ANSWER SCHEME PHYSICS MODULE 2022
CHAPTER 1 - MEASUREMENT 1
1 (a) Base // scalar 1
1
(b) T = 24/20 1
= 1.2 s 4M
1
(c) Time increases 1
1
2 (a) Length TOTAL
TOTAL 1
(b) Imperial unit 1
(c) mililiter – metric unit TOTAL 5M
TOTAL 3
inch – imperial unit TOTAL
TOTAL 1
gram - metric unit TOTAL 1
Miles – imperial unit 1
6M
3 (a) 1
1
Base quantity Derived quantity 1
1
Time Speed 4M
1
Length 1
1
(b) Physical quantity – length / height 1
Magnitude – 1.65 4M
Unit – meter // m 1
1
4 (a) The change in temperature of an object 1
(b) More sensitive 1
(c) Length // volume 4M
(d) Expand uniformly // high boiling point // opaque 1
1
5 (a) Temperature 1
(b) 0.1o C 1
(c) Digital thermometer // thermometer Q 4M
1
(d) Smaller scale 1
1
6 (a) Miliammeter // Ammeter 1
(b) Scalar
(c) 0.2 A
(d) Electric Current
7 (a) P inversely proportional to V // P is directly proportional to 1/V
(b) Draw a big triangle on the graph in Diagram 7.
(c) m= 2− 1 =
2− 1 1/
= PV
8 (a) Derived // scalar
(b) Draw a triangle
m = 2− 1
2− 1
=
(c) V decrease linearly to I 1
TOTAL 5M
CHAPTER 2 - FORCE & MOTION I 1
1
1 (a) P to Q 1
(b) Constant velocity
(c) Distance between consecutive dots are the same 1
(d) V = s/t 1
= 8.0 /(0.02 x 5) 5M
= 80.0 cm s-1 1
TOTAL
1
2 (a) Length is distance / path / route covered by an object from initial to final 1
position. 1
(b) (i) same 1
1
(ii) length in Diagram 2.2 is greater 1
(iii) average speed in Diagram 2.2 is greater
(c) The greater the length of ticker tape the greater the average speed 7M
(d) WX – the trolley move with increasing velocity 1
XY – the trolley moves with constant velocity
WX – trolley moves with constant acceleration 1
XY – trolley moves with zero acceleration 1
TOTAL 1
3 (a) Displacement is shortest path travelled by an object from the initial 1
5M
position and the final position in a specific direction 1
(b) Adam is at rest // not moving // stationary 1
1
(c) Total displacement = 70 + (-40) + 40 = 40 m 1
(d) V = s/t 4M
1
= 40/20 1
= 2 ms-1 1
1
TOTAL 4M
1
4 (a) Displacement 1
(b) (i) time
(ii) acceleration 1
(iii) force // momentum 1
TOTAL 4M
1
5 (a) BC – car moving with increasing velocity // the car accelerate 1
(b) V = (0 – 10)/(22-18) 1
= - 2.5 ms-1
(c) Velocity
TOTAL
6 (a) Velocity is rate of change of displacement
(b) (i) the car moves with zero acceleration at PQ
(ii) distance = ½ (u + v) t
= ½ (4 + 10) 2
= 14 m
TOTAL
7 (a) Rate of change of displacement
(b) (i) bus moves with decreasing velocity // constant deceleration
(ii) X to Y
(c) S = 2 x ½ (10)(10) 1
= 100 m 1
TOTAL 5M
8 (a) The motion of object affected by gravitational force // object move due to 1
gravitational field strength
(b) g 1
m
(c) North pole. 1
1
Distance nearer to the centre of the earth 4M
1
TOTAL
1
9 (a) Tendency of an object to remain at rest or, if moving to continue its 1
1
motion in a straight line at uniform velocity. 1
1
(b) A speeding car
6M
Car smaller inertia due to lower mass 1
1
(c) Divided tank will divide the mass of the oil
1
Inertia of the oil will be smaller
1
(d) Passenger in a car lurch forward when a moving car suddenly stops // 1
5M
souse come out when a shaking bottle suddenly stops // raindrops spun off 1
when a wet umbrella is rotated and suddenly stopped.
TOTAL
10 (a) Product of mass and velocity
(b) Use higher pressure inside the bottle // mass of the bottle is keep to
minimum
(c) The rocket will never tumble during flight // more stable as it flies
through the air
(d) P = mv
0.8 = 0.4 x v
V = 2 ms-1
TOTAL
11 (a) V2 V1
(b) Principle of conservation of momentum 1
(c) m1 v1 = m2 v2
TOTAL 1
0.25 x v = 0.01 x 6 1
v = 0.24 ms-1 1
(d) Rocket launching 5M
1
12 (a) Elastic collision 1
(b) m1 u1 + m2 u2 = m1 v1 + m2 v2 1
(c) u1 = (110 x 1000)/3600 = 30.56 ms-1 1
u2 = (100 x 1000)/3600 = 27.78 ms-1
m1 u1 + m2 u2 = (m1 + m2)v 1
(1000 x 30.56) + (800 x 27.78) = (1000 +800) v
v = 52 784 / 1800 1
= 29.323 ms-1 6M
TOTAL 1
1
13 (a) Rate of change of velocity
(b) States that the rate of change of momentum is directly proportional to the 1
force and acts in the direction of the applied force.
(c) Newton, N // kg m s-2 2
(d) F = ma 1
= (60 + 70 + 11.8) x 6 6M
= 850.8 N 1
TOTAL
1
14 (a) Rate of change of momentum in a collision or impact in a short period of 1
time 1
(b) In a collision the crumple zone increase the time impact 1
This will reduce the impulsive force 1
6M
(c) u = (100 x 1000)/3600 = 27.78 ms-1 1
F = (mv – mu)/t 1
= [(1500 x 0) – (1500 x 27.78)]/15 1
= - 2778 N 1
TOTAL 1
1
15 (a) Impulse is change of momentum 1
(b) The mattress is soft, time impact is high 1
Impulsive force decreases and reduce injuries 8M
(c) (i) same 1
(ii) height of high jump bar in Diagram 15.2 is greater 1
(iii) the final velocity in Diagram 15.2 is greater
(d) (i) the greater the height the greater the final velocity 1
(ii) the greater the final velocity the greater the impulse. 1
TOTAL 1
1
16 (a) Weight 6M
(b) (i) W = mg
250 = 20 g 1
g = 250/20
= 12.5 N kg-1 1
1
(ii) yes 1
(c) W = mg 1
5M
= 68 x 1/6 x 9.81 1
= 111.18 N
TOTAL
CHAPTER 3 - GRAVITATION
1 (a) Universal force // force acts between two bodies in the universe.
(b) F = GMm / r2
= ( 6.67 x 10 -11 x 5.97 x 1024 x 1.2 x 103 ) / (7.87 x 106)2
= 7714.929 N
(c) Satellite and moon
Lower mass // distance between the object is greater
TOTAL
2 (a) Universal force // force acts between two bodies in the universe.
