FizikPhysics

Print questions onlyCetak soalan sahaja Print with answersCetak dengan skema Print answer key onlyCetak kunci jawapan sahaja Practice shows answers as you go. Exam times you and hides answers until you submit.Latihan menunjukkan jawapan semasa menjawab. Peperiksaan menetapkan masa dan menyembunyikan jawapan sehingga dihantar.

1

QuestionSoalan 1

Impuls ialah hasil darab daya dengan masa, J = Ft. Apakah unit impuls dalam unit asas S.I.?Impulse is the product of force and time, J = Ft. What is the unit of impulse in S.I. base units?

• Akg m s⁻²
• Bkg m s⁻¹
• Ckg m² s⁻¹
• Dkg m² s⁻²

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B

The unit of impulse is N s. With N = kg m s⁻², N s = kg m s⁻² × s = kg m s⁻¹, which is also the unit of momentum.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 1, standard 1.1.3

TextbookBuku teks p.ms. 7

2

QuestionSoalan 2

Sebuah kereta dilanggar dari belakang oleh sebuah kereta lain. Mengapakah penyandar kepala dipasang pada setiap tempat duduk kereta?A car is hit from behind by another car. Why is a headrest fitted to every car seat?

• AUntuk menaikkan kedudukan duduk penumpang supaya jalan raya kelihatan lebih jelas.To raise the sitting position of the passenger so that the road is seen more clearly.
• BUntuk mengurangkan momentum kepala penumpang semasa perlanggaran dari hadapan.To reduce the momentum of the passenger's head in a collision from the front.
• CUntuk menghalang kepala tersentak ke belakang kerana inersia kepala itu.To stop the head from jerking backwards because of the inertia of the head.
• DUntuk menambah daya impuls yang bertindak pada kepala penumpang.To increase the impulsive force acting on the passenger's head.

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C

When the car is hit from behind, the body moves forward with the seat but the head, because of its inertia, tends to stay at rest and so is thrown backwards relative to the body. The headrest supports the head and prevents injury to the neck.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.4

TextbookBuku teks p.ms. 56

3

QuestionSoalan 3

Dalam sebuah taman permainan, kereta bom P berjisim 150 kg bergerak pada 4.0 m s⁻¹ dan melanggar kereta bom Q berjisim 100 kg yang bergerak pada 1.0 m s⁻¹ dalam arah yang sama. Selepas perlanggaran, kedua-dua kereta bergerak bersama-sama. Berapakah halaju bersama kereta itu?In a playground, bumper car P of mass 150 kg moving at 4.0 m s⁻¹ collides with bumper car Q of mass 100 kg moving at 1.0 m s⁻¹ in the same direction. After the collision the two cars move together. What is their common velocity?

• A2.8 m s⁻¹
• B2.4 m s⁻¹
• C2.5 m s⁻¹
• D5.0 m s⁻¹

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A

Conservation of momentum: 150(4.0) + 100(1.0) = (150 + 100)v, so v = 700 ÷ 250 = 2.8 m s⁻¹. 2.4 m s⁻¹ ignores the momentum of Q; 2.5 m s⁻¹ is the average of the two speeds.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 2, standard 2.5

TextbookBuku teks p.ms. 62

4

QuestionSoalan 4

Tikar getah yang tebal diletakkan di atas tanah di bawah buaian di sebuah taman permainan. Bagaimanakah tikar itu mengurangkan kecederaan apabila seorang kanak-kanak jatuh ke atasnya?A thick rubber mat is placed on the ground beneath a playground swing. How does the mat reduce injury when a child falls on it?

• AIa mengurangkan perubahan momentum kanak-kanak itu.It decreases the change in momentum of the child.
• BIa menambah impuls yang bertindak pada kanak-kanak itu.It increases the impulse acting on the child.
• CIa memendekkan masa hentaman supaya daya menjadi lebih kecil.It shortens the time of impact so that the force is smaller.
• DIa memanjangkan masa hentaman, maka daya impuls menjadi lebih kecil.It lengthens the time of impact, so the impulsive force is smaller.

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D

The child's change of momentum is the same with or without the mat. F = (mv − mu) ÷ t, so a longer impact time on the soft mat gives a smaller impulsive force.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.7

TextbookBuku teks p.ms. 68

5

QuestionSoalan 5

Rajah 1 menunjukkan graf daya-masa bagi tendangan seorang pemain bola sepak pada sebiji bola pegun berjisim 0.40 kg. Berapakah laju bola sejurus selepas ditendang?Diagram 1 shows the force-time graph of a footballer's kick on a stationary ball of mass 0.40 kg. What is the speed of the ball just after the kick?

• A30 m s⁻¹
• B15 m s⁻¹
• C7.5 m s⁻¹
• D750 m s⁻¹

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B

Impulse = area under the force-time graph = ½ × 0.040 × 300 = 6.0 N s. Impulse = change of momentum = mv, so v = 6.0 ÷ 0.40 = 15 m s⁻¹. 30 m s⁻¹ treats the area as a rectangle; 750 m s⁻¹ is F ÷ m.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 2, standard 2.7

TextbookBuku teks p.ms. 68

6

QuestionSoalan 6

Sebuah batu mempunyai berat 49 N di Bumi, tempat medan graviti, g = 9.8 N kg⁻¹. Batu itu dibawa ke sebuah planet yang mempunyai g = 3.7 N kg⁻¹. Berapakah jisim dan berat batu itu di planet tersebut?A rock has a weight of 49 N on the Earth, where the gravitational field strength is g = 9.8 N kg⁻¹. The rock is taken to a planet where g = 3.7 N kg⁻¹. What are the mass and the weight of the rock on that planet?

