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PERFECT SCORE MODULE Sekolah Berasrama Penuh Kementerian Pelajaran Malaysia 201 4 NAME: …………………………..…………………………………………………. SCHOOL……………………………………………………………………………….. PHYSICS Beyond A+ TEACHERS EDITION

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PHYSICS PERFECT SCORE

PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

PERFECT SCORE MODULESekolah Berasrama Penuh Kementerian Pelajaran Malaysia2014

NAME: ...

SCHOOL..PHYSICSBeyond A+

TEACHERS EDITION

MAKLUMAT MODUL

Modul ini mengandungi 2 bahagian: Section A dan Section B Section A soalan aneka pilihan untuk menguji penguasaan konsep pelajar mengikut topik. Section B soalan konstruk kefahaman dan penyelesaian masalah kuantitatif sebagai pengukuhan dan pengayaan konsep yang dikenalpasti lemah berdasarkan ujian penguasaan konsep dalam Section A Section B kemahiran asas matematik / sains

Keperluan Bahan

1. Modul Physics Perfect Score Beyond A+ 2014 (menguji penguasaan konsep dan pemantapan kemahiran)2. Modul Physics Perfect Score 2013 (pengayaan)3. Flip board/white board kecil/ /kertas mahjong4. Marker pen5. Label kumpulan (cadangan: mengikut topik sebagai expert group) 6. Alat radas (jika perlu)

CARTA ALIR PELAKSANAAN PROGRAM (Minimum 10 Jam)

1 jam 30 minit (pemilihan item adalah mengikut kelemahan pelajar dan dijalankan sebagai pra ujian)15 minit2 jam 15 minit Minimum 6 jam (mengikut kelemahan pelajar)Ujian Diagnostik (Section A)Semak JawapanAnalisis Skor IndividuPerbincangan soalan Diagnostik bersama Guru berdasarkan topik yang dikenalpasti lemahBerdasarkan Analisis Skor, pelajar mengenalpasti tajuk yang belum dikuasai Pelajar dibahagikan kepada kumpulan mengikut topik yang belum dikuasaiPEMANTAPANPerbincangan di dalam kumpulan soalan pada Section B (mengikut topik paling lemah yang dikenalpasti melalui Analisis Skor)Sessi pembentangan / Perkongsian konsep/kemahiranPengayaanLatihan menggunakan Modul Perfect Score 2013 mengikut kemahiran(mengikut kesesuaian sekolah)

SECTIONCONTENT Page

ADiagnostic Test Answer & Analysis5 - 6

BAnswer for Enhancement Question

1. Force & Motion7

2. Force and Pressure8

3. Heat10

4. Light12

5. Waves15

6. Electricity16

7. Electromagnetism17

8. Electronics18

9. Radioactivity20

PHYSICS PERFECT SCORE 2014 PANELS

NOR SAIDAH BT HASSAN - Kolej Tunku Kurshiah (TKC)( Head of Panels )

HASLINA BT ISMAIL - SMS Hulu Selangor (SEMASHUR)

JENNYTA BT NOORBI SMS TUANKU MUNAWIR (SASER)

SECTION A:

DIAGNOSTIC TEST (ANSWER AND ANALYSIS)

QuestionAnswerNumber of Wrong ResponseTopicRemarks

1. BForce and Motion

2. B

3. A

4. B

5. B

6. B

7. DF&P

8. D

9. C

10. C

11. A

12. C

13. C

14. B

15. D

16. CHeat

17. D

18. C

19. C

20. D

21. C

22. A

23. B

24. B

25. DLight

26. A

27. C

28. C

29. B

30. B

31. C

32. C

33. D

34. C

35. A

36. AWaves

37. B

38. B

39. B

40. D

41. C

42. D

43. AElectricity

44. B

45. A

46. D

47. C

48. A

49. D

50. C

51. A

52. C

53. C

54. B

55. D

56. C

57. AElectromagnetism

58. D

59. B

60. D

61. C

62. D

63. B

64. AElectronics

65. B

66. D

67. A

68. CRadioactivity

69. A

70. C

SECTION B

1. Force & Motion4 - 13

NoAnswersPhysics Concept/Principle/Law

1

Att= 0sand object is stationary at some position and remains stationary untilt= 2swhen it begins accelerating. It accelerates in a positive direction for 2 seconds untilt= 4sand then travels at a constant velocity for a further 2 seconds.