(b) (i) kg m s-2 1
(ii) F = ( 6.67 x 10 -11 x 70 x 53 ) / 12 1
= 2.475 x 10-7 N 1
(c) Mass too small. 1
3 (a) (i) v = 2πr / T TOTAL 5M
= 2π (3.84 x 108 ) / (27.3 x 24 x 60 x 60)
1
v = 1022.904 ms-1 1
(ii) F = ( 6.67 x 10 -11 x 5.97 x 1024 x 7.35 x 1022 ) / (3.84 x 108 )2 1
= 1.985 x 1020 N 1
(b) When the moon orbits the earth there is a centripetal force acting 1
between moon and earth.
Fg = Fc so the moon stays at same distance from the earth. 1
TOTAL 6M
4 (a) States that the gravitational force between two bodies is directly 1
proportional to the product of the masses of both bodies and inversely
proportional to the square of the distance between the centres of the two
bodies.
(b) (i) W = mg W = mg
= 8 x 9.81 = 20 x 9.81
= 78.48 N = 196.2 N
Weight P = 78.48 N 1
Weight Q = 196.2 N 1
(ii) F = GMm / r2 1
= ( 6.67 x 10 -11 x 8 x 20 ) / 1.752 1
= 3.485 x 10-9 N
(iii) No 1
(iv) The gravitational force between two small masses is small 1
(c) T = F = mv2/ r 1
= ( 0.075 x 1.82 ) / 0.8
= 0.304 N 1
TOTAL 9M
5 (a) Accelerate 1
(b) (i) unchanged 1
(ii) the mass of object does not affect the gravitational acceleration. 1
From formula g = GM/r2, g is affected by radius and mass of Earth 1
(c) (i) 1
1
(ii) g = 9.81 ms-2 1
(d) Blood circulation system – blood pump to heart decreases, reduce the 1
ability of regulate // blood tend to collect in the upper part of the body// 1
facial edema // decrease the ability to absorb oxygen. 9M
Bone fragility – bone loss, more fragile and more susceptible to fractures 1
1
// production of osteoblasts decreases 1
1
TOTAL 1
1
6 (a) Base // scalar 1
1
(b) (i) same mass 1
1
(ii) orbital period T2 is greater
1
(iii) orbital radius r2 is greater
1
(iv) the greater the radius the greater the period
(c) Kepler’s Third Law 1
1
(d) (i) unchanged
(ii) from formula T2 = (4 π 2 ) r3 / GM 1
1
period not affected by mass of satellite. 1
7 (a) Satellite that appears to be located at a fixed point in space when viewed 1
from the earth’s surface.
9M
(b) Position of satellite is relative to the earth antenna // move with same 1
1
rotation of earth 1
1
Period of rotation is 24 hours //Always at the same geographical location
(c) (i) v = √(GM)/r 1
1
= √ (6.67 x 10-11 x 5.97 x 1024 ) / (4.23 x 107) 1
= 3068.173 ms-1 1
1
(ii) Fg = GMm / r2 9M
= ( 6.67 x 10-11 x 5.97 x 1024 x 1200 ) / (4.23 x 107)2
1
= 267.055 N 1
1
(d) Fall into lower orbit and continue to revolve towards the earth until enters 1
the atmosphere
Movement of satellite of high linear speed against air resistance will
generate heat and causes satellite to burnt.
TOTAL
8 (a) Non – geosynchronous orbit
(b) Period shorter than 24 hours // lower height from earth
(c) Yes
Direction of its velocity constantly changing.
(d) v = (2πr)/T
= [2π (6400000 + 603000)] / (97 x 60 )
= 7560.334 m s-1
(e) Centripetal force
(f) Fc = (m v2 ) / r
= [10000 x (7560.334)2] / (6400000 + 603000)
= 81620.234 N
TOTAL
CHAPTER 4 - HEAT
1 (a) the change in temperature of an object
(b) More sensitive
(c) Length // volume
(d) Opaque // expand uniformly // high boiling point
TOTAL 4M
2 (a) Temperature 1
(b) 0.1o C 1
(c) Q 1
(d) Smaller scale 1
TOTAL 4M
3 (a) Kelvin // K 1
(b) No net heat flow between the liquid in the column and substance // 1
thermal equilibrium achieved 1
(c) T = 7−0 x 100 1
12−0
= 58.33o C
TOTAL 4M
4 (a) Heat energy required to increase the temperature of 1 kg substance by 1o 1
C
(b) To prevent heat loss to surrounding 1
(c) Pt = mcθ
20 x 10 x 60 = 0.5 x c x 26.6 2
C = 902.256 J kg-1 oC-1 1
(d) Rise in temperature is small
TOTAL 5M
5 (a) Scalar quantity 1
(b) (i) copper 1
Increase temperature faster 1
(ii) small 1
lighter 1
(c) (i) Pt = mcθ
500 t = 2.2 x 387 x 50 1
t = 85.14 s 1
(ii) Pt = mcθ
500 t = 2.5 x 500 x 50 1
t = 125 s 1
TOTAL 9M
6 (a) A form of energy 1
(b) Water bigger specific heat capacity 1
Larger mass contain greater heat energy 1
(c) mcθ = mcθ
2(T-30) = 9 (95-T) 1
2T – 60 = 855 -9T 1
T = 83.182 oC
TOTAL 5M
7 (a) When two object in thermal contact, temperature of both object are the 1
same and net heat transfer between the two object are zero
(b) 45oC 1
(c) Q = mcθ 1
= 0.3 x 900 x (100 – 45 )
= 14850 J 1
1
(d) Q = mcθ
14850 = 0.2 x 4200 x (45 – T)
T = 27.321 oC 1
TOTAL 6M
1
8 (a) B
1
(b) (i) E = Pt 1
= 16 x 25 x 60
= 24000 J 1
1
(ii) Pt = mcθ 5M
24000 = 0.5 x 4200 x θ 1
θ = 20 oC 1
TOTAL 1
1
9 (a) Heat energy required to increase temperature of 1 kg substance by 1oC 1
(b) (i) change in temperature in Diagram 9.2 is greater 1
(ii) specific heat capacity of Diagram 9.2 is lower
2
(iii) amount of heat supplied is the same
(iv) mass of both are the same 8M
(c) The smaller the specific heat capacity the greater the change in 2
temperature 1
(d) • Sand has smaller specific heat capacity, it is heated faster 1
1
• At daytime temperature of sand increase faster than water 1
• Hot air above the land rises upwards 1
• Cool air above the sea surface move towards land 1
1
TOTAL 9M
10 (a) P and R 1
1
(b) Q : Boiling
R : Freezing 1
1
(c) Ethanol has lower specific latent heat of vaporisation 4M
(d) (i) Q = mcθ + mlf + mcθ 1
1