• A5.0 kg; 49 N
• B13 kg; 49 N
• C13 kg; 18.5 N
• D5.0 kg; 18.5 N

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D

Mass does not change with place: m = 49 ÷ 9.8 = 5.0 kg. Weight on the planet W = mg = 5.0 × 3.7 = 18.5 N.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.8

TextbookBuku teks p.ms. 70

7

QuestionSoalan 7

Satelit S mengorbit Bumi dalam orbit bulatan berjejari r pada laju orbit v. Satelit kedua, T, mengorbit Bumi dalam orbit bulatan berjejari 4r. Berapakah laju orbit satelit T?Satellite S orbits the Earth in a circular orbit of radius r at an orbital speed v. A second satellite, T, orbits the Earth in a circular orbit of radius 4r. What is the orbital speed of satellite T?

• Av/4
• B2v
• Cv/2
• D4v

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C

The gravitational force provides the centripetal force, so v = √(GM/r) and v ∝ 1/√r. For radius 4r, v' = v ÷ √4 = v/2. v/4 wrongly takes v ∝ 1/r; 2v inverts the relationship.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 3, standard 3.3.3

TextbookBuku teks p.ms. 107

8

QuestionSoalan 8

Planet Y mempunyai jisim lapan kali jisim Bumi dan jejari dua kali jejari Bumi. Kekuatan medan graviti di permukaan Bumi ialah 9.8 N kg⁻¹. Berapakah kekuatan medan graviti di permukaan planet Y?Planet Y has a mass eight times the mass of the Earth and a radius twice the radius of the Earth. The gravitational field strength at the surface of the Earth is 9.8 N kg⁻¹. What is the gravitational field strength at the surface of planet Y?

• A19.6 N kg⁻¹
• B39.2 N kg⁻¹
• C78.4 N kg⁻¹
• D4.9 N kg⁻¹

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A

g = GM/R², so g ∝ M/R². The ratio is 8 ÷ 2² = 2, so g = 2 × 9.8 = 19.6 N kg⁻¹. 39.2 N kg⁻¹ divides by 2 instead of 2² and 78.4 N kg⁻¹ ignores the radius.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 3, standard 3.1.2

TextbookBuku teks p.ms. 82

9

QuestionSoalan 9

Seketul batu berjisim 0.50 kg dipusing dalam bulatan mengufuk berjejari 0.80 m di atas sebuah meja licin dengan seutas tali yang boleh menahan tegangan maksimum 20 N. Berapakah laju maksimum batu itu?A stone of mass 0.50 kg is whirled in a horizontal circle of radius 0.80 m on a smooth table by a string that can withstand a maximum tension of 20 N. What is the maximum speed of the stone?

• A4.0 m s⁻¹
• B7.1 m s⁻¹
• C5.7 m s⁻¹
• D32 m s⁻¹

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C

The tension provides the centripetal force: F = mv²/r, so v² = Fr/m = (20 × 0.80) ÷ 0.50 = 32 and v = 5.7 m s⁻¹. 32 m s⁻¹ is v² (the square root was not taken).

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 3, standard 3.1.3

TextbookBuku teks p.ms. 83

10

QuestionSoalan 10

Sebuah termometer diletakkan dalam secawan air panas. Selepas beberapa ketika, bacaan termometer itu menjadi tetap. Apakah yang ditunjukkan oleh keadaan ini?A thermometer is placed in a cup of hot water. After some time the reading of the thermometer becomes steady. What does this show?

• ATermometer dan air itu mempunyai jisim yang sama.The thermometer and the water have the same mass.
• BTermometer dan air itu mengandungi kuantiti haba yang sama.The thermometer and the water contain the same quantity of heat.
• CSemua haba air telah berpindah kepada termometer.All the heat of the water has been transferred to the thermometer.
• DTermometer dan air berada dalam keseimbangan terma, tanpa pengaliran haba bersih antara kedua-duanya.The thermometer and the water are in thermal equilibrium, with no net flow of heat between them.

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D

Thermal equilibrium is reached when the net rate of heat flow between two objects in contact is zero; they are then at the same temperature. Objects in equilibrium need not have the same mass or contain the same quantity of heat.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.1

TextbookBuku teks p.ms. 121

11

QuestionSoalan 11

Udara di dalam sebuah tayar kereta mempunyai tekanan 220 kPa pada 27 °C. Selepas perjalanan yang jauh, suhu udara di dalam tayar itu menjadi 57 °C. Anggap isi padu tayar tetap. Berapakah tekanan udara itu?The air in a car tyre has a pressure of 220 kPa at 27 °C. After a long journey the temperature of the air in the tyre becomes 57 °C. Assume the volume of the tyre is constant. What is the pressure of the air?

