F T 4 - yuhuaphysics.comFSY4p).pdf · • Ini adalah disebabkan titisan air pada permukaan payung itu bergerak secara serentak apabila payung itu dipusingkan. This is because the

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MODUL • Fizik TINGKATAN 4

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.

• Didapatibaldikosongitu 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-dua baldi diayun dan diberhentikan, baldi yang diisi dengan pasir lebih susah untuk 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.

• Inimenunjukkanbaldidenganjisimyanglebihbesarmenghasilkanrintanganyanglebihuntukberubahdarikeadaan 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

MEMAHAMI INERSIAUNDERSTANDING INERTIA2.3

Fizik Tg4 B2A 2015(FSY4p).indd 64 10/20/15 2:01 PM

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Aktiviti yang melibatkan inersiaActivities involving inertia

Apabila sekeping duit syiling 20 sen dikuis ke arah timbunan duitsyiling20senpadapermukaanyanglicin,duitsyilingdi 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.

Apabilakadbodditarikkeluardengancepat,duitsyilingituterusjatuh

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.

Letakkan segelas air di atas sekeping kertasA4. Dengan cepat tarikkeluar kertas itu secaramendatar.Apakah yang akan berlaku kepadagelas 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

Blok kayuWooden block

Sebuahblokkayudiletakkandiatassebuahtroliyangbergerakmenurunilandasan. Apabila gerakan troli itu dihalang oleh suatu penghalang,blok kayu itu akan kekal dalam keadaan 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 inersiaExamples 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?Apabila bas itu bergerak ke depan secara tiba-tiba dari rehat, 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 ituberhentisecaratiba-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 sedang bergerak. Apabila bas itu berhenti 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 inersiaActivities involving inertia

Sebuahbukuditarikkeluardarikedudukantengahnya.Bukudiatasnya

akan jatuh ke bawah secara terus . Inersia cuba menentang perubahannya dari keadaan rehat, iaitu, apabila buku ditarik keluar,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

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

• Sosdalambotol bergerak bersama-samadenganbotolsemasapergerakankebawah.

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

Kepalatukuldicantumdenganketatkepadapemegangnyadenganmengetukpenghujungpemegangnya,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.

• Inimenyebabkankepalatukulmeneruskangerakan ke bawah apabila gerakan pemegang itudiberhentikan.Dengan ini, hujung ataspemegang itu akandimasukkan lebihdalamkedalam 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.

• Apabilapayungituberhentimemusing, 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.Apabilakeretaberhentisecaramendadak,

tali pinggang itu mengelakkan pemandu daripada terhumban ke 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.Inersiakepalacenderunguntuk

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) Beg udara dipasang di dalam stereng. Ia membekalkan kusyen

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 kepada bahagian-bahagian lori yang tertentu supaya apabila loribergerak atau berhenti dengan tiba-tiba, perabot itu tidak akan jatuhatau 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 disebabkan oleh inersiajikaloritangkiituberhentidengantiba-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


Sebuahkapalminyakyangbesarmengambilmasayanglebihpanjanguntukmemecutkepadalajumaksimumnyadan 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 dengan 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 dimanipulasikanManipulated variable Jisim plastisin

Mass of plasticine

Pemboleh ubah bergerak balasResponding variable Tempoh ayunan

Period of oscillation

Pemboleh ubah dimalarkanConstant variable: Panjang bilah Hacksaw

Length of the Hacksaw blade

ProsedurProcedure

1 Letakkan sejumlah plastisin (berbentuk sfera) dengan jisim 30 g pada hujung bilahHacksaw.Place a lump of plasticine (sphere-shaped) with a mass of 30 g at the free end of the Hacksaw blade.

2 SesarkansedikitbilahHacksawdanlepaskannyasupayaiaberayunsecaramengufuk.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 = t

10 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 =

t10

st1 t2 tmin

30

40

50

60

70

Pengapit-G/G-clamp

Bilah HacksawHacksaw blade

Plastisin/Plasticine

Inersia dan Jisim / Inertia and MassEksperimenExperiment


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AnalisisAnalysis

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

T/s

m/g

PerbincanganDiscussion

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.

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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 .

