Softcopy of Fizik-T4standard

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Matapelajaran Tingkatan 1 2 3 4 5 SPM

MO

DU

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ChemistryKimiaFizikBiologiMatematikMatematik TambahanBahasa MalaysiaEnglishSainsSejarahGeografi

Edisi Guru

Edisi Guru

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4TINGKATAN

Dwibahasa

4TINGKATAN

Dwibahasa

l Modul P&P yang berkesan memudahkan dan mempercepatkan pengajaran guru.

l Menggunakan kaedah pengurusan grafik yang memudahkan pelajar memahami dan mengingati konsep-konsep fizik dalam Ingatan Jangka Panjang.

l Modul yang efisien mengandungi nota, latihan dan eksperimen yang ringkas, lengkap dan padat menyerupai format SPM.

l Mengikut Sukatan Pelajaran Tingkatan 4 yang terkini.

TINGKATAN 4

DWIBAH

ASAFIZ

IK

FIZIK • SIZE CLOSE: W8.5’’ x H11‘’

Sebabnya Modul Ini Menitikberatkan Penggunaan Unit S.I. Pada Setiap Langkah Penghitungan Dalam Fizik

• Sebagai ahli Fizik, penggunaan unit amat penting. Unit memberikan makna kepada nombor dalam setiap penghitungan dalam Fizik.

• Nilai nombor menjadi berbeza apabila unit yang berlainan digunakan, (misalnya, 9.2 m dan 9.2 mm mewakili panjang yang berbeza).

• Unit merupakan sebahagian penting dalam 'bahasa Fizik' yang kita sering gunakan. Unit mesti disebut dengan jelas apabila kuantiti fizik dikemukakan. Unit dapat 'menceritakan kisah Fizik'.

• Unit menggambarkan konsep fizik dengan lebih jelas. Unit merupakan blok-blok pembinaan dalam Fizik.

• Penggunaan unit dalam pengiraan membolehkan murid mengelakkan kesilapan secara automatis.

• Dengan menggunakan unit secara teliti dan lengkap dalam pengiraan, seseorang murid akan mencapai kejayaan dalam peperiksaan Fizik dan juga membina batu asas dalam pendidikan Fizik.

Walau bagaimanapun, cikgu-cikgu yang mengajar berhak untuk memilih sama ada untuk menggunakan unit dalam pengiraan setiap langkah atau mengikuti format peperiksaan SPM di mana murid hanya perlu meletakkan unit yang betul pada akhir jawapan sahaja.

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Pengenalan kepada FizikIntroduction to Physics 1

Daya dan GerakanForces and Motion 19

Daya dan Tekanan Forces and Pressure 112

HabaHeat 136

CahayaLight 159

KANDUNGANCONTENTS

00 FIZ Form 4 pre 1LP.indd 1 9/10/13 4:42 PM

MODUL • Fizik TINGKATAN 4

© Nilam Publication Sdn. Bhd.

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UNITPENGENALAN KEPADA FIZIKINTRODUCTION TO PHYSICS1

1.1 KUANTITI ASAS DAN KUANTITI TERBITAN BASE QUANTITIES AND DERIVED QUANTITIES

•MenerangkankuantitiasasdankuantititerbitanExplain what base quantities and derived quantities are

•MenyenaraikankuantitiasasdanunitnyaList base quantities and their units

•MenyenaraikankuantititerbitandanunitnyaList some derived quantities and their units

•MengungkapkankuantititerbitandanunitnyadalambentukkuantitiasasdanunitasasExpress derived quantities as well as their units in terms of base quantities and base units

1.2 IMBUHAN DAN NOTASI SAINTIFIK (BENTUK PIAWAI) PREFIX AND SCIENTIFIC NOTATION (STANDARD FORM)

•MengungkapkankuantitidenganmenggunakanimbuhandannotasisaintifikExpress quantities using prefixes and scientific notation

•MenyelesaikanmasalahyangmelibatkanpertukaranunitSolve problems involving conversion of units

1.3 KUANTITI SKALAR DAN KUANTITI VEKTOR SCALAR AND VECTOR QUANTITIES

•MenyatakantakrifkuantitiskalardankuantitivektorState the definitions of scalar quantity and vector quantity

•MemberikancontohkuantitiskalardankuantitivektorGive examples of scalar quantity and vector quantity

1.4 MEASUREMENTS PENGUKURAN

•Memilihperalatanyangsesuaiuntukmengukurkuantiti fizikChoose appropriate instruments to measure physical quantities

•Menerangkankepersisan,kejituandankepekaanExplain consistency, accuracy and sensitivity

•Menerangkanjenis-jenisralatdalameksperimenExplain types of experimental errors

•MenggunakanteknikyangsesuaiuntukmengurangkanralatdalameksperimenUsing appropriate techniques to reduce experimental errors

01 FIZ Form 4 Ch1 2F.indd 1 9/9/13 5:44 PM



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1 Kuantiti fizik ialah kuantiti yang boleh diukur.A physical quantity is a quantity that can be measured.

2 Kuantiti asas: ialah kuantiti fizik yang tidak boleh ditakrifkan dalam istilah kuantiti asas yang lain.

A base quantity: is a physical quantity which cannot be defined in terms of other base quantities.

3 Terdapat lima kuantiti asas dalam Unit Sistem Antarabangsa (unit S.I.).There are 5 base quantities in the International System of units (S.I. units).

Kuantiti asas Base quantity

Simbol untuk kuantiti asasSymbol for base quantity

Unit S.I. S.I. Unit

Simbol untuk Unit S.I.Symbol for S.I. Unit

Panjang Length

lmetermetre

m

Mass Jisim

m kilogram kg

MasaTime

tsaat

seconds

Arus elektrik Electric current

I ampere A

Suhu Temperature

T Kelvin K

4 Kuantiti terbitan: ialah kuantiti yang diterbitkan daripada kuantiti asas melalui pendaraban atau

pembahagian atau kedua-duanya.

A derived quantity: is a physical quantity which is derived from base quantities through multiplication or

division or both.

Kuantiti fizikPhysical quantity

Kuantiti asas

Base quantity

Kuantiti terbitan

Derived quantity

1.1 KUANTITI ASAS DAN KUANTITI TERBITANBASE QUANTITIES AND DERIVED QUANTITIES




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5 Tentukan unit terbitan untuk kuantiti terbitan yang berikut.Determine the derived unit for the following derived quantities.

Kuantiti terbitan Derived quantity

Hubungan dengan kuantiti asasRelationship with base quantity

Kuantiti terbitandari unit asas

Derived unit frombase units

LuasArea

Luas = panjang × panjang Area = length × length m × m = m2

Isi paduVolume

Isi padu = panjang × panjang × panjang

Volume = length × length × length

m × m × m = m3

KetumpatanDensity

Ketumpatan

= jisim

panjang × panjang × panjang

Density=

masslength × length × length

kg

m3 = kg m-3

HalajuVelocity Halaju =

sesaranmasa

Velocity = displacementtime

ms = m s–1

PecutanAcceleration

Pecutan

= perubahan halajumasa

Acceleration

= change in velocity

time

m s–1

s = m s–2

Berat Weight

Berat = jisim × pecutan gravitiWeight = mass × gravitational acceleration

kg m s-2

Momentum Momentum

Momentum = jisim × halaju Momentum = mass × velocitykg m s-1

TekananPressure

Tekanan = dayaluas

Pressure = forcearea

kg m–1 s–2 ; N m–2 ; pascal (Pa)

DayaForce

Daya = jisim × pecutan Force = mass × acceleration kg m s-2




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Kuantiti terbitan Derived quantity

Hubungan dengan kuantiti asasRelationship with base quantity

Kuantiti terbitandari unit asas

Derived unit frombase units

KerjaWork

Kerja = daya × sesaran Work = force × displacement kg m2 s-2 ; joule (J)

KuasaPower

Kuasa = kerjamasa

Power = worktime

kg m2 s-3 ; watt (W)

Tenaga kinetikKinetic energy

Tenaga kinetik

= 12

× jisim × (halaju)2

Kinetic energy

= 12

× mass × (velocity)2 kg m2 s–2 ; joule (J)

Tenaga keupayaan gravitiGravitational potential energy

Tenaga keupayaan graviti= jisim × pecutan graviti × tinggi

Gravitational potential energy= mass × gravitational

acceleration × heightkg m2 s–2 ; joule (J)

CasCharge

Cas = arus × masa Charge = current × time A s ; coulomb (C)

VoltanVoltage

Voltan = kerjacas

Voltage = work

chargeJ C–1 ; volt (V)

RintanganResistance

Rintangan = voltanarus

Resistance = voltagecurrent V A-1 ; ohm (Ω)




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1 Untuk menukar dari imbuhan kepada nombor biasa, darabkan dengan nilai imbuhan itu.To convert from prefixes to normal number, multiply with the value of the prefixes.

(a) 4 Gm = m Giga (G) = 109

Jadi/Therefore, 4 Gm = 4 × 109 m

= 4 000 000 000 m

(b) 260 mg = g mili / milli (m) = 10-3

Jadi/Therefore, 260 mg = 260 × 10-3 g = 0.26 g

2 Untuk menukar dari nombor biasa kepada imbuhan, bahagikan dengan nilai imbuhan itu.

To convert from normal numbers to prefixes, divide by the values of the prefixes.

(a) 325 s = Ms Mega (M) = 106

Jadi/Therefore, 325 s = 325 ÷ 106 Ms

= 0.000325 Ms

(b) 12 800 000 m = km kilo (k) = 103

Jadi/Therefore, 12 800 000 m = 12 800 000 ÷ 103 km = 12 800 km

3 Untuk menukar dari satu imbuhan kepada satu imbuhan yang lain, tukarkan kepada unit yang asal, kemudian baru tukar kepada imbuhan yang diminta.To change from one prefix to another prefix, change the prefix to the original unit, then only change it to the requested prefix.

(a) 3 060 kg = Tg 3 060 kg = 3 060 × 103 g = 3 060 000 g = 3 060 000 ÷ 1012 Tg

= 0.000 003 06 Tg

(b) 2 430 µm = cm 2 430 µm = 2 430 × 10-6 m = 0.002 43 m = 0.002 43 ÷ 10-2 cm = 0.243 cm

Contoh/Examples

1 Beberapa kuantiti fizikal mempunyai nilai yang sangat kecil atau sangat besar.Some physical quantities have very small or very large values.

2 Untuk mengendalikan nombor tersebut dengan mudah, imbuhan dan notasi saintifik diwujudkan.To handle such numbers more easily, prefixes and scientific notations have been developed.

3 Imbuhan dikaitkan dengan unit S.I. untuk mengungkapkan beberapa nilai yang tertentu. Prefixes are attached to S.I. units to express these values.

4 Jadual imbuhan/Table of prefixes:

1.2 IMBUHAN DAN BENTUK PIAWAIPREFIXES AND STANDARD FORM

Imbuhan Prefixes

SimbolSymbol

NilaiValue

tera T 1012

giga G 109

mega M 106

kilo k 103

desi/deci d 10-1

Imbuhan Prefixes

SimbolSymbol

NilaiValue

senti/centi c 10-2

mili/milli m 10-3

mikro/micro µ 10-6

nano n 10-9

piko/pico p 10-12




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5 Bentuk piawai boleh diungkapkan dalam A × 10n di mana 1 A 10 dan n ialah integer.Standard form can be expressed in the form A × 10n where 1 A 10 and n is an integer.

Tuliskan kuantiti fizik yang berikut dalam bentuk piawai. Write the following physical quantities in standard form.

Kuantiti fizik/Physical quantities Bentuk piawai/Standard form

Ketumpatan aluminium = 2 700 kg m-3

Density of aluminum = 2 700 kg m-3 2.7 × 103 kg m–3

Jarak planet Merkuri dari Matahari = 57 850 000 kmDistance of Mercury planet from the Sun = 57 850 000 km 5.785 × 107 km

Kelajuan cahaya = 380 000 000 m s-1

Speed of light = 380 000 000 m s-1 3.8 × 108 m s–1

Unit jisim atom = 0.000 000 000 000 000 000 000 000 001 66 kgAtomic mass units = 0.000 000 000 000 000 000 000 000 001 66 kg 1.6 × 10–27 kg

Cas satu elektron = 0.000 000 000 000 000 000 16 CCharge of an electron = 0.000 000 000 000 000 000 16 C 1.6 × 10–19 C

6 Kuantiti fizik yang ditulis dalam suatu unit boleh ditulis dalam unit yang lain.A physical quantity that is written in a certain unit can be rewritten in another unit.

1 Beberapa kuantiti fizik digunakan untuk menggambarkan pergerakan sesebuah objek.Some physical quantities are used to describe the motion of objects.

2 Kuantiti-kuantiti ini dapat dibahagikan kepada dua kategori: These quantities can be divided into two categories:

Kuantiti vektor

Vector quantity

Kuantiti skalar

Scalar quantity


1.3 KUANTITI SKALAR DAN KUANTITI VEKTORSCALAR QUANTITIES AND VECTOR QUANTITIES

(a) 1 m2 = 1 × 104 cm2

(b) 5 m3 = 5 × 106 cm3

(c) 1 000 cm3 = 1 × 10–3 m3

(d) Kelajuan sebuah kereta/Speed of a car

= 120 km j–1 = 33.33 m s–1

(e) Ketumpatan ais/Density of ice

= 0.9 g cm–3 = 900 kg m–3

(a) 1 cm2 = m2

1 cm

1 cm

1 cm2 = 1 cm × 1 cm = (1 × 10–2)m × (1 × 10–2)m = 1 × 10–4 m2

(b) 80 km j–1 = m s–1

80 km j–1 = 80 km

1 j

= 80 × 103 m

3 600 s

= 22.22 m s–1

1 j = 3 600 s

Contoh/Examples

Tuliskan kuantiti fizik berikut dengan unit yang diberikan./Write the following physical quantities in the unit given.




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1 Contoh-contoh bagi kuantiti skalar adalah/Examples of scalar quantities are:

Suhu, masa, laju, jarak, jisim/Temperature, time, speed, distance, mass.

2 Contoh-contoh bagi kuantiti vektor adalah/Examples of vector quantities are:

Halaju, pecutan, sesaran, momentum, daya/Velocity, acceleration, displacement, momentum, force.

Pertimbangkan kuantiti-kuantiti yang disenaraikan di bawah. Kategorikan setiap kuantiti sama ada kuantiti vektor atau kuantiti skalar.Consider the following quantities listed below. Categorise each quantity as being either a vector quantity or a scalar quantity.

Kuantiti/Quantity Kategori/Category

5 m Kuantiti skalar (tiada arah disertakan pada jarak) Scalar quantity (there is no direction listed for the distance)

30 cm s–1, Timur/EastKuantiti vektor (terdapat arah disertakan pada kelajuan)Vector quantity (there is direction listed for the speed)

5 km, Utara/NorthKuantiti vektor (terdapat arah disertakan pada jarak)Vector quantity (there is direction listed for the distance)

20 °C Kuantiti skalar (arah tidak dikaitkan)Scalar quantity (there is no direction involved)

256 bytes Kuantiti skalar (arah tidak terlibat)Scalar quantity (there is no direction involved )

4 000 kalori/caloriesKuantiti skalar (arah tidak terlibat)Scalar quantity (there is no direction involved)

Contoh/Examples

3 Dua kategori ini boleh dibezakan antara satu sama yang lain melalui definisi yang jelas: These two categories can be distinguished from one another by their distinct definitions:

i. Kuantiti skalar/Scalar quantity • Seorangbudakperempuanberjalansejauh4meter./A girl walks 4 meters.

• Magnitud: 4 meter Arah: Tiada arah

Magnitude: 4 metres Direction: No direction

ii. Kuantiti vektor/Vector quantity • Seorangbudakperempuanberjalansejauh4meterkeTimur. A girl walks 4 meters East.

• Magnitud: 4 meter Arah: Timur

Magnitude: 4 metres Direction: East

4 Kuantiti skalar merupakan kuantiti fizik yang mempunyai magnitud sahaja.

A scalar quantity is a physical quantity which has magnitude only.

Kuantiti vektor merupakan kuantiti fizik yang mempunyai kedua-dua magnitud dan arah .

A vector quantity is a physical quantity which has both magnitude and direction .




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1 Pengukuran adalah proses untuk menentukan nilai sesuatu kuantiti fizik dengan menggunakan alat saintifik.Measurement is the process of determining the value of a physical quantity using scientific instruments.

2 Meter (m), kilogram (kg) dan saat (s) adalah tiga unit asas untuk mengukur panjang , jisim

dan masa masing-masing.

The metre (m), the kilogram (kg) and .the second (s) are three basic units for measuring length ,

mass and time respectively.

Mengukur panjang / Measuring length

1 Panjang sesuatu objek boleh diukur dengan menggunakan pembaris meter, angkup vernier atau tolok skru mikrometer.The length of an object can be measured by using a metre rule, vernier callipers or micrometer screw gauge.

2 Kesesuaian alat ini bergantung kepada panjang yang diukur dan ketepatan yang diperlukan. The suitability of the instrument depends on the length to be measured and the accuracy required.

1.4 PENGUKURANMEASUREMENT

Alat pengukurMeasuring instrument

Senggatan terkecilSmallest scale division

Pembaris meter / Meter rule 0.1 cm

Angkup vernier / Vernier callipers 0.01 cm

Tolok skru mikrometer / Micrometer screw gauge 0.01 mm

(Senggatan terkecil pada alat pengukur menunjukkan kepekaan sesuatu alat)(The smallest scale division on the measuring instruments shows the sensitivity of the instruments.)

(a) Pengukuran menggunakan pembaris meterMeasurement using a metre rule

Pembaris meter memberikan bacaan dalam ketepatan 0.1 cm.

A metre rule gives readings to an accuracy of 0.1 cm.

Latihan/Exercises

Ukur dan tuliskan setiap yang berikut./Measure and write each of the following.

(i)

(ii)

Berdasarkan rajah, panjang objek = 6.2 cm – 1.0 cm = 5.2 cm

Based on the diagram, the length of the object = 6.2 cm – 1.0 cm = 5.2 cm

Lebar/Width = 1.6 cm

Panjang/Length = 7.0 cm

cm 21 3 4 5 6




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(b) Pengukuran dengan menggunakan angkup vernier/Measurement by using vernier callipers

1 Angkup vernier digunakan untuk mengukur/A pair of Vernier callipers is used to measure: (a) kedalaman/depth (b) dimensi luar /outer dimensions (c) dimensi dalam/ inner dimensions

2 Angkup vernier memberikan bacaan dalam ketepatan 0.01 cm

A pair of Vernier callipers gives readings to an accuracy of 0.01 cm.

0

cm

1 2 3

0 5 10

Rahang dalamInside jaw

Skru pemutarScrew

Skala utamaMain scale

Skala vernierVernier scale

Rahang luarOutside jaw

EkorTail

3 Angkup vernier mempunyai dua skala: skala utama dan skala vernier

A pair of Vernier callipers has two scales: main scale and vernier scale

4 (i) Panjang skala vernier = 0.9 cm/Length of vernier scale = 0.9 cm

(ii) Skala vernier boleh dibahagikan kepada 10 bahagian

The vernier scale is divided into 10 divisions

(iii) Panjang untuk senggatan terkecil oleh skala vernier = 0.09 cm

Length of the smallest scale division of the vernier scale = 0.09 cm

(iv) Panjang untuk senggatan terkecil pada skala utama = 0.1 cm

Length of the smallest scale division of the main scale = 0.1 cm

(v) Perbezaan antara senggatan terkecil oleh skala utama dan senggatan terkecil oleh skala vernier

= 0.1 cm – 0.09 cm = 0.01 cm

The difference between the smallest scale division of the main scale and the smallest scale division of the

vernier scale = 0.1 cm – 0.09 cm = 0.01 cm

(iii)

Berdasarkan gambar rajah, diameter pensel = 5.0 cm – 3.5 cm = 1.5 cm

Based on the diagram, the diameter of the pencil = 5.0 cm – 3.5 cm = 1.5 cm

1 2 3 4 5 6 7 8 9cm

Pensel/PencilSesikuSet square

EkorTailSkala utamaMain scaleSkala vernierVernier scaleSkru pemutarScrewRahang luarOutside jawRahang dalamInside jaw




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Latihan/Exercises

Apakah bacaan yang ditunjukkan oleh angkup tersebut?/What are the readings shown by the callipers?1 2

0 1 2 3

0 5 10

1cm cm

0 5 10

2

(c) Ukuran dengan menggunakan tolok skru mikrometer Measurement using a micrometer screw gauge

SpindalSpindle

KunciLock

AnvilAnvil

RacetRatchet knob

Bidal (skala vernier)Thimble (vernier scale)

Sleeve (skala utama)Sleeve (main scale)

mm

Bidal (skala vernier)Thimble (vernier scale)

KunciLock

RacetRatchet knob

Sleeve (skala utama)Sleeve (main scale)

SpindalSpindle

AnvilAnvil

0 13540

3025

1 Tolok skru mikrometer digunakan untuk mengukur objek yang sangat kecil sehingga 0.01 mm

(Ketepatan dalam 0.01 mm)

A micrometer screw gauge is used to measure very small objects to 0.01 mm. (Accuracy of 0.01 mm)

2 (a) Skru dalam spindal mempunyai jarak benang atau senggatan 0.01 mm.

The screw in the thimble has a thread distance or pitch of 0.01 mm.

(b) Apabila bidal dipusing satu pusingan lengkap, rahang bergerak dalam jarak 0.50 mm.

When the thimble is turned one complete rotation, the sliding jaw moves a distance of 0.50 mm.

3 (a) Skala bidal mempunyai 50 bahagian (senggatan).

The scale on the thimble has 50 divisions.

(b) Satu bahagian di atas skala bidal menunjukkan 0.01 mm .

One division on the thimble scale represents 0.01 mm .