Motion graph

2a97.2o

b

c4.54 N

3Constant speed, resultant force = 0F - 40 - 600 sin 25 = 0F = 293.57 N

2. Forces and Pressure

NoAnswersPhysics Concept/Principle/Law

1(a) Air pressure in the sticker decrease. Have the different between pressure in the pump and the air pressure surrounding. The force is produce Force > mirror weight(b) mirror weight= Vg= 2.5 x 10 3 x 1.5 x 0.5 x 0.01 x10 = 187.5 NAtmospheric pressureDifference in pressure

21. Spinning ball moving in the opposite direction with air flow at the upper surface 12. Spinning ball moving in the same direction with air flow at the lower surface13. Lower surface spins more faster than the upper surface of the ball1

Bernoullis principle

3(a) 1. Column of mercury in Diagram (b) is lower 2. At higher altitude, number of air molecules are smaller 3. Pressure exerted by the air molecules is smaller

(b) 1. Mercury column become lower 2. Gas pressure inside the tube push the mercuryAtmospheric pressureSimple mercury barometre

4(a) 1. Rubber tube is filled with water 2. Place the end tube Q lower than P 3. Pressure at P bigger than Q 4. Water flows from Q because there is difference in pressure(b) Q is at same level with P Or Q higher than P

Difference in pressureAtmospheric pressure

5(a)1. Measure the mass of the necklace2. Measure the volume of the necklace; 3. Place the necklace in the water. Volume of water displaced is measured by measuring cylinder; 4. volume of necklace = volume of water displaced5. Density of the necklace = mass/volume

(b)1. density = = = 13.25 g cm-3

2. Percentage = x 100% = 48.5%3. The necklace diamond is not genuineArchimedes principledensity

61. The best time is early morning2. The cool air is denser3. More air molecules can be displaced4. Produced more buoyant forceThe balloon can rise higher

Buoyant forcedensity

71. When force is exerted on Piston A, pressure is produced (P=F/A)2. Pressure will be transmitted uniformly and equally in all parts of the enclosed oil3. It obeys Pascal Law4. The same pressure exerted on bigger area, Piston B will produce bigger force (F=P x A)

(b)FB = ( FA AB) (AA) = (50 15) (2) = 375 N

AA DA = AB DB 2 21 = 15 DB DB = 28 cm

Pascal principleForce multification

F1/A1 = F2/A2

81. When the catch is still in the water, the buoyant force is bigger2. When the catch is getting out from the water, the volume of object immerse is smaller3. The volume of water displaced also smaller, thus the weight of water displaced is getting smaller4. The buoyant force is equal to the weight of water displaced5. The buoyant force is smaller and the catch feels heavier

Relationship between Bouyant force and depth of object immersed

91. Gas flows out through the jet with high velocity2. According to Bernoullis Principle, high velocity willproduce low pressure at the nozzles of the jet3. Higher atmospheric pressure pushes the air inside the cylinder trough the orifice4. The air will mix with the gas and complete combustion will occur

Bernoullis principle

3. Heat14 - 24

NoANSWERConcept/Principle

11. When temperature increases, the average kinetic energy increases2. Rate of collision between the air molecules and wall of the tire also increases.3. Rate of change of momentum increases4. Force exerted per unit are a increase, so the air pressure increases.Pressure Law

2(a) Pgas = 75 + 25 = 100 cm Hg(b) (i) When the gas is cooled down, the kinetic energy ofthe gas decreases, reducing the rate of collision between the gas molecules and the container, there for e pressure reduced.(ii)T1 = 127 + 273 = 300 K P1 = 100 cm Hg P2 = 75 cm HgTO = 300 x 75 = 75 K100(iii) Pressure Law

Pressure Law

P1 = P2T1 T2

331.25oC

4At lower lan, the density of air is higher.Hence it is more difficult to vaporizeSpecific Heat Capacity