• A242 kPa
• B250 kPa
• C464 kPa
• D200 kPa

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A

Gay-Lussac's law: P₁/T₁ = P₂/T₂ with T in kelvin: T₁ = 300 K, T₂ = 330 K. P₂ = 220 × 330 ÷ 300 = 242 kPa. 464 kPa uses the Celsius values (57 ÷ 27); 250 kPa adds the 30 °C rise to the pressure.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 4, standard 4.4.4

TextbookBuku teks p.ms. 161

12

QuestionSoalan 12

Pada waktu siang, daratan menjadi lebih panas daripada laut walaupun kedua-duanya menerima haba yang sama daripada Matahari. Mengapakah keadaan ini berlaku?During the day, the land becomes hotter than the sea although both receive the same amount of heat from the Sun. Why does this happen?

• ALaut mempunyai muatan haba tentu yang lebih rendah daripada daratan.The sea has a lower specific heat capacity than the land.
• BLaut mempunyai muatan haba tentu yang lebih tinggi daripada daratan.The sea has a higher specific heat capacity than the land.
• CLaut mempunyai haba pendam tentu pelakuran yang lebih rendah daripada daratan.The sea has a lower specific latent heat of fusion than the land.
• DLaut memantulkan semua haba Matahari.The sea reflects all the heat from the Sun.

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B

Water has a high specific heat capacity, so a large amount of heat is needed to raise its temperature by 1 °C. The land, with a lower specific heat capacity, heats up faster and becomes hotter during the day.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.2

TextbookBuku teks p.ms. 136

13

QuestionSoalan 13

Rajah 2 menunjukkan graf sesaran-masa bagi sebuah gabus yang terapung di atas air kolam apabila riak air melaluinya. Riak itu bergerak dengan laju 0.60 m s⁻¹. Berapakah panjang gelombang riak itu?Diagram 2 shows the displacement-time graph of a cork floating on the water of a pond as ripples pass it. The ripples travel at a speed of 0.60 m s⁻¹. What is the wavelength of the ripples?

• A0.40 m
• B1.8 m
• C1.2 m
• D0.90 m

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D

From the graph, two oscillations take 3.0 s, so the period T = 1.5 s. λ = vT = 0.60 × 1.5 = 0.90 m. 0.40 m is v ÷ T; 1.8 m uses the whole 3.0 s; 1.2 m multiplies v by 2 s.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 5, standard 5.1

TextbookBuku teks p.ms. 177

14

QuestionSoalan 14

Muncung sebuah picagari ditutup rapat dan penyorongnya ditolak masuk perlahan-lahan. Mengapakah tekanan udara yang terperangkap di dalamnya bertambah?The nozzle of a syringe is sealed and its plunger is pushed in slowly. Why does the pressure of the air trapped inside increase?

• AMolekul udara bergerak lebih laju kerana suhu udara naik.The air molecules move faster because the temperature of the air rises.
• BMolekul udara berlanggar dengan dinding lebih kerap kerana isi padu berkurang.The air molecules collide with the walls more often because the volume decreases.
• CBilangan molekul udara di dalam picagari bertambah.The number of air molecules in the syringe increases.
• DMolekul udara menjadi lebih besar apabila dimampatkan.The air molecules become larger when compressed.

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B

At constant temperature and mass (Boyle's law), a smaller volume means more molecules per unit volume, so the molecules hit the walls more frequently and the pressure is greater. The speed and the number of molecules do not change.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.4.2

TextbookBuku teks p.ms. 151

15

QuestionSoalan 15

Gelombang air bergerak dari air dalam ke air cetek berhampiran dinding pelabuhan. Dalam air dalam, lajunya 4.0 m s⁻¹ dan panjang gelombangnya 10 m. Panjang gelombang dalam air cetek ialah 6.0 m. Berapakah frekuensi gelombang dalam air cetek?Water waves move from deep water into shallow water near a harbour wall. In the deep water the speed is 4.0 m s⁻¹ and the wavelength is 10 m. The wavelength in the shallow water is 6.0 m. What is the frequency of the waves in the shallow water?

• A0.24 Hz
• B0.67 Hz
• C0.40 Hz
• D1.7 Hz

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C

The frequency does not change when waves are refracted. f = v ÷ λ = 4.0 ÷ 10 = 0.40 Hz in the deep water, and it is the same in the shallow water. 0.24 Hz is the speed in shallow water (2.4 m s⁻¹) divided by 10 m; 0.67 Hz divides the deep-water speed by the shallow-water wavelength.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 5, standard 5.4

TextbookBuku teks p.ms. 198

16

QuestionSoalan 16

Sebuah radar di atas kapal menggunakan gelombang mikro berpanjang gelombang 3.0 cm. Berapakah frekuensi gelombang mikro itu? [Laju gelombang elektromagnet = 3.0 × 10⁸ m s⁻¹]A radar on a ship uses microwaves of wavelength 3.0 cm. What is the frequency of the microwaves? [Speed of electromagnetic waves = 3.0 × 10⁸ m s⁻¹]

• A1.0 × 10¹⁰ Hz
• B9.0 × 10⁶ Hz
• C1.0 × 10⁸ Hz
• D1.0 × 10¹² Hz

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A

λ = 3.0 cm = 0.030 m. f = v ÷ λ = 3.0 × 10⁸ ÷ 0.030 = 1.0 × 10¹⁰ Hz. 1.0 × 10⁸ Hz comes from using 3.0 instead of 0.030 m.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.7

TextbookBuku teks p.ms. 221

17

QuestionSoalan 17

Gentian optik digunakan dalam telekomunikasi untuk membawa isyarat cahaya merentasi jarak yang jauh. Mengapakah cahaya tidak keluar daripada gentian itu?Optical fibres are used in telecommunication to carry light signals over long distances. Why does the light not escape from the fibre?

