Perfect Score 2014 Ans

21
http://spmsoalan.wordpress.com/ PERFECT SCORE MODULE Sekolah Berasrama Penuh Kementerian Pelajaran Malaysia 2014 NAME: …………………………..…………………………………………………. SCHOOL……………………………………………………………………………….. PHYSICS Beyond A+ TEACHERS EDITION

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Physics SPM

Transcript of Perfect Score 2014 Ans

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    PERFECT SCORE MODULE

    Sekolah

    Berasrama Penuh

    Kementerian

    Pelajaran Malaysia

    2014 NAME: ...

    SCHOOL..

    PHYSICS Beyond A+

    TEACHERS EDITION

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    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)

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    CARTA ALIR PELAKSANAAN PROGRAM (Minimum 10 Jam)

    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)

    Semak Jawapan

    Analisis Skor Individu

    Perbincangan soalan Diagnostik

    bersama Guru berdasarkan topik yang dikenalpasti

    lemah

    Berdasarkan Analisis Skor, pelajar mengenalpasti

    tajuk yang belum dikuasai

    Pelajar dibahagikan kepada kumpulan mengikut topik

    yang belum dikuasai

    PEMANTAPAN

    Perbincangan di dalam kumpulan soalan pada

    Section B (mengikut topik paling lemah yang

    dikenalpasti melalui Analisis Skor)

    Sessi pembentangan / Perkongsian

    konsep/kemahiran

    Pengayaan

    Latihan menggunakan Modul Perfect Score 2013

    mengikut kemahiran

    (mengikut kesesuaian sekolah)

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    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 CONTENT Page

    A Diagnostic Test Answer & Analysis 5 - 6

    B

    Answer for

    Enhancement

    Question

    1. Force & Motion 7

    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

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    SECTION A:

    DIAGNOSTIC TEST (ANSWER AND ANALYSIS)

    Question Answer Number of Wrong

    Response Topic Remarks

    1. B

    Force and Motion

    2. B

    3. A

    4. B

    5. B

    6. B

    7. D

    F&P

    8. D

    9. C

    10. C

    11. A

    12. C

    13. C

    14. B

    15. D

    16. C

    Heat

    17. D

    18. C

    19. C

    20. D

    21. C

    22. A

    23. B

    24. B

    25. D

    Light

    26. A

    27. C

    28. C

    29. B

    30. B

    31. C

    32. C

    33. D

    34. C

    35. A

    36. A

    Waves

    37. B

    38. B

    39. B

    40. D

    41. C

    42. D

    43. A

    Electricity

    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

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    Question Answer Number of Wrong

    Response Topic Remarks

    57. A

    Electromagnetism

    58. D

    59. B

    60. D

    61. C

    62. D

    63. B

    64. A

    Electronics 65. B

    66. D

    67. A

    68. C

    Radioactivity 69. A

    70. C

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    SECTION B

    No Answers Physics Concept/Principle/Law

    1

    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.

    Motion graph

    2

    a 97.2o

    b

    c 4.54 N

    3

    Constant speed, resultant force = 0

    F - 40 - 600 sin 25 = 0

    F = 293.57 N

    1. Force & Motion 4 - 13

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    No Answers Physics 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 N

    Atmospheric pressure

    Difference in pressure

    2

    1. Spinning ball moving in the opposite direction with

    air flow at the upper surface 1

    2. Spinning ball moving in the same direction with air

    flow at the lower surface 1

    3. Lower surface spins more faster than the upper

    surface of the ball 1

    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 mercury

    Atmospheric pressure

    Simple 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 pressure

    Atmospheric pressure

    5

    (a)1. Measure the mass of the necklace

    2. 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

    displaced

    5. Density of the necklace = mass/volume

    (b)1. density =V

    m =

    cm3 20

    g 265 = 13.25 g cm-3

    2. Percentage = 3.27

    25.13 x 100% = 48.5%

    3. The necklace diamond is not genuine

    Archimedes principle density

    6

    1. The best time is early morning

    2. The cool air is denser

    3. More air molecules can be displaced

    4. Produced more buoyant force

    The balloon can rise higher

    Buoyant force

    density

    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 oil

    3. It obeys Pascal Law

    4. The same pressure exerted on bigger area, Piston

    B will produce bigger force (F=P x A)

    Pascal principle

    Force multification

    F1/A1 = F2/A2

    2. Forces and Pressure

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    No Answers Physics Concept/Principle/Law

    (b)

    FB = ( FA AB) (AA)

    = (50 15) (2)

    = 375 N

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

    8

    1. When the catch is still in the water, the buoyant

    force is bigger

    2. When the catch is getting out from the water, the

    volume of object immerse is smaller

    3. The volume of water displaced also smaller, thus

    the weight of water displaced is getting smaller

    4. The buoyant force is equal to the weight of water

    displaced

    5. 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 velocity

    2. According to Bernoullis Principle, high velocity will produce low pressure at the nozzles of the jet

    3. Higher atmospheric pressure pushes the air inside

    the cylinder trough the orifice

    4. The air will mix with the gas and complete

    combustion will occur

    Bernoullis principle

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    No ANSWER Concept/Principle

    1

    1. When temperature increases, the average kinetic energy increases

    2. Rate of collision between the air molecules and wall of the tire also

    increases.