Nyatakan prinsip keabadian momentumState 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

MENGANALISIS MOMENTUMANALYSING MOMENTUM2.4

Pemain IPlayer I

Pemain IIPlayer II

Dalampermainanragbi,seorangpemainberjisim70kgbergerakdenganhalaju4ms-1 dan seorang pemain yang lain yang berjisim 75kgbergerakdengan3ms-1 menghala antara satu sama seperti yang ditunjukkan.Hitungkanmomentumkedua-dua pemain itumasing-masing.In a rugby game, a player of mass 70 kg is moving with velocity of 4 m s–1 and the other player of mass 75 kg is moving 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

Contoh/Examples 1

Nenek (m = 80 kg) menggelongsor sekeliling gelanggang gelongsordengan halaju 6m s–1. Tiba-tiba dia berlanggar denganBobby (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)

NenekGranny Bobby

Contoh/Examples 2


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Aktiviti/Activity 1

Aktiviti/Activity 2 Aktiviti/Activity 3

Rajahdisebelahmenunjukkanduaorangadik-beradikyangsedangmenggelongsor.Abang bergerak dan berlanggar dengan adiknya yang berada dalam keadaan rehat.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.

Menjentik sekeping duit syiling 20 sen,A secara terus kepadasekepingduitsyiling20sen,B yang lain.Flick a 20-cent coin, A, directly to another 20-cent coin, B.

A B

(a) Apakah yang berlaku kepada gerakan kedua-duaduit syiling selepas perlanggaran?

What happens to the motion of both coins after collision?

DuitsyilingAberhenti,duitsyilingB bergerak.

Coin A stops, coin B moves.

(b) Apakahyangberlakukepadamomentumduitsyiling 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.

Menjentik sekeping duit syiling 20 sen,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) ApakahyangberlakukepadamomentumduitsyilingA 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.


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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

Seorang budak perempuan berdiri dalam keadaan rehat di atas papan luncur. Diamembalingkanbolakehadapan.Bolaitubergerakkekiri.Budakperempuanbergerakke 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.


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Perlanggaran kenyalElastic collision

Perlanggaran tak kenyalInelastic 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

LetupanExplosion

Troli pegunStationary trolleys

m2 m1

m2 m1

v2 v1

pin

Sebelum letupan/Before explosion

Selepas letupan/After explosion

Sebelumletupan,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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(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 keabadian 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 rifle = m1 Jisim peluru/Mass of bullet = m2

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

• Apabilasepucuksenapangditembak,peluruyangberjisimm2 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 keabadianmomentum, suatumomentumyang 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 roketThe launching of rocket• Campuran bahan api hidrogen dan oksigen terbakar dengan letupan

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.

• Kelajuantinggigaspanasinimenghasilkanmomentumyangbesar ke

bawah .

This high-speed hot gas produces a large momentum downwards .• 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 keabadian momentum linearDescribe applications of the principle of conservation of linear momentum

Gas panasHot gas

Jet

Aplikasienjinjet:Application in the jet engine:• Suatugaspanasyangberkelajuantinggidipancutkeluardaribelakang

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

yangtinggi.Airmasukmelaluipembukaanyangbesardankeluarmelalui

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

Sebijibolayangberjisim5kgdibalingkanpadahalaju20kmj–1kepadaLilyyangberjisim60kgpadakeadaanrehat 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)

Contoh/Examples 1

Contoh/Examples 2


2

UNIT



Sebuahtrakyangberjisim1200kgbergerakpada30ms–1berlanggardengansebuahkeretayangberjisim1000kgyangbergerakdalamarahbertentanganpada20ms–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

Seorangmenembaksepucukpistolyangberjisim1.5kg.Jikapeluruituberjisim10gdanmempunyaihalaju300m s–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

Contoh/Examples 3

Contoh/Examples 4


2

UNIT



Bayangkan anda berlegar di sebelah kapal angkasapada orbit bumi dan rakan anda yang sama jisim bergerak dengan 4 km j–1 (dengan merujuk kepada kapal angkasa) melanggar anda. Jika dia memegang anda, berapakah kelajuan anda bergerak(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

Penyelesaian/SolutionJumlah momentum sebelum perlanggaranTotal momentum before collision

=

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, berapakahhalaju 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

Penyelesaian/SolutionJumlah momentum sebelum perlanggaranTotal momentum before collision

=

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.

•Bola ituakanbergerak 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.

• Ini adalah disebabkan seorang pemain bola sepak yang berlari mempunyai momentum yang 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.