• ATeras mempunyai indeks biasan lebih rendah daripada selongsong, maka cahaya dibengkokkan menghampiri garis normal.The core has a lower refractive index than the cladding, so the light bends towards the normal.
• BCahaya diserap oleh selongsong dan dipancarkan semula di sepanjang gentian.The light is absorbed by the cladding and re-emitted along the fibre.
• CCahaya bergerak lurus di tengah teras tanpa menyentuh selongsong.The light travels straight along the middle of the core without touching the cladding.
• DTeras mempunyai indeks biasan lebih tinggi daripada selongsong, maka cahaya mengalami pantulan dalam penuh berulang kali.The core has a higher refractive index than the cladding, so the light undergoes total internal reflection again and again.

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D

The core (higher refractive index) is surrounded by cladding (lower refractive index). Light meets the boundary at an angle greater than the critical angle and is totally internally reflected repeatedly along the fibre.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.2.3

TextbookBuku teks p.ms. 249

18

QuestionSoalan 18

Seekor ikan di dalam air pelabuhan dilihat dari atas pada kedalaman ketara 1.2 m. Indeks biasan air ialah 1.33. Berapakah kedalaman sebenar ikan itu?A fish in the water of a harbour is seen from directly above at an apparent depth of 1.2 m. The refractive index of water is 1.33. What is the real depth of the fish?

• A0.90 m
• B1.6 m
• C1.2 m
• D2.4 m

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B

n = real depth ÷ apparent depth, so real depth = 1.33 × 1.2 = 1.6 m. 0.90 m divides by n instead of multiplying.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 6, standard 6.1.5

TextbookBuku teks p.ms. 237

19

QuestionSoalan 19

Sebuah kanta cembung membentuk imej nyata pada skrin yang tiga kali ganda lebih besar daripada objek. Skrin itu berada pada jarak 24 cm dari kanta. Berapakah jarak fokus kanta itu?A convex lens forms a real image on a screen that is three times as large as the object. The screen is 24 cm from the lens. What is the focal length of the lens?

• A6.0 cm
• B8.0 cm
• C16 cm
• D32 cm

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A

m = v/u = 3, so u = 24 ÷ 3 = 8.0 cm. 1/f = 1/u + 1/v = 1/8.0 + 1/24 = 4/24, so f = 6.0 cm. 8.0 cm is the object distance; 16 cm is v − u.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 6, standard 6.4.2

TextbookBuku teks p.ms. 261

20

QuestionSoalan 20

Pernyataan manakah yang betul tentang kanta dan imej akhir bagi sebuah mikroskop majmuk?Which statement about the lenses and the final image of a compound microscope is correct?

• AKanta objektif mempunyai jarak fokus lebih panjang daripada kanta mata; imej akhir ialah imej nyata dan songsang.The objective lens has a longer focal length than the eyepiece lens; the final image is real and inverted.
• BKanta objektif mempunyai jarak fokus lebih panjang; imej akhir ialah imej maya, tegak dan dibesarkan.The objective lens has a longer focal length; the final image is virtual, upright and magnified.
• CKanta objektif mempunyai jarak fokus lebih pendek daripada kanta mata; imej akhir ialah imej maya, songsang dan dibesarkan.The objective lens has a shorter focal length than the eyepiece lens; the final image is virtual, inverted and magnified.
• DKedua-dua kanta mempunyai jarak fokus yang sama; imej akhir ialah imej nyata, songsang dan dikecilkan.The two lenses have the same focal length; the final image is real, inverted and diminished.

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C

In a compound microscope the focal length of the objective, fo, is less than that of the eyepiece, fe. The objective forms a real, inverted, magnified first image, which the eyepiece (acting as a magnifying glass) turns into a final image that is virtual, magnified and inverted compared with the object.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, Chapter 6, standard 6.5.2

TextbookBuku teks p.ms. 266

21

QuestionSoalan 21

Seorang murid mengkaji gerakan sebiji bola yang dilepaskan dari keadaan pegun menuruni sebuah model gelongsor taman permainan. Dia mengukur masa, t, bola itu bergerak sejauh s di sepanjang gelongsor. Jadual 1 menunjukkan keputusannya dan Rajah 3 menunjukkan graf t² melawan s.

s / m	0.20	0.40	0.60	0.80	1.00
t / s	0.63	0.89	1.10	1.26	1.41
t² / s²	0.40	0.79	?	1.59	1.99

Jadual 1A pupil investigates the motion of a ball released from rest down a model of a playground slide. She measures the time, t, taken by the ball to travel a distance s along the slide. Table 1 shows her results and Diagram 3 shows the graph of t² against s.

s / m	0.20	0.40	0.60	0.80	1.00
t / s	0.63	0.89	1.10	1.26	1.41
t² / s²	0.40	0.79	?	1.59	1.99

Table 1

[2 marksmarkah] (a)Nyatakan pemboleh ubah dimanipulasikan dan pemboleh ubah bergerak balas.State the manipulated variable and the responding variable.

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• Manipulated: distance travelled along the slide, s [1]1
• Responding: time taken, t (or t²) [1]1

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[1 markmarkah] (b)Hitung t² apabila s = 0.60 m.Calculate t² when s = 0.60 m.