Bacaan/Reading = (1.2 + 0.06) cm

= 1.26 cm

Bacaan/Reading

= (1.6 + 0.08) cm

= 1.68 cm




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Latihan/Exercises

Latihan/Exercises

Tuliskan bacaan tolok skru mikrometer yang ditunjukkan di rajah bawah.Write down the readings of the micrometer screw gauges shown in the diagrams below.

(a) (b) (c)

1.01 mm 1.96 mm 1.21 mm

Lengkapkan jadual di bawah./Complete the table below.

(a) Kuantiti fizik/Physical quantity Alat pengukur/Measuring instrument

Lebar sebuah mejaWidth of the table

Pembaris meterMetre rule

Diameter dalam paip air kuprumThe inner diameter of copper water pipe

Angkup vernierVernier callipers

Ketebalan wayarThickness of a piece of wire

Tolok skru mikrometerMicrometer screw gauge

Jarak seorang atlit melemparThe distance of a javelin throw

Pita pengukurMeasuring tape

Diameter sebatang paipThe diameter of a pipe

Angkup vernierVernier callipers

Ketebalan duit syiling 50 senThickness of a 50 cent coin

Tolok skru mikrometerMicrometer screw gauge

(b) Panjang/Length Alat pengukur/Measuring instrument

4.5 cm Pembaris meter/Metre rule

1.94 cm Angkup vernier/Vernier callipers

6.72 cm Angkup vernier/Vernier callipers

3.55 mm Tolok skru mikrometer/Micrometer screw gauge

0mm

1 5

0

45

0 1 0

5

45

40

0 1 25

20

15

mm mm

Mengukur Jisim/Measuring Mass

Jisim suatu objek boleh diukur dengan menggunakan neraca tuas atau neraca tiga alur .

The mass of an object can be measured by using a lever balance or a beam balance .

Mengukur Masa/Measuring Time

Masa diukur dengan menggunakan jam randik ./Time is measured by using a stopwatch .




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Kejituan Accuracy

1 Kejituan dalam ukuran ialah betapa hampir sesuatu nilai ukuran itu nilai sebenar .

Accuracy in measurement is the degree of closeness of a measurement value to the actual value .

2 Ukuran yang mempunyai peratusan ralat yang sangat kecil mempunyai kejituan yang tinggi .

A measurement with a smaller percentage error has a higher accuracy .

Kepersisan Consistency

1 Kepersisan ialah kemampuan sesuatu alat pengukur untuk mengukur kuantiti dengan sedikit atau tiada sisihan relatif dalam bacaan yang diperoleh.Consistency is the ability of a measuring instrument to measure in a consistent manner with little or no relative deviation in the readings obtained. The repeated measurement will produce almost the same reading.

2 Pengukuran dengan sisihan relatif yang lebih kecil mempunyai kepersisan yang lebih tinggi. A measurement with a smaller relative deviation has a higher consistency.

1.5 KEJITUAN, KEPERSISAN DAN KEPEKAANACCURACY, CONSISTENCY AND SENSITIVITY

Dalam pertandingan menembak, tiga orang peserta A, B dan C masing-masing melepaskan enam tembakan pada sasaran. Bandingkan kejituan dan kepersisan tiga penembak ini.In a shooting competition, three participants A, B and C each takes six shots at a target. Compare the accuracy and the consistency of the three shooters.

A B C

Jitu dan persis Persis tetapi tidak jitu Tidak jitu dan tidak persis

Accurate and consistent Consistent but not accurate Not accurate and not consistent

Latihan/Exercises

Kepekaan Sensitivity

1 Kepekaan ialah kebolehan sesebuah alat untuk mengesan perubahan yang kecil dalam sebarang kuantiti fizik yang diukur.

Sensitivity is the ability of an instrument to detect small changes in the physical quantity being measured.




1

UNIT

13

Semua ketidaktentuan eksperimen (ralat) adalah disebabkan oleh sama ada ralat rawak atau ralat sistematik .All experimental uncertainties (errors) are due to either random errors or systematic errors .

Ralat Rawak Random Errors

1 Ralat rawak ialah ketidakpastian dalam pengukuran yang disebabkan oleh:Random errors are uncertainties in the measurement due to:

(a) pemerhati/the observer (b) keadaan sekeliling/the surroundings (c) alat-alat/the instruments

2 Apabila kuantiti diukur untuk beberapa kali, ralat rawak akan menyebabkan bacaan menjadi lebih besar atau lebih kecil daripada bacaan sebenar.When a quantity is measured a few times, random errors will cause the readings either to be larger or smaller than the actual values.

RALAT RAWAK DAN RALAT SISTEMATIKRANDOM ERRORS AND SYSTEMATIC ERRORS

2 Alat pengukur yang lebih sensitif dapat A more sensitive measuring instrument is able to

(a) mengesan perubahan yang sangat kecil dalam kuantiti fizik yang diukur detect very small changes in the physical quantity that is being measured

(b) bertindak balas dengan cepat kepada perubahan dalam kuantiti fizik yang diukur respond more quickly towards changes in the physical quantity that is being measured

3 Alat pengukur yang mempunyai senggatan lebih kecil adalah lebih peka.

A measuring instrument which has smaller scale divisions is more sensitive.

Senaraikan perbezaan antara dua alat pengukur yang berikut.List the differences between these two measuring instruments.

Pembaris meter/A meter rule Pita pengukur/A measuring tape

21cm0 43 21 3 4 5

cm0

Pembaris meterA meter rule

Pita pengukurA measuring tape

Bilangan senggatan per cmNumber of divisions per cm

10 2

Senggatan skala terkecilSmallest scale division

0.1 cm 0.5 cm

KepekaanSensitivity

Tinggi/High Rendah/Low

Contoh/Examples




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14

3 Contoh bagi ralat rawak:Examples of random errors:

(a) kesilapan pengukuran yang berlaku apabila seseorang membaca bacaan

skala dari kedudukan mata pada alat yang salah ( ralat paralaks )

measurement error that happens when one reads a scale’s reading from a wrong

position of the eye or instrument ( parallax error )

Kedudukan mata yang betul sepatutnya berada berserenjang dengan skala.

The correct position of the eye should be perpendicular to the scale .

(i)

(ii) (iii)

(b) kesilapan yang dibuat apabila membaca skala suatu alatthe error made when reading the scale of an instrument

(c) kiraan bilangan ayunan yang salah dalam sistem yang bergetara wrong count of the number of oscillations in a vibrating system

(d) tekanan yang tidak konsisten semasa merapatkan rahang tolok skru mikrometer (Contoh: apabila mengukur diameter dawai)inconsistent pressures applied when closing the gap of a micrometer screw gauge (Example: when measuring the diameter of a wire)

(e) perubahan dalam persekitaran semasa eksperimen (Contoh: perubahan suhu yang mana ia telah dianggap malar)changes in the surroundings during an experiment (Example: the change of temperature which has been assumed to be constant)

4 Ralat rawak boleh dielakkan (atau dikurangkan) dengan mengulangi eksperimen

(mengambil beberapa bacaan) dan mengira nilai purata .

Random errors can be eliminated (or reduced) by repeating measurements (taking

several readings) and calculating the average value .

Ralat paralaksParallax errorPenjelasan/Explanation:(pandangan pemerhati tidak berserenjang dengan skala instrumen yang sedang dibaca)(the view of the observer is not perpendicular to the scale of the instrument that is being read)

Cara yang betul untuk membaca skala pembaris ditunjukkan dalam rajah. Mata mesti diletakkan tegak di atas tanda pada skala untuk mengelakkan ralat paralaks.The correct way to read the scale of a ruler is shown in the diagram. The eye must be placed vertically above the mark on the scale to avoid parallax error.

X

0

Y Z

cm

Ralat paralaks boleh menghasilkan bacaan yang lebih besar atau lebih kecil daripada nilai sebenar.Apabila kedudukan mata berada padaParallax errors can produce readings that are bigger or smaller than the actual value.When the position of the eye is at X – bacaan akan menjadi lebih besar daripada nilai sebenar. the reading will be greater

than the actual value.Z – bacaan akan menjadi lebih kecil daripada nilai sebenar. the reading will be smaller than the actual value.Y – tiada apa-apa kesilapan

paralaks. there will not be any

parallax errors.

Seorang pemerhati yang cekap akan mendapat bacaan tanpa atau dengan ralat paralaks yang kecil.An efficient observer will get readings without or with small parallax errors.

Ketiadaan ralat paralaks semasa menggunakan ammeter/voltmeter – menggunakan cermin di bawah penunjuk akan memastikan bahawa imej jarum tidak boleh dilihat semasa ukuran.No paralax when using an ammeter/voltmeter – using a mirror under the pointer will ensure that the image of the pointer cannot be seen during measurement.

1cm

MataEye

MataEye

MerkuriMercuryAir

Water

2




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15

Ralat Sistematik Systematic Errors

1 Ralat sistematik ialah ketidakpastian dalam pengukuran disebabkan oleh:Systematic errors are uncertainties in the measurements due to:

(a) pemerhatithe observer

Contoh/Example: (i) masa tindak balas (menggunakan jam randik, dll.)

reaction time (using stopwatch, etc.)

(ii) rabun jauh atau rabun dekat (eksperimen cahaya, dll) short-sightedness or long-sightedness (light experiments, etc)

(b) persekitaranthe surroundings

Contoh/Example:Andaian nilai pecutan disebabkan oleh graviti, 10 m s–2, adalah tidak tepat jika nilai g di tempat di mana eksperimen dijalankan berbeza dari 10 m s–2.Assuming the value of the acceleration due to gravity, 10 m s–2, is inaccurate if the value of g at the place where the experiment is carried out differs from 10 m s–2.

(c) alat-alatthe instruments

(i) ralat sifar zero error

Contoh/Example: • pembarismeteryangtelahrosakatauhausdihujungnya

a metre rule which has worn ends

1cm 2 3

• bacaanangkupvernieratautolokskrumikrometer bukan sifar walaupun rahang telah ditutup.

the reading of vernier callipers or micrometer screw gauge is not zero even when the jaws are closed.

• ammeterdanvoltmeteryangtidakmenunjukkanbacaansifarwalaupunterputusdarilitar.an ammeter and a voltmeter which do not show zero reading even when disconnected from a circuit.

• penimbang yang jarumnya tidak menunjukkan bacaan sifar walaupun tiada objek yangdiletakkan di atas penimbang itu.a balance which does not show zero reading even when no object is being placed on it.

(ii) kesalahan dalam alat fault in the instrument

Alat-alat ditentukurkan dalam kilang dalam keadaan suhu dan tekanan atmosfera tertentu. Kadangkala alat-alat ini digunakan dalam keadaan fizikal yang berbeza.Instruments are calibrated in the factory under specific temperature and atmospheric pressure. Sometimes these instruments are used under different physical conditions.

Contoh/Example:panjang angkup vernier keluli akan berubah apabila suhu berubah.the length of a steel vernier callipers will change when temperature changes.




1

UNIT

16

2 Ralat sistematik tidak boleh dielakkan dengan mengambil beberapa bacaan menggunakan alat yang sama, pemerhati yang sama atau alat yang sama.Systematic errors cannot be eliminated by taking several readings using the same instrument, same observer or same instrument.

3 Ralat sistematik boleh dielakkan dengan cara/Systematic errors can be eliminated by

(i) menggunakan alat yang berbeza /using different instruments

(ii) mengambil ukuran dengan teliti / taking measurements carefully

Ralat Sifar untuk Angkup Vernier Zero Error for Vernier Callipers

(a) (c)

Ralat sifar/Zero error = +0.04 cm (i) Ralat sifar/Zero error = 0.04 cm

(b)

Ralat sifar/Zero error = –0.03 cm (ii) Bacaan sebenar = 1.03 cm – 0.04 cm = 0.99 cm

The correct value = 1.03 cm – 0.04 cm = 0.99 cm

Ralat Sifar untuk Tolok Skru Mikrometer Zero Error for Micrometer Screw Gauge

(a) (c)

Ralat sifar/Zero error = +0.01 mm (i) Ralat sifar/Zero error = +0.01 mm

(b)

Ralat sifar/Zero error = –0.04 mm (ii) Bacaan sebenar = 1.46 mm – 0.01 mm = 1.45 mm

The correct value = 1.46 mm – 0.01 mm = 1.45 mm

0 1

0 5 10

cm

0 1

0 5 10

cm

0 1

0 5 10

cm

0 1

0 5 10

cm

0mm 5

045

04540

0mm

5045

0mm

10 04540

mm

PeneranganExplanation :

Ralat sifarZero error = –0.10 cm + 0.07 cm

= –0.03 cm




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17

1.5 KAJIAN SAINTIFIKSCIENTIFIC INVESTIGATIONS

Proses kajian saintifik adalah seperti ditunjukkan di bawah:The process of scientific investigation is shown as below:

InferensInference

Panjang bandul ringkas mempengaruhi tempoh bandul ringkas

The length of a simple pendulum affects the period of the simple pendulum

HipotesisHypothesis

Jika panjang bandul ringkas bertambah, maka tempohnya juga akan bertambah

If the length of the simple pendulum increases, its period will also increase

Tujuan eksperimenAim of the experiment

Untuk menyiasat hubungan antara tempoh bandul ringkas dengan panjangnya

To investigate the relationship between the period of a simple pendulum and its length

Pemboleh ubahVariables

Pemboleh ubah dimanipulasikan/Manipulated variable:Panjang bandul, l/The length of the pendulum, l

Pemboleh ubah bergerak balas/Responding variable:Tempoh bagi bandul ringkas, T/the period of the simple pendulum, T

Pemboleh ubah dimalarkan/Constant (fixed) variable:Jisim bandul/Mass of pendulum bob

Bahan dan radasMaterials and apparatus

Bandul, benang berpanjangan 70 cm, kaki retort dan pengapit, dua keping syiling kecil, jam

randik, pembaris meter

Pendulum bob, 70-cm length of thread, retort stand and retort clamps, two small coins, stopwatch, metre rule

ProsedurProcedure

Senarai bahan dan radasList of materials and apparatus

Kesimpulan Conclusion

PerbincanganDiscussion

AnalisisAnalysis

Pernyataan masalahProblem statement

PemerhatianObservation

InferensInference

HipotesisHypothesis

Pemboleh ubah/Variables• Pembolehubahdimanipulasikan Manipulated variable• Pembolehubahbergerakbalas Responding variable• Pembolehubahdimalarkan Constant variable(s)

EksperimenExperiment

Bandul ringkas / Simple pendulum




1

UNIT

18

Susunan radas dan prosedurArrangement of the apparatus and procedure

(a) Semua radas disusun seperti yang ditunjukkan dalam rajah itu. All the apparatus are set up as shown in the diagram.(b) Panjang benang dilaraskan supaya panjang, l = 10.0 cm. The thread is adjusted so that its length, l = 10.0 cm.(c) Bandul diayun untuk memulakan ayunan melalui sudut

yang kecil (tidak lebih daripada 10o). A gentle push is given to the pendulum bob to start swinging

through a small angle (not more than 10o).(d) Masa yang diambil untuk 10 ayunan lengkap, t, diambil

menggunakan jam randik. The time taken for 10 complete oscillations, t, is taken by

using the stopwatch.(e) Tempoh bandul ringkas, T (iaitu masa yang diambil untuk satu ayunan lengkap) dikira dengan menggunakan: The period of the simple pendulum, T (i.e. the time taken for one complete oscillation) is

calculated by using:

Tempoh/Period, T = Masa untuk 10 ayunan lengkap10

/ Time for 10 complete oscillations

10

(f) Langkah (c) hingga (e) diulangi untuk l = 20.0 cm, 30.0 cm, 40.0 cm, 50.0 cm dan 60.0 cm. Steps (c) to (e) are repeated for l = 20.0 cm, 30.0 cm, 40.0 cm, 50.0 cm and 60.0 cm.(g) Graf T melawan l diplotkan.

A graph of T against l is plotted.

Pengumpulan dataData collection

Panjang, l (cm)Length,l (cm)

Masa yang diambil untuk 10 ayunan lengkap, t (s)The time taken for 10 complete oscillations, t (s)

Tempoh bandul, T (s)

The period of the pendulum, T (s) t1 t2 tpurata/average

10.0

20.0

30.0

40.0

50.0

60.0

Graf (menggunakan kertas graf)Graph (using graph paper)

T (s)

l (cm)

KesimpulanConclusion

Bandul yang lebih panjang mengambil masa yang lebih panjang untuk membuat satu

ayunan lengkap (atau apabila panjang bandul ringkas bertambah, tempohnya bertambah).

The longer pendulum takes longer time to make a complete oscillation (or the period of a simple

pendulum increases with its length).

BenangThread

Bandul Pendulum bob

Kaki retortRetort stand

0



© Nilam Publication Sdn. Bhd.19

2

UNIT

UNITDAYA DAN GERAKANFORCES AND MOTION

2

2.1 MENGANALISIS GERAKAN LINEAR/ANALYSING LINEAR MOTION

•Menyatakanmaksudjarakdansesaran/Define distance and displacement

•Menyatakanmaksudlajudanhalajudanmenyatakanv= st/Define speed and velocity and state that v = s

t•Menyatakanmaksudpecutandannyahpecutandanmenyatakana = v – u

tDefine acceleration and deceleration and state that a = v – u

t•MenghitunglajudanhalajuCalculate speed and velocity

•MenghitungpecutandannyahpecutanCalculate acceleration and deceleration

•MenyelesaikanmasalahgerakanlineardenganpecutanseragamdenganSolve problems on linear motion with uniform acceleration using

(i) v = u + at

(ii) s = ut + 12

at2

(iii) v2 = u2 + 2as

2.2 MENGANALISIS GRAF GERAKAN/ANALYSING MOTION GRAphS

•Melakardanmentafsirkangrafsesaran-masadanhalaju-masaPlot and interpret displacement-time and velocity-time graphs

•Membuatkesimpulandaripadabentukgrafsesaran-masaapabila jasaddalamkeadaan:Deduce from the shape of a displacement-time graph when a body is:

(i) rehat/at rest(ii) bergerakdenganhalajuseragam/moving with uniform velocity(iii) bergerakdenganhalajutidakseragam/moving with non-uniform velocity

•Menentukanjarak,sesarandanhalajudaripadagrafsesaran-masaDetermine distance, displacement and velocity from a displacement-time graph

•Membuatkesimpulandaripadabentukgrafhalaju-masaapabila jasaddalamkeadaan:Deduce from the shape of a velocity-time graph when a body is:

(i) rehat/at rest

(ii) bergerakdenganhalajuseragam/moving with uniform velocity

(iii) bergerakdenganpecutanseragam/moving with uniform acceleration

•Menentukanjarak,sesaran,halajudanpecutandaripadagrafhalaju-masaDetermine distance, displacement, velocity and acceleration from a velocity-time graph

•MenyelesaikanmasalahgerakanlineardenganpecutanseragamSolve problems on linear motion with uniform acceleration

2.3 MEMAhAMI INERSIA/UNdERSTANdING INERTIA

•MenerangkanapaituinersiaExplain what inertia is

•MenghubungkaitkanjisimdenganinersiaRelate mass to inertia

•MembericontohsituasiyangmelibatkaninersiaGive examples of situations involving inertia

•Mencadangkancarauntukmengurangkankesannegatif inersiaSuggest ways to reduce the negative effects of inertia

02 FIZ Form 4 Ch2 A 2F.indd 19 9/10/13 4:26 PM


© Nilam Publication Sdn. Bhd. 20

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2.4 MENGANALISIS MoMENtuM/ANALYSING MOMENTUM

•MenyatakanmaksudmomentumsuatuobjekDefine the momentum of an object

•Menyatakanmaksudmomentumsebagaihasildarabjisim(m)danhalaju(v), iaitumomentum= mvDefine momentum as the product of mass (m) and velocity (v), i.e. momentum = mv

•MenyatakanPrinsipKeabadianMomentumState the Principle of Conservation of Momentum

•MenghuraikanaplikasiPrinsipKeabadianMomentumDescribe applications of Conservation of Momentum

•MenyelesaikanmasalahyangmelibatkanmomentumSolve problems involving momentum

2.5 MEMAhAMI KESAN DAyA/UNdERSTANdING ThE EffEcT Of fORcE

•MenghuraikankesandayaseimbangyangbertindakkeatasobjekDescribe the effects of balanced forces acting on the object

•MenghuraikankesandayayangtidakseimbangyangbertindakkeatasobjekDescribe the effects of unbalanced forces acting on an object

•MenyelesaikanmasalahdenganmenggunakanF = maSolve problems using F = ma

2.6 MENGANALISIS IMpuLS DAN DAyA IMpuLS/ANALYSING IMpULSE ANd IMpULSIvE fORcE

•MenerangkanapaituImpulsExplain what an impulsive force is

•MembericontohsituasiyangmelibatkandayaimpulsGive examples of situations involving impulsive forces

•Menyatakanmaksudimpulssebagaiperubahanmomentum,iaituFt = mv – muDefine impulse as a change of momentum, i.e. Ft = mv – mu

•Menyatakanmaksuddayaimpulssebagaikadarperubahanmomentumdalamperlanggaranatauletupan, iaitu Define impulsive force as the rate of change of momentum in a collision or explosion, i.e.