= 0.9 (4.2 x 103(39) + 0.9 (3.34 x 105 ) + 0.9 ( 2 x 103)(8)
= 462420 J 1
(ii) no heat loss to the surrounding 1
1
TOTAL 1
11 (a) A measure of degree of hotness
1
(b) 80 oC 1
(c) Pt = mlf 1
100 x (90-50) = 0.025 lf
lf = 160000 J kg-1
TOTAL
12 (a) Melting
(b) Energy absorbed to overcome attractive forces between molecules
(c) Q = mlf
3360 = m (3.36 x 105)
m = 0.01 kg
(d) (i) high
Not easy to increase in temperature // absorbed heat slower from
surrounding
(ii) heat insulator // wood // high specific heat capacity
No heat allow into the container // temperature increase slower
(iii) lid with clip // container 2 layer // layer is vacuum
Reduce heat conduction // reduce heat absorbed TOTAL 1
TOTAL 10M
13 (a) 1 : volume
2 : pressure 1
1
(b) Constant // unchanged 1
(c) Boyle’s Law 1
4M
14 (a) (i) the volume increases 1
(ii) at the surface the pressure decreases 1
1
(b) The lower the pressure the greater the volume of trapped gas 1
(c) Boyle’s Law 2
(d)
Shape -1M
Lable axis – 1M
TOTAL 6M
1
15 (a) 1 : Volume 1
1
2 : Temperature
1
(b) Charles’ Law
1
(c) = 5M
1
10 12 1
27+273 = 1
1
T = 360 K 1
1
= 360 – 273 = 87 oC 1
1
TOTAL
8M
16 (a) A form of energy transfer from hot to cold object 1
1
(b) (i) length of trapped air in Diagram 16.2 is greater 1
1
(ii) temperature of trapped air in Diagram 16.2 is greater 1
1
(iii) volume 1
1
(c) The greater the temperature the greater the volume
(d) Charles’ Law 8M
(e) (i) Absolute zero
(ii) molecule of substance static // molecule of gas at rest // molecules not
moving
TOTAL
17 (a) Kelvin // K
(b) (i) volume of trapped gas in Diagram 17.2 is greater
(ii) temperature of tapped gas in Diagram 17.2 is greater
(iii) pressure exerted is the same // constant
(c) The greater the temperature the greater the volume
(d) Charles’ Law
(e) 35 =
40+273 70+273
V = 38.355 m3
TOTAL
18 (a) Degree of hotness 1
1
(b) (i) pressure decreases 1
(ii) -273 oC 1
(c) At rest // static // not moving 1
(d) = 1
6M
1.25 100000 = 1
27+273 42+273 1
1
P = 131250 Pa 1
1
TOTAL 5M
19 (a) Increases 1
1
(b) Pressure increases 1
1
(c) When kinetic energy increases the pressure increases at constant volume
1
(d) The greater the temperature the greater the pressure 1
(e) Gay – Lussac’s Law 1
TOTAL 1
CHAPTER 5 - WAVES
1 (a) Maximum displacement from equilibrium position
(b) B
(c) Same length
(d) Resonance
(e) T = 1/f
=½
= 0.5 s
(f) The frequency of external force is equal to the natural frequency of the
bridge
Amplitude of oscillation become maximum
(g)
2
TOTAL 10M
1
2 (a) Phase 1
1
(b) (i) Depth same 1
1
(ii) frequency in Diagram 2.1 (a) is greater 1
1
(iii) wavelength in Diagram 2.2(b) is greater
(iv) the greater the frequency the lower the wavelength
(c) When depth decreases, the speed of wave decreases
Hence the wavelength decreases
(d) =
1.2/ 1.5 = 0.6/λ 1
1
λ = 0.75 m
TOTAL 9M
3 (a) Longitudinal wave 1
(b) (i) Rarefaction 1
(ii) 1.2 m 1
(c) v = λf
f = 330 / 1.2 1
= 275 Hz 1
(d) Speed increases 1
Distance between particles are closer hence energy easily transferred 1
TOTAL 7M
4 (a) Reflection of sound waves 1
(b) Time for the reflected sound is recorded 1
Using formula s = vt /2 the depth is calculated 1
(c) 1600 = (v x 9.4)/2 1
v = 340.426 m s-1 1
(d) (i) sound waves // longitudinal waves 1
High frequency //sound waves transferred parallel to wave 1
propagation
(ii) frequency high 1
High energy // sound move further 1
TOTAL 9M
5 (a) Reflection of sound waves 1
(b) Longitudinal waves 1
(c) (i) the roughness of surface in Diagram 5.1 is greater 1
(ii) amplitude in Diagram 5.2 is greater 1
(d) (i) the greater the roughness of surface the lower the amplitude of 1
reflected wave
(ii) the higher the amplitude the higher the energy 1
(iii) the higher the roughness the lower the energy of reflected waves 1
(e) (i) microwaves 1
(ii) high frequency 1
TOTAL 9M
6 (a) Refraction of water waves 1
(b) Energy is converged to the cape 1
(c) (i) 1
(ii) correct wave pattern TOTAL 2
5M
7 (a) Transverse waves 1
(b) Perpendicular 1
(c) (i) velocity decreases 1
(ii) = 1
1
100/ 40 = v2/20
v = 50 cm s-1
TOTAL 5M
8 (a) Diffraction of water waves 1
(b) Amplitude / energy decreases 1
(c) (i) size of slit in Diagram 8.2 is smaller 1
(ii) Diagram 8.2 the shape is more curve / more circular 1
(d) The smaller the slit the more curve the shape of wave after passing through 1
the slit
TOTAL 5M
9 (a) A phenomena where waves bend /spread out when passing through an 1
opening or around an obstacle.
(b) Unchanged 1
(c)
2
(d) v = fλ 1
1
50 = λx 23
λ = 2.17 m
TOTAL 6M
10 (a) (i) spread // bend 1
(ii) amplitude decreases 1
Energy spread out in a wider area 1
(b) Diffraction of waves 1
(c) Unchanged 1
(d) (i) move downwards // at trough 1
(ii) at rest // static // at calm water 1
(e) Reflection 1
TOTAL 8M
11 (a) Interference of water waves 1
(b) Distance between two consecutive antinodal line increases 1
(c) (i) cork move upwards // cork at maximum amplitude // cork at the peak 1
(ii) crest meet crest // at maximum amplitude 1
(d) Depth increases, wavelength increases 1
According to equation λ = ax/D, the greater the λ the greater the distance 1
between two consecutive antinodal line.
