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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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  • 1. PERFECTSCOREMODULETEACHERSEDITIONSekolahBerasrama PenuhKementerianPelajaran Malaysia2014NAME: ...SCHOOL..PHYSICSBeyond A+

2. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014MAKLUMAT MODUL Modul ini mengandungi 2 bahagian: Section A dan Section B Section A soalan aneka pilihan untuk menguji penguasaan konsep pelajar mengikuttopik. Section B soalan konstruk kefahaman dan penyelesaian masalah kuantitatif sebagaipengukuhan dan pengayaan konsep yang dikenalpasti lemah berdasarkan ujianpenguasaan konsep dalam Section A Section B kemahiran asas matematik / sainsKeperluan Bahan1. Modul Physics Perfect Score Beyond A+ 2014 (menguji penguasaan konsep danpemantapan 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)TOGETHER we must succeed, TOGETHER we will succeedPage 2 3. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014CARTA ALIR PELAKSANAAN PROGRAM (Minimum 10 Jam)TOGETHER we must succeed, TOGETHER we will succeed1 jam 30 minit (pemilihanitem adalah mengikutkelemahan pelajar dandijalankan sebagai praujian)15 minit2 jam 15 minitMinimum 6 jam(mengikutkelemahanpelajar)Page 3Ujian Diagnostik (Section A)Semak JawapanAnalisis Skor IndividuPerbincangan soalan Diagnostikbersama Guru berdasarkan topik yang dikenalpastilemahBerdasarkan Analisis Skor, pelajar mengenalpastitajuk yang belum dikuasaiPelajar dibahagikan kepada kumpulan mengikut topikyang belum dikuasaiPEMANTAPANPerbincangan di dalam kumpulan soalan padaSection B (mengikut topik paling lemah yangdikenalpasti melalui Analisis Skor)Sessi pembentangan / Perkongsiankonsep/kemahiranPengayaanLatihan menggunakan Modul Perfect Score 2013mengikut kemahiran(mengikut kesesuaian sekolah) 4. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014SECTION CONTENT PagePHYSICS PERFECT SCORE 2014 PANELSNOR 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:TOGETHER we must succeed, TOGETHER we will succeedPage 4A Diagnostic Test Answer & Analysis 5 - 6BAnswer forEnhancementQuestion1. Force & Motion 72. Force and Pressure 83. Heat 104. Light 125. Waves 156. Electricity 167. Electromagnetism 178. Electronics 189. Radioactivity 20 5. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014DIAGNOSTIC TEST (ANSWER AND ANALYSIS)Question Answer Number of WrongResponse Topic Remarks1. B44. B45. A46. D47. C48. A49. D50. C51. A52. C53. C54. B55. D56. C57. A ElectromagnetismTOGETHER we must succeed, TOGETHER we will succeedForce and Motion2. B3. A4. B5. B6. B7. DF&P8. D9. C10. C11. A12. C13. C14. B15. D16. CHeat17. D18. C19. C20. D21. C22. A23. B24. B25. DLight26. A27. C28. C29. B30. B31. C32. C33. D34. C35. A36. AWaves37. B38. B39. B40. D41. C42. D43. AElectricityPage 5 6. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Question Answer Number of WrongResponse Topic Remarks58. D59. B60. D61. C62. D63. B64. A65. B Electronics66. D67. A68. C69. A Radioactivity70. CSECTION BTOGETHER we must succeed, TOGETHER we will succeedPage 6 7. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers Physics Concept/Principle/Law1At t = 0s and object is stationary at someposition and remains stationary until t =2s when it begins accelerating. Itaccelerates in a positive direction for 2seconds until t = 4s and then travels at aconstant velocity for a further 2 seconds.TOGETHER we must succeed, TOGETHER we will succeedMotion graph2a 97.2obc 4.54 N3Constant speed, resultant force = 0F - 40 - 600 sin 25 = 0F = 293.57 NPage 71. Force & Motion 4 - 13 8. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers Physics Concept/Principle/Law1(a) Air pressure in the sticker decrease.Have the different between pressure in the pumpand 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 Nm265 g13.25TOGETHER we must succeed, TOGETHER we will succeedAtmospheric pressureDifference in pressure21. Spinning ball moving in the opposite direction withair flow at the upper surface 12. Spinning ball moving in the same direction with airflow at the lower surface 13. Lower surface spins more faster than the uppersurface of the ball 1Bernoullis principle3(a) 1. Column of mercury in Diagram (b) is lower2. At higher altitude, number of air molecules aresmaller3. Pressure exerted by the air molecules is smaller(b) 1. Mercury column become lower2. Gas pressure inside the tube push the mercuryAtmospheric pressureSimple mercury barometre4(a) 1. Rubber tube is filled with water2. Place the end tube Q lower than P3. Pressure at P bigger than Q4. Water flows from Q because there isdifference in pressure(b) Q is at same level with POr Q higher than PDifference in pressureAtmospheric pressure5(a)1. Measure the mass of the necklace2. Measure the volume