    3. Rate of change of momentum increases

    4. 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 of

    the 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 Hg

    TO = 300 x 75 = 75 K

    100

    (iii) Pressure Law

    Pressure Law

    P1 = P2

    T1 T2

    3 31.25oC

    4 At lower lan, the density of air is higher.

    Hence it is more difficult to vaporize

    Specific 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 C

    m1 C1 1 = m2 C2 2

    6

    Q = mc

    7

    (a) (i) - The rate of heat transfer between two bodies are

    The 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 water

    m1 C1 1 = m2 C2 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-1C-1

    (ii) Heat released by water is absorb by the metal //

    no heat loss to surrounding

    m1 C1 1 = m2 C2 2

    3. Heat 14 - 24

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    No ANSWER Concept/Principle

    8 A BoylesLaw

    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

    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 original

    position.

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    Num. Answer Concept

    1.

    1. Light rays and reflection

    2. Extrapolate and draw the image

    3. Incident angle = reflected angle; Object distance = image

    distance

    4. Characteristics of image: Virtual, inverted, same size

    The 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 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 index

    Critical 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

    4. Light

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    Num. Answer Concept

    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 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

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    Num. Answer Concept

    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

    11.

    Real, inverted, diminished

    v = 15 cm

    m = 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

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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 singers

    sound

    Both systems are in resonance

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

    3 170 m S = 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)

    5. Waves

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    um Answer Concept

    1 (i) V2 = 4 V

    (ii) I =5

    4 = 0.8 A

    (ii) R = 8.0

    2 = 2.5

    V = IR

    2 (a) Total resistance in the circuit (b) If one bulb is blown the other still can be used Lower the total resistance Maintain the potential difference same as the supply through

    The 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 = 60/50

    r = 1.2 1/r = 1/20 + 1/20 @ 1/r = 2/20 @ 1/r = 1/10

    r = 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

    6. Electric

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    No Answers Physics

    Concept/Principle/Law

    1. 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 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 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 emf

    Induced current

    Faradays Law Factors affected

    induced emf

    Lenzs law

    5. 1. rotate the coil in clock wise direction

    2. the coil cut across the magnetic field

    3. current is induced in the coil

    4. the commutator change the direction in the coil so that the

    direction of current in external circuit I always the same.

    generators

    7. Electromagnet

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    No Answers Physics Concept/Principle/Law

    1 (a)

    1. When someone speaks at the microphone, the

    current

    produced flows to the circuit

    2. The capacitor is used to avoid direct current from

    battery

    to flow through the microphone

    3. The current will give changes to the magnitude of

    base-

    current// IB become bigger

    4. 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

    (b)

    (c)

    Logic gates

    Truth table

    3

    1. 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 k

    Potential difference across P = 3500 x 6 =

    2.2 V

    (3500 + 6000)

    2 Potential difference across Q = 6000 x 6 =

    Potential divider

    Transistor as switching circuit

    8. Electronic 9. 49 61

    Q Q

    R

    Q

    P

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

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    No Answers Physics Concept/Principle/Law

    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

    5

    (a)

    (b) During hot weather

    1. resistance at termistor decrease ,

    potential difference across thermistor will

    decrease

    2. Potential difference across R will increase

    3. This will produce bigger base current , and will increase

    the collector current

    4. 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 V

    Effect to VBE; effect to output

    6

    (a) 7.5 V

    (b)

    Rt

    Rt

    600

    95.1 OR AIb 0125.0

    600

    5.7

    120Rt 1200125.0

    5.1Rt

    (c) 33 105.1210100 Ic

    = A3105.87

    VBE

    Potential divider

    IE = IB + IC

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    No Answers Physics

    Concept/Principle/Law

    1 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

    3. The substance should emit particles (the radiation can be 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 detector

    Characteristic of

    radiation

    2

    1. 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 decay

    5. 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 = 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 graph

    2- T1/2 = 4 days

    Half life

    10. Radioactivity

  • PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014

    Page 21 TOGETHER we must succeed, TOGETHER we will succeed

    No Answers Physics

    Concept/Principle/Law

    (b)

    1. Shape of graph

    2. One point is correct

    3. Two or more point

    6

    1. Put the radioactive source opposite the detector

    2. Detector is connected to the thickness indicator

    3. Detector detect the reading of the changes in counts

    4. 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

    More SPM papers at: http://spmsoalan.wordpress.com/