Perfect Score 2014 Ans
description
Transcript of Perfect Score 2014 Ans
-
http://spmsoalan.wordpress.com/
PERFECT SCORE MODULE
Sekolah
Berasrama Penuh
Kementerian
Pelajaran Malaysia
2014 NAME: ...
SCHOOL..
PHYSICS Beyond A+
TEACHERS EDITION
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 2 TOGETHER we must succeed, TOGETHER we will succeed
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)
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 3 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 4 TOGETHER we must succeed, TOGETHER we will succeed
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
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 5 TOGETHER we must succeed, TOGETHER we will succeed
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
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 6 TOGETHER we must succeed, TOGETHER we will succeed
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
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 7 TOGETHER we must succeed, TOGETHER we will succeed
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
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 8 TOGETHER we must succeed, TOGETHER we will succeed
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
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 9 TOGETHER we must succeed, TOGETHER we will succeed
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
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 10 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 11 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 12 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 13 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 14 TOGETHER we must succeed, TOGETHER we will succeed
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
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 15 TOGETHER we must succeed, TOGETHER we will succeed
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
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 16 TOGETHER we must succeed, TOGETHER we will succeed
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
More SPM papers at: http://spmsoalan.wordpress.com/
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 17 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 18 TOGETHER we must succeed, TOGETHER we will succeed
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
Page 19 TOGETHER we must succeed, TOGETHER we will succeed
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
-
PHYSICS PERFECT SCORE SEKOLAH BERASRAMA PENUH 2014
Page 20 TOGETHER we must succeed, TOGETHER we will succeed
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/