F = mv – mut

•MenerangkansituasidimanaataumengurangkanmasaperlanggaranterhadapnilaidayaimpulsExplain the effect of increasing or decreasing the time of impact on the magnitude of the impulsive force

•MenghuraikansituasidimanadayaimpulsperludikurangkandancarauntukmengurangkannyaDescribe situations where an impulsive force needs to be reduced and suggest ways to reduce it

•MenghuraikansituasidimanadayaimpulsmendatangkanmanfaatDescribe situations where an impulsive force is beneficial

•MenyelesaikanmasalahyangmelibatkandayaimpulsSolve problems involving impulsive forces

2.7 CIRI-CIRI KESELAMAtAN yANG DIpERLuKAN DALAM KENDERAANBEING AwARE Of ThE NEEd fOR SAfETY fEATURES IN vEhIcLES

•Menghuraikankepentinganciri-cirikeselamatandalamkenderaanDescribe the importance of safety features in vehicles

2.8 MEMAhAMI GRAvItI/UNdERSTANdING GRAvITY

•Menerangkanpecutanyangdisebabkanolehgraviti/Explain acceleration due to gravity•Menyatakanapaitumedangraviti/State what a gravitational field is•Menyatakanmaksudkekuatanmedangraviti/Define gravitational field strength•Menentukannilaipecutanyangdisebabkanolehgraviti/Determine the value of acceleration due to gravity•MenyatakanmaksudBerat(W)sebagaihasildarabjisim(m)danpecutanyangdisebabkanoleh graviti (g), iaitu W = mg

Define weight (W) as the product of mass (m) and acceleration due to gravity (g), i.e W = mg

•MenyelesaikanmasalahyangmelibatkanpecutangravitiSolve problems involving acceleration due to gravity




2

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2.9 MENGANALISIS KESEIMbANGAN DAyA ANALYSING fORcES IN EqUILIBRIUM

•MenghuraikansituasikeseimbangandayaDescribe situations where forces are in equilibrium

•MenyatakanapaitudayapaduanState what a resultant force is

•TambahkanduadayauntukmenentukandayapaduanAdd two forces to determine the resultant force

•MeleraikandayakepadaduakomponendayaResolve a force into two effective component forces

•MenyelesaikanmasalahyangmelibatkankeseimbangandayaSolve problems involving forces in equilibrium

2.10 MEMAhAMI KERjA, tENAGA, KuASA DAN KECEKApAN UNdERSTANdING wORK, ENGERY, pOwER ANd EffIcIENcY

•MenyatakanmaksudKerja(W)sebagaihasildarabdaya(F)yangdikenakandansesaran(s)objekdalamarahdayatersebut, iaituW = Fs Define work (W) as the product of an applied force (F) and displacement (s) of an object in the direction of the applied force, i.e W = Fs

•MenyatakanapabilaKerjadilakukan,tenagadipindahkandaripadasatuobjekkeobjeklainState that when work is done, energy is transferred from one object to another

•Menyatakanmaksudtenagakinetik,Ek,danmenyatakanbahawaEk=12

mv2

Define kinetic energy, Ek, and state that Ek = 12

mv2

•Menyatakanmaksudtenagakeupayaangraviti,Ep, danmenyatakanbahawaEp = mghDefine gravitational potential energy, Ep, and state that Ep = mgh

•MenyatakanPrinsipKeabadianTenaga/State the Principle of Conservation of Energy

•MenyatakanmaksudKuasa,P,danmenyatakanbahawaP= TenagaMasa

Define power, P, and state that P = EnergyTime

•MenerangkanapaitukecekapanperalatanExplain what efficiency of a device is

•Menyelesaikanmasalahyangmelibatkankerja,tenaga,kuasadankecekapanSolve problems involving work, energy, power and efficiency

2.11 MEMAhAMI KEKENyALAN UNdERSTANdING ELASTIcITY

•MenyatakanmaksudkekenyalanDefine elasticity

•MenyatakanHukumHookeState Hooke’s Law

•Menyatakanmaksudtenagakeupayaankenyal,Ep,danmenyatakanbahawaEp=12

kx2

Define elastic potential energy, Ep, and state that Ep = 12

kx2

•Menentukanfaktor-faktoryangmempengaruhikekenyalanDetermine the factors that affect elasticity

•MenghuraikanaplikasikekenyalanDescribe applications of elasticity

•MenyelesaikanmasalahyangmelibatkankekenyalanSolve problems involving elasticity




2

UNIT

1 Jarak ialah jumlah laluan yang dilalui dari satu lokasi ke satu lokasi yang lain

Distance is the total path length travelled from one location to another

Kuantiti/Quantity: Skalar/Scalar Unit SI/SI unit: meter (m)

2 Sesaran ialah/Displacement is

(a) jarak dalam arah tertentu./the distance in a specific direction .

(b) jarak antara dua lokasi yang diukur sepanjang laluan yang paling pendek yang menghubungkannya dalam arah tertentu.

the distance between two locations measured along the shortest path connecting them in a specific direction. (c) jarak kedudukan akhir dari kedudukan awal dalam arah tertentu.

the distance of its final position from its initial position in a specified direction.

Kuantiti/Quantity: Vektor/Vector Unit SI/SI unit: meter (m)

3 Rajah di sebelah kanan menunjukkan beza antara jarak dan sesaran. Apabila Ah Chong berjalan kaki di sepanjang Jalan Baik dari rumah A ke sekolah B,The diagram on the right shows the difference between distance and displacement. When Ah Chong walked from House A to School B, along Good Road,

Jarak/Distance = Panjang laluan di sepanjang Jalan Baik Length of the road along Good Road

Sesaran/Displacement = Panjang garis lurus AB / Length of the straight line ABRumahHouse

SekolahSchool

Jalan BaikGood Road

A

B

Rahim berjalan dari rumahnya ke simpang sejauh 1.5 km dari rumahnya. Kemudian dia berpatah balik dan berhenti di warung Pak Din yang sejauh 0.5 km dari rumahnya. Rahim walked from his house to the junction which is 1.5 km from his house. Then he turned back and stopped at Pak Din’s stall which is 0.5 km from his house.(a) Berapakah sesaran Rahim dari rumahnya/What is Rahim’s displacement from his house (i) apabila dia sampai di simpang?/when he reached the junction?

1.5 km ke timur/1.5 km to the east

(ii) apabila dia berada di warung Pak Din?/when he was at Pak Din’s stall?

0.5 km ke barat/0.5 km to the west

(b) Selepas bersarapan pagi, Rahim berjalan pulang ke rumahnya. Apabila dia sampai di rumahnya,After breakfast, Rahim walked back to his house. When he reached home,

(i) berapakah jumlah jarak yang dilalui oleh Rahim?/what was the total distance travelled by Rahim?

1.5 km + 1.5 km + 0.5 km + 0.5 km = 4 km

(ii) berapakah jumlah sesaran Rahim dari rumahnya?/what was Rahim’s total displacement from his house?

0 km

Contoh/ExampleRumah RahimRahim’s house

0.5 km 1.5 km

Warung Pak DinPak Din’s stall

B O

Definisi jarak dan sesaranDefine distance and displacement

2.1 MENGANALISIS GERAKAN LINEARANALYSING LINEAR MOTION

Utara/North




2

UNIT

1 Laju ialah kadar perubahan jarak

Speed is the rate of change of distance

Laju, v = Jarak dilalui

Masa yang diambilSpeed, v = Distance travelled

Time taken

Kuantiti/Quantity: Skalar/Scalar Unit SI/SI unit: m s–1

2 Halaju ialah kadar perubahan sesaran

Velocity is the rate of change of displacement

Halaju, v = Sesaran

Masa yang diambilVelocity, v = Displacement

Time taken

Kuantiti/Quantity: Vektor/Vector Unit SI/SI unit: m s–1

3 Arah halaju adalah arah sesaran. Direction of velocity is the direction of the displacement.

Definisi Laju dan HalajuDefine Speed and Velocity

Laju purata,

v = Jumlah jarak dilalui, s

Jumlah masa yang diambil, t

Average speed,

v = Total distance travelled, sTotal time taken, t

Halaju purata:

v = Sesaran, s

Jumlah masa yang diambil, t

Average velocity:

v = Displacement, sTotal time taken, t

Laju sekata Laju yang magnitudnya kekal sama tanpa mempertimbangkan arahnya.

Uniform speed Speed that remains the same in magnitude regardless of its direction.

Halaju sekata Halaju yang magnitud dan arahnya kekal sama.

Uniform velocity Velocity that remains the same in magnitude

and direction .

Suatu objek mempunyai halaju tak sekata jika:An object has a non-uniform velocity if:

(a) arah gerakan berubah atau gerakan tidak linear.

the direction of motion changes or the motion is not linear.

(b) magnitud halaju berubah.

the magnitude of its velocity changes.




2

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Sebuah kapal terbang menuju ke utara selama 1 jam dengan halaju300 km j–1. Kemudian, kapal terbang bergerak ke timur selama 1 jam dengan halaju 400 km j–1. An aeroplane flies towards the north for 1 hour with a velocity of 300 km h–1. Then, the plane moves to the east for 1 hour with a velocity of 400 km h–1. (a) Berapakah laju purata kapal terbang itu? What is the average speed of the plane?

Laju purata = Jumlah jarakJumlah masa

Average speed = Total distance

Total time

Jarak OA = LajuOA × MasaOA Distance OA = SpeedOA × TimeOA

= 300 km j–1 × 1 jam = 300 km h–1 × 1 hour = 300 km = 300 km

Jarak AB = LajuAB × MasaAB Distance AB = SpeedAB × TimeAB

= 400 km j–1 × 1 jam = 400 km h–1 × 1 hour = 400 km = 400 km

∴ Laju purata = (300 km + 400 km)2 jam

∴ Average speed = (300 km + 400 km)

2 hours

= 700 km2 jam

= 350 km j–1 = 700 km2 hours

= 350 km h–1

(b) Berapakah halaju purata kapal terbang itu?/What is the average velocity of the plane?

Dari (a); From (a);

JarakOA = 300 km DistanceOA = 300 kmJarakAB = 400 km DistanceAB = 400 km

∴ SesaranOB = (300 km)2 + (400 km)2 ∴ DisplacementOB = (300 km)2 + (400 km)2 = 500 km = 500 km

∴ Halaju purata = SesaranMasa

∴ Average velocity = Displacement

Time

= 500 km

2 j =

500 km2 h

= 250 km j–1 = 250 km h–1

400 km j–1

300 km j–1

A

O

B

Contoh/Example 1

Bacaan meter had laju bagi sebuah kereta yang bergerak ke arah utara menunjukkan 80 km j–1. Sebuah kereta yang lain bergerak pada 80 km j–1 menuju ke selatan. Adakah kelajuan kedua-dua kereta itu sama? Adakah halaju kedua-dua kereta itu sama? Terangkan jawapan anda.The speedometer reading for a car travelling north shows 80 km h–1. Another car is travelling at 80 km h–1 towards the south. Is the speed of both cars the same? Is the velocity of both cars the same? Explain your answer.

Kelajuan kedua-dua kereta itu adalah sama iaitu 80 km j–1 tetapi halaju adalah tidak sama kerana arah

kedua-dua kereta itu berbeza. / The speed of both cars is the same, that is, 80 km h–1 but the velocity is not the

same because the cars are in different directions.

Contoh/Example 2




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Pecutan sifar bermaksud suatu objek berada dalam keadaan pegun atau bergerak pada halaju malar, a = 0

Zero acceleration means an object is at rest or is moving at a constant velocity, a = 0

‘a positif’ (pecutan): Halaju suatu objek bertambah dari halaju awal, u, kepada halaju akhir, v yang lebih tinggi.

‘positive a’ (acceleration): The velocity of an object increases from an initial velocity, u, to a higher final velocity, v.

‘a negatif’ (nyahpecutan): Halaju suatu objek berkurang dari halaju awal, u, kepada halaju akhir, v yang lebih rendah.

‘negative a’ (deceleration): The velocity of an object decreases from an initial velocity, u, to a lower final velocity, v.

Seorang penunggang basikal bermula dari keadaan rehat dan menambahkan halajunya pada kadar seragam sehingga dia mencapai halaju 4.0 m s–1 dalam 5.0 s. Berapakah purata pecutannya?A cyclist starts from rest and increases his velocity at a constant rate until he reaches a velocity of 4.0 m s–1 in 5.0 s.What is his average acceleration?

Halaju awal/Initial velocity = 0

Halaju akhir/Final velocity = 4.0 m s–1

Masa yang diambil/Time taken = 5.0 s

Contoh/Example

pecutan, a = v – u

t acceleration, a =

v – ut

a = (4.0 – 0) m s–1

5.0 s

= 0.8 m s–2

1 Pecutan, a, didefinisikan sebagai kadar perubahan halaju .

Acceleration, a, is defined as the rate of change of velocity .

2 Formula dan Unit SI:/Formula and SI unit:

Pecutan, a = Perubahan halaju

Masa yang diambil Acceleration, a =

Change in velocityTime taken

= Halaju akhir, v – Halaju awal, u

Masa yang diambil, t = Final velocity, v – Initial velocity, u

Time taken, t

= v – u

t = v – u

t

Unit SI/SI unit: m s–2

Definisi pecutan dan nyahpecutanDefinition of acceleration and deceleration




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Perhatikan gerakan sebuah kereta dalam Rajah (a) dan Rajah (b) yang merupakan gambar stroboskop.Observe the motion of the car in Diagram (a) and Diagram (b) which are stroboscopic pictures.

(a) (b) Rajah (a)/Diagram (a) Rajah (b)/Diagram (b)

(i) Huraikan perubahan halaju sebuah kereta dalam Rajah (a). Adakah kereta memecut atau menyahpecut? Describe the changes in velocity of the car in Diagram (a). Is the car accelerating or decelerating?

Jarak antara dua gambar berturut-turut bertambah. Halaju kereta itu bertambah. Kereta itu memecut.

The distance between two consecutive images increases. Velocity of the car is increasing. The car accelerates.

(ii) Huraikan perubahan halaju sebuah kereta dalam Rajah (b). Adakah kereta memecut atau menyahpecut? Describe the changes in velocity of the car in Diagram (b). Is the car accelerating or decelerating?

Jarak antara dua gambar berturut-turut berkurang. Halaju kereta itu berkurang. Kereta itu mengalami

nyahpecutan.

The distance between two consecutive images decreases. Velocity of the car is decreasing. The car decelerates.

Contoh/Example

1 Lengkapkan jadual berikut untuk membuat perbandingan antara dua istilah yang diberi.Complete the following tables to do comparisons between the terms given.

Jarak/Distance Sesaran/Displacement Laju/Speed Halaju/Velocity

Jarak ialah jumlah panjang laluan yang dilalui dari satu lokasi kesatu lokasi yang lain.Total path length travelled from one location to another.

Sesaran ialah jarak dalam arah tertentu.Displacment is the distance in a specified direction.

Laju ialah kadar perubahan jarak.Speed is the rate of change of distance.

Halaju ialah kadar perubahan sesaran.Velocity is the rate of change of displacement.

Kuantiti asasBase quantity

Kuantiti asasBase quantity

Quantiti terbitanDerived quantity

Quantiti terbitanDerived quantity

Kuantiti skalar Scalar quantity

Kuantiti vektorVector quantity

Kuantiti skalarScalar quantity

Kuantiti vektorVector quantity

2 Seorang budak berjalan sepanjang laluan PQ./A boy walks along path PQ. Cari/Find (a) jumlah jarak dilalui./total distance travelled.

(5 + 7 + 5 + 10 + 10 + 10 + 10)m = 57 m

(b) sesaran./displacement.

(7 + 10 + 10)m = 27 m

Latihan/Exercises

5 m

7 m

10 m

10 m

10 m

PQ




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Hubung kait Sesaran, Halaju, Pecutan dan MasaRelating Displacement, Velocity, Acceleration and Time

Jangka masa detik:/Ticker timer:

• Iadisambungkebekalankuasa arus ulang-alik 50 Hz. Apabila dihidupkan, bilah pengetuk akan

bergetar 50 kali sesaat.

It is connected to an alternating current supply of 50 Hz. When it is switched on, the iron strip will vibrate 50 times per second.

• Masadiambiluntukmembuat50titikpadapitadetik

ialah 1 saat. Jadi, selang masa antara dua titik yang berturutan ialah 150

s = 0.02 s.

The time taken to make 50 dots on the ticker tape is 1 second. Hence, the time interval between 2 consecutive

dots is 150

s = 0.02 s.

• 1detikdidefinisikansebagai selang masa antara 2 titik .

1 tick is defined as the time interval between 2 dots .

6 – 12 V a.c.

Jangka masa detikTicker timerTroli dinamik

Dynamic trolley

Aktiviti 1 Kaedah pengiraanActivity 1 Method of calculation

Menyiasat gerakan dalam makmal iaitu menentukan jarak / sesaran, kelajuan / halaju, masa dan pecutan /nyahpecutanTo investigate motion in laboratory to determine distance / displacement, speed / velocity, time and acceleration /deceleration

10 ticks/detik

8.0 cm BA

3 Isikan tempat kosong/Fill in the blanks:

(a) Laju malar 10 m s–1 Jarak sejauh 10 m dilalui setiap saat

A constant speed of 10 m s–1 A distance of 10 m is travelled every second

(b) Halaju malar –10 m s–1 Sesaran sejauh 10 m yang dilalui setiap saat dalam arah bertentangan

A constant velocity of –10 m s–1 A displacement of 10 m is travelled every second in the opposite direction.

(c) Pecutan malar 4 m s–2 Halaju meningkat dengan 4 m s–1 setiap saat

A constant acceleration of 4 m s–2 Velocity increases by 4 m s–1 every second

(d) Nyahpecutan malar 4 m s–2 Halaju berkurang dengan 4 m s–1 setiap saat

A constant deceleration of 4 m s–2 Velocity decreases by 4 m s–1 every second




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LANGKAH 1: Menentukan masa diambil untuk 1 detik.STEP 1: Determination of time taken for 1 tick.

(a) Masa yang diambil untuk 50 detik = 1 saat

(b) Masa yang diambil untuk 1 detik = 0.02 s

(c) Masa yang diambil dari A ke B = 10 detik

= 0.2 s

(a) Time taken for 50 ticks = 1 second

(b) Time taken for 1 tick = 0.02 s

(c) Time taken from A to B = 10 ticks

= 0.2 s

LANGKAH 2: Menentukan sesaranSTEP 2: Determination of displacement

Sesaran suatu objek ditentukan dengan mengukur panjang pita detik yang ditarik melalui jangka masa detikThe displacement of the object is determined by measuring the length of the ticker tape that is pulled through the ticker time

Sesaran A ke B = 8.0 cm/Displacement from A to B = 8.0 cm

LANGKAH 3: Menentukan halajuSTEP 3: Determination of velocity

Halaju, v = SesaranMasa

= 8.0 cm0.2 s

= 40.0 cm s–1

Velocity, v = DisplacementTime

= 8.0 cm0.2 s

= 40.0 cm s–1

LANGKAH 4: Menentukan pecutanSTEP 4: Determination of acceleration

1.5 cm 3.5 cm 5.5 cm 7.5 cm

A υ νB C D E

Dari jalur pertama: Halaju pertama, u, pada AB/From the first strip: Initial velocity, u, at AB

uAB = 1.5 cm0.2 s

= 7.5 cm s–1

Dari jalur terakhir: Halaju terakhir, v, pada DE/From the final strip: Final velocity, v, at DE

vAB = 7.5 cm0.2 s

= 37.5 cm s–1

Selang masa, bagi perubahan halaju, t/The time interval, t, for the change in the velocity

t = (4 – 1) × 0.2 s = 0.6 s

Pecutan/Acceleration,

a = (37.5 – 7.5) cm s–1

0.6 s = 50.0 cm s–2




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1

Arah pergerakanDirection of motion

0.2 cm 1.4 cm

Berdasarkan rajah di atas, hitungkan pecutan objek itu.Based on the diagram above, calculate the acceleration of the object.

Penyelesaian/Solution

u = 0.2 cm0.02 s

= 10 cm s–1

v = 1.4 cm0.02 s

= 70 cm s–1

t = (5 – 1) × 0.02 s = 0.08 s

a = (70 – 10) cm s–1

0.08 s = 750 cm s–2 = 7.5 m s–2

2

BA C D E2.0 cm 4.0 cm 6.0 cm 8.0 cm

Arah pergerakanDirection of motion

Rajah di atas menunjukkan suatu pita detik yang mengandungi 5 detik untuk setiap selang AB-BC-CD dan DE. Hitungkan pecutan objek itu.The diagram above shows a ticker tape contains 5 ticks for every interval AB-BC-CD and DE. Calculate the acceleration of the object.


u = 8.0 cm0.1 s

= 80.0 cm s–1

v = 2.0 cm0.1 s

= 20.0 cm s–1

t = (4 – 1) × 0.1 s = 0.3 s

a = (20.0 – 80.0)0.3 s

= –200 cm s–2 = –2.0 m s–2

3 Rajah di sebelah menunjukkan carta pita detik bagi sebuah troli yang bergerak. Frekuensi bagi jangka masa detik ialah 50 Hz. Setiap jalur pita merupakan panjang 10 detik. The diagram on the right shows a ticker tape chart for a moving trolley. The frequency of the ticker-timer used is50 Hz. Each strip of the tape is a 10-ticks length.

(a) Berapakah selang masa antara dua titik?What is the time interval between two dots?

0.02 s

(b) Berapakah selang masa untuk satu jalur?What is the time interval for one strip?

0.02 × 10 = 0.2 s

(c) Berapakah halaju awal?What is the initial velocity?

u = 2.0 cm0.2 s

= 10.0 cm s–1

(d) Berapakah halaju akhir?/What is the final velocity?

v = 12.0 cm0.2 s

= 60.0 cm s–1

(e) Berapakah selang masa yang diambil untuk berubah dari halaju awal kepada halaju akhirnya?What is the time interval to change from its initial velocity to its final velocity?

t = (11 – 1) × 0.2 s = 2.0 s

(f) Berapakah pecutan objek itu?/What is the acceleration of the object?

a = (60.0 – 10.0) cm s–1

2.0 s = 25.0 cm s–2

Latihan/Exercises

12.0

Panjang 10 detik/cm10-tick length/cm

0

2.0

4.0

6.0

8.0

10.0

Detik/Ticks

Jalur pitaStrip of the tape




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Aktiviti 2 Untuk mengenal pasti jenis gerakanActivity 2 To identify the types of motion

PITA DETIK DAN CARTATICKER TAPE AND CHARTS

Panjang/Length (cm)

Arah gerakanDirectionof motion

Detik/Ticks

Jenis gerakan: Type of motion:

Laju malar/Constant speed

Jarak dilalui antara dua titik berturutan:Distance between two consecutive dots:

Sama/equal

1 2 3 4 5 6

4.0

0.5

1.2

1.9

2.6

3.3

Panjang/Length (cm)


Detik/Ticks

• Jarakantaraduatitikberturutan bertambah secara seragam.

