading of voltmeter to the reading of the
ammeter. 3

Procedure: Ammeter A
1. An electrical circuit is set up as shown in
Rheostat PQ
Figure 3.28.
Constantan wire
2. A constantan wire is connected across V
terminals P and Q. The length of the wire Voltmeter
across P and Q is adjusted to be 30 cm long.
Figure 3.28
3. The switch is connected and the rheostat is
adjusted so that the ammeter gives a reading
of 0.5 A. The reading of voltmeter is taken
and recorded in Table 3.6.

4. The constantan wire is removed. Steps 2 and
3 are repeated for copper wire and tungsten
wire.

Results: Table 3.6

Wire Reading of ammeter, Reading of voltmeter, Resistance,
—V
I/A V/V R = I / Ω

Constantan 0.5 6.3 12.6

Copper 0.5 0.2 0.4

Tungsten 0.5 0.6 1.2

Discussions:
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

Fo5rm 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
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 Figure 3.29 Heating element

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
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 Figure 3.30 Copper wires in
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.

(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

Fo4rm 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.

Water waves Gap
• Diffraction of water waves when moving through Region of
calm water
a gap produces a region of calm water suitable for
berthing of ships and water recreational activities.

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 Barrier Gap

1. Figure 5.48 shows water waves approaching a gap in a Ripple
ripple tank. tank
(a) Draw the wave pattern formed to show the wave
phenomenon that occurs.  C3
(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 BA

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

Fo5rm Physics   Chapter 2 Pressure

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

Flask Gas

20 cm

12 cm

SPM Figure 1 Figure 3
What is the physics principle or law that is applied What is the value of the atmospheric pressure?

to move the arm of the excavator? C3 C3
A Pascal’s principle A 71 cm Hg
B Archimedes’ principle B 75 cm Hg
C Bernoulli’s principle C 76 cm Hg
D Newton’s Third Law of Motion D 91 cm Hg
Clo2n.e Figure 2 shows the supply of water from a tank.
The reading of a pressure meter at X is 2.8 × 105 4. Figure 4 shows a hydraulic system to lift a car.
Pa and the density of water is 1 000 kg m–3.
[g = 9.81 m s–2]

Tank Area, Pressure, P2 Area,
A1 Pressure, P1 A2
h
Figure 4
X The multiplying factor of the hydraulic system is
Figure 2
C3
What is the value of h in the situation above?
C3 A —AA–12–
B —AA–21–
A 2.85 m
B 28.0 m C —PP–21–
C 28.5 m D —PP–21–
D 285 m

298

Chapter 2 Pressure  Physics Fo5rm

SPM Clo5n.e 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
Figure 5 A the surface area of the water increases
Which relationship is correct? C4 B the volume of air in the container increases
[F = buoyant force; W = weight of the boy; C the tap is nearer to the floor
D the tap is at a deeper level below the water
B = weight of water displaced] surface
A F = W and W = B
B F . W and W , B 2 8. Figure 8 shows two syringes used to study CHAP
C F , W and W = B Pascal’s principle. A 5 N force applied at the
D F = W and W , B small syringe produces a 20 N force at the large
6. Figure 6 shows a Venturi tube. piston. The cross-sectional area of the large
syringe is 7.2 × 10–4 m2.
Air
STU Small Large
syringe syringe

Figure 8
What is the cross-sectional area of the small

syringe? C4

Figure 6 A 9.0 × 10–5 m2 C 3.6 × 10–4 m2
B 1.8 × 10–4 m2 D 2.88 × 10–3 m2
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
hh[hUTS===hhheeeiiiggghhhtttooofffwwwaaattteeerrrcccooolluluummmnnniiinnntttuuubbbeeeTUS;;] on a row of glass tubes. When the pupil blows
C4 in between the two cans, both the cans move
towards each other.

A hS . hT . hU
B hT . hU . hS
C hT . hS . hU
D hU . hT . hS

7. Figure 7 shows water flowing out of a container. Glass tube

Glass tube

Figure 9 C4
This situation can be explained by
Figure 7
A Newton’s second law of motion
B Archimedes’ principle
C Bernoulli’s principle
D Pascal’s principle

299

Chapter 3 Gravitation  Physics Fo4rm

The shortest distance between the probe and A C

the planet is r. The speed v of the probe at theGM
nearest position is 1.5 r , where G is the B D

universal gravitational constant. Which

diagram shows the correct path of the probe? 