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• 1.10² = 1.21 s²1

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[1 markmarkah] (c)Berdasarkan Rajah 3, nyatakan hubungan antara t² dengan s.Based on Diagram 3, state the relationship between t² and s.

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• t² is directly proportional to s (a straight line through the origin)1

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[2 marksmarkah] (d)Hitung kecerunan graf itu dan nyatakan unitnya.Calculate the gradient of the graph and state its unit.

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• Gradient = (2.0 − 0) ÷ (1.00 − 0) [1]1
• = 2.0 s² m⁻¹ [1]1

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[2 marksmarkah] (e)Bagi gerakan dari keadaan pegun, s = ½at². Tunjukkan bahawa kecerunan graf ialah 2/a dan hitung pecutan bola itu.For motion from rest, s = ½at². Show that the gradient of the graph is 2/a and calculate the acceleration of the ball.

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• t² = (2/a)s, so gradient = 2/a [1]1
• a = 2 ÷ 2.0 = 1.0 m s⁻² [1]1

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[1 markmarkah] (f)Ramalkan masa yang diambil oleh bola itu untuk bergerak sejauh 1.50 m.Predict the time taken by the ball to travel 1.50 m.

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• t² = 2.0 × 1.50 = 3.0 s², so t = 1.73 s (accept 1.7 s)1

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[1 markmarkah] (g)Mengapakah masa bagi setiap jarak diukur beberapa kali dan puratanya digunakan?Why is the time for each distance measured several times and the average used?

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• To reduce the effect of random error (reaction time in starting and stopping the stopwatch) and obtain a more accurate value1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.2.2

TextbookBuku teks p.ms. 12

22

QuestionSoalan 22

Dalam sebuah pameran sains, seorang murid menggunakan sebuah kanta cembung untuk membentuk imej nyata sebatang lilin setinggi 2.5 cm pada skrin. Dia mengubah jarak objek, u, dan mengukur jarak imej, v, pada skrin. Jadual 2 menunjukkan keputusannya.

u / cm	18	24	36	60
v / cm	36	24	18	15

Jadual 2At a science fair, a pupil uses a convex lens to form real images of a candle 2.5 cm tall on a screen. She changes the object distance, u, and measures the image distance, v, on the screen. Table 2 shows her results.

u / cm	18	24	36	60
v / cm	36	24	18	15

Table 2

[2 marksmarkah] (a)Nyatakan dua ciri imej pada skrin apabila u = 18 cm.State two characteristics of the image on the screen when u = 18 cm.

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Any 2 pointsMana-mana 2 isi

• Real1
• Inverted1
• Magnified (any two) [1 each]1

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[2 marksmarkah] (b)Hitung jarak fokus kanta itu menggunakan u = 18 cm dan v = 36 cm.Calculate the focal length of the lens using u = 18 cm and v = 36 cm.

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• 1/f = 1/u + 1/v = 1/18 + 1/36 [1]1
• f = 12 cm [1]1

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[2 marksmarkah] (c)Hitung pembesaran linear dan tinggi imej apabila u = 18 cm.Calculate the linear magnification and the height of the image when u = 18 cm.

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• m = v/u = 36 ÷ 18 = 2.0 [1]1
• Height of image = 2.0 × 2.5 = 5.0 cm [1]1

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[2 marksmarkah] (d)Ramalkan jarak imej apabila u = 16 cm.Predict the image distance when u = 16 cm.

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• 1/v = 1/f − 1/u = 1/12 − 1/16 [1]1
• v = 48 cm [1]1

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[2 marksmarkah] (e)Lilin itu diletakkan 8.0 cm dari kanta. Mengapakah imej tidak dapat dibentuk pada skrin?The candle is placed 8.0 cm from the lens. Why can no image be formed on the screen?

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• The object is inside the focal length (u < f = 12 cm) [1]1
• The refracted rays diverge, so the image formed is virtual (upright and magnified) and cannot be formed on a screen [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.4.2

TextbookBuku teks p.ms. 261

23

QuestionSoalan 23

Jadual 3 menunjukkan jisim dan jejari Bulan. Sebuah pengorbit bulan berjisim 2 000 kg mengorbit Bulan dalam orbit bulatan pada ketinggian 1.00 × 10⁵ m dari permukaan Bulan.

Jasad	Jisim / kg	Jejari / m
Bulan	7.35 × 10²²	1.74 × 10⁶

Jadual 3Table 3 shows the mass and the radius of the Moon. A lunar orbiter of mass 2 000 kg orbits the Moon in a circular orbit at a height of 1.00 × 10⁵ m above the surface of the Moon.

Body	Mass / kg	Radius / m
Moon	7.35 × 10²²	1.74 × 10⁶

Table 3

[1 markmarkah] (a)Apakah yang dimaksudkan dengan kekuatan medan graviti?What is meant by gravitational field strength?

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• The gravitational force acting on a mass of 1 kg (g = F/m, unit N kg⁻¹)1

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[2 marksmarkah] (b)Hitung kekuatan medan graviti di permukaan Bulan.Calculate the gravitational field strength at the surface of the Moon.

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• g = GM/R² = (6.67 × 10⁻¹¹)(7.35 × 10²²) ÷ (1.74 × 10⁶)² [1]1
• g = 1.62 N kg⁻¹ [1]1

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[1 markmarkah] (c)Hitung berat seorang angkasawan berjisim 65 kg di permukaan Bulan.Calculate the weight of an astronaut of mass 65 kg on the surface of the Moon.