of the necklace;3. Place the necklace in the water. Volume ofwater displaced is measured by measuring cylinder;4. volume of necklace = volume of waterdisplaced5. Density of the necklace = mass/volume(b)1. density =V=20 cm3= 13.25 g cm-32. Percentage =27.3x 100% = 48.5%3. The necklace diamond is not genuineArchimedes principledensity61. The best time is early morning2. The cool air is denser3. More air molecules can be displaced4. Produced more buoyant forceThe balloon can rise higherBuoyant forcedensity7 1. When force is exerted on Piston A, pressure isproduced (P=F/A)2. Pressure will be transmitted uniformly and equallyin all parts of the enclosed oil3. It obeys Pascal Law4. The same pressure exerted on bigger area, PistonB will produce bigger force (F=P x A)(b)Pascal principleForce multificationF1/A1 = F2/A2Page 82. Forces and Pressure 9. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers Physics Concept/Principle/LawFB = ( FA AB) (AA)= (50 15) (2)= 375 NAA DA = AB DB2 21 = 15 DBDB = 28 cm81. When the catch is still in the water, the buoyantforce is bigger2. When the catch is getting out from the water, thevolume of object immerse is smaller3. The volume of water displaced also smaller, thusthe weight of water displaced is getting smaller4. The buoyant force is equal to the weight of waterdisplaced5. The buoyant force is smaller and the catch feelsheavierTOGETHER we must succeed, TOGETHER we will succeedRelationship between Bouyantforce and depth of objectimmersed91. 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 insidethe cylinder trough the orifice4. The air will mix with the gas and completecombustion will occurBernoullis principlePage 9 10. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No ANSWER Concept/Principle11. When temperature increases, the average kinetic energy increases2. Rate of collision between the air molecules and wall of the tire alsoincreases.3. Rate of change of momentum increases4. Force exerted per unit are a increase, so the air pressure increases.TOGETHER we must succeed, TOGETHER we will succeedPressure Law2(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 collisionbetween the gas molecules and the container, therefor e pressure reduced.(ii) T1 = 127 + 273 = 300 K P1 = 100 cm Hg P2 = 75 cm HgTO = 300 x 75 = 75 K100(iii) Pressure LawPressure LawP1 = P2T1 T23 31.25oC4At lower lan, the density of air is higher.Hence it is more difficult to vaporizeSpecific Heat Capacity5(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 26Q = mcq7(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 12C1 = 420 J kg-1C-1(ii) Heat released by water is absorb by the metal //no heat loss to surroundingm1 C1 1 = m2 C2 2Page 103. Heat 14 - 24 11. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No ANSWER Concept/Principle8 A BoylesLaw9(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)55.6oCTOGETHER we must succeed, TOGETHER we will succeedm1 C1 1 = m2 C2 210The 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.Page 11 12. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Num. Answer Concept1.1. Light rays and reflection2. Extrapolate and draw the image3. Incident angle = reflected angle; Object distance = imagedistance4. Characteristics of image: Virtual, inverted, same sizeTOGETHER we must succeed, TOGETHER we will succeedThe law of reflectionPlane mirror2.1. Bring each mirror one by one close to an object and observe theimage 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 animage in a convexmirror3.1. A convex mirror always forms an upright image of an object2. It also forms a diminished image3. As a result images of large number of objects can be seen in themirror at the same time4. The mirror can be tilted // use photosensors mounted in the mirrorto detect light and dim the mirrorcharacteristics of animage in a convexmirror4.Use n = 1/sin x to get n (critical angle equation)Use n = sin i / sin r to get yy = 27.4 oRefractive indexCritical anglePage 124. Light 13. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Num. Answer Concept5. 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, totalinternal reflection occurs To the observer, light is appearing to come in a straight linecreating the form of image on the road.TOGETHER we must succeed, TOGETHER we will succeedThe Laws Of Refraction6.Situation B Light travels in straight line. In A, when the cup is empty, the edge of the cup stops observerseeing the coin. When the water is poured into the cup, the light travels fromoptical denser medium (water) to less dense medium (air).