5(i) 100C(ii) m=V = (1) (100) = 100g(iii) .2 x 379 ( 100-T) = 0.1 x 4200 x (T-28) T = 39 Cm1 C1 1 = m2 C2 2

6

Q = mc

7(a) (i)- The rate of heat transfer between two bodies areThe same- The temperature of the two bodies are the same

(ii)40C

(iii)Prevent heat loss to surrounding

(b) (i)Heat supplied by hot metal = heat received by waterm1 C1 1 = m2 C2 20.4 xC1 x (100-40) = 0.2 x 4200x (40 28)0.4 x C1 x 60= 0.2 x 4200x 12 C1 = 420 J kg-1C-1

(ii)Heat released by water is absorb by the metal // no heat loss to surrounding

m1 C1 1 = m2 C2 2

8ABoylesLaw

9(a) (i) The degree of hotness of an object(i) 1 x 103 (1.0 x 60) = 0.05 c (78 20) 2.069 x 105 Jkg-1oC-1

(b) 0.05 (2.069 x 105)(78 ) = 2.0 (4 200) ( 28) 55.6oC

m1 C1 1 = m2 C2 2

10The heat is transferred from hot water to the dented ping pong ball.The air temperature in the dented ping pong ball increased. The air pressure of dented ping pong increased. The air pressure pushed the wall of the ball back to its originalposition.

4. Light

Num.AnswerConcept

1.

1. Light rays and reflection2. Extrapolate and draw the image3. Incident angle = reflected angle; Object distance = image distance4. Characteristics of image: Virtual, inverted, same sizeThe law of reflection Plane mirror

2. 1. Bring each mirror one by one close to an object and observe the image formed in it. 2. If the image is of the same size as that of the object and upright, the mirror is a plane mirror.3. If the image is highly diminished and upright, it is a convex mirror4. If the image is large and upright, it is a concave mirror.characteristics of an image in a convex mirror

3. 1. A convex mirror always forms an upright image of an object2. It also forms a diminished image 3. As a result images of large number of objects can be seen in the mirror at the same time4. The mirror can be tilted // usephotosensorsmounted in the mirror to detect light and dim the mirrorcharacteristics of an image in a convex mirror

4. Use n = 1/sin x to get n (critical angle equation)Use n = sin i / sin r to get y

y = 27.4 oRefractive indexCritical angle

5. The layers of air nearer the road warmer The density of air decrease nearer to the road surface. The light travel from denser to less dense area. The light refract away from the normal When the angle of incidence exceed the critical angle, total internal reflection occurs To the observer, light is appearing to come in a straight line creating the form of image on the road.The Laws Of Refraction

6. Situation B Light travels in straight line.

In A, when the cup is empty, the edge of the cup stops observer seeing the coin. When the water is poured into the cup, the light travels from optical denser medium (water) to less dense medium (air). (diagram) The light refracted away and it bends over the edge so the observer can see the coin. (diagram)Real Depth And Apparent Depth

7. When a coin is placed under an empty beaker, the light travels from the air glass air the wall of the beaker air, before it enters the observers eye. Therefore, making it possible for the observer to see the coin. (ray diagram) When the water is poured in the beaker, the light travels from the air glass water the wall of the beaker and through the air to the eye. The index of refraction is too great ; the light refracted and bends and change in angle, so the observer cannot be able to see the coin. (ray diagram)

8. The instructor I appear to be at higher position due to refractionLight refracts towards normal as it travels from less dens medium (air) to water (denser medium)Light appears to travel in straight line to the scuba diverArrow: from instructor to the observer

9. Increase the angle of incidence, i, then angle of refraction, r will also increase Keep on increasing the angle of incidence until angle of refraction is 90 The angle of incidence is called critical angle Increase the angle of incidence more than the critical angle The ray will be reflected. Critical Angle and total internal reflection

10. 1. The convex lens is aimed/focused to a distant object (infinity) 2. The screen is adjusted until a sharp image is formed on the screen3.The distance between the screen and the lens is measuredl4.Focal length = distance between the screen and the lensFocal Point And Focal Length Of A Lens

11.