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• W = mg = 65 × 1.62 = 105 N (accept 105 to 106 N)1

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[2 marksmarkah] (d)Hitung halaju lepas dari permukaan Bulan.Calculate the escape velocity from the surface of the Moon.

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• v = √(2GM/R) = √[2(6.67 × 10⁻¹¹)(7.35 × 10²²) ÷ (1.74 × 10⁶)] [1]1
• v = 2.37 × 10³ m s⁻¹ (2.37 km s⁻¹) [1]1

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[2 marksmarkah] (e)Hitung laju linear pengorbit itu.Calculate the linear speed of the orbiter.

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• r = 1.74 × 10⁶ + 1.00 × 10⁵ = 1.84 × 10⁶ m [1]1
• v = √(GM/r) = √[(6.67 × 10⁻¹¹)(7.35 × 10²²) ÷ (1.84 × 10⁶)] = 1.63 × 10³ m s⁻¹ [1]1

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[2 marksmarkah] (f)Hitung tempoh orbit pengorbit itu dalam minit.Calculate the orbital period of the orbiter in minutes.

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• T = 2πr / v = 2π(1.84 × 10⁶) ÷ (1.63 × 10³) = 7.09 × 10³ s [1]1
• = 118 minutes (accept 117 to 119) [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, Chapter 3, standard 3.3.3

TextbookBuku teks p.ms. 107

24

QuestionSoalan 24

Rajah 4 menunjukkan dua pencelup, S₁ dan S₂, berjarak 4.0 cm dalam sebuah tangki riak. Kedua-duanya bergetar dengan fasa yang sama pada frekuensi 15 Hz dan menghasilkan gelombang bulatan berlaju 30 cm s⁻¹. Setiap bulatan menunjukkan kedudukan sebuah puncak gelombang pada satu ketika; bulatan biru berasal daripada S₁ dan bulatan jingga daripada S₂. Titik P berada pada garis pembahagi dua serenjang S₁S₂. Titik Q berjarak 6.0 cm dari S₁ dan 8.0 cm dari S₂. Titik R berjarak 6.0 cm dari S₁ dan 7.0 cm dari S₂.Diagram 4 shows two dippers, S₁ and S₂, 4.0 cm apart in a ripple tank. Both vibrate in phase at a frequency of 15 Hz and produce circular waves of speed 30 cm s⁻¹. Each circle shows the position of a crest at one instant; the blue circles come from S₁ and the orange circles from S₂. Point P lies on the perpendicular bisector of S₁S₂. Point Q is 6.0 cm from S₁ and 8.0 cm from S₂. Point R is 6.0 cm from S₁ and 7.0 cm from S₂.

[1 markmarkah] (a)Nyatakan fenomena yang berlaku apabila gelombang daripada S₁ dan S₂ bertindih.State the phenomenon that occurs when the waves from S₁ and S₂ overlap.

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• Interference (superposition of waves)1

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[1 markmarkah] (b)Hitung panjang gelombang gelombang itu.Calculate the wavelength of the waves.

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• λ = v/f = 30 ÷ 15 = 2.0 cm1

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[1 markmarkah] (c)Hitung beza lintasan di Q dan nyatakan nilainya dalam bentuk λ.Calculate the path difference at Q and express it in terms of λ.

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• 8.0 − 6.0 = 2.0 cm = 1λ1

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[2 marksmarkah] (d)Nyatakan jenis interferens di Q dan terangkan bagaimana ia berlaku.State the type of interference at Q and explain how it occurs.

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• Constructive interference [1]1
• A crest from S₁ meets a crest from S₂ (path difference is a whole number of wavelengths), so the waves are in phase and add up to a larger amplitude [1]1

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[2 marksmarkah] (e)Nyatakan jenis interferens di R dan terangkan jawapan anda.State the type of interference at R and explain your answer.

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• Destructive interference [1]1
• Path difference = 1.0 cm = ½λ, so a crest from S₁ meets a trough from S₂ and the waves cancel out (zero / minimum amplitude) [1]1

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[2 marksmarkah] (f)Jarak antara S₁S₂ dan satu garis selari ialah D = 16 cm. Hitung jarak, x, antara dua garis antinod yang bersebelahan pada garis itu. [λ = ax/D]A line parallel to S₁S₂ is at a distance D = 16 cm from them. Calculate the distance, x, between two adjacent antinodal lines on this line. [λ = ax/D]

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• x = λD / a = (2.0)(16) ÷ 4.0 [1]1
• x = 8.0 cm [1]1

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[1 markmarkah] (g)Frekuensi pencelup ditambah kepada 30 Hz dengan laju gelombang tetap. Hitung nilai x yang baharu.The frequency of the dippers is increased to 30 Hz and the speed of the waves stays the same. Calculate the new value of x.

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• λ becomes 1.0 cm, so x = 4.0 cm (x is halved)1

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For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, Chapter 5, standard 5.6

TextbookBuku teks p.ms. 216

25Choose 1Pilih 1

QuestionSoalan 25

Dua orang murid menjalankan kajian perlanggaran antara troli P dan troli Q di atas landasan licin yang geseran telah dikompensasikan. Jadual 4 menunjukkan data bagi dua eksperimen.

