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Page 1: Modul perfect score sbp physics spm 2014 skema

PERFECT SCORE MODULE

Sekolah Berasrama Penuh Kementerian Pelajaran Malaysia

2014 NAME: …………………………..………………………………………………….

SCHOOL………………………………………………………………………………..

PHYSICSBeyond A+

TEACHERS EDITION

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‘PHYSICS PERFECT SCORE’ SEKOLAH BERASRAMA PENUH 2014

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 mahjong

4. ‘Marker pen’

5. Label kumpulan (cadangan: mengikut topik sebagai ‘expert group’)

6. Alat radas (jika perlu)

CARTA ALIR PELAKSANAAN PROGRAM (Minimum 10 Jam)

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SECTION CONTENT Page

A Diagnostic Test – Answer & Analysis 5 - 6

1. Force & Motion 7

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1 jam 30 minit (pemilihan item adalah mengikut kelemahan pelajar dan dijalankan sebagai pra ujian)

15 minit

2 jam 15 minit

Minimum 6 jam (mengikut kelemahan pelajar)

Ujian Diagnostik (Section A)Ujian Diagnostik (Section A)

Semak JawapanAnalisis Skor IndividuSemak Jawapan

Analisis Skor Individu

Perbincangan soalan Diagnostik bersama Guru berdasarkan topik yang dikenalpasti

lemah

Perbincangan soalan Diagnostik bersama Guru berdasarkan topik yang dikenalpasti

lemah

Berdasarkan Analisis Skor, pelajar mengenalpasti tajuk yang belum dikuasai Berdasarkan Analisis Skor, pelajar mengenalpasti

tajuk yang belum dikuasai

Pelajar dibahagikan kepada kumpulan mengikut topik yang belum dikuasaiPelajar dibahagikan kepada kumpulan mengikut topik

yang belum dikuasai

PEMANTAPANPerbincangan di dalam kumpulan soalan pada Section B (mengikut topik paling lemah yang

dikenalpasti melalui Analisis Skor)

PEMANTAPANPerbincangan di dalam kumpulan soalan pada Section B (mengikut topik paling lemah yang

dikenalpasti melalui Analisis Skor)

Sessi pembentangan / Perkongsian konsep/kemahiranSessi pembentangan / Perkongsian

konsep/kemahiran

PengayaanLatihan menggunakan Modul Perfect Score 2013

mengikut kemahiran(mengikut kesesuaian sekolah)

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B

Answer for Enhancement Question

2. Force and Pressure 8

3. Heat 10

4. Light 12

5. Waves 15

6. Electricity 16

7. Electromagnetism 17

8. Electronics 18

9. Radioactivity 20

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)

Question

AnswerNumber of Wrong

ResponseTopic Remarks

1. B Force and Motion2. B3. A4. B

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Question

AnswerNumber of Wrong

ResponseTopic Remarks

5. B6. B7. D

F&P

8. D9. C10. C11. A12. C13. C14. B15. D16. C

Heat

17. D18. C19. C20. D21. C22. A23. B24. B25. D

Light

26. A27. C28. C29. B30. B31. C32. C33. D34. C35. A36. A

Waves

37. B38. B39. B40. D41. C42. D43. A

Electricity

44. B45. A46. D47. C48. A49. D50. C51. A52. C53. C54. B55. D56. C57. A

Electromagnetism

58. D59. B60. D61. C62. D63. B64. A

Electronics65. B66. D67. A

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Question

AnswerNumber of Wrong

ResponseTopic Remarks

68. CRadioactivity69. A

70. C

SECTION B

1. Force & Motion 4 - 13

No Answers Physics Concept/Principle/Law1

Motion graph

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No Answers Physics Concept/Principle/Law

At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds.

2

a 97.2o

b

c 4.54 N

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

2. Forces and Pressure

No Answers Physics Concept/Principle/Law1 (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

Atmospheric pressureDifference in pressure

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No Answers Physics Concept/Principle/Law(b) mirror weight= ρVg= 2.5 x 10 3 x 1.5 x 0.5 x 0.01 x10 = 187.5 N

2 1. 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 surface 13. Lower surface spins more faster than the upper surface of the ball 1

Bernoulli’s 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 mercury

Atmospheric 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 genuine

Archimedes’ principledensity

6

1. 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

7 1. 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

Pascal principleForce multification

F1/A1 = F2/A2

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No Answers Physics Concept/Principle/Law 2 21 = 15 DB

DB = 28 cm

8

1. 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

9

1. Gas flows out through the jet with high velocity2. According to Bernoulli’s 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

Bernoulli’s principle

3. Heat 14 - 24

No ANSWER Concept/Principle

1 1. When temperature increases, the average kinetic energy Pressure Law

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No ANSWER Concept/Principleincreases

2. 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.