(e) (i) infront of microphone 1
No disturbance of sound from microphone 1
(ii) further / greater 1
To produce more loud sound // greater number of x 1
TOTAL 10M
12 (a) Sound with same frequency and same phase difference 1
1
(b) Longitudinal waves 1
Vibration of medium particle parallel to wave propagation 1
1
(c) (i) Constructive interference
Amplitude is maximum 1
(ii) λ = ax / D 1
= (1.5) (0.4) / 2 7M
= 0.3 m 1
1
TOTAL 1
1
13 (a) Light with one colour or 1 wavelength 1
1
(b) (i) a in Diagram 13.2 is greater
1
(ii) same 1
(iii) same 1
9M
(iv) x in Diagram 13.1 is greater 1
1
(c) The greater the a the lower the x (can use symbol because in question given
symbol) 1
1
(d) λ = ax/D 4M
450 x 10-9 = (0.2 x 10-3)(x) / 6 1
x = 0.0135 m 1
y = 4 x 0.0135 1
= 0.054 m
1
TOTAL 1
14 (a) X-ray 5M
1
(b) To detect broken bone 1
(c) v = λf 1
3 x 108 = λ x 2 x 1013 1
λ = 1.5 x 10-5 m 1
TOTAL 1
1
15 (a) Electromagnetic waves 1
1
(b) 1. transverse waves // can reflect, refract, diffract and interfere 9M
2. transfer energy // speed of light
(c) =
( 2.27 x 108) / λ = ( 3 x 108) / 750 nm
λ = 567.5 nm
TOTAL
16 (a) Equal to
(b) (i) Frequency controller B / Diagram 16.2 is greater
(ii) √ toy car B still can be controlled ….. 80.0 m from the controller
(c) Controller A
λ = (3 x 108) / 2.7 x 107
= 11.11 m
Controller B
λ = (3 x 108) / 4.9 x 107
= 6.122 m
(d) Wavelength of controller A is greater
(e) (i) the greater the frequency the greater the maximum distance
(ii) the greater the frequency the lower the wavelength
TOTAL
17 (a) Number of complete oscillation in I second 1
(b) (i) wavelength of gamma ray is shorter 1
(ii) frequency of gamma ray is bigger 1
(iii) energy of gamma ray is higher 1
(c) The lower the wavelength the greater the frequency 1
(d) Same 1
(e) λ = v/f
= 3 x 108 / 2.5 x 108 1
= 1.2 m 1
TOTAL 8M
CHAPTER 6 – LIGHT AND OPTICS
1 (a) Refraction of light 1
1
(b) Speed of light 2
(c)
TOTAL 4M
2 (a) A phenomena of velocity of light changes when propagates in medium of 1
different optical densities.
(b) Light travel from low optical density to higher optical density 1
(c) (i) angle of refraction in Diagram 2.1 is greater 1
(ii) refractive index in Diagram 2.2 is greater 1
(iii) optical density in Diagram 2.2 is greater 1
(d) (i) the greater the refractive index the lower the angle of refraction 1
(ii) the greater the refractive index the greater the optical density 1
(e) nair sin i = ndiamond sin r
1 x sin 70 = 2.42 sin r 1
r = 22.85 0 1
TOTAL 9M
3 (a) Refraction of light 1
(b) (i) the speed of light in Diagram 3.1 is greater 1
(ii)angle r in Diagram 3.2 is smaller 1
(iii) refractive index in Diagram 3.2 is greater 1
(c) (i) the greater the speed the greater the angle r 1
(ii) the greater the refractive index the lower the speed 1
(d) 2
TOTAL 8M
4 (a) Process of transmission of light by matter // The ratio of the speed of light 1
in medium
(b) (i) speed of light is Perspex / Diagram 4.2 is higher 1
(ii) the refractive index of glass is greater. 1
(c) (i) The greater the speed of light the lower the refractive index 1
(ii) the greater the refractive index the greater the optical density 1
(d) Light travel from dense to less dense medium 1
Speed of light change from low to high, light refract away from normal. 1
Light travel in straight line so object seen above the real object.
(e) Depth of image decreases 1
TOTAL 8M
5 (a) (i) Diagram 5.2 is more bend 1
(ii) density of sea water is greater 1
(iii) the greater the density the greater the bending 1
(b) From side the size of pencil is bigger 1
(c) Refraction of light 1
(d) 1
1
M1 – 2 ray drawn correctly
M2 – image position above the object with dotted line
6 (a) n = 1 TOTAL 7M
sin
1.51 = 1 1
sin 1
C = 41.472 0
1
(b) Critical angle is smaller than incidence angle. 1
1
(c) Total internal reflection
(d) Light propagate from dense to less dense
Angle of incidence is greater than critical angle 1
TOTAL 6M
7 (a) Prism periscope
(b) (i) 1
1
M1 – prism draw correctly
M2 – 1 ray into prism 1 with normal and direction
M3 – 1 ray into prism 2 with normal and direction
(ii) so that total internal reflection can occur 1
(iii) refer Diagram in (b)(i) 2
(c) Same size // upright // virtual 1
TOTAL 6M
8 (a) A phenomenon occurs when light propagate from denser medium to less 1
dense medium which i > c
(b) n = 1/sin c
sin c = 1 sin c = 1 1
2.42 1.5 1
c = 24.41o c = 41.81o 1
(c) (i) refractive index in Diagram 8.2 is greater 1
(ii) incident angle in Diagram 8.2 greater 1
(iii) critical angle in Diagram 8.2 smaller 1
(d) The greater the refractive index the lower the critical angle. 1
TOTAL 8M
9 (a) Total internal reflection 1
(b) i > c 1
Light propagate from dense to less dense. 1
(c) Sharper image formed 1
(d) (i) na sin 30 = n sin 18 1
n = sin 30 = 1.62 1
sin 18
(ii) n = 1/ sin c
1.62 = 1/ sin c 1
c = 38.12o 1
(iii) total internal reflection occur 1
TOTAL 9M
10 (a) Convex lens 1
(b) 1
(c) P = 1 / 10 x 10-2 TOTAL 1
= 10 D 1
4M
11 (a) Distance between optical centre and focal point 1
(b) (i) thickness of the lens is greater in Diagram 11.2 1
(ii) focal length in Diagram 11.2 is shorter 1
(c) The higher the thickness the shorter the focal length 1
(d) Refraction of light 1
(e) (i)
3
(ii) virtual // upright // diminished 1
TOTAL 9M
1
12 (a) Real and inverted 1
(b) (i) object distance in Diagram 12.1 is greater 1
(ii) image distance in Diagram 12.1 is shorter 1
(iii) size image in Diagram 12.1 is smaller 1
(c) (i) The greater the object distance the smaller the image distance 1
(ii) the greater the object distance the smaller the size of image 1
(d) Image is bigger and virtual 1
Image formed at the same side of the object 8M
TOTAL 1
13 (a) Light change in velocity when travel in different optical density 1
(b) (i) P = 1/f
= 1 / 0.02 1
= 50 D 1
(ii) 1/f = 1/u + 1/v
1/5 = 1/15 + 1/v
V = 7.5 cm
(c)
3
(d) (i) f <u< 2f 1
To magnified the image // to produced real and magnified image 1
(ii) > fo + fe 1
to produce image of objective lens greater than 2f 1
(iii) L 1
TOTAL 12M
14 (a) 2
(b) Real, inverted and magnified TOTAL 1
(c) (i) long 1
1
High magnification (Reject : larger) 1
(ii) big 1
Allow more light to enter (Reject : clearer) 1
(d) Lens S 8M
1
15 (a) Reflection 2
(b) Laterally inverted / upright / same size / virtual 3
(c) (i)
(ii) make up mirror // dentist mirror 1
TOTAL 7M
1
16 (a) Reflection of light 1
(b) (i) object distance in Diagram 16.2 is greater 1
(ii) image distance in 16.1 is greater 1
(iii) size of image in Diagram 16.1 is greater 1
(iv) the greater the object distance the lower the image distance 1
(v) the greater the object distance the lower the size of image 3
(c)
17 (a) Convex mirror TOTAL 9M
(b) Wider view TOTAL 1
(c) (i) concave 1
Reflected light is converged 1
(ii) Big 1
Collect more light 1
(iii) Smooth // shiny 1
To reflect greater light 1
(d) Q 1
1
18 (a) Reflection of light 9M
(b) Upright / virtual / diminished 1
(c) 1
(d) (i) concave mirror 3
Image is magnified
(ii) u < f TOTAL 1
To produced upright image 1
1
CHAPTER 7 – FORCE AND MOTION II 1
1 (a) Resultant force is zero 9M
(b) (i) net force, F = 0 1
(ii) resistive force, Fr = 5000 N 1
1
(c) F = ma
9000 – 5000 = 1000 a TOTAL 1
a = 4 m s-2 1
5M
2 (a) Tension 1
(b) (i) F = ma
W – Fr = ma 1
(3 x 9.81 ) – 6 = (3 + 1 ) a 1
a = 5.858 m s-2
(ii) F = ma TOTAL 1
T – Fr = ma 1
T – 6 = 1 x 5.858 5M
T = 11.858 N 1
3 (a) F = 0 1
(b) (i) R – W = ma
R = W + ma
= 68 (9.81) + 68 (2.5)
= 837.08 N 1
(ii) W – R = ma
1
W–R=0 1
R = W = 68 x 9.81
TOTAL 1
= 667.08 N 1
(iii) W – R = ma 7M
1
R = W – ma
=W–mg=0
4 (a) Tension
(b)
Fy 1
(c) Fy = F sin 30 TOTAL 1
F = (50 x 9.81) / sin 30 1
= 981 N 4M
1
5 (a) A push or a pull on an object
(b) Fy = F sin 40 1
= 20 sin 40 1
= 12.86 N
(c) F = ma TOTAL 1
12.86 – 2 = 10 a 1
a = 1.332 m s-2 5M
1
6 (a) Resultant force = 0
(b) (i)
1
(ii) W = mg = 7 x 9.81 = 68.67 N 1
(c) F – w = ma
TOTAL 1
F = ma + W 1
= 7(2) + 68.67 5M
= 82.67 N 1
7 (a) Force acting upwards on the rope.