C3

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

rmoon = 60.1 R

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 × 1024 kg and its radius is 6.37 × 106 m. Calculate
(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 Quantum Physics

7

7.1 Quantum Theory Of Important Learning Standard Page
Light • Explain the initiation of the quantum theory. 472
• Describe quantization of energy. 476
7.2 Photoelectric Effect • Explain wave-particle duality. 477
• Explain concept of photon. 478
7.3 Einstein's • Solve problems using photon energy, E = hf and power, P = nhf. 478
Photoelectric Theory 479
• Explain photoelectric effect. 482
• Identify four characteristics of photoelectric effect that cannot be
483
explained using classsical theory.
485
• State minimum energy required for a photoelectron to be emitted 485
from a metal surface using Einstein’s equation, hf = W + —12 mvm 2ax 486
488
• Explain threshold frequency, f0 and work function, W. 489
• Determine work function of metal, W = hf0 = —hλc0–
• Solve problems using Einstein’s equation for photoelectric effect.
• Explain production of photoelectric current in a photocell circuit.
• Describe applications of photoelectric effect in daily life.

• 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
ℎf = W + —21 mvm2 ax in which n is the number of photons emitted per
Wwis htoherreketbdhyor,nefse0h,iosWltdh=rwehasfhv0eo=lled–nhλ—fgrc0tehquoef nmceytoafl
• metal and λ0 second
• de Broglie's wavelength, λ = —Ph

• Photon energy, E = hf = –hλ—c • Momentum of particle, p = AB2BmBKB

whereby, K = —12 mv2

470

Fo5rmQuantum Physics

Quantum theory471 Photoelectric explained by Einstein's
of light effect
photoelectric theory
Explanation of black Four characteristics of photoelectric
body spectrum effect • Einstein's photoelectric equation,
• The higher the light photon hf = W + —12 mv2
Max Planck • Work function, W = hf0 = –hλ—c0
frequency, the higher the
Particle with Wave with particle kinetic energy of the emitted Photoelectric current
wave properties properties photoelectrons. production
• The minimum frequency to emit
de Broglie's Discrete energy a photoelectron is the threshold
hypothesis, λ = —hp packet, E = hf • fTrheequkeinnectyi,cf0energy of
photoelectrons is independent of
the intensity of the light
• The emission of photoelectrons is
instantaneous

Wave-particle • Photon energy, E = –hλ—c Applications
duality • Power, P = nhf • Solar cells
• Light detector of automatic door
Application • Image sensor
Electron • ISS solar panel

microscope

Fo5rm 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
10–11 10–10 10–9 10–8 10–7 10–6 10–5 10–4 10–3 10–2 10–1 1 101 102 103 104

Gamma X-rays Ultraviolet Infrared Microwave Radio waves
rays radiation radiation

1020 1019 1018 1017 1016 1015 1014 1013 1012 1011 1010 109 108 107 106 105 104 Frequency / Hz

Increasing frequency

Visible light

400 500 600 700 750
Wavelength / nm

Figure 7.1

3. Any object with a temperature above absolute The usual meaning of 'black' is absorbs visible light.
zero emits electromagnetic radiation at all But in Physics, a black body is one of that absorbs
wavelengths. all the radiation that falls on it, at all wavelengths.
It is the perfect absorber and radiator. Our Sun is a
4. Objects with low temperature or cold objects back body.
emit waves with low frequencies such as radio
waves or microwaves. 8. The emitted radiaton forms a continuous
spectrum and is unaffected by the nature of the
5. Meanwhile, objects with high temperature or black body surface.
hot objects emit waves with high frequencies
such as visible light and ultraviolet radiation. 9. When the temperature of an object rises, the
object (black body radiator) emits thermal
6. A black body is an idealised body that is able to radiation at all wavelengths.
absorb all electromagnetic radiation that falls
on it. The light rays that

Simulation of black body enter the ear cavity
spectrum
https://bit.ly/35FlCtc will undergo repeated

7. A black body also emit thermal radiation reflections on the inner Ear
depending on its temperature and is known as walls of the ear cavity. cavity
black body radiator. At each reflection,

CHAP Thermal radiation is electromagnetic radiation parts of the rays are

7 consist of visible light and radiation that cannot be absorbed by the inner
seen by the human eye such as infrared radiation.
walls of the ear until all the rays are absorbed. Thus,

the ear cavity acts like a black body.