Contoh/Examples 5 Contoh/Examples 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,pitaselofan,troli,landasan,bekalankuasaa.u.12V.Ticker timer, ticker tape, plasticine, cellophane tape, trolleys, runway, 12 V ac power supply.

ProsedurProcedure

1 Dirikansatulandasandenganmengubahsuaikecerunannyasupayalandasanterpampasgeserandi 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.

Pita detikTicker tape

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-langkah2–5diulangidenganjisimP dan jisim Qyangberbezasepertiditunjukkandalam 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

Keabadian Momentum / Conservation of MomentumEksperimenExperiment


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

Kuantiti fizikPhysical quantity



Panjang10detik10-tick length

18.4cm 9.2cm

Masadiambiluntuk10detikTime taken for 10 ticks

0.2s 0.2s

HalajuVelocity

18.4cm0.2s =0.92ms–1 9.2cm

0.2s =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

PerbincanganDiscussion

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)

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 ,iaakanmembatalkanantara

satusamalain.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]ataubergerakpada 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

Cawanituberadadalamkeadaanrehat.Dayabersih

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

Kapalterbangbergerakdenganhalajumalar.Dayabersih

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:Seretan/Drag

MEMAHAMI KESAN DAYAUNDERSTANDING THE EFFECTS OF A FORCE2.5


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,bagisuatuobjekberkadarsongsangdenganjisimnya,m/The 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

Duaorangpemudamenolakjisimyangsamatetapi 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 mengenakan daya yangsama. 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

Semakin besar daya, semakin besar pecutan.

The larger the force, the greater the

acceleration.

Semakin besar jisim, semakin kecil pecutan.

The greater the mass, the smaller the

acceleration.

InferensInference

Pecutan,a,bergantungkepadadayayangdikenakan,FThe acceleration, a, depends on applied force, F

Pecutan,a,bergantungkepadajisimyangdikenakan,mThe acceleration, a, depends on applied mass, m

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

EksperimenExperiment


2

UNIT



Pemboleh ubah dimanipulasikanManipulated 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

Pita detikTicker tape

Jangka masa detikTicker timer

Landasan terpampasgeseran

Friction-compensatedrunway

Tali tak kenyalInelastic string

Blok kayuWooden block

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 berjisim1.0 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,pecutantrolidihitung denganmenggunakan formula, a

= (v – u)

tFrom 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 Sebuah troli dengan jisim , m = 1.0 kgdiletakkan 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 pemberat iniialah10N)The ticker timer is switched on and the trolley is pulled by a weight of constant force, 10 N

4 Daripitadetikyangdiperoleh,pecutantrolidihitung denganmenggunakan 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-langkah 2 – 4 diulangi denganmenambahkan 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-langkah 2 – 4 diulangi denganmelekat pemberat berslot pada troli supaya jisim troli,m=1.5kg,2.0kg,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/kg

Mass, m/kg

Jisim songsang, 1

m /kg–1

Inverse of mass, 1m /kg–1

Pecutan, a/cm 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

1jisim ,

1m /kg–1

1mass ,

1m /kg–1

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 =

8.0N2kg

a = (18–2)N10kg

= 16N10kg

=4ms-2/4Nkg-1 =1.6ms-2/1.6Nkg-1

(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-2/2.5Nkg-1 = 0ms-2


2

UNIT



Menyelesaikan masalah menggunakan F = maSolve problems using F = ma

2 Seoranglelakimenolaktroliyangberisikotak(jumlahjisim5kg)diataspermukaanyanglicin.Jikadiamenggunakandaya30Nuntukmenolaktroliitu,apakahmagnituddan arah pecutan troli itu?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 Sebuah bas berjisim 2 000 kg bergerak dengan halaju seragam 40 m s-1 sejauh 2 500 m sebelum berehat.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

KBAT


2

UNIT



Menerangkan daya impulsExplain what an impulsive force is

Dayayangbesaryangbertindakdalam tempohmasayang singkat semasaperlanggaranatau letupandikenali

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 – u

t)

F = mv – mu

t = 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 impulsDefine 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 – mu

t, Ft = mv – mu = perubahan momentum = impuls

From F = mv – mu

t, 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 momentum

Time.

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

MENGANALISIS IMPULS DAN DAYA IMPULSANALYSING IMPULSE AND IMPULSIVE FORCE2.6


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