(b) 3
M1 – label all 3 forces 1
M2 – direction correct 1
M3 – angle correct 1
(c) Using cosine rule
a2 = b2 + c2 – 2 bc cos A @
T2 = 4502 + 490.52 – 2(450)(490.5) cos 40 1
1
T = 323.913 N 1
Using sin rule TOTAL 7M
1
Fy1 = 450 sin 50 2
= 344.72 N
Fy2 = W – Fy1 = 490.5 – 344.72 = 145.78 N
Fy2 = T sin 27
T = 145.78 / sin 27 = 321.11 N
8 (a) Resultant force = 0
(b)
M1 – 2 forces label 1
M2 – direction force correct 1
TOTAL 5M
(c) W = mg
= 15 x 9.81
= 147.15 N
9 (a)
2
M1 – 3 forces label correctly 1
M2 – direction of force is correct 1
(b) W = mg 1
= 150 x 9.81
= 1471.5 N 1
(c) 1
3 forces with correct label and direction 1
(d) W + T1 + T2 = 0 8M
1
TOTAL
10 (a) Elasticity is property of material that enables an object to return to its 1
1
original shape and size after the force applied on it is removed.
(b) k = F/x = 5 /2 1
1
= 2.5 N cm-1 @ 250 N m-1 1
(c) 5 N ---- 2 cm
1
12 N --- x = (12 /5) x 2 1
= 4.8 cm 1
9M
Total extension = 4.8 x 2 = 9.6 cm @ 0.096 m 1
(d) 5N ---- 2 cm
1
Parallel = 12/2 = 6 N 1
x = (6/5 ) x 2
1
= 2.4 cm 1
TOTAL 1
1
11 (a) Maximum force that can be applied to the spring before it lost its 1
elasticity 1
(b) Arrange in parallel
The mass will be divided equally between two spring and it will become
stiffer.
(c) (i) short
Bigger spring constant
(ii) low
Not easy to extend // stiffer
(iii) steel
Strong
(iv) strong 1
1
Not break easily. 11M
1
TOTAL 1
1
12 (a) Diameter // spring constant 1
1
(b) (i) Number of spring in Diagram 12.2 1
1
(ii) compression in Diagram 12.2 is lower 1
8M
(iii) Diagram 12.2 is stiffer
1
(iv) the greater the number of spring, the lower the compression 1
(v) the lower the compression the greater the stiffness 1
1
(c) Diagram 12.2 1
5M
The weight can be divided and the extension will be lower 1
1
TOTAL
1
CHAPTER 8 - PRESSURE 1
1
1 (a) P = = 5M
1
ℎ 1
= = 1
1
P = ρhg 4M
1
(b) N m-1 // Pa 1
1
(c) Shape // area // volume // mass 1
1
TOTAL 1
1
2 (a) Mass per unit volume 1
8M
(b) Patm + PA = Patm + PB 1
ρhg = ρhg 1
1
1000 x 15 = ρ x 8 1
1
ρ = 1875 kg m-3 5M
(c) Decreases
Density of mercury is greater
TOTAL
3 (a) Depth of liquid
(b) D increases
(c) D decreases
Alcohol is less dense
TOTAL
4 (a) Force per unit area
(b) (i) depth of Q is greater
(ii) horizontal distance of water spurting at Q is greater
(iv) The pressure of water at point Q is greater
(c) (i) the greater the horizontal distance the greater the pressure
(ii) the greater the depth of water the greater the pressure
(d) Distance decreases
The depth of water in the storage tank decreases thus pressure decreases
TOTAL
5 (a) Mercury barometer // fortin barometer
(b) 76 cm Hg
(c) P = ρ h g = 13600 x 0.76 x 9.81
= 101396.16 Pa
(d) Length h decreases
TOTAL
6 (a) Aneroid barometer 1
1
(b) Altitude N is greater 1
1
(c) (i) volume Diagram 6.3 is greater 1
1
(ii) atmospheric pressure at M is higher
1
(iii) the greater the altitude the greater the volume of vacuum 1
(iv) The bigger the volume of vacuum the lower the atmospheric 8M
1
pressure
1
(d) Pressure atmospheric outside increases cause the chamber to contract 1
1
This movement drive mechanical levers inside which move the needle 1
1
Small changes in external air pressure causes its metal to extend and 1
7M
contract 1
1
TOTAL 1
7 (a) Pressure due to the weight of the layer of air acting on the surface of the 1
1
earth 5M
(b) High altitude lack of oxygen 1
1
Increase in the breathing rate
1
(c) (i) bigger turbine engine // bigger outflow valve 1
1
Suck in more into the fuselage / cabin 5M
1
(ii) use emergency oxygen mask 1
1
Direct supply of oxygen
1
TOTAL 1
1
8 (a) (i) Pressure gas is greater than atmospheric pressure 6M
1
(ii) Pg = Patm + PH2O 1
1
= 10.3 + 0.36 = 10.66 m H2O 1
(iii) P = ρ h g
= 1000 x 10.66 x 9.81
= 104574.6 Pa
TOTAL
9 (a) Pg = Patm + PHg
= 76 + 40
= 116 cm Hg
(b) P = ρ h g
= 13600 x 1.16 x 9.81
= 154762.56 Pa
(c) No
TOTAL
10 (a) Pascal’s Principle
(b) (i) P = = 5
4
= 1.25 N cm-2 // 12500 Pa
(ii) F = PA
= 12500 x 8 x 10-4