Real, inverted, diminishedv = 15 cmm = v/u m = 0.5Relationship Between u, v and f Lens equation

12. By using a convex lens, f = 20 cm(ray diagram)The Use Of Lenses In Optical Devices

13. Objective lens: Y Eyepiece lens: X

The diagram shows the microscope in normal adjustment, that is, with the final image at the near point (25 cm from the eye) (distance D from the eye lens). (This setting gives the maximum angular size of image without eye strain.)

The Use Of Lenses In Optical Devices

5. Waves

Num.AnswerConcept

120 cm s-1

2When the singer sings, she produces a high frequency soundThe frequency of the glass equal with the frequency of the singers soundBoth systems are in resonanceSo the glass will oscillates at its maximum aplitude and it breaks.

3170 mS = vt 2

4When the prongs of the tuning fork move outward, it produce a region of compressionWhen the prongs of the tuning fork move inward, it produce a Region of rarefactionCandle flame in front of a loud speaker that emits sound waveCandle flame vibrates forward and backward

5(a) Transverse / Plane waves(b) Show the path is not bended when enter the shallow area and is bended away from the normal line when enter the deep areaShow the wavelength is decreased in shallow area And is equal in deep area

(c) =

= 4.5 m (answer with correct unit)

6. Electric

umAnswerConcept

1(i) V2 = 4 V

(ii) I = = 0.8 A

(ii) R = = 2.5

V = IR

2(a) Total resistance in the circuit(b) If one bulb is blown the other still can be usedLower the total resistanceMaintain the potential difference same as the supply throughThe household appliances(c) (i) Control the speed of the fan (ii) 1/r = 1/20 + 1/(20+10) @ 1/r = 1/20 + 1/30 @ 1/r = 50/60 @r = 60/50r = 1.2 1/r = 1/20 + 1/20 @ 1/r = 2/20 @ 1/r = 1/10r = 10

3(a) Note : The flame flatten and spread out more toward negative plate

(b) The heat of burning candle produces positive and negative ions.2 The positive ions which are heavier is pulled towards negative plate with a large proportion flame

7. Electromagnet

NoAnswersPhysics Concept/Principle/Law

1. Adevice thattransferselectrical energyintosound The wire from theamplifiercarries analternating current The interaction between magnetic field of the current carrying conductor and the permanent magnet produces force Thecoilwhich can slidebackwardsandforwardsover thecentral pole of acircularpermanent magnet makes thecoil(and thepapercone)move backwardsandforwards at thesamefrequency as thechanging current. Thepaper conethenmoves theairbackwardsandforwardswhich creates thesound

2. When the switch is on, the current flows through the copper wire The interaction between magnetic field of the current carrying conductor and the permanent magnet produces force The catapult field is produced (diagram) the magnetic lines of force are close together near the wire on the left so forcing it to the right.

3. The diaphragm is attached to the coil. When the diaphragm vibrates in response to incoming sound waves, the coil moves backwards and forwards past the magnet. This creates an induced current in the coil which is channeled from the microphone along wires

4. (a) 1. Magnet pushed inside, magnetic flux is cut by the wire 2. According to Faradays Law; ## 3. emf is induced in the solenoid 4. so, the current is induced

(b) 1. The bigger number of turns, the bigger magnetic flux is cut by the wire 2. According to Faradays Law; ## 3. The bigger emf is induced in the solenoid 4. so, the bigger current is induced

(c) 1. The bigger speed, the rate of cutting of magnetic field is bigger 2. According to Faradays Law; ## 3. bigger emf is induced in the solenoid 4. so, the bigger current is induced, pointer of the galvanometer will deflected more

(d) 1. When the N pole is pushed into the solenoid, cutting of magnetic field occur 2. The current induced produces north pole on the left side, 3. so as to oppose the oncoming magnet, obeying the Lenzs Law 4. I will flows in anti clock wise direction

Induced emfInduced currentFaradays LawFactors affected induced emfLenzs law

5. 1. rotate the coil in clock wise direction2. the coil cut across the magnetic field3. current is induced in the coil4. the commutator change the direction in the coil so that the direction of current in external circuit I always the same.