(diagram) The light refracted away and it bends over the edge so theobserver can see the coin. (diagram)Real Depth AndApparent Depth7. When a coin is placed under an empty beaker, the light travelsfrom the air glass air the wall of the beaker air, beforeit enters the observers eye. Therefore, making it possible for the observer to see the coin. (raydiagram) When the water is poured in the beaker, the light travels from theair glass water the wall of the beaker and through the airto the eye. The index of refraction is too great ; the light refracted andbends and change in angle, so the observer cannot be able tosee 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 observerPage 13 14. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Num. Answer Concept9. Increase the angle of incidence, i, then angle of refraction, r willalso increase Keep on increasing the angle of incidence until angle ofrefraction 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.TOGETHER we must succeed, TOGETHER we will succeedCritical Angle and totalinternal reflection10.1. 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 FocalLength Of A Lens11.Real, inverted, diminishedv = 15 cmm = v/u m = 0.5Relationship Between u,v and fLens equation12.By using a convex lens, f = 20 cm(ray diagram) The Use Of Lenses InOptical Devices13.Objective lens: Y Eyepiece lens: X The Use Of Lenses InOptical DevicesPage 14 15. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Num. Answer ConceptThe diagram shows the microscope in normal adjustment, that is, withthe final image at the near point (25 cm from the eye) (distance Dfrom the eye lens). (This setting gives the maximum angular size ofimage without eye strain.)Num. Answer Concept1 20 cm s-12When the singer sings, she produces a high frequency soundThe frequency of the glass equal with the frequency of the singerssoundBoth systems are in resonanceSo the glass will oscillates at its maximum aplitude and it breaks.3 170 m S = vtTOGETHER we must succeed, TOGETHER we will succeed24When the prongs of the tuning fork move outward, it produce aregion of compressionWhen the prongs of the tuning fork move inward, it produce aRegion of rarefactionCandle flame in front of a loud speaker that emits sound waveCandle flame vibrates forward and backward5(a) Transverse / Plane wavesPage 155. Waves 16. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014Num. Answer Concept(b) Show the path is not bended when enter the shallow area and isbended away from the normal line when enter the deep areaShow the wavelength is decreased in shallow area And is equal indeep area(c) =l = 4.5 m (answer with correct unit)um Answer Concept1 (i) V2 = 4 V(ii) I =45= 0.8 A(ii) R =20.8= 2.5 WTOGETHER we must succeed, TOGETHER we will succeedV = IR2 (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 = 60/50Page 166. Electric 17. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014r = 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 negativeplate(b) The heat of burning candle produces positive and negativeions.2 The positive ions which are heavier is pulled towardsnegative plate with a large proportion flameNo Answers PhysicsTOGETHER we must succeed, TOGETHER we will succeedConcept/Principle/Law1. A device that transfers electrical energy into sound The wire from the amplifier carries an alternating current The interaction between magnetic field of the current carryingconductor and the permanent magnet produces force The coil which can slide backwards and forwards over the centralpole of a circular permanent magnet makes the coil (andthe papercone) move backwards and forwards at the same frequency as the changing current. The paper cone then moves the air backwards and forwards whichcreates the sound2. When the switch is on, the current flows through the copper wire The interaction between magnetic field of the current carryingconductor and the permanent magnet produces force The catapult field is produced (diagram) the magnetic lines of force are close together near the wire on theleft so forcing it to the right.Page 177. Electromagnet 18. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 20143. The diaphragm is attached to the coil. When the diaphragm vibrates in response to incoming soundwaves, the coil moves backwards and forwards past the magnet. This creates an induced current in the coil which is channeled fromthe microphone along wires4. (a) 1. Magnet pushed inside, magnetic flux is cut by the wire2. According to Faradays Law; ##3. emf is induced in the solenoid4. so, the current is induced(b) 1. The bigger number of turns, the bigger magnetic flux is cut by thewire2. According to Faradays Law; ##3. The bigger emf is induced in the solenoid4. so, the bigger current is induced(c) 1. The bigger speed, the rate of cutting of magnetic field is bigger2. According to Faradays Law; ##3. bigger emf is induced in the solenoid4. so, the bigger current is induced, pointer of the galvanometer willdeflected more(d) 1. When the N pole is pushed into the solenoid, cutting of magneticfield occur2. The current induced produces north pole on the left side,3. so as to oppose the oncoming magnet, obeying the Lenzs Law4. I will flows in anti clock wise directionTOGETHER we must succeed, TOGETHER we will succeedInduced emfInduced currentFaradays LawFactors affected inducedemfLenzs law5. 