= 10 N
(c) Use smaller area of small piston
TOTAL
11 (a) Not easy to compress // no air bubble
(b) (i) same area
(ii) area of big piston in Diagram 11.2 is bigger
(iii) force exerted on big piston in Diagram 11.2 is bigger
(c) (i) the greater the area of large piston the greater the force 1
(ii) Pascal’s Principle 1
6M
TOTAL 1
1
12 (a) Bernoulli’s Principle 1
1
(b) FL = W = 2000 x 9.81 1
= 19620 N 1
1
(c) (i) high 1
1
Increase speed 9M
1
(ii) high
1
Greater acceleration
1
(iii) big 1
1
Greater lift force 5M
1
TOTAL 1
1
13 (a) Bernoulli’s Principle 1
(b) (i) P = 1500 – 500
1
= 1000 Pa 5M
1
(ii) F = PA
1
= 1000 x 0.25
= 250 N
(c) So the car will not be lifted when move with greater speed
TOTAL
14 (a) Bernoulli’s Principle
(b) Y
(c) At Y air is high speed
(d) At Z greater pressure, the different pressure at Y and Z produced a force
to push the liquid into the narrow tube
The greater speed of air from X produce fine spray
TOTAL
15 (a) Bernoulli’s Principle
(b)
(c) (i) aerofoil TOTAL 1
Greater lift force 1
1
(ii) low 1
Lighter 1
(iii) high 1
Greater different in pressure 8M
1
16 (a) Gravitational force 1
(b) (i) Weight of boat in Diagram 16.2 is > 1
(ii) volume of water displaced in Diagram 16.2 is > 1
(iii) buoyant force in Diagram 16.2 is greater 1
(c) The greater the volume displaced the greater the buoyant force
(d) Fb = W 1
= 629 x 9.81
= 6170.49 N 1
1
(e) The boat will displaced more water 1
9M
The density of river is lower 1
1
TOTAL 1
1
17 (a) Weight of bottle Q is the same as buoyant force 1
(b) (i) weight of Q is > 1
(ii) volume of Q immersed is > 1
1
(iii) weight of water displaced by Q is > 1
9M
(c) (i) the greater the weight of water displaced the greater the volume of 1
bottle immersed 1
1
(ii) the greater the weight of bottle the greater the weight of water
1
displaced 1
1
(d) (i) throw some of the load 6M
1
(ii) to reduce the weight of the boat // to reduce weight of boat immersed
(iii) Archimedes’ Principle 1
1
TOTAL 1
1
18 (a) Archimedes’ Principle 1
1
(b) (i) FB = W 1
= 1.5 x 107 x 9.81 8M
= 1.4715 x 108 N 1
1
(ii) ρ V g = m g
1.03 x 103 x V x 9.81 = 1.4715 x 108 1
V = 14563.12 m3 1
1
(c) Weight of submarine is greater than buoyant force 1
1
TOTAL 1
1
19 (a) Force acting upwards on an object immersed in a liquid when pressure
1
different between lower and upper surface area
(b) (i) Diagram 19.2 volume displaced is >
(ii) Buoyant force in Diagram 19.2 is >
(iii) Pulling force in Diagram 19.2 is greater
(c) The greater the volume of air displaced the greater the buoyant force
(d) The greater the buoyant force the greater the pulling force
(e) Yes
Density of air is lower at the top od mountain
TOTAL
20 (a) Archimedes’ Principle
(b) FB = W
(c) (i) FB = W
= 0.05 x 9.81
= 0.4905 N
(ii) length immersed greater
Density of petrol is lower than water
(d) (i) aerodynamic
Reduce water friction
(ii) FB = W A = 0.5 – 0.12
= 0.38 m2
V = Ah = 0.38 x 2
ρVg=mg
1000 x 0.38 x 2 = (100 + x) 1
x = 660 kg 1
TOTAL 12M
(iii) Q
1
CHAPTER 9 - ELECTRICITY
1 (a) A region where charge particles experience electric force 2
(b)
(c) (i) Q = It 1
= 2 x 2 x 60 1
= 240 C
TOTAL 1
(ii) Q = n e 1
240 = n x 1.6 x 10-19 7M
n = 1.5 x 1021 electron 1
2 (a) Coulomb // C // As
(b)
1
(c) Ion positive greater mass attracted to negative plate 1
Ion positive move slower so the division of flame is greater towards 1
negative plate.
TOTAL 4M
3 (a) (i) high 1
Greater strength of electric field 1
(ii) small distance 1
Stronger electric field 1
(iii) low mass 1
Lighter 1
(iv)metallic 1
Electric conductor 1
(b) J 1
TOTAL 9M
4 (a) Parallel 1
(b) A1 : 1.0 A 1
A3 : 0.5 A 1
(c) V = IR
= 0.5 x 3 1
= 1.5 V 1
(d)
2
TOTAL 7M
5 (a) Series 1
(b) V = IR
3 = 0.75 x (2R) 1
R=2Ω 1
(c) increases 1
TOTAL 4M
6 (a) Ohm’s Law states that the current through a conductor between two point 1
is directly proportional to the voltage across the two points provided all
physical condition and temperature remain constant.