472 7.1.1

Fo5rm 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
randomly in any direction and produce light. 5. When the temperature of the black body
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
emitted. 6. However, the experimental results of black-
body radiation are inconsistent with classical
3. As stated in classical theory, the electron that physics theory.
vibrates at the same frequency should have the
same energy content. 7. Several physicists that involve in the quantum
theory development are shown as the following.

Classical Theory Isaac Newton
(a) Isaac Newton (1643-1727) https://bit.ly/37FU9KI

• Described light as a single stream of Thomas Young
particles or corpuscles in 1704. https://bit.ly/3ozx6aH

• Unsuccessful in explaining the
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
light is a wave.
• Unable to explain the radiation
spectrum produced by black bodies.

(c) John Dalton (1766-1844) John Dalton
• Matter consists of basic particles https://bit.ly/35BN6zU
that cannot be further divided
called atoms. J.J. Thomson
• Same elements have the same type https://bit.ly/3kpyFW1
of atoms.
• Unable to explain the light
spectrum produced by atoms.

(d) J.J. Thomson (1856-1940)
CHAP • Discovered negatively charged
subatomic particles called electrons
7 in 1897.
• Designed experiment to study the
behaviour of electrons.
• Unable to explain the line spectrum
of light produced by atoms.

474 7.1.1

Chapter 7 Quantum Physics  Physics Fo5rm

Example 1 6. The relationship between the momentum of
Calculate the photon energy for light with particle, p and its wavelength, λ is
wavelengths 450 nm and 700 nm. Compare both
photons. λ = —hp–

Solution which h is Planck’s constant (6.63 × 10–34 J s)

Planck’s constant, h = 6.63 × 10–34 J s
Speed of light in vacuum, c = 3.00 × 108 m s–1
Wavelength, = 450 10–9 m
Wavelength, λ1 = 700 × 10–9 m K = —21 mv2 × —mm– ➞ K = —(m2—mv—)2 = —2pm—2
λ2 ×
which p = mv, p = AB2mBBKB
E = hf = —hλ—c
7. The larger the momentum of particle, the shorter
Photon energy of 450 nm: the wavelength produced. The momentum of
particle is p = mv, then —2pm—2 = K
1 2E1 = 6.63 × 10–34 —435.—000—××—110—0–89–
λ = —mh–v– = —AB—2Bmh—BK—B
= 4.42 × 10–19 J
where m is the mass of particle, v is the velocity
Photon energy of 700 nm: of particle and K is the kinetic energy of particle.
8. Since the value of h is too small, the particle
1 2E2 = 6.63 × 10–34 —730.—000—××—110—0–89– with large mass will have too short of de Broglie
wavelength to be detected. Thus, the wave
= 2.84 × 10–19 J characteristics cannot be observed.
9. In 1927, the presence of wave properties of
The shorter the wavelength of light, the higher electrons was confirmed through the electron
the photon energy. diffraction experiments.
10. Photograph 7.1 shows the diffraction pattern of
Wave-Particle Duality electrons through a thin layer of graphite. The
pattern in Photograph 7.2 resembles the light
Wavelength simulation diffraction pattern through an aperture.
http://bit.ly/39xNYqu
Photograph 7.1 Photograph 7.2
1. Light has wave properties because it shows the
phenomena of diffraction and interference. 7 11. This observation proved de Broglie’s hypothesis. CHAP

2. Object has particle properties because it has
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
characteristics can be shown by light particles
such as electron, proton and neutron.

7.1.3 477

Fo5rm 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
beam, electrons will be emitted from the
Photocell metal surface. These electrons are called
photoelectrons.
Source of e- e- _ Light sensitive (b) The emitted photoelectrons are attracted to
light e- e- metal the anode which has positive potential.
Cathode (c) The movement of the photoelectrons
+ e- e- from the cathode to the anode produces
Anode a photocurrent inside the circuit. The
Vacuum milliammeter shows the value of this

current.