The distance between two consecutive dots increases

uniformly.

• Halajuobjekitu bertambah secara seragam.

The velocity of the object increases uniformly.

• Objekitubergerakpada pecutan seragam.

The object moves at a uniform acceleration .

Panjang/Length (cm)


Detik/Ticks

• Jarakantaraduatitikberturutan berkurang secara seragam.

The distance between two consecutive dots decreases

uniformly.

• Halajuobjekitu berkurang secara seragam.

The velocity of the object decreases uniformly.

• Objekitubergerakpada nyahpecutan seragam.

The object moves at a uniform deceleration .

Aktiviti 3 Untuk menentukan sesaran, halaju purata dan pecutanActivity 3 To determine displacement, average velocity and acceleration

RadasApparatus

Jangka masa detik, troli, bekalan kuasa 12 V, landasan, pita detik, pembarisTicker timer, trolley, 12 V power supply, runway, ticker tape, ruler

ProsedurProcedure

1 Naikkan satu hujung landasan kepada ketinggian yang munasabah.Raise one end of runway to a reasonable height.

2 Lalukan pita detik melalui jangka masa detik dan sambungkan kepada troli di atas landasan.Pass the ticker tape through the ticker timer and attach it to a trolley at the top of the runway.

(i)

(ii)

(iii)




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ProsedurProcedure

3 Hidupkan jangka masa detik dan lepaskan troli.Switch on the ticker timer and release the trolley.

4 Apabila troli itu sampai ke penghujung landasan, hentikan troli dan potongkan pita itu.When the trolley reaches the end of the runway, stop it and cut the tape.

Pita detikTicker tape

Jangka masa detikTicker timer Troli

Trolley

LandasanRunway

Blok kayuWooden block

Sumber kuasaPower supply

5 Tanda dan potongkan pita kepada jalur-jalur 10 detik dari permulaan titik pertama jelas dilihat.Mark and cut the tape into 10-tick strips from the start of the first clear dot.

6 Lekatkan jalur-jalur 10 detik bersebelahan pada kertas untuk membuat satu carta pita.Paste the 10-tick strips side-by-side on a paper to make a tape chart.

Pengumpulan dataCollecting data

CariFind

Jalur pertama1st strip

Jalur terakhirLast strip

Sesaran bagi 10 detikThe displacement of the 10-tick

Masa yang diambil bagi jalur 10 detikThe time interval for the 10-tick strip

Halaju purata jalur 10 detikAverage velocity of the 10-tick strip

Perubahan halaju antara dua jalur 10 detikChange in velocity between the two 10-tick strips

Masa yang diambil untuk perubahan halajuTime taken for the change in velocity

PecutanAcceleration

KesimpulanConclusion

Apabila halaju suatu objek bertambah, ia mengalami pecutan.When the velocity of an object increases, it experiences acceleration.

Selesaikan masalah pada gerakan linear dengan pecutan seragamSolve problems on linear motion with uniform acceleration

(1) v = u + at (2) s = u + v2 t (3) s = ut + ½ at2 (4) v2 = u2 + 2as

di mana/where:s: sesaran/displacementu: halaju awal/initial velocity

v: halaju akhir/final velocitya: pecutan/acceleration

t: masa/time




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1 Sebuah kereta memecut dari keadaan rehat ke 25 m s–1 dalam 4 s. Cari pecutan kereta itu.A car accelerates from rest to 25 m s–1 in 4 s. Find the acceleration of the car.


u = 0, v = 25 m s–1, t = 4 s, a = ?

v = u + at a = v – ut

= 25 m s–1 – 04 s

= 6.25 m s–2

2 Sebuah kereta memecut dari keadaan rehat pada 3 m s–2

sepanjang suatu jalan lurus. Berapakah sesaran yang dilalui oleh kereta itu selepas 4 s?A car accelerates from rest at 3 m s–2 along a straight road. How far has the car travelled after 4 s?

Penyelesaian/Solutionu = 0, a = 3 m s–2, t = 4 s, sesaran/displacement = ?

s = ut + 12

at2

= 0 + 12

(3 m s–2)(4 s)2 = 24 m

3 Sebuah kereta bergerak dengan halaju 20 m s–1 sepanjang jalan lurus. Pemandu itu menekan brek selama 5 s. Ia

menyebabkan nyahpecutan 3 m s–2, berapakah halaju akhir kereta itu?A car is travelling at 20 m s–1 along a straight road. The driver brakes for 5 s. This causes a deceleration of 3 m s–2. What is the final velocity of the car?

Penyelesaian/Solutionu = 20 m s–1, t = 5 s, a = –3 m s–2, v = ?v = u + at = 20 m s–1 + (–3 m s–2)(5 s) = 5 m s–1

4 Sebuah kereta bergerak dengan halaju malar 40 m s–1. Pemandu ternampak suatu penghalang di hadapannya dan dia segera menekan brek. Dia dapat memberhentikan keretanya dalam masa 8 s. Jarak

antara penghalang itu dari kereta apabila pemandu ternampak penghalang itu ialah 180 m. Berapakah jarak penghalang itu dari kereta selepas ia berhenti?A car was moving at a constant velocity of 40 m s–1. The driver saw an obstacle in front and he immediately stepped on the brake pedal. He managed to stop the car in 8 s. The distance of the obstacle from the car when the driver spotted it was 180 m. How far was the obstacle from the car after it stopped?


u = 40 m s–1, v = 0, t = 8 s, s = ?

s = u + v2 t = (40 m s–1 + 0)(8 s)

2 = 160 m

Jarak penghalang dari kereta itu selepas berhenti / The distance of the obstacle from the car after it stopped = 180 m – 160 m= 20 m

Latihan/Exercises

t = 4 s

0 m s–1

Sesaran / Displacement

a = 3 m s–2

t = 5 s

20 m s–1 v

a = –3 m s–2




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2.2 MENGANALISIS GRAF GERAKANANALYSING MOTION GRAPH

Daripada graf gerakan, anda akan memahami:From motion graphs, you will understand:(i) berapa jauh yang telah dilalui oleh sesuatu objek – (jarak, sesaran) how far an object has travelled – (distance, displacement)(ii) berapa cepat ia bergerak – (laju, halaju) how fast it is moving – (speed, velocity)(iii) perubahan halaju terhadap masa – (pecutan / nyahpecutan) change of velocity with time – (acceleration / deceleration)

Lakar dan tafsir graf sesaran-masaPlot and interpret displacement-time graph

Rajah di sebelah kanan menunjukkan gerakan sebuah kereta pada masa yang berlainan.The diagram on the right shows the motion of a car with respect to time.(a) Lakarkan graf sesaran-masa kereta itu. Plot a displacement-time graph for the car.(b) Hitungkan kecerunan graf. Calculate the gradient of the graph.

Kecerunan/Gradient = (50 – 0) m5 s

= 10 m s–1

(c) Apakah unit bagi kecerunan? What is the unit of this gradient?

m s–1

(d) Apakah kuantiti fizik yang ditunjukkan oleh unit ini?What is the physical quantity shown by this unit?

Halaju/velocity

0 mKedudukan:Position

10 m 20 m 30 m 40 m 50 m

t = 0 s 1 s 2 s 3 s 4 s 5 s

00

10

20

30

40

50

1 2 3

Masa/Time (s)

Kedu

duka

n/Po

sitio

n (m

)

4 5

Kecerunan graf = halajuThe gradient = velocityof the graph

Membuat kesimpulan daripada bentuk graf sesaran-masaDeduction from the shape of a displacement-time graph

Objek pada keadaan rehat

Object at rest

Sesaran/Displacement,s/m

t/s0

Objekberadadalamkeadaan rehat kerana ia

berada pada kedudukan yang sama pada bila-bila masa.

The object is at rest because it is at the same position at any time.

Halaju = kecerunan grafVelocity = gradient of the graph

= 0 m s–1

Graf sesaran-masaDisplacement-time graph




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Objek bergerak dengan halaju

malarObject moving at constant velocity

t/s0

20

5


• Objekbergerakpadahalaju malar kerana sesaran bertambah secara seragam dalam setiap saat.

Object travels at constant velocity because displacement increases constantly in every second.

• Garislurusgrafmempunyaikecerunanyang tetap .

The straight line of the graph has a constant gradient. • Halaju =kecerunan

Velocity = gradient

= 4 m s–1

Objek bergerak dengan pecutan

Object moving with acceleration

t/s0


• Objekbergerakdenganpecutankeranakadar

perubahan sesaran bertambah . Object moves with acceleratiom because the rate of change

of displacement is increasing .

• Kecerunanlengkungan bertambah , dan ini

menunjukkan halaju bertambah.

The gradient of the curve is increasing showing that

the velocity is increasing.

• Objekmengalami pecutan .

The object experiences acceleration .

Menentukan jarak, sesaran dan halaju dari graf sesaran-masaDetermine distance, displacement and velocity from the displacement-time graph

Rajah di bawah menunjukkan graf sesaran-masa bagi suatu objek.The diagram below shows a displacement-time graph of an object.

Berdasarkan graf,Based on the graph,

(a) Hitungkan halaju objek antara calculate the velocity of the object between (i) A dan/and B (ii) B dan/and C (iii) C dan/and D

v = 20 m10 s

v = 0 v = – 20 m5 s

= 2 m s–1 = –4 m s–1

(b) Gambarkan gerakan objek itu antaraDescribe the motion of the object between

(i) A dan B: halaju malar (ii) B dan C: dalam keadaan rehat

A and B: constant velocity B and C: at rest

(iii) C dan D: halaju malar tetapi objek bergerak pada arah yang bertentangan

C and D: constant velocity but in opposite direction

Sesaran/mDisplacement / m

B C

DA0

20

10 30 35Masa/sTime / s




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Lakar dan tafsir graf halaju-masaPlot and interpret velocity-time graph

Rajah menunjukkan kereta bergerak dari rehat dalam suatu garis lurus.The diagram shows the car move from rest in a straight line.(a) Lakarkan graf halaju-masa bagi

kereta ini. Plot the velocity-time graph for the car.

(b) Hitungkan kecerunan graf. Calculate the gradient of the graph.

KecerunanGradient

= (10 – 0) m s–1

(5 – 0) s = 2 m s–2

(c) Apakah unit bagi kecerunan ini? What is the unit of this gradient?

m s–2

(d) Apakah kuantiti fizik yang ditunjukkan oleh unit ini?What is the physical quantity shown by this unit?

pecutan/acceleration

t = 0 s 1 s 2 s 3 s 4 s 5 s

0 m s–1 2 m s–1 4 m s–1 6 m s–1 8 m s–1 10 m s–1Halaju:Velocity

Graf halaju-masaVelocity-time graph• Kecerunangraf=pecutanataunyahpecutan The gradient of graph = acceleration or deceleration• Luasdibawahgraf=sesaran The area under the graph = displacement

00

2

4

6

8

10

1 2 3 4 5

Masa/Time(s)

Halaju/Velocity(m s–1)

Graf halaju-masaVelocity-time graph

(c) Cari/Find (i) jumlah jarak/total distance (ii) jumlah sesaran/total displacement

= (20 + 0 + 20)m = 40 m = (20 + 0 – 20)m = 0

(d) Hitungkan/Calculate (i) laju purata (ii) halaju purata gerakan zarah itu. the average speed the average velocity of the moving particle.

= jumlah jarakmasa

= 40 m35 s

= 1.14 m s–1

= total distance

time

= 40 m35 s

= 1.14 m s–1

= sesaranmasa

= 0

= displacement

time

= 0




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Membuat kesimpulan daripada bentuk graf halaju-masaDeduce from the shape of velocity-time graph

ObjekberadadalamkeadaanrehatObject at rest

Objekbergerakdenganhalaju malar

Object moves with constant velocity

ObjekbergerakdenganpecutanmalarObject moving with constant

acceleration

Objekbergerakdengannyahpecutan malar

Object moves with constant deceleration

Nota/Notes:• Dalamanalisisgerakanlinear,jikahalajusuatuzarahadalahsifar,inibermaknazarahituberhentibergerak. In linear motion analysis, if the velocity of a particle is zero, it means that the particle has stopped moving. • Jikahalajuzarahitumenjadinegatif,makazarahitubergerakdalamarahbertentangandenganarahgerakanawalnya. If the velocity of the particle becomes negative, then the particle is moving opposite to its earlier direction of motion.

• Kecerunan=pecutan= 0

Gradient = acceleration = 0

• Luasdibawahgraf=sesaran= 0

Area under the graph = displacement = 0

• Objekberadadalamkeadaan rehat .

Object is at rest .

• Kecerunan/Gradient = pecutan/acceleration = 0

• Luasdibawahgraf=sesaran= 10 m s–1 × 2 s = 20 m

Area under the graph = displacement = 10 m s–1 × 2 s = 20 m

• Sesarandilalui= 20 m

Displacement travelled = 20 m

• Objekbergerakdenganhalaju malar

Object moves with constant velocity

• Kecerunan/Gradient = malar/constant =

18 m s–1

3 s = 6 m s–2

• Pecutan/Acceleration = 6 m s–2

• Luasdibawahgraf/Area under the graph

= 12

(18 m s–1)(3 s) = 27 m

• Sesarandilalui/Displacement travelled = 27 m

• Objekbergerakdenganpecutan malar

Object moves with constant acceleration

• Kecerunangrafadalah malar dan negatif

The gradient of the graph is constant and negative

• Objekitubergerakdengan nyahpecutan seragam

The object is moving with uniform deceleration

v/m s-1

t/s0

v/m s-1

t/s0 2

10

v/m s-1

t/s0

18

3

v

t0




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Menentukan jarak, sesaran, halaju dan pecutan daripada graf halaju-masaDetermine distance, displacement, velocity and acceleration from a velocity-time graph

Berikut menunjukkan graf halaju-masa bagi sebuah kereta. Berdasarkan graf, The following shows the velocity-time graph of a car. Based on the graph,

Halaju/Velocity / m s-1

K

L M

N

OJ0

20

–10

10

10 20 30 4035 45 50Masa/sTime / s

(a) hitungkan pecutan kereta itu bagi calculate the acceleration of the car for (i) JK

a = 20 m s–1

10 s = 2 m s–2

(ii) KL

a = – (20 – 10) m s–1

(20 – 10) s = –1 m s–2

(iii) LM a = 0

(iv) MN

a = – 20 m s–1

10 s = –2 m s–2

(b) Nyatakan jenis gerakan kereta itu bagi State the types of motion of the car for

(i) JK : pecutan malar (ii) KL : nyahpecutan malar (tetap)

constant acceleration constant deceleration

(iii) LM : halaju tetap (iv) MN : nyahpecutan tetap

constant velocity constant deceleration

(c) Hitungkan jumlah sesaran yang dilalui oleh kereta itu semasa Calculate the total displacement travelled by the car during (i) gerakan bagi 10 s yang pertama (ii) kereta bergerak dengan halaju seragam the first 10 s of motion the car moves with uniform velocity

Sesaran Displacement

= 12

(20 m s–1)(10 s) Sesaran Displacement

= 10 m s–1 × 10 s = 100 m

= 100 m

(iii) gerakan bagi 10 s yang terakhir the last 10 s of motion Sesaran / Displacement = 0

(d) Hitungkan / Calculate (i) jumlah jarak bagi keseluruhan perjalanan (ii) jumlah sesaran bagi keseluruhan perjalanan. the total distance for the whole journey the total displacement for the whole journey. Sesaran / Displacement Sesaran / Displacement

= (100 + 150 + 100 + 25 + 50) m = 425 m – 50 m= 425 m = 375 m

(e) Hitungkan / Calculate (i) laju purata / the average speed (ii) halaju purata / the average velocity

Laju purata / Average speed Halaju purata / Average velocity

= 425 m45 s

= 9.44 m s–1 = 375 m45 s

= 8.33 m s–1




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Menyelesaikan masalah gerakan linear dengan pecutan seragamSolve problems on linear motion with uniform acceleration

P Q

R

S

O0

20

10

–10

2 4 6 8 t/s


1 Hitungkan/Calculate (i) halaju bagi OP, QR dan RS

velocity of OP, QR and RS (ii) sesaran/displacement Penyelesaian/Solution:

(i) OP: halaju/velocity = 20 m2 s

= 10 m s–1

QR: halaju/velocity = – 20 m2 s

= –10 m s–1

RS: halaju/velocity = – 10 m1 s

= –10 m s–1

(ii) s = (20 + 0 – 20 – 10)m = –10 m

2 Hitungkan/Calculate (i) pecutan bagi OA dan BC acceleration of OA and BC (ii) jumlah sesaran/total displacement

Penyelesaian/Solution:

(i) OA: pecutan acceleration

= 10 m s–1

10 s = 1 m s–2

BC: pecutan acceleration

= – 10 m s–1

5 s = –2 m s–2

(ii) Jumlah sesaran Total displacement

= 12

(25 + 10)s (10 m s–1)

= 175 m

Halaju / m s-1Velocity / m s-1

A B

CO0

10

5

5 10 15 20 25t/s

Kesimpulan/Conclusion 1 Kecerunan graf s melawan t memberi halaju suatu objek.

Gradient of the graph s against t gives the velocity of an object. 2 Kecerunan graf v melawan t memberi pecutan suatu objek.

Gradient of the graph v against t gives the acceleration of an object. 3 Luas di bawah graf v melawan t memberi sesaran yang dilalui oleh objek.

Area under the graph v against t gives the displacement travelled by the object.




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Ringkasan bagi bentuk graf gerakanSummary of shapes of motion graphs

GrafGraph

s melawan ts against t

v melawan tv against t

a melawan ta against t

Halaju sifarZero velocity

s

t0

v

t0

Halaju negatifNegative velocity

s

t0

v

t0

Halaju seragamUniform velocity

s

t0

v

t0

a

t0

Pecutan seragamUniform acceleration

s

t0

v

t0

a

t0

Nyahpecutan seragamUniform deceleration

s

t0

v

t0

a

0 t




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2.3 MEMAHAMI INERSIAUNDERSTANDING INERTIA

Jelaskan apakah inersia/Explain what inertia is

Inersia suatu objek ialah kecenderungan objek itu kekal dalam keadaan rehat atau terus bergerak

dalam keadaan gerakannya

The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its motion

• Suatuobjekberadadalamkeadaanrehatakancenderungkekaldalamkeadaan rehat .

An object in a state of rest tends to remain at rest .

• Suatuobjekyangberadadalamkeadaanbergerakcenderunguntukkekaldalamkeadaan gerakan .

An object in a state of motion tends to stay in motion .

Hukum Newton pertama/Newton’s first law:Setiapobjekakanterusberadadalamkeadaanrehatataukeadaangerakannyadenganhalajuseragam

kecuali ia dikenakan daya luar.

Every object continues in its state of rest or of uniform motion unless it is acted upon by an external force.

Semakinbesarjisim,semakinbesar inersia /The larger the mass, the larger the inertia

• Duabaldikosongdigantungdengantalidarisiling. Two empty buckets are hung with rope from the ceiling.• Sebuahbaldidiisidenganpasirmanakalabaldiyanglainadalahkosong. One bucket is filled with sand while the other bucket is empty.• Kemudian,kedua-duabaldiditolak. Then, both buckets are pushed.

• Didapatibaldikosong itu senang ditolak berbanding dengan baldi yang diisi dengan pasir.

It is found that the empty bucket is easy to push compared to the bucket with sand.

• Baldiyangdiisidenganpasiradalahlebih susah untuk bergerak.

The bucket filled with sand is more difficult to move.• Apabila kedua-duabaldi diayundandiberhentikan, baldi yangdiisi denganpasir lebih susahuntuk

diberhentikan. When both buckets are oscillating and an attempt is made to stop them, it is more difficult to stop the bucket

filled with sand.• Inimenunjukkanbaldidenganjisimyanglebihbesarmenghasilkanrintanganyanglebihuntukberubah

dari keadaan rehat atau dari keadaan gerakan. This shows that the bucket with a bigger mass offers a greater resistance to change from its state of rest or

from its state of motion.

• Olehitu,suatuobjekdenganjisimyangbesarmempunyaiinersiayanglebih besar .

So, an object with a larger mass has a larger inertia.

Hubung kait inersia dengan jisim/Relate mass to inertia

TaliRopes

BaldiBuckets

PasirSand

03 FIZ Form 4 Ch2 B 2F.indd 40 9/10/13 4:30 PM



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Aktiviti yang melibatkan inersia/Activities involving inertia

Apabilasekepingduitsyiling20sendikuiskearahtimbunanduitsyiling20senpadapermukaanyanglicin,duitsyiling di bawah dihentam keluar tanpa menggerakkan duit

syiling yang lain. Ini menunjukkan bahawa inersia bagi timbunan duit syiling di atas bercenderung untuk kekal

dalam keadaan rehat dan menentang gerakan.When a 20 cent coin is flicked towards a stack of 20 cent coins on a

smooth surface, the bottom coin is knocked off without moving

the rest of coins. This shows that the inertia of the stack of coins

above tends to remain at rest and resists motion.

Apabila kadbod ditarik keluar dengan cepat, duit syiling itu

terus jatuh ke dalam gelas. Inersia duit syiling itu

mengekalkannya dalam keadaan rehat walaupun kadbod itu ditarik keluar.When the cardboard is pulled away quickly, the coin drops straight into the glass.