(b) (i) Arrangement in Diagram 6.2 is series while Diagram 6.1 is parallel 1
(ii) brightness of bulb P and Q in Diagram 6.1 is greater 1
(iii) potential difference across each bulb in Diagram 6.1 is bigger 1
(c) (i) when the arrangement of bulbs are parallel, the brightness is greater 1
(ii) the voltage across each bulb is greater when the arrangement is 1
parallel
(d) (i) P > Q = R 1
(ii) voltage across P is greater // current across P is bigger 1
TOTAL 8M
7 (a) Work done to move one coulomb of charge from point to point 1
(b) (i) V = IR
9 = 2R 1
1
R = 4.5 Ω 1
1
(ii) 1 = 1 + 1
6 15 1
7
= 30 = 4.29 Ω
R = 4.5 – 4.29
= 0.214 Ω
(iii) V = IR
= 2 x 0.214 1
= 0.428 V 1
(iv)V = 9 – 0.428 = 8.572 1
V = IR
8.572 = I (6)
I1 = 1.43 A 1
(v) I = 2 – 1.43 1
I2 = 0.57 A 1
8 (a) Conductor which obeys Ohm’s Law TOTAL 12M
1
(b) (i) thickness in Diagram 8.2(a) is bigger 1
(ii) resistance in Diagram 8.1(b) is greater 1
(iii) the greater the thickness the lower the resistance 1
(c) (i) 1 = 1 + 1 1
20 20 1
R = 10 Ω
R = 10 + 20 = 30 Ω
(ii) V = IR
6 = I (30) 1
I = 0.2 A 1
(iii) V = IR
= 0.2 x 20 1
=4V 1
(d) Decreases because effective resistance increases 1
TOTAL 11M
9 (a) Rate of charge flow 1
(b) (i) length in Diagram 9.1(a) is > 1
(ii)resistance in Diagram 9.1(b) > 1
(iii) the greater the length the greater the resistance 1
1
(c) (i) 1 = 1 + 1 1
10 10 1
R=5Ω
R = 5 + 10 = 15 Ω
(ii) V = IR
6 = I (15) 1
I = 0.4 A 1
(iii) P = I2 R 1
= 0.22 x 10
= 0.4 W 1
(d) Current increases 1
TOTAL 12M
10 (a) Ratio of voltage to current 1
(b) (i) P and Q 1
(ii)voltage of each bulb is greater 1
1
(iii) 1 = 1 + 1 = 1Ω
2 2 1
1 = 1+1= 5
4 1 4
R = 0.8 Ω
V = IR
3 = I (0.8) = 3.75 A 1
TOTAL 6M
11 (a) Work done by an electrical source in moving one coulomb of charge in a 1
complete circuit
(b)
3
(c) (i) E = 1.5 V 1
(ii) 1.5 – 1.35 = 0.3 r 1
r = 0.5 Ω 1
(iii) V = IR
1.35 = 0.3 R 1
R = 4.5 Ω 1
TOTAL 9M
1
12 (a) The resistance cause by the electrolyte in the dry cell
(b) (i) The emf of both Diagram is 3.0 V 1
E = 2 x 1.5 1
= 3.0 V
(ii) Diagram 12.1 1
r = 1 +1 = 2 Ω
Diagram 12.2 1
1= 1+ 1 1
2 2 1
1
r=1Ω 1
(c) E = I (R + r) 9M
1
3 = I (3 + 1)
I = 0.75 A 1
1
TOTAL 1
13 (a) Work done by an electrical source to move one coulomb of charge in a 1
1
complete circuit 1
(b) (i) electromotive force is the same 1
1
(ii) reading of voltmeter in Diagram 13.1 is bigger 9M
(iii) ammeter reading in Diagram 13.1 is greater 1
(c) (i) the lower the ammeter reading the greater the voltage drop
(ii) the greater the voltage drop the greater the internal resistance 1
(d) (i) Voltmeter reading : decreases 1
1
Ammeter reading : increases 1
(ii) effective resistance decreases 1
1
TOTAL 1
14 (a) 1.5 Joule of energy is use by an electrical source to move one coulomb of 1
9M
charge in a complete circuit
(b) An open circuit no current flow, voltmeter measure emf 1
1
Switch is closed, current flow in external circuit
Voltmeter measure potential difference 1
(c) (i) tungsten 1
High resistance 1
(ii) coiled 1
Produced more heat // high resistivity 1
(d) R 1
TOTAL
15 (a) (i) E = Pt
= 2 x 6 x 30
= 360 kWh
(ii) cost = E x cost
= 360 x 0.2
= RM 72.00
(b) (i) coil wire
High resistance
(ii) nichrome
Heated faster // high resistivity // high resistance
(c) N 1
TOTAL 9M
1
16 (a) 2000 J energy is consume per second when connected to 240 V potential
difference 1
1
(b) (i) P = VI
2000 = 240 I 1
I = 8.33 A 1
1
(ii) P = 2 1
1
1
1
R = 2402 10M
1
2000
1
= 28.8 Ω 1
(c) (i) coiled
1
Long wire // high resistance 1
(ii) nichrome
1
Heated faster // produced more heat
(d) P 1
1
TOTAL 1
17 (a) 100 Joule per second energy is released when connected to 240 V 9M
1
potential difference.
(b) P = VI 1
1
100 = 240 I 1
I = 0.417 A 1
(c) (i) P : E = VIt 1
1
= 240 x 0.6 x 90 1
= 12960 J 1
Q : E = VIt 9M
= 240 x 0.5 x 100 = 12000 J
R : E = VIt 1
= 240 x 0.4 x 120 = 11520 J
(ii) R
(iii) use least energy to toast the bread
TOTAL
18 (a) 18 Joule per second energy released when connected to 6 V of potential
difference.
(b) V = IR
6 = 3R
R=2Ω
(c) (i) coil
Greater length of wire
(ii) small
High resistance
(iii) high
Not melt at high temperature
TOTAL
CHAPTER 10 - ELECTROMAGNETISM
1 (a) A region in which magnetic material experience a force
(b) (i)
1
(ii) Fleming’s left hand rule 1
(c) Increase electric current // stronger magnet // smaller distance between the 1
magnets // greater diameter of the rod
TOTAL 4M
2 (a) A resultant field from the combination of magnetic field of permanent 1
magnet and magnetic field of current carrying conductor.
(b) (i) angle in Diagram 2.2 is bigger 1
(ii) current in Diagram 2.2 is bigger 1
(iii) same 1
(c) Force // catapult force 1
(d) The greater the current the greater the force 1
(e) Fleming’s left hand rule 1
(f) Current flow in the copper frame produced catapult force 1
The force pushed the copper frame outwards 1
TOTAL 9M
3 (a) A region where magnetic material experience a magnetic force 1
(b) (i) number of dry cell in Diagram 3.2 is greater 1
(ii) angle of deflection in Diagram 3.2 is bigger 1
(iii) final position in Diagram 3.2 is bigger / further 1
(iv)Force 1
(v)the greater the number of cells the greater the electric current 1
(vi)the greater the current the greater the force 1
(c) Fleming’s left hand rule 1
(d) Stronger magnet // smaller distance between magnet 1
TOTAL 9M
4 (a) (i), (ii), (iii)
3
(b) Interaction between two magnetic field 1
Magnetic field of permanent magnet and magnetic field of current 1
carrying conductor
(c) (i) high 1
Greater current flow 1
(ii) curve / C shape magnet 1
Produced radial magnetic field 1
TOTAL 9M
5 (a) Swing outwards // deflect to the right 1
(b) Interaction of the two fields 1
will produced a resultant force that push the rod to the right. 1
(c) 2
(d) (i) curve / U shape / C shape TOTAL 1
Produced radial field 1
(ii) many / high 1
Stronger magnetic field of current carrying conductor 1
(iii) soft iron 1
Easy to magnetised and demagnetised 1
11M
6 (a) (i) Direct current 1
(ii) pointer deflect in one direction 1
(b) (i)
3
(ii) catapult force 1
(c) Fleming’s left hand rule 1
(d) (i) 1
(ii) magnetic field is more concentrate 1
(e) Bigger diameter of wire // greater number of turns 1
TOTAL 10M
7 (a) A region where magnetic particle experience a force 1
(b) (i) same 1
(ii) distance of copper wire in Diagram 7.3 is > 1
(iii) strength of magnetic field in Diagram 7.3 is > 1
(c) (i) the greater the strength of magnetic field the greater the distance 1
(ii) the greater the strength the greater the force 1
(d) (i) the wire vibrates // wire stationary 1
(ii) the change of cycle is too fast in changing the direction of current in 1
the wire // AC changes direction alternately in 0.02 s
TOTAL 8M
8 (a) Electric current produce from changing of magnetic flux 1
(b) (i) P = VI
= 0.5 x 6 1
=3W 1
1
(ii) 3 x 100 = 20 % 1
15
(c) (i) high 1
More cutting of magnetic flux 1
(ii) soft iron 1
Greater changing of magnetic flux 1
(iii) DC 1
Produced of induced current in one direction 1
(iv)S 1
TOTAL 12M
9 (a) A region where magnetic particle experience a force 1
(b) (i) same 1
(ii) Diagram 9.1 is South while Diagram 9.2 is North 1
(iii) Deflection of Diagram 9.1 is to the right while Diagram 9.2 is to the 1
left
(c) When south enters the solenoid the deflection is to the right when North 1
enters the solenoid the defection is to the right.