1 23 4
0 5
mA

+_ −+

Battery Milliammeter Video of demonstration of
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: P lanck’s constant kit (9 V battery, 1 kΩ potentiometer, LEDs of different colours, milliammeter and
voltmeter)

Instructions: 23
14
1. The Planck's constant kit is connected as shown 05

in Figure 7.6. 1 23 4 V
0 5
mA Voltmeter
2. The knob on the potentiometer is adjusted to
obtain the voltage, V = 0.2 V. The milliammeter −+ −+

reading is recorded in Table 7.2. Milliammeter LED
3. Step 2 is repeated for V = 0.4 V, 0.6 V, 0.8 V, and

3.0 V.

4. A graph of current against voltage is drawn. Potentiometer
V –+
Then, the activation voltage, a of the red LED
Battery
is determined from the intercept value on the
Figure 7.6
voltage axis.

5. Steps 2 to 4 are repeated using orange, green

and blue LEDs.

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.

7CHAP 1.0 1.5 2.0 2.5 3.0
Voltage, V / V

Figure 7.7

480 7.2.1

Chapter 7 Quantum Physics  Physics Fo5rm

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, —1λ /106 m–1 • Video to determine
colour λ / nm Va / V Planck’s Constant
White 793 1.35 1.261
1.605 http://bit.ly/395cDTW
Red 623 1.78 1.706
Orange 1.764 • Classroom
Green 586 1.90 2.141 fundamentals:
measuring the
Blue 567 2.00 Planck’s constant

467 2.45 http://bit.ly/2XafllD

2. Based on the Table 7.2, the graph of V against —1λ is plotted.
a

Va / V

2.6 Note:
For simplicity, we
2.4 can assume the
energy loses inside
2.2 the p-n junction of
LED to be equal for
all the LEDs.

2.0

1.8

1.6

1.4

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 λ1– / 106 m–1

Figure 7.8

3. From the plotted graph, the gradient of the graph, m and Planck’s constant, h are determined.
Given h = —mc—e ; e = charge of one electron (1.60 × 10–19 C) and c = speed of light in vacuum

(3.00 × 108 m s–1)

m = ——2.—50—–—1.—38—– and h = —(1—.19—×—31—.00–06—)(×1—.16—008—× 1—0—–19)–
(2.20 – 1.26)106

= 1.19 × 10–6 V m = 6.35 × 10–34 J s

Discussion: V V-intercept I V
a
The activation voltage, can be obtained through from the graph of against as shown in

Figure 7.9. The activation voltage, V has a linear relationship with —1λ as shown in Figure 7.10. Gradient of
a

a graph of V against —1λ , m is equal to the value of —he—c . Therefore, the value of Planck’s constant can be CHAP
a
7
determined from the expression —mc—e .

7.2.1 74.82.11

Fo5rm Physics   Chapter 7 Quantum Physics
I
Va

m = _he_c

0 Va V

Figure 7.9 Graph of I against V Figure 7.10 Graph of Va against —λ1–

The Characteristics of the Photoelectric Effect (d) fTrheequthenrecsyhroelqdufirreeqduteonpcryo, df0uicsetphheomtSoioneuliremcceturmic

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
surface by thermionic emission may take
4. Dim light has low energy so the electrons need some time.
a longer time to absorb sufficient amount of
energy to escape from the metal surface. Source
of light
5. However, the results of the photoelectric effect
experiments show that:
(a) The higher the frequency of the photon of
light, the higher the kinetic energy of the
photoelectrons emitted from the metal Metal

surface. Thermio nic emission Metal
(b) The minimum frequency of light needed for
a metal to emit electrons is known as the • The emission of electrons from a
metal surface by photoelectric effect is
tThhreeskhionledtifcreeqnueerngcyy,off0 for that metal. instantaneous.
(c) photoelectrons does
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

7CHAP 4. Why are photoelectrons emitted instantaneously from a metal surface when it is illuminated by a
light of certain frequency? C3

5. Does an increasing of light intensity increase the kinetic energy of the photoelectrons? Why? C4

482 7.2.2

Fo5rm 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 1020 Hz? C3 in hydrogen atom.