The inertia of the coin maintains it in its rest position even when the cardboard is withdrawn.

LetakkansegelasairdiatassekepingkertasA4.Dengancepattarik keluar kertas itu secara mendatar. Apakah yang akanberlaku kepada gelas air itu?Place a glass of water on a piece of A4 paper. Suddenly you pull the paper horizontally. What happens to the glass of water?

Gelas air itu kekal dalam keadaan rehat. Inersia gelas yang

berisi air itu cenderung mengekalkan gelas air dalam

kedudukan rehat.

The glass of water remains at rest. The inertia of the glass of water

still wants it to remain at rest position.

TroliTrolley

PenghalangObstracle


Sebuahblokkayudiletakkandiatassebuahtroliyangbergerakmenuruni landasan. Apabila gerakan troli itu dihalang olehsuatupenghalang,blokkayuituakankekaldalamkeadaan gerakan dan ia menggelongsor ke hadapan. Inersia blok kayu itu berkecenderung untuk mengekalkan keadaan gerakannya.A wooden block is placed on top of a moving trolley down a runway. When the motion of the trolley is stopped by an obstacle, the wooden

block will continue its state of motion and slide forward.

The inertia of the wooden block tends to keep its state of motion.




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Contoh situasi yang melibatkan inersia/Examples of situations involving inertia

Basyangpegun/The bus is stationary

Basbergeraksecaratiba-tibakedepanThe bus moves forward suddenly

Penumpang di dalam bas akan terhumban ke belakang apabila bas yang pegun memecut ke hadapan. Mengapa?Passengers in a bus will be thrown backwards when a stationary bus starts to accelerate. Why?Apabilabasitubergerakkedepansecaratiba-tibadarirehat, inersia badan penumpang masih kekal dalam keadaan rehat.

Ini menyebabkan badannya terhumban ke belakang .

When the bus moves forward suddenly from rest, the inertia of the passenger's body tends to keep him at rest. This causes his body

to be thrown backwards .

Bassedangbergerak/The bus is moving

Basberhentisecaratiba-tibaThe bus stops suddenly

Penumpang dalam bas yang bergerak terhumban ke hadapan apabila bas itu berhenti secara tiba-tiba. Mengapa?Passengers in a moving bus will be thrown forward when the bus comes to a halt suddenly. Why?

Penumpang berada dalam keadaan gerakan apabila bas itu sedangbergerak.Apabilabas ituberhenti secara tiba-tiba,

inersia badan penumpang cenderung untuk terus bergerak ke

hadapan . Ini menyebabkan badan penumpang terhumban ke

hadapan .

The passengers are in a state of motion when the bus is moving. When the bus stops suddenly, the inertia of the passenger tends to

continue in its forward motion. This causes his body to be thrown

forward .

Aktiviti yang melibatkan inersia/Activities involving inertia

Sebuah buku ditarik keluar dari kedudukan tengahnya. Buku

di atasnya akan jatuh ke bawah secara terus . Inersia

cuba menentang perubahannya dari keadaan rehat,iaitu,apabilabukuditarikkeluar,buku-bukudiatastidakakanbergerakbersama-sama.A book is pulled out from its central position. The books on top will

drop straight downwards . Inertia tries to resist the change from rest, that is, when the book is pulled out, the books on top do not follow suit.




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Gerakan ke bawah yang cepatFast downward motion

SosSauce

Soscilidalambotolbolehdituangkeluardengansenang jikabotoldigerakkan turundengancepatdanberhentisecaratiba-tiba.Jelaskan.Chili sauce in the bottle can be easily poured out if the bottle is moved down fast with a sudden stop. Explain.

• Sos dalambotol bergerak bersama-samadengan botol semasa pergerakan kebawah.

The sauce in the bottle moves with the bottle during the downward movement.

• Apabilabotolituberhentisecaratiba-tiba, inersia sos menyebabkan ia terus bergerak ke bawah dan mengakibatkan sos dituang keluar dari botol itu.

When the bottle is stopped suddenly, the inertia of the sauce causes it to continue in its downward movement and thus the sauce is poured out of the bottle.

Gerakan ke bawah yang cepatFast downward motion

Kepalatukuldicantumdenganketatkepadapemegangnyadengan mengetuk penghujungpemegangnya,secaramenegak,diataspermukaanyangkeras.

The head of hammer is secured tightly to its handle by knocking one end of the handle, held vertically, on a hard surface.

• Inimenyebabkan kepala tukulmeneruskan gerakan ke bawah apabila gerakan pemegangitudiberhentikan.Denganini,hujungataspemegangituakandimasukkanlebih dalam ke dalam kepala tukul.

This causes the hammer head to continue on its downward motion when the motion of the handle is stopped. So that the top end of the handle is slotted deeper into the hammer head.

Titisan air pada payung basah akan jatuh apabila budak itu memusingkan payung itu.The water droplets on a wet umbrella will fall when the girl rotates the umbrella.

• Iniadalahdisebabkantitisanairpadapermukaanpayungitu bergerak secara serentak apabila payung itu dipusingkan.

This is because the water droplets on the surface of the umbrella move simultaneously as the umbrella is rotated.

• Apabilapayung ituberhentimemusing, inersia titisan air akan terus mengekalkan pergerakannya.

When the umbrella stops rotating, the inertia of the of water droplets will continue in its original motion.

Seorangbudakmelarikandiridarilembudalamgerakanzig-zag.Mengapa?A boy runs away from a cow in a zig-zag motion. Why?

Lembuitumempunyaijisimyanglebihbesar,makainersianyajugalebih

besar.Jadi,lembuitusukaruntukmenukararahgerakannya.

A cow has a larger mass, so it has a larger inertia. So the cow has difficulty to

change its direction of motion.




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Cadangan untuk mengurangkan kesan negatif inersiaSuggestions to reduce the negative effects of inertia

1 Keselamatandalamkereta:Safety in a car:

(a) Talipinggangkeledarmengekalkanpemandupadatempatduduknya.Apabilakeretaberhentisecara

mendadak,talipinggangitumengelakkanpemandudaripadaterhumbanke hadapan .A safety belt secures a driver to his seat. When the car stops suddenly, the seat belt prevents the driver from

being thrown forward .

(b) Alaskepalamencegahkecederaanlehersemasaperlanggarandaribelakang.Inersiakepalacenderung

untuk mengekalkannya keadaan rehat apabilabadandigerakkansecaratiba-tibakedepan.A headrest prevents injuries to the neck during rear-end collisions. The inertia of the head tends to keep it

in its state of rest when the body is moved forward suddenly.

(c) Begudaradipasangdidalamstereng.Iamembekalkankusyen

untuk mengelakkan pemandu daripada terhentam pada stereng atau papan pesawat kereta semasa perlanggaran.An air bag is fitted inside the steering wheel. It provides a cushion

to prevent the driver from hitting the steering wheel or dashboard during a collision.

2 Perabot yang diangkat oleh lori biasanya perlu diikat dengan tali kepadabahagian-bahagian lori yang tertentu supaya apabilaloribergerakatauberhentidengantiba-tiba,perabotitutidakakanjatuh atau tidak akan terhumban ke depan.

Furniture carried by a lorry normally is tied by ropes to certain fixed parts of the lorry so that when the lorry moves or stops suddenly, the furniture will not fall or will not be thrown forward.

3 Empattangkikecildimanajisimmuatandibahagiantaratangki-

tangki tersebut akan mempunyai inersia yang lebih kecil.

Ini akan mengurangkan impak pada setiap tangki yang disebabkanolehinersiajikaloritangkiituberhentidengantiba-tiba.

Four small tanks with distributed mass will have smaller inertia .

This will greatly reduce the inertial impact on each tank if the tanker stops suddenly.

TaliRope

Treler dengan 4 tangki kecilTrailer with 4 small tanks

Kepala loriTractor Lori tangki

Tanker


Sebuah kapal minyak yang besar mengambil masa yang lebih panjang untuk memecut kepada lajumaksimumnya dan ia mengambil beberapa kilometer untuk berhenti walaupun propelernya telah diterbalikkan. Mengapa?A massive oil tanker (a very big ship) takes a long time to accelerate to its full speed and a few kilometers to come to a stop even though the engine has reversed its propeller to slow it down. Why?

Kapalminyakyangbesarmempunyaijisimyanglebihbesar,jadiinersianyajugalebihbesar.Olehitu,

ia adalah lebih sukar untuk memberhentikan kapal minyak.

The massive oil tanker has larger mass, so it has a larger inertia. So it is more difficult to stop the oil tanker.




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TujuanAim

Untuk mengkaji hubungan antara jisim dan inersia (tempoh ayunan)

To study the relationship between mass and inertia (period of oscillation)

RadasApparatus

BilahHacksaw,pengapit-G,jamrandikdanplastisin.Hacksaw blade, G-clamp, stopwatch and plasticine.

Pemboleh ubahVariables

Pemboleh ubah dimanipulasi/Manipulated variable Jisim plastisin/mass of plasticine

Pemboleh ubah bergerak balas/Responding variable Tempoh ayunan/period of oscillation

Pemboleh ubah dimalarkan/Fixed variable: Panjang bilah Hacksaw/length of the Hacksaw blade

ProsedurProcedure

1 Letakkansejumlahplastisin(berbentuksfera)denganjisim30gpadahujungbilahHacksaw.Place a lump of plasticine (sphere-shaped) with a mass of 30 g at the free end of the Hacksaw blade.

2 Sesarkan sedikit bilah Hacksaw dan lepaskannya supaya ia berayun secaramengufuk.Displace the Hacksaw blade slightly and release it so that it oscillates horizontally.

3 Tentukandanrekodkanmasayangdiambiluntuk10ayunanlengkap,t saat.Determine and record the time taken for 10 complete oscillations, t seconds.

4 Hitungkantempohayunan,T = t10

saat.

Calculate period of oscillation, T = t10

seconds.

5 Ulangilangkah1–4eksperimendenganjisim40g,50g,60gdan70g.Repeat steps 1 – 4 of the experiment with mass of 40 g, 50 g, 60 g and 70 g.

6 Lakarkan graf tempoh ayunan melawan jisim.Plot the graph of period of oscillation against mass.

KeputusanResults

Jisim / gMass / g

Masa untuk 10 ayunan, t/sTime for 10 oscillation, t/s T = t

10 s

t1 t2 tmin

30

40

50

60

70

Pengapit-G/G-clamp

Bilah HacksawHacksaw blade

Plastisin/Plasticine

EksperimenExperiment Inersia dan Jisim / Inertia and Mass




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AnalisisAnalysis

Lakarkan graf Tmelawanjisim,m.Plot the graph T against mass, m.

T/s

m/g


1 Nyatakan kuantiti yang digunakan untuk mewakili inersia dalam aktiviti ini.State the quantity used to represent inertia in this activity.

Tempoh ayunan/Period of oscillation

2 Apakahhubunganantaratempohayunansuatuobjekdenganinersianya?What is the relationship between the period of oscillation of an object and its inertia?

Semakinpanjangtempohayunan,semakinbesarinersia.

The longer the period of oscillation, the larger the inertia.

3 Daripadagraf,nyatakanhubunganantaraFrom the graph, state the relationship between

(a) tempoh ayunan dengan jisim objek.period of oscillation and mass of object.

Semakinbesarjisim,semakinpanjangtempohayunan.

The larger the mass, the longer the period of oscillation.

(b) inersia suatu objek dan jisimnya.inertia of an object and its mass.

Semakinbesarjisimobjek,semakinbesarinersianya.

The larger the mass of the object, the larger its inertia.

0




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Pemain IPlayer I

Pemain IIPlayer II

2.4 MENGANALISIS MOMENTUMANALYSING MOMENTUM

Definisi momentum sebagai hasil darab jisim dan halajuDefine momentum as the product of mass and velocity

Momentum = jisim × halaju Momentum = mass × velocity

UnitSI: kg m s-1 atau N s (Newton saat)

SI unit: kg m s-1 or N s (Newton second)

Momentum adalah suatu kuantiti vektor .Arahmomentummengikutarah halaju .

Momentum is a vector quantity. The direction of the momentum follows the direction of the velocity .

Dalampermainanbolasepak,seorangpemainberjisim70 kg bergerak dengan halaju 4 m s-1 dan seorang pemain yang lain yang berjisim 75 kg bergerakdengan 3 m s-1 menghala antara satu sama seperti yangditunjukkan.Hitungkanmomentumkedua-duapemainitumasing-masing.In a football game, a player of mass 70 kg is moving with velocity of 4 m s–1 and the other player of mass 75 kg ismoving with 3 m s-1 towards each other as shown. Calculate the momentum of the two players respectively.Penyelesaian/SolutionMomentum pemain I/Momentum player I = m1v1=(70kg)(4ms

–1)=280kgms-1

Momentum pemain II/Momentum player II = m2v2=(75kg)(–3ms–1)=–225kgms-1

Nenek (m = 80 kg) menggelongsor sekeliling gelangganggelongsordenganhalaju6ms–1.Tiba-tibadiaberlanggardenganBobby(m=40kg)yangberadadalamkeadaanrehat.HitungkanmomentumnenekdanBobbymasing-masing.Granny (m = 80 kg) whizzes around the ring with a velocity of 6 m s–1. Suddenly she collides with Bobby (m = 40 kg) who is at rest. Calculate the momentum of granny and Bobby respectively.

Penyelesaian/SolutionMomentum nenek/granny = m1v1=(80kg)(6ms

–1)=480kgms–1

MomentumBobby=m2v2=(40kg)×(0ms–1)=0kgms–1 (dalam keadaan rehat / at rest)

Contoh/Example 1

Contoh/Example 2

NenekGranny Bobby

Nyatakan prinsip keabadian momentum/State the principle of conservation of momentum

Tanpakehadirandayaluar,jumlahmomentumdalamsuatusistemkekaltidakberubah.In the absence of an external force, the total momentum of a system remains unchanged.Jumlah momentum sebelum perlanggaran/letupan = Jumlah momentum selepas perlanggaran/letupan Total momentum before collision/explosion = Total momentum after collision/explosion




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Aktiviti/Activity 1

Aktiviti/Activity 2 Aktiviti/Activity 3

Rajah di sebelah menunjukkan dua orang adik-beradik yang sedangmenggelongsor.Abangbergerakdanberlanggardenganadiknyayangberadadalamkeadaanrehat.Apakahgerakanmerekaselepasperlanggaran?The diagram on the right shows two brothers skating. The elder brother moves and collides with his younger brother who is at rest. What is their movement after the collision?

Selepasperlanggaran,/After collison,

Laju abang berkurang ./The speed of the elder brother decreases .

Laju adik bertambah ./The speed of the younger brother increases .

Momentum abang berkurang ./Momentum of the elder brother decreases .

Momentum adik bertambah ./Momentum of the younger brother increases .

Adakahjumlahmomentumsebelumperlanggaransamadenganjumlahmomentumselepasperlanggaran?Is the total momentum before collision equal to the total momentum after collision?

Ya/Yes.

Menjentiksekepingduitsyiling20sen,A secara terus kepadasekepingduitsyiling20sen,B yang lain.Flick a 20-cent coin, A, directly to another 20-cent coin, B.

A B

(a) Apakahyangberlakukepadagerakankedua-duaduit syiling selepas perlanggaran?

What happens to the motion of both coins after collision?

DuitsyilingAberhenti,duitsyilingB bergerak.

Coin A stops, coin B moves.

(b) Apakah yang berlaku kepada momentum duitsyiling A selepas perlanggaran?

What happens to the momentum of coin A after collision?

Momentum duit syiling A dipindahkan kepada

duit syiling B selepas perlanggaran.

Momentum of coin A is transferred to coin B after

collision.

Menjentiksekepingduitsyiling20sen,A secara terus kepadaduitsyiling20senB dan C.Flick a 20-cent coin A, directly to 20-cent coins B and C.

A B C (a) Gambarkan gerakan semua duit syiling selepas

perlanggaran. Describe the motion of all the coins after collision.

DuitsyilingA/Coin A:Berhenti/Stop

DuitsyilingB/Coin B:Rehat/At rest

DuitsyilingC/Coin C: Bergerakkekanan/Moves to the right

(b) Apakah yang berlaku kepada momentum duitsyiling A selepas perlanggaran?What happens to the momentum of coin A after collision?

Momentum duit syiling A dipindahkan ke duit

syiling B dan duit syiling C.

Momentum of coin A is transferred to coin B and to

coin C.




2

UNIT

Aktiviti/Activity 4

Aktiviti/Activity 5

Rajahdisebelahmenunjukkansebijibolakeluli,A ditarik dan dilepaskan. The diagram on the right shows a steel ball, A is pulled and released.(a) Bola itu akan berlanggar dengan empat biji bola yang lain. Ini akan

menyebabkanbolaterakhir,Ebergerakkeketinggianyang sama dengan ketinggian bola A.

The ball will collide with the other four balls. This will cause the last ball, E to

move to the same height as ball A.

Adakahmomentumdiabadikan?Is the momentum conserved?

Ya/Yes

(b) Apakahyangakanberlakujikakedua-duabolaAdanBditarikdankemudiandilepaskan? What will happen if two balls A and B are pulled and then released?

BolaD dan E akan bergerak ke ketinggian yang sama dengan bola A dan Bmasing-masing.BolaC akan

berada dalam keadaan rehat.

Balls D and E will rise to the same heights of balls A and B respectively. Ball C is at rest.

E D C B A

Seorangbudakperempuanberdiridalamkeadaanrehatdiataspapanluncur.Dia membalingkan bola ke hadapan. Bola itu bergerak ke kiri. Budakperempuan bergerak ke kanan.A girl is standing at rest on the skateboard. She throws the massive ball forward. The ball moves to the left. The girl moves to the right.

• Momentumbolasebelumbalingan=0

Momentum of the ball before the throw = 0

• Momentumbudakperempuansebelumbalingan/Momentum of the girl before the throw = 0

• Jumlahmomentumselepasbalingansamadenganjumlahmomentumsebelumbalingan = 0

Total momentum after the throw is equal to total momentum before the throw = 0

• Jumlahmomentumselepasbalingan=momentumbola+momentumbudakperempuan=0

Total momentum after the throw = momentum of the ball + momentum of the girl = 0

Jadi, selepasbalingan,magnitudmomentumbudakperempuanadalah sama dengan magnitud

momentum bola tetapi dalam arah bertentangan .

Therefore, after the throw, the magnitude of the momentum of the girl is equal to the magnitude of the momentum

of the ball but in the opposite direction.




2

UNIT

Perlanggaran kenyal/Elastic collision Perlanggaran tak kenyal/Inelastic collision

m1 m2 m1 m2

u1 u2

Sebelum perlanggaran Before collision

Selepas perlanggaranAfter collision

v1 v2

m1 m2 m1 m2

u1 u2 vm1 + m2



• Kedua-duaobjekbergeraksecara berasingan denganhalajumasing-masingselepasperlanggaran.

Both objects move separately at their respective velocities after the collision.

• Jumlah momentum diabadikan. Total momentum is conserved.

• Jumlah tenaga diabadikan. Total energy is conserved.

• Tenaga kinetik diabadikan. Kinetic energy is conserved.

• Kedua-duaobjekbergabungdanbergerakbersama

dengan satu halaju sepunya selepas perlanggaran.The two objects combine and move together with a

common velocity after the collision.

• Jumlah momentum diabadikan. Total momentum is conserved.

• Jumlah tenaga diabadikan. Total energy is conserved.

• Tenaga kinetik tidak diabadikan. Kinetic energy is not conserved.

m1 m2 m1 m2

u1 u2 v1 v2



Tuliskan persamaan yang menghubungkaitkan jumlah momentum sebelum perlanggaran dengan jumlah momentum selepas perlanggaran:Write equation which relates the total momentum before collision with the total momentum after collision:

m1u1 + m2u2 = m1v1 + m2v2

m1 m2 m1 m2

u1 u2 v



Tuliskan persamaan yang menghubungkaitkan jumlah momentum sebelum perlanggaran dengan jumlah momentum selepas perlanggaran:Write equation which relates the total momentum before collision with the total momentum after collision:

m1u1 + m2u2 = (m1 + m2)v

Letupan/Explosion

Troli pegunStationary trolleys

m2 m1

m2 m1

v2 v1

pin

Sebelum letupan/Before explosion

Selepas letupan/After explosion

Sebelum letupan, kedua-duaobjek bercantum bersama dan berada dalam keadaan rehat. Selepas letupan, kedua-dua objek

bergerak pada arah yang bertentangan .

Before explosion, both the objects stick together and are at rest. After

explosion, both objects move at opposite directions.

Jumlah momentum sebelum

letupan adalah sifar .The total momentum before

explosion is zero .

Jumlah momentum selepas

letupan = m1v1+m2v2

Total momentum after explosion

= m1v1 + m2v2




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UNIT

(Catatan: v2 bernilai negatif)(Remarks: V2 has a negative value)

Apakah yang dimaksudkan dengannilai negatif bagi v2?Why does v2 have a negative value?

Arahbertentangan

Opposite direction

Daripadaprinsipkeabadianmomentum:From the principle of conservation of momentum:

Jumlah momentum sebelum perlanggaranTotal momentum before collision

= Jumlah momentum selepas perlanggaranTotal momentum after collision

Terbitkan persamaan untuk letupan:Derive an equation for explosion:0=m1v1+m2v2m1v1 = –m2v2

Huraikan aplikasi prinsip kebadiaan momentumDescribe applications of the principle of conservation of momentum

m1 + m2

(a) Sebelum letupan Before explosion

(b) Selepas letupan After explosion

Pegun/Stationaryu = 0

m1

PeluruBullet

m2

v2

v1

Catatan/Remarks: Jisim senapang/Mass of riffle = m1 Jisim peluru/Mass of bullet = m2

Selepasletupan/After explosion: v1 = Halaju senapang/Velocity of riffle v2 = Halaju peluru/Velocity of bullet

• Apabila sepucuk senapang ditembak, peluru yang berjisimm2 bergerak dengan halaju tinggi, v2. Ini menghasilkan suatu

momentum ke arah hadapan . When a rifle is fired, the bullet of mass m2 moves with a high velocity, v2.