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 = P2

T1 T2

3 31.25oC

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

Specific Heat Capacity

5

(i) 100°C(ii) m=ρV = (1) (100) = 100g(iii) .2 x 379 ( 100-T) = 0.1 x 4200 x (T-28) T = 39° C

m1 C1 θ 1 = m2 C2 θ 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) 40°C(iii) Prevent heat loss to surrounding(b) (i)

Heat supplied by hot metal = heat received by waterm1 C1 θ 1 = m2 C2 θ 2

0.4 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-1°C-1

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

m1 C1 θ 1 = m2 C2 θ 2

8 A Boyle’sLaw

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

m1 C1 θ 1 = m2 C2 θ 2

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No ANSWER Concept/Principle

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

10

The 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

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Num.

Answer Concept

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 size

The law of reflection Plane mirror

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Num.

Answer Concept

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

mirror

4. 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 object

2. 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 time

4. The mirror can be tilted // use photosensors mounted in the

mirror to detect light and dim the mirror

characteristics 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 o

Refractive 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,

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Num.

Answer Concept

before it enters the observer’s 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 refraction

Light 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 diver

Arrow: 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

screen

3.The distance between the screen and the lens is measuredl

4.Focal length = distance between the screen and the lens

Focal Point And Focal Length Of A Lens

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Num.

Answer Concept

11.

Real, inverted, diminishedv = 15 cmm = v/u m = 0.5

Relationship 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

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Num.

Answer Concept

1 20 cm s-1

2

When the singer sings, she produces a high frequency sound

The frequency of the glass equal with the frequency of the

singer’s sound

Both systems are in resonance

So the glass will oscillates at its maximum aplitude and it breaks.

3 170 mS = vt 2

4

When the prongs of the tuning fork move outward, it produce a

region of compression

When the prongs of the tuning fork move inward, it produce a

Region of rarefaction

Candle flame in front of a loud speaker that emits sound wave

Candle 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 area

Show the wavelength is decreased in shallow area And is equal in deep area

(c) =

= 4.5 m (answer with correct unit)

6. Electric

um Answer Concept

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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 Ω1/r = 1/20 + 1/20 @ 1/r = 2/20 @ 1/r = 1/10r = 10 Ω

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

No AnswersPhysics

Concept/Principle/Law1. A device that transfers electrical energy into sound

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The wire from the amplifier carries an alternating current The interaction between magnetic field of the current carrying

conductor and the permanent magnet produces force The coil which can slide backwards and forwards over the central

pole of a circular permanent magnet makes the coil (and the papercone) move backwards and forwards

at the same frequency as the changing current. The paper cone then moves

the air backwards and forwards which creates the sound

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 wires4. (a) 1. Magnet pushed inside, magnetic flux is cut by the wire

2. According to Faraday’s 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 Faraday’s 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 Faraday’s 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 Lenz’s Law 4. I will flows in anti clock wise direction

Induced emfInduced currentFaraday’s LawFactors affected induced emfLenz’s 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. Electronic 9. 49 – 61

No Answers Physics Concept/Principle/Law1 (a)

1. When someone speaks at the microphone, the IE = IB + IC ; IC >> IBTransistor as an amplifier

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No Answers Physics Concept/Principle/Lawcurrent 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 Ω

2 (a) 0001 , AND

(b)

(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

4 1. When there is a fire burning, R at T = 3.5 kΩ 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

Potential dividerTransistor as switching circuit

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Q Q

R

Q

P

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No Answers Physics Concept/Principle/Law and the bell will rings

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

No AnswersPhysics

Concept/Principle/Law1 1. Small amount of radioisotope is put in the water reservoir

2. The substance must be in liquid state so it is easy to flow in the water

Radioactive detectorCharacteristic of

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No AnswersPhysics

Concept/Principle/Law3. The substance should emit γ 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.

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 carbon–14 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 10 7

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 = mc2

2.9 x 10 -11 = m x (3.0 x 108)2

m = 3.22 x 10-28 kg

Chain reaction

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

(b)

Half life

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No AnswersPhysics

Concept/Principle/Law

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

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

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