generators

8. Electronic9. 49 61

NoAnswersPhysics Concept/Principle/Law

1(a)1. When someone speaks at the microphone, the current produced flows to the circuit2. The capacitor is used to avoid direct current from battery to flow through the microphone3. The current will give changes to the magnitude of base- current// IB become bigger4. When IB changes IC also changes// IB bigger, IC also bigger The speaker will produce bigger audio/amplified

(b) Vzy = 1 V 1. VXY = 5 V

2, R1 x 6 = 5 R1 + 1000 3, R1 = 5000 IE = IB + IC ; IC >> IB

Transistor as an amplifier

2 (a) 0001, AND

QQ (b)

RQP (c)

Logic gatesTruth table

31. Connect the dry cell terminal to the Y-input of CRO. 2. The Y-gain is set to a value so that the direct current wave form displayed on the screen CRO. 3. Determine the distance / part of y-axis. 4. Potential different = ( Y-gain scale) x (Vertical distance of direct current wave)

CRO

41. When there is a fire burning, R at T = 3.5 k Potential difference across P = 3500 x 6 = 2.2 V (3500 + 6000) 2 Potential difference across Q = 6000 x 6 = 3.8 V or (6-2.2) = 3.8 V (3500 + 6000)

3. Potential difference across Q exceed / greater than 3.2 V, so the transistor is functioned

4. The solenoid become magnetised, G will swicth on and the bell will rings

Potential dividerTransistor as switching circuit

5(a)

(b) During hot weather1. resistance at termistor decrease , potential difference across thermistor will decrease 2. Potential difference across R will increase3. This will produce bigger base current , and will increase the collector current4. Electric relay will switch on the air conditioner.

(c) During cold weather , resistance at thermistor increase.1. Potential difference across thermistor will increase.2. Potential difference across R will decrease.3. This will produce smaller base-current and no current flow in collector circuit. 4. Electric relay will swith off the air conditioner.

Relationship R and VEffect to VBE; effect to output

6(a) 7.5 V

(b)

OR

(c)

= VBE

Potential divider

IE = IB + IC

10. Radioactivity

NoAnswersPhysics Concept/Principle/Law

11. Small amount of radioisotope is put in the water reservoir2. The substance must be in liquid state so it is easy to flow in the water3. The substance should emit particles (the radiation canbe detected above the ground ) 4. A Geiger-Muller counter is moved over the pipe according to the layout plan. 5. At a point where the Geiger-Muller counter detected high radiation level, indicating the point of leakage.

Radioactive detectorCharacteristic of radiation

21. Carbon-14 atom is a radioactive substance which is easily absorbed by living plants.2. After the plants dies, the activity of Carbon-14 will decline since no new carbon-14 is absorbed. (carbon-14 will decay to nitrogen-14)3. The difference between the concentration of carbon14 in the material to be dated and the 4. Concentration in the atmosphere provides gives the rate of carbon-14 decay5. By calculating the activity of carbon-14, the age of the dead plant/fossil can be determined (half-life of carbon-14 is 5,730 years)

Application of radioisotopes

Carbon dating

3 (a) Energy released E = mc2 = 3.5 x 10-9 x ( 3 x 108)2 = 3.15 x 107 J

(b) Power obtained P = E/t= 3.15 x 107 1.5 x 10-3 = 2.1 x 1010 W

Nuclear energy

E mc2

4(a)

1. Neutron bombarded a uranium nucleus //Diagram 2. Three neutrons produced // Diagram 3. The new neutron bombarded a new uranium nucleus // Diagram 4. For every reaction, the neutrons produced will generate a chain reaction // Diagram

(b) E = mc22.9 x 10 -11 = m x (3.0 x 108)2m = 3.22 x 10-28 kg

Chain reaction

5(a)1- Show the line in the graph2- T1/2 = 4 days

(b)

1. Shape of graph2. One point is correct3. Two or more point

Half life

6

1. Put the radioactive source opposite the detector2. Detector is connected to the thickness indicator3. Detector detect the reading of the changes in counts4. If the reading of the detector is less than the specified value, the thickness of the aluminium foil is too thick/ vice versa

Application of radioactive

TOGETHER we must succeed, TOGETHER we will succeedPage 21