A 10–15 J C 10–13 J E4
B 10–14 J D 10–12 J E3

2. The energy carried by one blue light photon of E2
450 nm is C3
E1
A 2.21 × 10–19 J C 6.63 × 10–19 J Figure 1
B 4.42 × 10–19 J D 8.84 × 10–19 J
The possible number of spectral lines produced
3. In the Einstein’s photoelectric equation from the transition of energy levels is C4
—21 mv2 = hf – W, for a metal, the symbol W A 3
represents the C4 B 6
A minimum energy electron escaped from the C 9
metal. D 12
B minimum energy for ionization of one atom
in the metal. 7. The consecutive light photons of wavelength
C minimum photon energy needed to remove 590 nm produced by a laser source were
one photoelectron from the metal surface. separated at a distance of 0.20 m. What is the
D minimum potential energy for one power of the laser source in nW? C4
photoelectron to be liberated from the metal A 0.1
surface. B 0.3
C 0.5
4. Photoelectron cannot be liberated from clean D 0.7
copper surface by visible light because the
C4 8. Photon is the name given to C2
A photon energy for visible light is less than A one unit of energy quanta.
the work function of copper B one quantum of electromagnetic radiation.
B threshold frequency of copper is less than C one electron emitted from a clean metal
the frequencies of visible light surface.
C light intensity of visible light is insufficient to D one electron liberated from a metal surface
remove photoelectron from copper surface by light radiation.
D kinetic energy of the free electron inside
copper is less than photon energy of visible 9. Which of the following experiment phenomena
light provide evidence for the existence of discrete
energy levels in atoms? C2
5. What is the de Broglie wavelength of a particle A Photoelectric effect
with mass m moving at velocity of v ? C1 B Line spectrum from sodium lamps
C Energy distribution of black-body radiation
A —mh—v C —hvm— D Visible light spectrum from tungsten
CHAP B —2π—hm—v D —2mπ—vh– filament lamps

7

490

Chapter 7 Quantum Physics  Physics Fo5rm

Table 12 shows nine electronic components and gadgets that may be connected to complete the
electronic eye system circuit.

Table 12

Capacitor Photodiode Thermistor

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 6 V d.c.
wavelengths were taken from the manufacturer’s –
catalogue. The threshold voltage, V is the potential
difference across the LED when it begins to conduct A V
is measured by adjusting the potentiometer to set the Figure 1.1 CHAP
value of ammeter to zero.
(ii) The values of the wavelengths for the five LEDs are 7
λ = 467 nm, 567 nm, 586 nm, 623 nm and 793 nm as
labelled on them.
(iii) Tabulate your readings in a table for all values of λ, V
and –λ—1 .

Table 1

λ

V
–λ—1

501

Fo5rm Physics   Chapter 7 Quantum Physics

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

(c) Draw a graph of V against –λ—1 .

(d) Based on your graph in (c), state the relationship between V and λ.

(e) Calculate the gradient, k of the graph V against –λ—1 in S.I. unit.

(f ) If k = –h—ec where e = charge of an electron (1.60 × 10–19 C) s–1)
c = speed of light in vacuum (3.00 × 108 m
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.

0 1 23 0 1 23 0 1 23

V VV

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

12 12
03 03

V V

CHAP λ = 623 nm λ = 793 nm V
V= V V=
7
Figure 1.2
502

SPM MODEL PAPER

Paper 1
[40 marks]
Instruction: Answer all questions.

1. Which of the following is a base unit? A The gradient of the graph is —ab SPM MODEL PAPER
A kJ B p increases linearly with q
B µm C If q = 1 then p = —ab + c
C cd D The equation of the graph is p = —ab q + —ab
5. Assume that the length, L, time, T and density,
2. Which of the following are derived quantities? ρ are chosen as the new base quantities of a
I Mass of a satellite physics, what is the relationship of the derived
II Proton charge quantity of mass in terms of ρ, L and T?
III Electrical power A ρL3
A I and II only B LT 3
B I and III only C ρ3L
C II and III only
D 3AB(ρBLBTB)
3. The following statement provides information
about a particular system. 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
the field of science, sports, trade, medicine P
and so on. QR S T

Which is the system mentioned above? Figure 2
A Metric System When ball P is pulled to the left, released and hit
B Imperial System
C Universal Measurement System ball Q with a displacement, the ball P becomes
D International System of Units stationary while the ball T moves forward with
the same displacement. Give a reason for this
4. Figure 1 shows the relationship between the situation.
physical quantities of p and q. A Impulsive force before collision = Impulsive

p force after collision
B Velocity of ball P before collision = Velocity
a
of ball T after collision
(0, c) b C Kinetic energy before collision = Kinetic

0q energy after collision
D Momentum before collision = Momentum
Figure 1
Which of the following statements is incorrect? after collision

503

Physics   SPM Model Paper

Paper 2

Section A

[60 marks]
Instructions: Answer all questions in this section.

1. Coulomb’s law can be expressed in the form of the equation F = —k—rQ 2—q where F is the force, Q and q are
the charges and r is the distance between the two charges.