Eksperimen	Jisim P / kg	Jisim Q / kg	Halaju awal P / m s⁻¹	Halaju awal Q / m s⁻¹	Halaju selepas perlanggaran
1	0.50	1.50	4.0	0	P dan Q bergerak bersama
2	0.50	1.50	4.0	0	P melantun balik pada 2.0 m s⁻¹

Jadual 4Two pupils carry out a study of collisions between trolley P and trolley Q on a runway whose friction has been compensated. Table 4 shows the data for two experiments.

Experiment	Mass of P / kg	Mass of Q / kg	Initial velocity of P / m s⁻¹	Initial velocity of Q / m s⁻¹	Velocity after the collision
1	0.50	1.50	4.0	0	P and Q move together
2	0.50	1.50	4.0	0	P rebounds at 2.0 m s⁻¹

Table 4

[1 markmarkah] (a)Nyatakan Prinsip Keabadian Momentum.State the Principle of Conservation of Momentum.

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• In a closed system (no external force), the total momentum before a collision is equal to the total momentum after the collision1

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[2 marksmarkah] (b)Hitung halaju bersama troli P dan Q dalam Eksperimen 1.Calculate the common velocity of trolleys P and Q in Experiment 1.

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• 0.50(4.0) + 0 = (0.50 + 1.50)v [1]1
• v = 1.0 m s⁻¹ [1]1

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[2 marksmarkah] (c)Hitung halaju troli Q dalam Eksperimen 2.Calculate the velocity of trolley Q in Experiment 2.

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• 0.50(4.0) = 0.50(−2.0) + 1.50v [1]1
• v = 2.0 m s⁻¹ in the original direction of P [1]1

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[2 marksmarkah] (d)Hitung impuls yang bertindak pada troli P dalam Eksperimen 2.Calculate the impulse acting on trolley P in Experiment 2.

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• Impulse = mv − mu = 0.50(−2.0) − 0.50(4.0) [1]1
• = −3.0 N s (magnitude 3.0 N s, opposite to the original motion) [1]1

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[1 markmarkah] (e)Masa sentuhan antara P dengan Q ialah 0.020 s. Hitung purata daya impuls pada P.The contact time between P and Q is 0.020 s. Calculate the average impulsive force on P.

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• F = 3.0 ÷ 0.020 = 150 N1

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[2 marksmarkah] (f)Dengan menggunakan Hukum Gerakan Newton Ketiga, terangkan mengapa impuls yang bertindak pada Q sama besar tetapi berlawanan arah dengan impuls pada P.Using Newton's Third Law, explain why the impulse acting on Q is equal in size and opposite in direction to the impulse on P.

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• The force of P on Q and the force of Q on P are equal in magnitude and opposite in direction (action and reaction acting on different bodies) [1]1
• The contact time is the same, so the impulses (Ft) are equal and opposite: Q gains 1.50 × 2.0 = 3.0 N s while P loses 3.0 N s [1]1

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[2 marksmarkah] (g)Dua buah troli, R berjisim 0.80 kg dan S berjisim 0.20 kg, pegun bersentuhan dengan satu spring termampat di antaranya. Apabila spring itu dilepaskan, troli S bergerak pada 2.0 m s⁻¹. Hitung halaju troli R.Two trolleys, R of mass 0.80 kg and S of mass 0.20 kg, are at rest in contact with a compressed spring between them. When the spring is released, trolley S moves at 2.0 m s⁻¹. Calculate the velocity of trolley R.

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• 0 = 0.80v + 0.20(2.0) [1]1
• v = −0.50 m s⁻¹ (0.50 m s⁻¹ in the opposite direction to S) [1]1

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[3 marksmarkah] (h)Selepas Eksperimen 1, troli P dan Q yang bergerak bersama menghadapi daya geseran 0.50 N apabila landasan tidak lagi dikompensasikan. Hitung nyahpecutan dan jarak yang dilalui sebelum berhenti.After Experiment 1, the trolleys P and Q moving together meet a friction force of 0.50 N when the runway is no longer compensated. Calculate the deceleration and the distance travelled before they stop.

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• a = F/m = 0.50 ÷ 2.0 = 0.25 m s⁻² [1]1
• v² = u² + 2as: 0 = (1.0)² + 2(−0.25)s [1]1
• s = 2.0 m [1]1

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[3 marksmarkah] (i)Seorang murid berkata, "Dalam Eksperimen 1, momentum telah hilang kerana halaju selepas perlanggaran, 1.0 m s⁻¹, lebih kecil daripada 4.0 m s⁻¹." Nilaikan pernyataan itu.A pupil says, "In Experiment 1 momentum has been lost because the velocity after the collision, 1.0 m s⁻¹, is smaller than 4.0 m s⁻¹." Evaluate the statement.

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• The statement is not correct [1]1
• Momentum before = 0.50 × 4.0 = 2.0 kg m s⁻¹ and after = 2.00 × 1.0 = 2.0 kg m s⁻¹, so the total momentum is conserved [1]1
• The velocity is smaller because the same momentum is now shared by a larger mass (2.00 kg instead of 0.50 kg) [1]1

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[2 marksmarkah] (j)Kapal yang berlabuh di jeti menggunakan tayar lama sebagai penghadang. Terangkan bagaimana tayar itu melindungi kapal dan jeti.Ships berthing at a jetty use old tyres as fenders. Explain how the tyres protect the ship and the jetty.