This creates a momentum in the forward direction.• Daripada prinsip keabadian momentum, suatu momentum yang

sama tetapi bertentangan arah dihasilkan supaya senapang itu

tersentak ke belakang . From the principle of conservation of momentum, an equal but opposite

momentum is produced to recoil the rifle backward .

RoketRocket

Gas panasHot gas

Pelancaran roket/The launching of rocket• Campuranbahanapihidrogendanoksigenterbakardenganletupan

dalam kebuk pembakaran. Gas panas dalam jet itu dipancutkan

dengan kelajuan yangsangattinggimelaluiekzos. A mixture of hydrogen and oxygen fuels burn explosively in the

combustion chamber. Jets of hot gases are expelled at very high speed through the exhaust.

• Kelajuan tinggi gas panas ini menghasilkan momentum yang

besar ke bawah .

This high-speed hot gas produces a large momentum .• Dengan prinsip keabadian momentum, suatu momentum yang sama tetapi bertentangan arah dihasilkan dan menggerakkan

roket itu ke atas . By the principle of conservation of momentum, an equal but opposite

momentum is produced and propels the rocket upwards .




2

UNIT

Huraikan aplikasi prinsip kebadiaan momentum linearDescribe applications of the principle of conservation of linear momentum

Gas panasHot gas

Jet

Aplikasienjinjet:Application in the jet engine:• Suatu gas panas yang berkelajuan tinggi dipancut keluar dari

belakang dengan momentum tinggi .

A high-speed hot gas is ejected from the back with high momentum .

• Inimenghasilkanmementumyang sama tetapi bertentangan

arah untuk menolak jet bergerak ke hadapan.

This produces an equal and opposite momentum to propel the jet plane forward.

Gerakan udara ke belakangMovement of air backwards

Bot berkipasFan boat

Dalamkawasanpaya,suatubotberkipasdigunakan.In a swamp area, a fan boat is used. • Kipasitumenghasilkangerakanudaraberkelajuantinggike

belakang . Ini menghasilkan suatu momentum yang besar ke belakang.

The fan produces a high speed movement of air backwards . This

produces a large momentum backwards.

• Dengankeabadianmomentum,suatumomentumyang sama tetapi bertentangan arah dihasilkan dan ditindakkan ke atas bot

itu.Jadi,botituakanbergerakke hadapan .

By conservation of momentum, an equal but opposite momentum

is produced and acts on the boat. So the boat will move forward .

SotongSquid

Seekorsotongbergerakdenganmengeluarkancecairpada halaju

yang tinggi.Airmasukmelalui pembukaanyangbesar dankeluar

melalui tiubyangkecil.Airdipaksakeluarpada kelajuan tinggi

ke belakang. Magnitud momentum air dan sotong adalah sama

tetapi pada arah yang bertentangan. Ini menyebabkan sotong itu

bergerak ke hadapan .

A squid propels by expelling a liquid at high velocity . Water enters through a large opening and exits through a small tube. The water is forced

out at a high speed backward. The magnitude of the momentum of

water and squid are equal but opposite in direction. This causes the

squid to jet forward .




2

UNIT

Menyelesaikan masalah melibatkan momentum linearSolve problems involving linear momentum

KeretaAyangberjisim1000kgbergerakpada20ms–1 berlanggar dengan kereta Byangberjisim1200kgdanbergerakpada10ms–1dalamarahyangsama.Akibatnya,keretaB,bergerakkehadapanpada15ms–1.Berapakahhalaju,v, bagi kereta A sebaik sahaja selepas perlanggaran?Car A of mass 1 000 kg moving at 20 m s–1 collides with car B of mass 1 200 kg moving at 10 m s-1 in the same direction. If car B is shunted forwards at 15 m s–1 by the impact, what is the velocity, v, of car A immediately after the crash?

uA = 20 m s-1

m1 = 1 000 kg

A

uB = 10 m s-1

m2 = 1 200 kg

B

Penyelesaian/Solution Jumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(1000kg)(20ms–1)+(1200kg)(10ms–1) =(1000kg)v+(1200kg)(15ms–1) 20000kgms–1+12000kgms–1 =(1000kg)(v)+18000kgms–1

(1000kg)(v) =14000kgms–1

∴ v =14ms–1

Latihan/Exercises 1

Sebijibolayangberjisim5kgdibalingkanpadahalaju20kmj–1kepadaLilyyangberjisim60kgpadakeadaan rehat di atas ais. Lily menangkap bola itu dan kemudian menggelongsor dengan bola di atas ais. Tentukan halaju Lily dengan bola selepas perlanggaran. A 5 kg ball is thrown at a velocity of 20 km h–1 towards Lily whose mass is 60 kg at rest on ice. Lily catches the ball and subsequently slides with the ball across the ice. Determine the velocity of Lily and the ball together after the collision.

u1 = 20 km j-¹m1 = 5 kg

u2 = 0 km j-¹m2 = 60 kg

m1 = 5 kg

v = ?

m2 = 60 kg

Penyelesaian/Solution Jumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(5kg)(20kmj–1)+(60kg)(0kmj–1) =(5+60)kg× v (100+0)kgkmj–1 =(65kg)v ∴ v =1.54kmj–1

(v =1.54kmh–1)

Latihan/Exercises 2




2

UNIT

Sebuahtrakyangberjisim1200kgbergerakpada30ms–1 berlanggar dengan sebuah kereta yang berjisim 1000kgyangbergerakdalamarahbertentanganpada20ms–1.Selepasperlanggaran,kedua-duakenderaanitubergerakbersama.Berapakahhalajukedua-duakenderaanitusebaiksahajaselepasperlanggaran?A truck of mass 1 200 kg moving at 30 m s–1 collides with a car of mass 1 000 kg which is traveling in the opposite direction at 20 m s–1. After the collision, the two vehicles move together. What is the velocity of both vehicles immediately after collision?

30 m s-¹ 20 m s-¹ v

(a) Sebelum perlanggaran Before collision

(b) Selepas perlanggaran After collision

Penyelesaian/SolutionJumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(1200kg)(30ms–1)+(1000kg)(–20ms–1)=(1200+1000)kg× v (36000–20000)kgms–1=(2200kg)v (2200kg)v=16000kgms–1

∴ v=7.27ms–1

Latihan/Exercises 3

Seorangmenembak sepucuk pistol yang berjisim 1.5 kg. Jika peluru itu berjisim 10 g danmempunyaihalaju300ms–1selepastembakan,berapakahhalajusentakanpistolitu?A man fires a pistol which has a mass of 1.5 kg. If the mass of the bullet is 10 g and it has a velocity of 300 m s–1 after shooting, what is the recoil velocity of the pistol?

1.5 kg 10 g

300 m s-¹v

Pegun/Stationary

(a) Sebelum tembakan Before shooting

(b) Selepas tembakan After shooting

Penyelesaian/Solution Jumlah momentum sebelum tembakan = Jumlah momentum selepas tembakan Total momentum before explosion = Total momentum after explosion

0kgms–1=(1.5kg)(v)+(0.010kg)(300ms–1) (1.5kg)(v)=–3.0kgms–1

∴ v=–2.0ms–1

Latihan/Exercises 4




2

UNIT

Bayangkanandaberlegardi sebelahkapalangkasapada orbit bumi dan rakan anda yang sama jisim bergerak dengan 4 km j–1 (dengan merujuk kepada kapal angkasa) melanggar anda. Jika dia memeganganda,berapakahkelajuanandabergerak(dengan merujuk kepada kapal angkasa itu)?Imagine that you are hovering next to a space shuttle in earth orbit and your buddy of equal mass who is moving at 4 km/hr (with respect to the ship) bumps into you. If she holds onto you, how fast do you move (with respect to the ship)?

Sebelum perlanggaranBefore collison

m

Dalam gerakanIn motion

u1 = 4 km j–1

Dalam keadaanrehatAt rest

u2 = 0 km j–1

m

Dalam gerakan bersamapada laju yang sama

In motion togetherat the same speed

mm

Selepas perlanggaranAfter collison



=

Jumlah momentum selepas perlanggaranTotal momentum after collision

(mkg)(4kmj–1)+(mkg)(0kmj–1) = (m+m) kg × v(4m) kg km j–1+0=(2m) kg × v

∴ v = (4m) kg km j–1

(2m) kg =2kmj–1

Seekorikanyangbesaryangberjisim3mbergerakdengan2ms–1 bertemu seekor ikan kecil yang berjisim m dalam keadaan rehat. Ikan besar itu menelan ikan kecil dan meneruskan gerakan dengan kelajuan yang berkurang. Jika jisim ikan besar adalah tiga kali ganda jisim ikan kecil,berapakah halaju ikan besar selepas menelan ikan kecil itu?A large fish of mass 3 m is in motion at 2 m s–1 when it encounters a smaller fish of mass m which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish after swallowing the smaller fish?

3m m

Dalam gerakan/In motionu1 = 2 m s–1

RehatAt rest

Sebelum perlanggaranBefore collison

Selepas perlanggaranAfter collison



=

Jumlah momentum selepas perlanggaranTotal momentum after collision

[(3m) kg ×(2ms–1)]+0=(3m +m) kg × v (6m) kg m s–1 =(4m) kg × v ∴ 4v =6ms–1

v =1.5ms–1

Huraikan apa yang dilakukan oleh penjaga gol sebelum dia menendang bola itu.Describe what the goalkeeper does before kicking the ball.•Penjagagolituakanmengambilbeberapalangkahkebelakangdankemudian

berlari ke hadapan untuk menendang bola itu.

The goalkeeper takes a few steps backwards and then runs forward to kick the ball.

•Bolaituakanbergerak lebih jauh/lebih cepat apabila ditendang semasa berlari berbanding dengan tendangan dari kedudukan pegun.

The ball goes further / faster when kicked while running compared to kicking from a standing position.

• Iniadalahdisebabkanseorangpemainbolasepakyangberlarimempunyaimomentumyang besar dan

momentumnya dipindahkan kepada bola.

This is because a running football player has a large momentum and his momentum is transferred to the ball.

Latihan/Exercises 5 Latihan/Exercises 6




2

UNIT

TujuanAim

Untuk menunjukkan jumlah momentum bagi suatu sistem tertutup adalah malar dalam perlanggaran tak kenyal.To show that the total momentum of a closed system is constant in an inelastic collision.

RadasApparatus

Jangkamasadetik,pitadetik,plastisin,pita selofan, troli, landasan,bekalankuasaa.u.12V.Ticker timer, ticker tape, plasticine, cellophane tape, trolleys, runway, 12 V ac power supply.

ProsedurProcedure

1 Dirikansatulandasandenganmengubahsuaikecerunannyasupayalandasanterpampasgeseran di mana troli boleh bergerak turun landasan dengan halaju malar.Set up a runway and adjust the slope to compensate for friction where the trolley moves down the runway with constant velocity.

2 Letakkan plastisin pada troli P dan Q supaya mereka akan melekat antara satu sama lain semasa perlanggaran.Fix plasticine on trolleys, P and Q so that they can stick together upon collision.


Jangka masa detikTicker timer Landasan

terpampasgeseranFriction-

compensatedrunway

Blok kayu/Wooden block

Bekalan kuasaPower supply

Troli PTrolley P Plastisin

Plasticine Troli QTrolley Q

3 Pita detik diletakkan melalui jangka masa detik dan dilekatkan pada troli P.A ticker tape is passed through the ticker timer and is attached to trolley P.

4 Mulakan jangka masa detik dan tolakkan troli P supaya ia bergerak menuruni landasan dan berlanggar dengan troli Q,yangberadadalamkeadaanrehat.Start the ticker timer and give trolley P a push so that it will move down the runway and collide with trolley Q, which is at rest.

5 Daripadapitadetikyangdiperoleh,tentudanukurkanhalajuberikut.From the ticker tape obtained, determine and measure the following velocities.

(a) Halaju troli Psebelumperlanggaran,uP / Velocity of trolley P before collision, uP

(b) Halaju troli Qsebelumperlanggaran,uQ / Velocity of trolley Q before collision, uQ

(c) Halaju troli (P+Q)selepasperlanggaran,vVelocity of trolley (P + Q) after collision, v

6 Langkah-langkah 2 – 5 diulangi dengan jisimP dan jisim Q yang berbeza sepertiditunjukkan dalam jadual di bawah.Steps 2.5 are repeated for different masses of P and Q as shown in the table below.

Sebelum perlanggaranBefore collision


Jisim troli PMass of trolley P

kg

Jisim troli QMass oftrolley Q

kg

Halaju PVelocity of P

m s–1

Jumlah momentum

Total momentum kg m s–1

Halaju sepunyaCommon velocity

m s–1

Jumlah momentum

Total momentum kg m s–1

1 1

2 1

1 2

2 2

3 2

EksperimenExperiment Keabadian Momentum / Conservation of Momentum




2

UNIT

KeputusanResult

Contoh: Perlanggaran tak kenyal antara dua troliExample: Inelastic collision between two trolleys

18.4 cm 9.2 cm



Perlanggaran berlaku di siniCollision

occurs here

Arah gerakan/Direction of motion




Panjang10detik10-tick length

18.4cm 9.2cm

Masadiambiluntuk10detikTime taken for 10 ticks

0.2s 0.2s

HalajuVelocity

18.4cm0.2s

=0.92ms–1 9.2cm0.2s

=0.46ms–1

Jisim troli (jisim 1 troli = 1 kg)Mass of trolley (mass of 1 trolley = 1 kg)

1 kg 2kg

MomentumMomentum

(1kg)(0.92ms–1)=0.92kgms–1

(2kg)(0.46ms–1)=0.92kgms–1


1 Bandingkanjumlahmomentumsebelumperlanggarandanselepasperlanggaran.Compare the total momentum before collision and after collision.

Jumlah momentum sebelum dan selepas perlanggaran adalah sama.

The total momentum before collision and after collision are equal.

2 Nyatakan satu kesimpulan.State a conclusion.

Tanpakehadirandayaluar,jumlahmomentumsebelumperlanggaranadalahsama

dengan jumlah momentum selepas perlanggaran.

In the absence of any external force, the total momentum before collision is equal to total

momentum after collision.

3 Apakahtujuanutamamengubahsuailandasansupayalandasanterpampasgeseran?What is the main purpose of adjusting the runway so that it is friction-compensated?

Troli bergerak dengan halaju malar.

The trolley moves with constant velocity.




2

UNIT

Huraikan kesan daya tidak seimbang yang bertindak ke atas suatu objekDescribe the effects of unbalanced forces acting on an object

Apabiladayayangbertindakkeatasobjek tidak seimbang , terdapatdaya bersih yang bertindak ke atasnya.

When the forces acting on an object are not balanced , there must be a net force acting on it.

Dayabersihdikenalisebagaidaya paduan yang bertindak ke atasnya.

The net force is known as the resultant force acting on it.

Dayatidakseimbang=dayabersih=dayayangdikenakan–dayageseranThe unbalanced force = net force = force applied – frictional force

Kesan:Bolehmenyebabkanbadanseseorang/Effect: Can cause a body to• bertukarkeadaanrehatnya(objekituakanmemecut)

change its state at rest (an object will accelerate)• bertukarkeadaangerakannya(suatuobjekyangbergerakakanmemecut/nyahpecutataumenukararahnya)

change its state of motion (a moving object will accelerate/decelerate or change its direction)

2.5 MEMAHAMI KESAN DAYAUNDERSTANDING THE EFFECTS OF A FORCE

Huraikan kesan daya seimbang yang bertindak ke atas objekDescribe the effect of balanced forces acting on an object

Daya seimbang/Balanced force

Apabiladaya-dayayangbertindakkeatassuatuobjekdalamkeadaan seimbang ,iaakanmembatalkan

antarasatusamalain.Dayabersihadalah sifar .

When the forces acting on an object are balanced , they cancel each other out. The net force is zero .

Kesan/Effect:

Objekberadadalamkeadaan rehat [halaju = 0] atau bergerak pada halaju malar [pecutan = 0]

The object is at rest [velocity = 0] or moves at constant velocity [acceleration = 0 ]

BeratWeight

Daya dikenakan oleh meja ke atas cawanForce exerted by table on the cup

Cawan itu berada dalam keadaan rehat. Daya

bersih yang bertindak ke atasnya adalah sifar . The cup stays at rest. The net force acting on it is

zero . W = R di mana/where W:Berat/Weight R : Tindak balas normal/Normal reaction

Daya angkat, U/Lift, U

Berat, WWeight, W

Tujahan, FThrust, F

Seretan, GDrag, G

Kapal terbang bergerak dengan halaju malar. Daya

bersih yang bertindak ke atasnya adalah sifar .The plane moves with constant velocity. The net force acting

on it is zero .W = U di mana/where W :Berat/Weight U :Dayaangkat/Lift

F = G di mana/where F : Tujahan/Thrust G:Seratan/Drag




2

UNIT

Menentukan hubungan antara daya, jisim dan pecutan (F = ma)Determine the relationship between force, mass and acceleration (F = ma)

Hukum Gerakan NewtonKeduaNewton’s Second Law of Motion

F = 5 N

m = 25 kg

Pecutan yang dihasilkan oleh daya ke atas suatu objek adalah berkadar langsung dengan magnitud daya bersih yang dikenakan dan berkadar songsang dengan jisim objek itu.

The acceleration produced by a force on an object is directly proportional to the

magnitude of the net force applied and is inversely proportional to the mass of the object.

Daya=Jisim×Pecutan F = ma

Force = Mass × Acceleration F = ma

Hubungan antara a dan FRelationship between

a and F

a α FPecutan,a,berkadarlangsungdengandayayangdikenakan,FThe acceleration, a, is directly proportional to the applied force, F

a

0 F

Hubungan antara a dan mRelationship between

a and m

a α 1m

Pecutan, a, bagi suatu objek berkadar songsang denganjisimnya,mThe acceleration, a, of an object is inversely proportional to its mass, m

1m

a

0

Hubungan antaraRelationship between a α F a α m

SituasiSituation

A

B

Dua orang pemuda menolak jisim yangsama tetapi pemuda A menolak dengan daya yang lebih besar. Jadi dia bergerak dengan lebih cepat.Both men are pushing the same mass but man A pushes with a greater force. So he moves faster.

A

B

Dua orang pemuda mengeluarkan dayayang sama. Tetapi pemuda B bergerak dengan lebih cepat daripada pemuda A.Both men exerted the same force. But man B moves faster than man A.

HipotesisHypothesis

Semakinbesardaya,semakinbesarpecutan.

The larger the force, the greater the

acceleration.

Semakinbesarjisim,semakinkecilpecutan.

The greater the mass, the smaller the

acceleration.

EksperimenExperiment

Mencari hubungan antara daya, jisim dan pecutanFind the relationship between force, mass and acceleration




2

UNIT

Pemboleh ubah dimanipulasiManipulated variable

Daya/Force Jisim/Mass

Pemboleh ubah bergerak balasResponding variable

Pecutan/Acceleration Pecutan/Acceleration

Pemboleh ubah dimalarkanConstant variable

Jisim/Mass Daya/Force

Bahan dan radasMaterials and apparatus

Jangkamasadetikdanpitadetik,bekalankuasa,landasanterpampasgeseran,

pembaris,troli,takallicin(denganpengapit),talitakkenyal,pemberatberslot

Ticker timer and ticker tape, power supply, friction-compensated runaway, ruler, trolley,

smooth pulley (with clamp), inelastic string, slotted weights

RajahDiagram


Jangka masa detikTicker timer

Landasan terpampasgeseran

Friction-compensatedrunway

Tali tak kenyalInelastic string


Bekalan kuasa a.u.a.c. power supply

Troli PTrolley P

Takal licinSmooth pulley

PemberatberslotSlottedweight

Susunanradasuntukmengkajihubunganantara(1) daya dan pecutan(2)jisimdanpecutan

Arrangement of apparatus to investigate the relationship between(1) Force and acceleration(2) mass and acceleration

ProsedurProcedure

1 Radas disusun seperti ditunjukkan dalam rajah di atas.The apparatus is set up as shown in the diagram above.

2 Sebuah troli berjisim 1.0 kg (jisimmalar) diletakkan di atas landasan. Pita detik dilekat pada troli itu.A trolley of mass 1.0 kg (constant mass) is placed on the runway. A length of ticker tape is attached to the trolley.

3 Jangka masa detik dihidupkan dan troli itu ditarik oleh pemberat yang mempunyai daya,F=10.0N. The ticker timer is switched on and the

trolley is pulled by a weight of force, F = 10.0 N.

4 Daripitadetikyangdiperoleh,pecutantroli dihitung dengan menggunakan

formula,a = (v – u)t

From the ticker tape obtained, the acceleration of the trolley is calculated by using the formula, a = (v – u)

t.

1 Radas disusun seperti ditunjukkan dalam rajah di atas.The apparatus is set up as shown in the diagram above.

2 Sebuahloridenganjisim,m=1.0kgdiletakkan di atas landasan. Pita detik dilekat pada troli itu.A trolley of mass, m = 1.0 kg is placed on the runway. A length of ticker-tape is attached to the trolley.

3 Jangka masa detik dihidupkan dan troli itu ditarik oleh pemberat (daya malar pemberatiniialah10N)The ticker timer is switched on and the trolley is pulled by a weight of constant force, 10 N

4 Daripitadetikyangdiperoleh,pecutantroli dihitung dengan menggunakan

formula,a = (v – u)t

.From the ticker tape obtained, the acceleration of the trolley is calculated by

using the formula, a = (v – u)t

.




2

UNIT

5 Langkah-langkah2–4diulangidenganmenambahkan pemberat berslot supaya F=15.0N,20.0N,25.0Ndan30.0N.Steps 2 – 4 are repeated by adding slotted weights to pull the trolley so that F = 15.0 N, 20.0 N, 25.0 N and 30.0 N.

