modul perfect score sbp physics spm 2014 skema

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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 oRefract