(a) Based on the equation, state one example of [2 marks]
(i) base quantity,
(ii) vector quantity.

(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.

Velocity / m s–1

SPM MODEL PAPER 20 [2 marks]
[1 mark]
15
[2 marks]
10

5

0 10 20 30 40 50 60 70 80 90 Time / s
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?
(c) Describe the movement of the MRT,

(i) from 10 s to 70 s
(ii) from 70 s to 90 s.

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.

Cargo ship 20° P
20° Tug boats

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

Mark 0.0 cm Crocodile clip Slide wire
A Meter rule
Bulb

SPM MODEL PAPER V
Figure 1.1

(a) The circuit has been set up for you. Carry out the following steps below:
(i) Turn on the switch.
(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

R = —VI–. 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 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0

A A

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

0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 SPM MODEL PAPER

A A

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

0 0.2 0.4 0.6 0.8 1.0

A

When V = 2.6 V,
I= A

Figure 1.2

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

FORM 4 (b) (i) Duration, T = —22—04
= 1.2 s
Chapter 1 Measurement Paper 1

1.1 1. D 2. A 3. A 4. A 5. D (ii) l = —4 —× —(13—0.14––2—)2 × (1.2)2
6. B 7. A 8. C 9. B 10. C = 0.36 m
1. (a) Length 11. B 12. D 13. C 14. B 15. D
(b) Base unit: Meter; 16. D 17. C 18. A 19. B 20. C Section B
Symbol of unit: m 21. A 22. B 23. A 24. B 25. D 4. (a)
Magnitude: 1.55;
Symbol of physical quantity: l Paper 2 S.I. unit Car Speed of car / m s–1
Section A meter A 30.00
2. (a) Scalar quantity has magnitude 1. (a) Base quantity second B 37.49
only and vector quantity has both C 25.20
magnitude and direction. Length kilogram D 33.30
E 24.20
(b) Scalar quantities: Distance, speed, Time
time
Mass
Vector quantities: Acceleration
(b) P = Force × —LTe—inm—gte—h (b) Both of them use walkie talkie
3. Base quantities: Time, mass, length = —ks—g×—ms— × —ms– or smart phone. The time taken
Derived quantities: Volume, area to pass through point O and P is
= kg m2 s–3 recorded and the time interval, t
FORM 4 ANSWERS 1.2 is determined or other reasonable
(c) S.I. unit is not same as metric method.
1. (a) Graph functions as a visual tool unit because the time unit cannot
to represent the relationship be expressed in terms of 10. (c) Malaysia speed limit
between two physical quantities. = 110 km h–1
For example, 1 hour = 60 minutes = 30.56 m s–1
(b) Steps to analyse graph: = 60 × 60 seconds Thus, car B exceeds the speed
(i) State the relationship
between two variables = 3 600 seconds. limit on Malaysia highway.
(ii) Determine the gradient of Metric unit is based upon power (d) Use the ultrasound detector,
graph
(iii) Determine the area under of tens or decimal-based but S.I. time tracker application or other
the graph units of time cannot be expressed reasonable method. Record
(iv) Determine the value of in terms of 10. several readings on speed within
a physical quantity from 5 km and calculate the average
interpolation 2. (a) (i) Base quantity: Mass values.
(v) Make a prediction through (ii) Derived quantity: Force
extrapolation (iii) Vector quantity: Force

2. (a) V increases (b) S.I. unit G 5. (a) In old system: F = mlt–2
= —F—r2—
(b) V= 0 cm3, θ = –250 °C so, FAω m = Ft2l–1 = A—12 ,
θ = 300 °C, V =22 cm3 Mm In new system: l
= —s —×kgs—×m—k—×g—m×—2kg–
V / cm3 = kg–1 m3 s–2 t = ω–1 m = Ft2l–1 = Fω–2A– —21

so,

3. (a) (i) Time taken, T = 26 – 2 (b) There is no suitable measuring

= 24 s device to measure force, area and

(ii) To get the value of one frequency accurately.

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 cm3 20
the average value of two sets – The unit for derived quantity

of readings for 20 complete becomes very complex and

– 250°C 10 oscillations. Use electronic or hinders the communications

θ/ °C digital stopwatch to measure between physicist.
100 200 300
–300 –200 –100 0 the time more accurately. – or other reasonable answers.

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