5 Langkah-langkah2–4diulangidenganmelekat pemberat berslot pada troli supaya jisim troli,m = 1.5 kg, 2.0 kg,2.5kgdan3.0kg.Steps 2 – 4 are repeated by taping up slotted weights to the trolley to give m = 1.5 kg,2.0 kg, 2.5 kg and 3.0 kg.

Merekodkan dataRecording data

Daya, F/NForce, F/N

Pecutan, a/cm s–2

Acceleration, a/cm s-2

10.0

15.0

20.0

25.0

30.0

Jisim, m/kgMass, m/kg

Pecutan, a/m s–2

Acceleration, a/cm s–2

1.0

1.5

2.5

2.5

3.0

Menganalisis dataAnalysing data

Pecutan, a/cm s–2


0 Daya, F/NForce, F/N

Pecutan, a/cm s–2


0 Jisim, m/kgMass, m/kg

Menyelesaikan masalah menggunakan F = maSolve problems using F = ma

1 Hitungkan pecutan bagi blok di bawah:/Calculate the acceleration of the block:

(a) m = 2 kg

F = 8.0 N (c) m = 10 kg

F = 18 NF = 2 N

a = Fm = 8.0N

2kg a = (18–2)N

10kg = 16N

10kg

=4m s-2 =1.6ms-2

(b) m = 8 kg

F = 14 NF = 6 N (d) m = 12 kg

F = 10 NF = 5 NR = 5 N

a = (14+6)N

8kg =

20N8kg

a = (10–5–5)N12kg

=0

=2.5ms-2 =0ms-2




2

UNIT

Menyelesaikan masalah menggunakan F = maSolve problems using F = ma

2 Seorang lelaki menolak troli yang berisi kotak (jumlah jisim 5 kg) di ataspermukaanyanglicin.Jikadiamenggunakandaya30Nuntukmenolaktroliitu,apakahmagnituddanarahpecutantroliitu?A man pushes a trolley with a box (total mass 5 kg) on a smooth surface. If he uses a force of 30 N to move the trolley, what is the magnitude and direction of the acceleration of the trolley?

Penyelesaian/Solution:F = ma

a = 30N5kg

=6ms-2 ke kanan / to the right.

3 Sebuahobjekyangberjisim2kgditarikdiatastanahdengandaya5Ndanhalajumalar.An object of mass 2 kg is pulled on the floor by a force of 5 N and has a constant velocity.

(a) Berapakahdayageseranantaraobjekdantanah?What is the frictional force between the object and the floor?

(b) Hitungkanpecutanobjekitujikaobjekituditarikdengandaya17N.Calculate the acceleration of the object if the object is pulled by a 17 N force.

Penyelesaian/Solution:(a) R ialah daya geseran/R is the frictional force F1 – R = ma ∴ R = F1 – ma Olehkeranahalajumalar/Because the velocity is constant, a=0 ∴ R = F1–0 = F1

=5N(b) F2 – R = ma 17N–5N=(2kg) (a)

a = 12N2kg

=6ms–2

4 Sebuahbasberjisim2000kgbergerakdenganhalajuseragam40ms-1sejauh2500msebelumberehat.Hitungkan A bus of mass 2 000 kg travels at a uniform velocity 40 m s-1 for a distance of 2 500 m before it comes to rest. Calculate

(a) purata nyahpecutan bas itu./the average deceleration of the bus. (b) purata daya yang dikenakan oleh brek itu untuk membolehkan bas itu berhenti bergerak.

the average force applied by the brakes to bring the bus to a standstill.

Penyelesaian/Solution:(a) v2 = u2+2as (b) F = ma 0=(40ms–1)2+2a(2500m) =(2000kg)(–0.32ms–2) ∴5000a=–1600ms–2 =–640N a=–0.32ms-2 (Negatif bermaksud daya untuk menentang

gerakan/Negative means force to resist the motion)

F = 30 N

5 kg

F1 = 5 NR




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2.6 MENGANALISIS IMPULS DAN DAYA IMPULSANALYSING IMPULSE AND IMPULSIVE FORCE

Menerangkan daya impuls/Explain what an impulsive force is

Daya yang besar yang bertindak dalam tempohmasa yang singkat semasa perlanggaran atau letupan

dikenali sebagai daya impuls .

A large force that acts over a short period of time during a collision or explosion is known as an impulsive force .

Daripadahubunganantaradaya,jisimdanpecutan:From the relationship between force, mass and acceleration:

F = ma = m ( v – ut

)

F = mv – mut

= perubahan momentum

t Unit = N = kg m s-2

m = jisim/mass u = halaju awal/initial velocity t = masa/time v = halaju akhir/final velocity

Dayaimpulsialahkadar perubahan momentum dalam perlanggaran atau letupan.

An impulsive force is the rate of change of momentum in a collision or explosion.

Mendefinisikan impuls/Define impulse

Impuls didefinisikan sebagai perubahan momentum /Impulse is defined as the change of momentum .

atau (momentum akhir – momentum awal) atau (mv – mu)or (final momentum – initial momentum) or (mv – mu)

Unit: kg m s-1 atau/or N s

DaripadaF = mv – mut

, Ft = mv – mu = perubahan momentum = impuls

From F = mv – mut

, Ft = mv – mu = change of momentum = impulse

Impuls ialah hasil darab antara daya dan masa .

The product of the force and the time is called the impulse.

Kesan peningkatan dan pengurangan masa perlanggaranThe effects of increasing and decreasing the time of collision

Daripadaformula,F = Perubahan momentumMasa

/From the formula, F = Change of momentumTime

.

Dayaimpulsberkadar songsang dengan masa sentuhan atau tindakan atau perlanggaran.

Impulsive force is inversely proportional to the time of contact or impact or collision.

Tempoh masa yang panjang/Longer period of time Dayaimpuls kecil / Impulsive force is small

Tempoh masa yang pendek/Shorter period of time Dayaimpuls besar / Impulsive force is large




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UNIT

2.7 KESELAMATAN KENDERAANVEHICLE SAFETY

Komponen/Component Fungsi/Function

Pelapik kepalaHeadrest Untuk mengurangkan kesan inersia terhadap kepala pemandu. Mengurangkan

kecederaan leher apabila kereta dilanggar daripada belakang.

To reduce the inertia effect on the driver’s head. Reduce neck injury when the car is

hit from behind.

Beg udaraAir bag Menyerap hentakan dengan menambahkan masa perlanggaran apabila kepala

pemandu terhentak ke stereng. Oleh itu daya impuls dikurangkan.

Absorbing impact by increasing the collision time when the driver’s head is thrown

towards the steering. So the impulsive force is reduced.

Cermin hadapan kereta/Cermin keselamatanWindscreen/safety glass

Kaca tahan pecah yang tidak akan mudah pecah kepada serpihan yang kecil dengan

mudah semasa perlanggaran. Mengurangkan kecederaan disebabkan oleh serpihan

kaca yang berselerak.

Shatterproof glass that will not break into small pieces easily during collision. This will

reduce injuries caused by scattered glass.

Pelapik kepalaHeadrest

Zon kemek belakangRear crumple zone

Cermin hadapan keretaWindscreen

Beg udaraAir bag

Zon remuk depanFront crumple zone

Bumper depanFront bumper Tali pinggang keselamatan

Safety beltBar hentaman sisiSide impact bar

Sistem brek anti kunciAnti-lock braking system (ABS)

SteringSteering

04 FIZ Form 4 Ch2 C 2F.indd 67 9/10/13 4:28 PM



2

UNIT

Komponen/Component Fungsi/Function

Zone remuk (depan dan belakang)Crumple zone (front and rear)

Boleh dimampatkan ketika kemalangan. Jadi ia akan meningkatkan masa yang

diperlukan kereta untuk berhenti sepenuhnya. Maka ia akan kaca yang

mengurangkan daya impuls.

Can be compressed during an accident. So it can increase the time taken by the car to

come to a complete stop. So it can reduce the impulsive force.

Bumper depanFront bumper Menyerap hentakan akibat daripada kemalangan. Diperbuat daripada keluli,

aluminium, plastik, getah dan fiber komposit.

Absorb the shock from the accident. Made from steel, aluminium, plastic, rubber and

composite fibres.

Sistem brek anti kunciAnti-lock braking system (ABS)

Membolehkan pemandu memberhentikan kereta dengan segera tanpa

menyebabkan roda terkunci apabila brek ditekan secara tiba-tiba. Mengelakkan

kereta daripada menggelongsor.

Enables drivers to quickly stop the car without the wheels locking when the brake is

applied suddenly. Prevents the car from skidding.

Bar hentaman sisiSide impact bar Boleh dimampatkan ketika kemalangan. Jadi ia akan meningkatkan masa yang

diperlukan kereta untuk berhenti sepenuhnya. Maka ia akan mengurangkan

daya impuls.

Can be compressed during accident. So it can increase the time the car takes to come

to a complete stop. So it can reduce the impulsive force.

Tali pinggang keselamatanSafety belt

Untuk mengurangkan kesan inersia dengan mengelakkan pemandu daripada

tercampak ke hadapan.

To reduce the inertia effect by preventing the driver from being thrown forward.

Papan pesawatDashboard Semasa perlanggaran, papan pesawat meroboh. Ini akan menyerap kesan

hentaman dengan meningkatkan masa perlanggaran antara kepala pemandu

dan stereng. Jadi ia mengurangkan daya impuls.

During collision, the dashboard collapses. This will absorb the impact by increasing the

time of collision between the driver's head and the steering. This will reduce the

impulsive force.




2

UNIT

2.8 MEMAHAMI GRAVITIUNDERSTANDING GRAVITY

Kekuatan medan graviti/Gravitational field strength

• Objekjatuhbebaskeranaditarikkearahpusatbumiolehdayatarikan graviti .

Objects experience free fall because they are pulled towards the centre of the Earth by the force of gravity .

• Kekuatanmedangraviti= daya tarikan gravitijisim

. / Gravitational field strength = gravitational force

mass.

• DipermukaanBumi, At the surface of the Earth, Kekuatan medan graviti = 10 N kg–1 gravitational field strength = 10 N kg–1

= 10 m s–2 = 10 m s–2

• SetiapkilogramjisimpadapermukaanBumimengalamidayagravitisebanyak10Nyangbertindakkeatasnya. Earth kilogram of mass at the Earth's surface has a gravitational force of 10 N acting on it.

Aktiviti 1: Pecutan disebabkan gravitiActivity 1: Acceleration due gravity

Rajah di sebelah menunjukkan gambarfoto stroboskop bagi bola yang jatuh bebas dan graf halaju lawan masa bagi gerakannya. The diagram on the right shows a stroboscopic photograph of a free falling ball and its velocity-time graph.(a) Perhatikan gambarfoto dan terangkan halaju bola. Observe the photograph and describe the velocity of the ball.

Halaju bola itu meningkat dengan seragam.

The velocity of the ball increases uniformly.

(b) Apakah yang boleh anda simpulkan daripada graf kecerunan v – t? What can we deduce from the gradient of the v – t graph?

Kecerunan ialah pecutan bola itu.

The gradient is the acceleration of the ball.

(c) Terangkan gerakan bola tersebut. Describe the motion of the ball.

Bola tersebut bergerak dengan pecutan seragam.

The ball moves with constant acceleration.

Terangkan pecutan yang disebabkan oleh graviti, g/Explain acceleration due to gravity, g

Pecutan disebabkan oleh graviti, g, ialah pecutan bagi objek yang disebabkan oleh daya tarikan graviti.

Acceleration due to gravity, g, is the acceleration of an object due to the pull of the gravitational force .

Nilai piawai bagi pecutan graviti, g, ialah 9.81 m s-2. Nilai g yang sering digunakan ialah 10 m s-2.

Magnitud bagi pecutan yang disebabkan oleh graviti bergantung pada kekuatan medan graviti .The standard value of the gravitational acceleration, g, is 9.81 m s-2. The value of g is often taken to be 10 m s-2

for simplicity. The magnitude of the acceleration due to gravity depends on the gravitational field strength .

V

t0




2

UNIT

Apakah jatuh bebas?/What is free fall?

Objek dikatakan 'jatuh bebas' apabila ia jatuh di bawah daya tarikan graviti sahaja.

An object is falling freely when it is falling under the gravitational force only.

Sehelai kertas tidak jatuh bebas kerana kejatuhannya dipengaruhi oleh rintangan udara .

A piece of paper does not fall freely because its fall is affected by air resistance .

Objek hanya jatuh bebas di dalam vakum . Ketiadaan udara bermaksud tiada rintangan udara yang

menentang pergerakan objek.

An object falls freely only in vacuum . The absence of air means there is no air resistance to resist the motion of the object.

Di dalam vakum, kedua-dua objek yang ringan dan berat jatuh bebas. Ia jatuh dengan pecutan graviti iaitu pecutan disebabkan oleh graviti, g.

In vacuum, both light and heavy objects fall freely. They fall with the gravitational acceleration, that is the acceleration due to gravity, g.

Aktiviti 2: Pecutan disebabkan graviti/Activity 2: Acceleration due gravity

Pegang dua biji batu yang berbeza saiz pada ketinggian yang sama, kemudian kedua-dua batu itu dijatuhkan serentak daripada ketinggian yang sama. Hold two stones of different sizes at the same height and then drop both stones simultaneously from the same height.(a) Huraikan bagaimana halaju berubah. Describe how the velocity changes.

Halaju meningkat dengan seragam.

The velocity increases uniformly.

(b) Bandingkan masa yang diambil untuk batu mencecah lantai. Compare the time taken for the stones to reach the floor.

Sama/same

(c) Adakah pecutan batu dipengaruhi oleh jisimnya? Is the acceleration of each stone influenced by its mass?

Jisim tidak mempengaruhi pecutan.

Mass does not affect the acceleration.




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UNIT

Aktiviti 3: Yang mana satukah mencecah tanah dahulu?Activity 3: Which one reaches the ground first?

Bola golf dan sehelai kertas dipegang pada ketinggian yang sama dan dijatuhkan serentak.Hold a golf ball and a piece of paper at the same height and drop them simultaneously.(a) Objek yang manakah mencecah tanah dahulu? Which object reaches the floor first?

Bola golf.

The golf ball.

(b) Terangkan mengapa./Explain why.

Kertas mempunyai luas pemukaan yang besar.

Jadi lebih banyak rintangan udara yang

bertindak ke atasnya.

The paper has large surface area. As such, the air

resistance acting on it is big.

Ulangi dengan bola golf dan sehelai kertas yang direnyukkan.Repeat with a golf ball and a piece of paper which is crumpled.(a) Objek yang manakah mencecah tanah dahulu? Which object reaches the floor first?

Kedua-duanya mencecah tanah pada masa

yang sama.

Both reach the floor at the same time.

(b) Terangkan mengapa./Explain why.

Kedua-dua objek mempunyai saiz dan luas

pemukaan yang sama. Jisim tidak memberi

kesan kepada pecutan graviti.

Both objects have same size and surface area.

Mass does not affect gravitational acceleration.

Bola golfGolf ball

KertasPaper

Bola golfGolf ball

Kertas yang direnyukkanPaper which is crumpled

Aktiviti 4: Perbezaan antara jatuh bebas di atmosfera (udara) dan jatuh bebas di dalam vakum bagi duit syiling dan bulu pelepah.

Activity 4: The difference between free fall in atmosphere and free fall in a vacuum of a coin and a feather.

Duit syiling dan bulu pelepah dilepaskan daripada ketinggian yang sama serentak di dalam makmal.A coin and a feather are released from the same height simultaneously in the laboratory.

Pemerhatian/Observation

Duit syiling jatuh lebih cepat daripada bulu pelepah.

The coin falls faster than the feather.

Duit syiling dan bulu pelepah yang sama diletakkan di dalam satu tiub vakum dan kemudian dijatuhkan serentak pada ketinggian yang sama.The same coin and feather are put into a vacuum tube and then dropped simultaneously from the same height.

Pemerhatian/Observation

Kedua-dua objek mencecah ke bawah silinder

pada masa yang sama .Both objects reach the bottom of the cylinder at the

same time.




2

UNIT

g sebagai kekuatan medan graviti. g sebagai pecutan yang disebabkan gravitig as gravitational field strength. g as acceleration due to gravity

1 Sebuah objek yang jatuh bebas berdekatan dengan permukaan bumi akan memecut pada 10 m s-2.

An object falling freely near the earth's surface will accelerate at 10 m s-2.

2 Setiap kilogram jisim yang berdekatan dengan permukaan bumi mempunyai daya graviti 10 N yang bertindak ke atasnya.Each kilogram of mass near the earth’s surface has a gravitational force of 10 N acting on it.

Kekuatan medan graviti , g = 10 N kg-1/ Gravitational field strength, g = 10 N kg-1

Pecutan disebabkan graviti/Ac celeration due to gravity, g = 10 m s-2

Nilai g boleh ditulis sebagai 10 m s-2 atau 10 N kg-1. The approximate value of g can be written either as 10 m s-2 or as 10 N kg-1.

Penjelasan/Explanation

Rintangan udara yang besar bertindak ke atas

bulu pelepah kerana ia mempunyai luas permukaan yang besar.

A bigger air resistance acts the feather because

it has a large surface area .

Daya graviti pada duit syiling mampu untuk

mengatasi rintangan udara lebih baik daripada bulu pelepah.

The gravitational force on the coin is able to

overcome air resistance better than the feather.

Penjelasan/Explanation

Di dalam keadaan vakum, tiada rintangan udara.Hanya terdapat satu daya yang bertindak ke atas

objek iaitu daya graviti .

In vacuum, there is no air resistance . The only

force acting on both objects is the force of gravity .

Kedua-dua objek jatuh bebas dengan pecutan yang disebabkan graviti walaupun berbeza dari

segi jisim dan bentuk .

Both objects free fall with acceleration due to

gravity despite the differences in their mass and

shapes .

Aktiviti 5: Pecutan yang disebabkan gravitiActivity 5: Acceleration due to gravity

Tujuan/Aim

Menentukan pecutan disebabkan graviti/To determine the acceleration due to gravity.

Radas/Apparatus

Jangka masa detik, bekalan kuasa 12 V, bangku, pengapit-G, pemberat, pita detikTicker timer, 12 V ac power supply, stool, G-clamp, slotted weight, ticker tape.

Prosedur/Procedure

1 Potong sekeping pita detik lebih kurang 2.5 m panjang dan lalukan melalui jangka masa detik yang diapit kepada kerusi oleh pengapit-G.Cut a piece of ticker tape about 2.5 m long and pass through the ticker timer which is clamped to a stool using G-clamp.




2

UNIT

2 Sambungkan satu hujung pita pada pemberat 100 g.Attach one end of the tape to the 100 g slotted weight.

3 Hidupkan jangka masa detik dan pemberat dilepaskan supaya ia jatuh bebas.Switch on the ticker timer and release the slotted weight so that it falls freely.

4 Kaji pita itu untuk menentukan nilai bagi pecutan disebabkan oleh graviti, g.Analyse the tape to determine the value of the acceleration due to gravity, g.

Perbincangan/Discussion

1 Mengapakah sukar untuk menentukan pergerakan objek yang jatuh dengan hanya memerhatikannya jatuh?Why is it difficult to describe the motion of a falling object by just observing it fall?

Objek bergerak sangat laju.

The object moves very fast.

2 Apakah jenis pergerakan objek jika ia jatuh di bawah tarikan graviti?What is the type of motion of an object falling under the pull of gravity?

Pecutan seragam./Constant acceleration.

3 Mengapakah pergerakan pemberat boleh diandaikan sebagai jatuh bebas?Why is it that the motion of the slotted weight can be assumed to be a free fall?

Ringtangan udara yang kecil boleh diabaikan.

The small air resistance is negligible.

4 Apakah langkah yang akan anda ambil untuk mengurangkan geseran antara pita dan jangka masa detik?What steps did you take to minimise the friction between the ticker tape and the ticker timer?

Pegang pita detik dalam keadaan menegak dan lepaskannya. Pastikan ia jatuh mealui jangka masa

detik dengan lancar.

Hold the ticker tape vertically when releasing it. Make sure it slips through the ticker timer smoothly.

5 Terangkan mengapa perlu menjatuhkan pemberat daripada kedudukan yang tinggi.Explain the need for the slotted weight to be dropped from a high position.

Pengiraan akan menjadi lebih tepat kerana ralat eksperimen dikurangkan.

The calculation will be more accurate because experimental errors are reduced.

6 Tunjukkan bagaimana anda mengira g daripada pita.Show how you would calculate g from the tape.

u = s1

t1 v = s2

t2a = v – u

t

Pengapit-GG-clamp

Jangka masadetikTicker timer


Bekalankuasa, 12 VA.C. Powersupply, 12 V

BangkuStool

PemberatWeight

Kepingan polisterenaPolystyrene sheet




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7 Bandingkan nilai pecutan disebabkan oleh graviti daripada aktiviti ini dengan nilai yang sebenar. Berikan alasan yang munasabah bagi perbezaan di antara dua nilai tersebut.Compare the value of the acceleration due to gravity from this activity with the actual value. Give possible reasons for any difference in these two values.

Nilai daripada eksperimen adalah lebih rendah berbanding dengan nilai sebenar. Sebabnya ialah

rintangan akibat jangka masa detik. / The value from the experiment is lower than the actual value.

The reason is the resistance caused by the ticker timer.

8 Bandingkan nilai bagi g daripada eksperimen jika anda mengulangi eksperimen dengan menggunakan pemberat 200 g dan 300 g.Compare the values of g from the experiment if you repeat the experiment using 200 g and 300 g weights.

Keputusan sepatutnya sama.

The result should be the same.

9 Apakah yang boleh anda simpulkan tentang hubungan antara g dan jisim bagi objek yang jatuh?What can you conclude about the relationship between g and the mass of the falling object?