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 thedirection of current in external circuit I always the same.generatorsNo Answers Physics Concept/Principle/Law1 (a)1. When someone speaks at the microphone, thecurrentproduced flows to the circuit2. The capacitor is used to avoid direct current frombatteryto flow through the microphone3. The current will give changes to the magnitude ofbase-current//IB become bigger4. When IB changes IC also changes// IB bigger, ICalsobiggerThe speaker will produce bigger audio/amplifiedIE = IB + IC ; IC >> IBTransistor as an amplifierPage 188. Electronic 9. 49 61 19. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers Physics Concept/Principle/Law(b)Vzy = 1 V1. VXY = 5 V2, R1 x 6 = 5R1 + 10003, R1 = 5000 2 (a) 0001 , AND(b)(c)QTOGETHER we must succeed, TOGETHER we will succeedLogic gatesTruth table31. Connect the dry cell terminal to the Y-input ofCRO.2. The Y-gain is set to a value so that the directcurrent waveform displayed on the screen CRO.3. Determine the distance / part of y-axis.4. Potential different =( Y-gain scale) x (Vertical distance of direct currentwave)CRO4 1. When there is a fire burning, R at T = 3.5 kPotential 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 / greaterthan3.2 V, so the transistor is functioned4. The solenoid become magnetised, G will swicth onand the bell will ringsPotential dividerTransistor as switching circuit5(a)Relationship R and VEffect to VBE; effect to outputPage 19Q QRP 20. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers Physics Concept/Principle/Law(b) During hot weather1. resistance at termistor decrease ,potential difference across thermistor willdecrease2. Potential difference across R will increase3. This will produce bigger base current , and willincreasethe collector current4. Electric relay will switch on the air conditioner.(c) During cold weather , resistance at thermistorincrease.1. Potential difference across thermistor will increase.2. Potential difference across R will decrease.3. This will produce smaller base-current and nocurrentflow in collector circuit.4. Electric relay will swith off the air conditioner.6(a) 7.5 V(b)( )1.5 9 OR I A b 0.0125RtRt+=600= 7.5 =600Rt =120WRt 1.5= =120W0.0125(c)Ic =10010-3 -12.510-3= 87.510-3 ATOGETHER we must succeed, TOGETHER we will succeedVBEPotential dividerIE = IB + ICNo Answers PhysicsConcept/Principle/Law1 1. Small amount of radioisotope is put in the water reservoir2. The substance must be in liquid state so it is easy to flow inthe water3. The substance should emit particles (the radiation canbe detected above the ground )4. A Geiger-Muller counter is moved over the pipeaccording to the layout plan.5. At a point where the Geiger-Muller counter detectedhigh radiation level, indicating the point ofleakage.Radioactive detectorCharacteristic ofradiation2Page 2010. Radioactivity 21. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers PhysicsTOGETHER we must succeed, TOGETHER we will succeedConcept/Principle/Law1. Carbon-14 atom is a radioactive substance which iseasily absorbed by living plants.2. After the plants dies, the activity of Carbon-14 willdecline since no new carbon-14 is absorbed.(carbon-14 will decay to nitrogen-14)3. The difference between the concentration of carbon14in the material to be dated and the4. Concentration in the atmosphere provides gives the rateof carbon-14 decay5. By calculating the activity of carbon-14, the age of thedead plant/fossil can be determined(half-life of carbon-14 is 5,730 years)Application ofradioisotopesCarbon dating3(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 10 71.5 x 10-3= 2.1 x 1010 WNuclear energyE mc24 (a)1. Neutron bombarded a uranium nucleus //Diagram2. Three neutrons produced // Diagram3. The new neutron bombarded a new uranium nucleus //Diagram4. For every reaction, the neutrons produced will generate achain reaction // Diagram(b) E = mc22.9 x 10 -11 = m x (3.0 x 108)2m = 3.22 x 10-28 kgChain reaction5 (a)1- Show the line in the graph2- T1/2 = 4 days(b)Half lifePage 21 22. PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014No Answers PhysicsTOGETHER we must succeed, TOGETHER we will succeedConcept/Principle/Law1. Shape of graph2. One point is correct3. Two or more point61. 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, thethickness of the aluminium foil is too thick/ vice versaApplication ofradioactivePage 22