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• The soft rubber fender lengthens the time of impact when the ship touches the jetty [1]1
• The change in momentum is the same, so the impulsive force F = (mv − mu) ÷ t is smaller and less damage is caused [1]1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, Chapter 2, standard 2.5

TextbookBuku teks p.ms. 62

26Choose 1Pilih 1

QuestionSoalan 26

Rajah 5 menunjukkan graf suhu melawan masa bagi 0.50 kg pepejal X yang dipanaskan oleh pemanas rendam 250 W. Anggap semua haba daripada pemanas diserap oleh X.Diagram 5 shows the temperature-time graph of 0.50 kg of a solid X heated by a 250 W immersion heater. Assume that all the heat from the heater is absorbed by X.

[1 markmarkah] (a)Nyatakan takat lebur X.State the melting point of X.

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• 80 °C1

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[2 marksmarkah] (b)Terangkan mengapa suhu X tidak berubah semasa X melebur walaupun haba masih dibekalkan.Explain why the temperature of X does not change while X is melting although heat is still supplied.

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• The heat absorbed is used to weaken / break the bonds between the molecules (latent heat) [1]1
• It does not increase the average kinetic energy of the molecules, so the temperature stays constant [1]1

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[2 marksmarkah] (c)Hitung muatan haba tentu X dalam keadaan pepejal.Calculate the specific heat capacity of X in the solid state.

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• Pt = mcΔθ: 250 × 120 = 0.50 × c × (80 − 20) [1]1
• c = 1.0 × 10³ J kg⁻¹ °C⁻¹ [1]1

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[2 marksmarkah] (d)Hitung haba pendam tentu pelakuran X.Calculate the specific latent heat of fusion of X.

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• Melting takes 420 − 120 = 300 s; Pt = ml: 250 × 300 = 0.50 × l [1]1
• l = 1.5 × 10⁵ J kg⁻¹ [1]1

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[2 marksmarkah] (e)Hitung muatan haba tentu X dalam keadaan cecair.Calculate the specific heat capacity of X in the liquid state.

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• 250 × (600 − 420) = 0.50 × c × (120 − 80) [1]1
• c = 2.25 × 10³ J kg⁻¹ °C⁻¹ [1]1

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[1 markmarkah] (f)Muatan haba tentu ikan ialah 3 200 J kg⁻¹ °C⁻¹. Apakah maksud pernyataan ini?The specific heat capacity of fish is 3 200 J kg⁻¹ °C⁻¹. What does this statement mean?

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• 3 200 J of heat is needed to raise the temperature of 1 kg of fish by 1 °C1

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[3 marksmarkah] (g)Seorang nelayan menyimpan 4.0 kg ikan pada 28 °C bersama 1.0 kg ais pada 0 °C di dalam kotak bertebat. Hitung suhu akhir ikan itu apabila semua ais telah melebur. [Haba pendam tentu pelakuran ais = 3.34 × 10⁵ J kg⁻¹; muatan haba tentu air = 4 200 J kg⁻¹ °C⁻¹]A fisherman stores 4.0 kg of fish at 28 °C together with 1.0 kg of ice at 0 °C in an insulated box. Calculate the final temperature of the fish when all the ice has melted. [Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹; specific heat capacity of water = 4 200 J kg⁻¹ °C⁻¹]

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• Heat lost by fish = 4.0 × 3 200 × (28 − θ) [1]1
• Heat gained = 1.0 × 3.34 × 10⁵ + 1.0 × 4 200 × (θ − 0) [1]1
• 12 800(28 − θ) = 334 000 + 4 200θ, so θ = 1.4 °C [1]1

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[2 marksmarkah] (h)Sebuah belon di sebuah pesta di taman permainan berisi udara 3.0 × 10⁻³ m³ pada 27 °C. Belon itu diletakkan dalam peti sejuk pada −13 °C. Hitung isi padu udara dalam belon itu. Anggap tekanan tetap.A balloon at a playground fair holds 3.0 × 10⁻³ m³ of air at 27 °C. The balloon is placed in a freezer at −13 °C. Calculate the volume of the air in the balloon. Assume the pressure is constant.

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• T₁ = 300 K, T₂ = 260 K [1]1
• V₂ = 3.0 × 10⁻³ × 260 ÷ 300 = 2.6 × 10⁻³ m³ [1]1

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[2 marksmarkah] (i)Terangkan mengapa isi padu udara berkurang apabila suhu menurun pada tekanan tetap, menggunakan teori kinetik gas.Explain why the volume of the air decreases when the temperature decreases at constant pressure, using the kinetic theory of gas.

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• The molecules move more slowly (less kinetic energy) and hit the walls less often and with less force [1]1
• To keep the pressure constant the volume must decrease so that the rate of collisions with the walls per unit area stays the same [1]1

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[3 marksmarkah] (j)Nelayan lain mencadangkan kotak aluminium dan bukannya kotak polistirena untuk menyimpan ais dan ikan. Nilaikan cadangan itu.Another fisherman suggests an aluminium box instead of a polystyrene box to keep the ice and the fish. Evaluate the suggestion.

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• The suggestion is not suitable [1]1
• Aluminium is a good conductor of heat, so heat from the surroundings flows into the box quickly [1]1
• The ice melts faster and the fish are not kept cold for long; polystyrene is a poor conductor (good insulator) and keeps the ice longer [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 4, standard 4.3

TextbookBuku teks p.ms. 138

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