Jisim tidak mempengaruhi pecutan graviti, g.

Mass does not affect the gravitational acceleration, g.

Definisi beratDefinition of weight

Berat, W, bagi sesuatu objek ialah daya graviti yang dikenakan ke atasnya.The weight, W, of an object is the gravitational force acting on it.

Berat ialah daya dan diukur dalam unit Newton, N . Berat ialah kuantiti vektor .

Weight is a force and is measured in Newton, N . Weight is a vector quantity.

Sebiji batu yang berjisim m, dilepaskan jatuh bebas pada pecutan graviti, g.A stone of mass, m, is released and free falls with a gravitational acceleration of g.

Daya yang bertindak ke atas batu hanyalah berat, W, di mana ianya menuju ke arah bawah.The only force acting on the stone is its weight, W, which is downward.

Hukum gerakan Newton kedua:Newton’s second law of motion:F = ma di mana/where F = W , a = g

Oleh itu/Therefore:

W = mg

W = berat/weightm = jisim/massg = pecutan graviti/acceleration due to gravity unit bagi g ialah = m s-2

the unit of g is = m s–2

Berat/Weight = m × 10 N

Bumi/Earth

Jisim/Mass = m kg

Pecutan = gAcceleration = g




2

UNIT

Berat berubah, jisim tetapWeight changes but mass is fixed

Di Bulan, berat kita lebih ringan daripada di Bumi, ini kerana medan graviti di Bulan adalah lebih kecil .

On the Moon, our weight is less than that on Earth because

the Moon’s gravitational field is smaller than that of the Earth.Malahan di Bumi, berat kita berbeza sedikit dari suatu tempat ke tempat yang lain, kerana kekuatan medan graviti yang berbeza. Semakin jauh dari bumi, berat kita

semakin berkurang .Even on Earth, our weight can vary slightly from place to place, because the Earth’s gravitational field strength varies.

Moving away from the Earth, our weight decreases .Jika kita boleh pergi lebih jauh ke dalam ruang angkasa lepas dan bebas daripada sebarang tarikan graviti, berat

kita akan menjadi sifar .If we could go deep into space, and be free from any

gravitational pull, our weight would be zero .Sama ada di atas Bumi, Bulan atau di angkasa, jisim kita tetap tidak akan berubah.Whether on the Earth, Moon or deep in space, our mass does not change.

Nilai bagi pecutan graviti, gB, di bulan ialah 16

daripada nilai gE di bumi.The value of the gravitational acceleration, gB, on

the Moon is 16

the value of gE on the Earth.

Di angkasa lepas

Deep in space

Di permukaan

BulanOn Moon’s

surface

Di permukaan

BumiOn Earth’s

surface

JisimMass

100 kg 100 kg 100 kg

BeratWeight

0 N 16

(1 000) N 1 000 N

Perbezaan antara berat dan jisimDifference between weight and mass

Berat/Weight Jisim/Mass

Definisi Definition

Daya graviti yang bertindak ke atas objek.

The force of gravity acting on the object.

Jumlah jirim di dalam objek.

The amount of matter in the object.

Perubahan / Tiada perubahanChange / unchanged

Berat sesuatu objek berubah dengan kekuatan medan graviti pada sesuatu tempat.

The weight of an object changes with the gravitational field strength at the location.

Jisim sesuatu objek tidak berubah walau di mana-mana.

The mass of an object is unchanged anywhere.

Kuantiti asas atau kuantiti terbitanBase quantity or derived quantity

Kuantiti terbitan

A derived quantity

Kuantiti asas

A base quantity

Kuantiti vektor atau kuantiti skalarVector of scalar quantity

Kuantiti vektor

A vector quantity

Kuantiti skalar

A scalar quantity

Unit SI/SI unit Newton, N/Newton, N Kilogram, kg/Kilogram, kg




2

UNIT

Untuk objek yang jatuh dengan pecutan, g, berikut adalah persamaan-persamaan yang berkaitan:For an object falling with acceleration g, the following equations apply: 1 v = u + at di mana/where s = sesaran/displacement 2 s = ut + ½ at2 u = halaju awal/initial velocity 3 s = ½ (u + v)t v = halaju akhir/final velocity 4 v2 = u2 + 2as t = masa/time

a = g, pecutan graviti/acceleration due to gravity,

Nota/Notes: 1 Apabila sebuah objek jatuh bebas: a = g = 10 m s-2 (pecutan)

When an object fall freely: a = g = 10 m s-2 (acceleration) 2 Apabila sebuah objek dilambung ke atas: a = –g = –10 m s-2. (nyahpecutan)

When an object is thrown upwards: a = –g = –10 m s-2. (deceleration) 3 Pada kedudukan yang tertinggi, v = 0.

At the highest point, v = 0. 4 Jatuh ke bawah, v adalah positif./Downward direction, v is positive. 5 Arah ke atas, v adalah negatif./Upward direction, v is negative.

Contoh/Example

1 Sebiji batu jatuh daripada ketinggian 45 m.A rock falls from a height of 45 m.

(a) Berapa lamakah masa yang diambil oleh batu itu untuk mencecah ke tanah?How long does it take to reach the ground?

(b) Berapakah halaju batu itu semasa ia menghentam lantai?What is its velocity as it hits the ground?

Penyelesaian/Solution(a) u = 0, s = 45 m, g = 10 m s–2, t = ? s = ut + ½ gt2

45 m = 0 + ½ (10 m s–2)(t2) t2 = 9 s2

t = 3 s(b) v = u + gt = 0 + (10 m s–2)(3 s) = 30 m s–1

2 Sebiji bola dilambung ke atas daripada tanah dengan halaju 30 m s–1. Selepas beberapa lamakah bola itu akan menyentuh tanah semula?A ball is thrown upwards from the ground with a velocity of 30 m s–1. After how many seconds will it strike the ground again?

g = –10 m s–2

v = 0

u = 30 m s–1

Penyelesaian/SolutionUntuk gerakan ke atas,/For the upward motion, u = 30 m s–1, v = 0, g = –10 m s–2, v = u + gt ∴0 = 30 m s–1 + (–10 m s–2)(t) 10t = 30 s t = 3 s (gerakan ke atas/upward motion)Maka, masa untuk gerakan ke bawah juga mengambil 3 s. Oleh itu, ia mengambil masa 6 saat.The time taken for the downward motion is also 3 s.So it takes a total of 6 seconds.

Andaikan g = 10 m s-2 dan tiada rintangan udara. / Assume g = 10 m s-2 and there is no air resistance.

PecutanAcceleration

gTinggi/Height

hMasaTime

t

Objek dilambung ke atasObject thrown upwards

titik tertinggi, v = 0highest point, v = 0




2

UNIT

Latihan/Exercises

Andaikan nilai g = 10 m s–2./Assume the value of g = 10 m s–2.

1 Amir menjatuhkan batu ke dalam perigi. Jika jarak antara bahagian atas perigi dan permukaan air ialah 20 m,Amir releases a stone into a well. If the distance between the top of the well and the water surface is 20 m,

(a) berapakah masa yang diambil oleh batu itu untuk sampai ke permukaan air?what is the time required for the stone to reach the surface of the water?

(b) berapakah halaju batu itu apabila ia terkena permukaan air?what is the velocity of the stone when it strikes the surface of the water?

Penyelesaian/Solution(a) u = 0, s = 20 m , g = 10 m s–2 , t = ? s = ut + ½ gt2 20 m = 0 + ½ (10 m s–2)(t2) t2 = 4 s2 t = 2 s(b) v2 = u2 + 2gs v2 = 0 + 2(10 m s–2)(20 m) ∴v = 20 m s–1

2 Suatu objek yang berjisim 5 kg dilepaskan dari sebuah bangunan setinggi 500 m. BerapakahAn object of mass 5 kg is released from a tall building of height 500 m. What is the

(a) berat objek itu?/weight of the object? (b) kekuatan medan graviti?/gravitational field strength? (c) masa yang diambil untuk sampai ke tanah?/time taken to reach the ground?

Penyelesaian/Solution(a) W = 5 kg × 10 m s–2 (c) u = 0, s = 500 m, g = 10 m s–2

= 50 N s = ut + ½ gt2

(b) g = 10 N kg–1 atau/or 10 m s–2 500 m = 0 + ½(10 m s–2)(t2) t2 = 100 s2 t = 10 s

2.9 MENGANALISIS KESEIMBANGAN DAYAANALYSING FORCES IN EQUILIBRIUM

Keseimbangan daya/Forces in equilibrium

1 Apabila suatu daya dikenakan terhadap objek dan ia mengekalkan keadaan pegun atau

bergerak dengan halaju seragam, maka objek itu dikatakan berada di dalam keadaan keseimbangan.

When forces act upon an object and it remains stationary or moves at a constant velocity , the object is said to be in a state of equilibrium.

2 Apabila keadaan keseimbangan berlaku, daya paduan yang bertindak ke atas objek itu

adalah sifar iaitu tiada daya bersih bertindak ke atasnya.

When equilibrium is reached, the resultant force acting on the object is zero , i.e. there is no net force acting upon it.

Menerangkan situasi di mana daya berada dalam keseimbanganDescribe situations where forces are in equilibrium

Perhatian/Note:Untuk (b), mengapa v2 = u2 + 2gs digunakan dan bukan v = u + gt?For (b), why v2 = u2 + 2gs is used and not v = u + gt?

Jawapan/Answer:Semua nilai yang diperlukan diberi dalam soalan.All the values required are given in the question.




2

UNIT

Menyatakan maksuddaya paduan

State what a resultant force is

Daya paduan: daya tunggal yang menunjukkan kesan daripada gabungan

dua atau lebih daya dalam magnitud dan arah

Resultant force: a single force that represents the combined effect of two or

more forces in magnitude and direction

Daya paduan ialah hasil tambah vektor bagi dua atau lebih daya yang bertindak ke atas objek. Dalam kes-kes berikut, jika F ialah daya paduan, maka,The resultant force is the vector sum of two or more forces which act on the object. In the cases below, if F is the resultant force, hence,

1F

2F

1F2F

Daya paduan/Resultant force = F = F1 + F2 Daya paduan/Resultant force = F = F1 – F2

Latihan/Exercises

1 Hitungkan daya paduan. Ke arah manakah objek itu bergerak?Calculate the resultant force. Which direction does the object move?

5 N12 N

Daya paduan, F/Resultant force, F = 5 N + 12 N = 17 N Arah ke kanan/To the right.

2 Hitungkan daya paduan. Ke arah manakah objek itu bergerak?Calculate the resultant force. Which direction does the object move?

5 N12 N

Daya paduan, F/Resultant force, F = 12 N – 5 N = 7 N Arah ke kanan/To the right.

3 Seekor kuda menarik kereta dengan daya 500 N. Seorang petani membantu kuda itu menolak

kereta itu dengan daya 200 N. Berapakah daya paduan?A horse pulls a cart with a force of 500 N. A farmer helps the horse by pushing the cart with a force of 200 N. What is the resultant force?Daya paduan, F/Resultant force, F = 500 N + 200 N = 700 N ke kanan/to the right

4 Seekor kuda menarik kereta dengan daya 500 N. Seorang petani menarik kereta itu pada arah

bertentangan dengan daya 200 N. Berapakah daya paduan?A horse pulls a cart with 500 N force. A farmer pulls the cart with a force of 200 N but in opposite direction. What is the resultant force?Daya paduan, F/Resultant force, F= 500 N – 200 N = 300 N ke kanan/to the right

200 N500 N

500 N200 N




2

UNIT

5 Rajah 5.1 menunjukkan sebiji ladung berjisim 0.3 kg digantung dari siling.Diagram 5.1 shows a pendulum bob of mass 0.3 kg hung from the ceiling.Benang itu ditarik secara mengufuk dengan daya, F, supaya sudut antara benang dengan garis mencancang adalah 40°seperti ditunjukkan dalam Rajah 5.2.The thread is then pulled horizontally by a force, F, so that the thread makes an angle of 40° with the vertical line as shown in Diagram 5.2.

(a) Dalam ruang di bawah, lukis sebuah segi tiga keseimbangan daya bagi T, F dan berat ladung itu.In the space below, draw a triangle of forces in equilibrium for T, F and the weight of the bob.

T

F

BeratWeight3.0 N

40°

Perhatian/Note:Arah bagi tiga daya itu adalah berkitar.The directions of the three forces are cyclic.

(b) Hitung Daya, FCalculate the force, FDari segi tiga di atas,/From the triangle above,

F3.0 N

= tan 40°

F = 3.0 N tan 40° = 2.52 N

(c) Hitung tegangan, T, dalam benang ituCalculate the tension, T, in the string.Dari segi tiga di atas,/From the triangle above,

3.0 NT

= kos 40°

T = 3.0 Nkos 40°

= 3.92 N

Rajah 5.1/Diagram 5.1

SilingCeiling Benang

Thread

Ladung Pendulum Bob

Perhatian/Note: Soalan 5 di atas boleh dijawab dengan kaedah lukisan berskala. Question 5 above can be answered by scale-drawing.

40°

m = 0.3 kg

BeratWeight

T

F

Rajah 5.2/Diagram 5.2




2

UNIT

Dua daya yang betindak pada sudut tertentu antara satu sama lainTwo forces acting at an angle to each other

Daya paduan bagi dua daya yang bertindak ke atas dua objek pada dua arah berbeza, boleh ditentukan

dengan menggunakan kaedah segi tiga daya atau kaedah segi empat selari daya.

The resultant force of two forces, which act on an object in two different directions, can be determined by a triangle

of forces or a parallelogram of forces.

Dua daya yang bertindak pada satu titik melalui satu sudut [kaedah segi empat selari]Two forces acting at a point at an angle [Parallelogram method]

Skala/Scale: 1 cm = N

LANGKAH 1/ STEP 1: Dengan menggunakan pembaris dan protaktor, lukis dua daya F1 dan F2 bermula dari satu titik, X.Using ruler and protractor, draw the two forces F1 and F2 from a point, X.

60 °C

F1

X

F2

LANGKAH 2/ STEP 2: Lengkapkan rajah segi empat selari.Complete the parallelogram.

F1

X

F2

LANGKAH 3/ STEP 3: Lukis pepenjuru (dari X) bagi segi empat selari bagi menunjukkan magnitud dan arah bagi daya paduan, F. Draw the diagonal (from X) of the parallelogram. The diagonal representsthe resultant force, F in magnitude and direction.

Daya paduan

Resultant force

F1

X

F2

F1

F2

60 °C

Contoh/Example

Rajah di bawah menunjukkan dua daya yang bertindak ke atas objek P.The diagram below shows two forces acting on object P.

5 N

12 N60°

F1

F2P

Tentukan daya paduan yang terhasil.Determine the resultant force produced.




2

UNIT

Jawapan/Answer:Kaedah I/Method IKaedah segi empat selari dayaParrallelogram of forces method 1 Tentukan skala. Dengan menggunakan pembaris dan protaktor, lukis dua daya F1 dan F2 bermula dari

titik O.Set a scale. Using a ruler and protactor, draw the two forces, F1 and F2 from a point O.

2 Lengkapkan rajah segi empat selari./Complete the parallelogram.

3 Lukis pepenjuru bagi segi empat selari. Pepenjuru menunjukkan magnitud dan arah daya paduan, F.

Draw the diagonal of the parallelogram. The diagonal represents the resultant force, F in magnitude and

direction .

5 cm = 5 N

12 cm = 12 N

60°

O

Kaedah II/Method IIKaedah segi tiga dayaTriangle of forces method 1 Tentukan skala. Dengan menggunakan pembaris dan protaktor, lukis daya yang pertama, F1 dari titik O.

Set a scale. Using a ruler and protractor, draw the first force, F1, from a point O.

2 Lukis daya yang kedua F2 dari hujung atas F1.Draw the second force, F2 from the head of F1.

3 Lengkapkan segi tiga daya dengan melukis garisan dari pangkal F1 ke hujung F2.Complete a triangle of forces by drawing a line from the tail of F1 to the head of F2.

4 Pepenjuru menunjukkan magnitud dan arah daya paduan, F.The diagonal represents the resultant force, F, in magnitude and direction.

F2 = 12 N

F1 = 5 N

F = Daya paduan Resultant force

60°

O




2

UNIT

1 Dengan menggunakan skala dan kaedah yang sesuai, tentukan daya paduan. (Perhatian: magnitud dan arah diperlukan).By using suitable scale and method, determine the resultant force (Note: Magnitudde and direction are required).

(a) (b) (c)

5 N

5 N

8 N

8 N

6 N

10 N

600 1200

2 Lengkapkan rajah untuk menunjukkan daya paduan. Complete the diagram to show the resultant force.

2 N

5 N

120°

Jawapan/Answers:

1 Skala/Scale: 1 cm : 1 N

(a)

8 N

6 N

37°

Daya paduan = 10.0 N

Resultant force

Latihan/Exercises

10 N pada sudut 37° dengan daya 8 N10 N at angle of 37° with the 8 N-force




2

UNIT

(b)

8 N

10 N

260600

Daya paduan = 15.6 N

Resultant force

(c)

5 N

5 N

600

600

Day

a pa

duan

= 5

.0 N

Resu

ltant

forc

e

2 Skala/Scale: 2 cm : 1 N

(a)

2 N

5 N

120°16°

Daya paduan/Resultant force = 6.2 N

15.6 N pada sudut 26° dengan daya 10 N15.6 N at an angle of 26° with the 10 N-force

5 N pada sudut 60° dengan daya 5 N5 N at an angle of 60° with 5 N-force

6.2 N pada sudut 16° dengan daya 5 N6.2 N at an angle of 16° with 5 N-force




2

UNIT

Daya dileraikan kepada komponen berkesanResolve a force into the effective components

Leraian Daya/Resolution Of Forces:

Daya F boleh dileraikan kepada dua komponen yang berserenjang/bersudut tegak antara satu sama lain:

A force F can be resolved into two components which are perpendicular/at 90° to each other:

(a) Fx: komponen mengufuk / horizontal component,

(b) Fy: komponen menegak / vertical component,

sin θ = Fy

F kos θ =

Fx

FFy = F sin θ Fx = F kos θ

Fy

θ

θ

F

Fx

Latihan/Exercises

1 Dapatkan komponen mengufuk dan komponen menegak daya tersebut. (Perhatian: Pertimbangkan magnitud daya yang positif sahaja.)Find the horizontal component and the vertical component of the force. (Note: Consider only the positive magnitudes of the forces.)(a) (b)

(c) (d)

Fx

75 N = sin 70°

∴ Fx = 75 N sin 70° = 70.48 N

FY

75 N = kos 70°

∴ FY = 75 N kos 70° = 25.65 N

Fx

6 N = kos 60°

∴ Fx = 6 N kos 60° = 3.0 N

FY

6 N = sin 60°

∴ FY = 6 N sin 60° = 5.20 N

Fx

200 N = kos 50°

∴ Fx = 200 N kos 50° = 128.6 N

FY

200 N = sin 50°

∴ FY = 200 N sin 50° = 153.2 N

Fx

5 N = sin 40°

∴ Fx = 5 N sin 40° = 3.21 N

FY

5 N = kos 40°

∴ FY = 5 N kos 40° = 3.83 N

Fy

Fx

200 N

500

500

400400

Fx

Fy

F = 5 N

0

Fy

Fx

700700

75 N

600

600

Fy

Fx

F = 6 N

0




2

UNIT

2 Rajah menunjukkan troli yang berjisim 2 kg di atas permukaan licin, ditarik oleh daya 6 N pada sudut 60° dengan ufuk.The diagram shows a trolley of mass 2 kg on a smooth surface being pulled by a force of 6 N at an angle of 60° with the horizontal.

(a) Berapakah komponen mengufuk daya itu?What is the horizontal component of the force?

(b) Berapakah pecutan troli itu?/What is the acceleration of the trolley?


(a) Fx

6 N = kos 60° (b) F = ma, a = 3 N

2 kg = 1.5 m s-2

∴ Fx = 6 N kos 60° = 3.0 N

3 Rajah menunjukkan sebuah kereta sedang ditunda. Kabel itu mempunyai daya F, 5 000 N. The diagram shows a car being towed. The cable has a force F of 5 000 N.

(a) Tunjukkan dan labelkan/Indicate and label: • dayaF/the force F • dayakomponenmengufuk, Fx /the horizontal component force Fx

• dayakomponenmenegak,Fy /the vertical component force Fy

(b) Cari/Find • dayamengufukkabelyangmenggerakkankeretakehadapan.

the horizontal force of the cable which moves the car forward. • dayamenegakkabel./the vertical force of the cable.

Penyelesaian/Solution(a)

Fy

Fx

60°

F = 5 000 N

(b) Fx

5 000 N = kos 60°

∴ Fx = 5 000 N kos 60° = 2 500 N

FY

5 000 N = sin 60°

∴ FY = 5 000 N sin 60° = 4 330 N

4 Seorang pelancong menarik begnya dengan daya 100 N pada sudut 55° dari garis mengufuk. A tourist pulls his bag with a force of 100 N at an angle of 55° with the horizontal.

(a) Tunjukkan dan labelkan/Indicate and label: • dayaF/the force F • dayakomponenmengufuk, Fx /the horizontal component force Fx

• dayakomponenmenegak,Fy /the vertical component force Fy

(b) Cari/Find • dayamengufukbegyangmenggerakkannyakehadapan.

the horizontal force of the cable which moves it forward. • dayamenegakbeg./the vertical force of the bag.

Penyelesaian/Solution(a)

Fy

Fx

55°

F = 100 N

(b) Fx

100 N = kos 55°

∴ F = 100 N kos 55° = 57.36 N

FY

100 N = sin 55°

∴ FY = 100 N sin 55° = 81.92 N

600

6 N

Fx

2 kg

Permukaan licin/Smooth surface

600


Documents

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