(d) (i) South 1
(ii) Lenz’s Law 1
(e) Deflection increases 1
The higher the height the greater the speed, the greater the cutting of 1
magnetic flux
TOTAL 9M
10 (a) Mechanical energy to electric energy // kinetic energy to electrical energy 1
(b) Electromagnetic induction 1
(c) (i) high strength 1
Greater cutting of magnetic flux 1
(ii) many 1
Greater induced emf 1
(iii) copper 1
Low resistance 1
(d)
1
TOTAL 9M
1
11 (a) Induced electrical current 1
1
(b) There is changing / cutting of magnetic flux 1
(c) North
(d) Lenz’s Law
(e) (i)
1
(ii) no changing of magnetic flux 1
TOTAL 6M
12 (a) North
(b) (i) 1
1
(ii) refer diagram in (b)(i) 1
(c) Move the solenoid with greater speed 1
4M
TOTAL 1
13 (a) A region where magnetic particle experience a force 1
1
(b) (i) height in Diagram 13.1 is greater 1
(ii) angle of deflection in Diagram 13.1 is greater 1
(iii) same 1
(iv)when height increases the deflection increases 1
(v)height increases the induced current increases 7M
1
(c) Faraday’s Law 1
TOTAL 1
1
14 (a) Electric current produced from cutting of magnetic flux 1
(b) (i) number of turns in Diagram 14.1 is > 1
(ii) same 1
(iii) induced current in Diagram 14.1 is > 1
(c) The greater the strength of magnetic field the greater the induced current 8M
(d) Faraday’s Law 1
(e) Magnet move away from solenoid // magnet push to the right 1
Solenoid become North pole, same pole repel 1
TOTAL 1
1
15 (a) Electric current produced from cutting of magnetic flux 1
(b) (i) diagram 15.2 is > 1
(ii) same 1
(iii) Diagram 15.2 is > 8M
(c) (i) the greater the height the greater the deflection 1
(ii) the greater the height the greater the induced current
(d) The galvanometer will deflect and back to zero continuously
On and off will produced changing of magnetic field
TOTAL
16 (a) Step up transformer
(b) The alternating current will produce a changing magnetic field at primary 1
coil 1
1
This will induced a changing magnetic field in secondary coil 1
1
(c) Easy to magnetised and demagnetised
(d) (i) 20 x 100 = 85 1
1
240 8M
1
Ip = 0.098 A
1
(ii) use greater diameter of wire // laminated soft iron core // wound 1
secondary coil on top of primary coil // use copper wire 1
1
TOTAL 1
1
17 (a) Step down transformer 1
1
(b) VEF = 60 – 48 9M
1
= 12 V 1
1
= 240 = 1000
1
12 1
1
Ns = 50 turns
1
(c) (i) Np > NDE > NEF
(ii) same 1
(iii) Vp > VDE > VEF 8M
1
(d) The greater the number of turns the greater the voltage 1
(e) zero 1
18 (a) Step down TOTAL 1
TOTAL 1
(b) = 240 = 1500
1
1000
V = 160 V
(c) (i) P = VI
= 240 x 0.25
= 60 W
(ii) Poutput = 3 x 18 = 54 W
n = 54 100
60
= 90 %
(iii) bigger diameter of wire
19 (a) To increase or decrease output voltage
(b) (i) 240 = 500
12
Ns = 25 turns
(ii) P = VI
24 = 12 I
I=2A
(c) (i)
(ii) to convert AC to DC C 1
20 (a) Step down transformer TOTAL 7M
1
(b) There is changing of magnetic flux in secondary coil 1
1
Induced emf produce in secondary coil and induced current flow to light
1
up the bulb
1
(c) (i) 5V
1
(ii) 240 = 960 6M
5 1
1
Ns = 20 turns 1
(d) Not light up 1
4M
TOTAL 1
1
CHAPTER 11 - ELECTRONICS 1
1
1 (a) Fast moving electron // ray produced from thermionic emission 1
1
(b) (i) light blocked by the maltase cross
(ii) electron beam that reached the fluorescent strike the atom to change
kinetic energy to light energy
(c) Fleming’s left hand rule
TOTAL
2 (a) Thermionic emission
(b) To produce greater strength of electric field
(c) (i) EHT in Diagram 2.2 is greater
(ii) angle of deflection in Diagram 2.2 is greater
(iii) same direction
(d) The greater the voltage the greater the deflection
(e)
1
(i) 1
(ii) Fleming’s left hand rule
TOTAL 8M
3 (a) Diode 1
(b) (i) current not flow // connection reversed biased 1
(ii) reversed the battery terminal // reversed the diode connection 1
(c) (i)
1
(ii)
1
4 (a) Diode TOTAL 5M
1
(b) One way TOTAL 1
(c) Lights up TOTAL 1
1
There is flow of current // connection is forward biased 4M
1
5 (a) Diode 1
(b) (i) capacitance in Diagram 5.3 is bigger 1
(ii) smoothness in Diagram 5.3 is greater 1
(iii) peak voltage is the same 1
(iv)the greater the capacitance the greater the smoothness 1
(c) Full-wave rectification 1
(d) Each cycle the capacitor stored charge 1
When the voltage decreases, capacitor released charge 8M
1
6 (a) Diode 1
(b) Convert AC to DC
(c)
1
(d) Second cycle become forward biased 1
TOTAL 4M
7 (a) npn 1
(b) To switch on the secondary circuit 1
(c) (i) V = 6-4 = 2V 1
1
(ii) 2 =( )6
+7000 1
2R + 14000 = 6R
R = 3500 Ω
(d) (i) m = 62 1
0.4 1
m = 155
(ii) amplification factor 1
(e) amplifier 1
TOTAL 9M
8 (a) npn 1
(b) (i) VB in Diagram 8.2 is greater 1
(ii) bulb in Diagram 8.2 lights up 1
(iii) IB in Diagram 8.2 is greater 1
(c) (i) the greater the base voltage the greater the base current 1
(ii) the greater the base current the greater the collector current 1
(d) No 1
Not enough voltage 1
TOTAL 8M
9 (a) As an automatic switch 1
(b) NPN 1
(c) (i) LDR 1
(ii) resistance of X will increase 1
10 (a) Npn TOTAL 4M
(b) (i) thermistor TOTAL 1
Detect heat 1
(ii) alarm 1
To switch on when collector current flow 1
(iii) emitter connection forward biased to the battery terminal 1
Charge carrier electron can flow to collector 1
(c) X 1
1
11 (a) Npn 8M
(b) (i) 6 V 1
(ii) to limit base current 1
(c) 3 = ( +1 000) 6 1
R = 0.5(1000 + R) 1
R = 1000 Ω
(d) Current amplifier TOTAL 1
12 (a) Transistor 1
(b) 6M
1
1
(c) (i) resistor at the base circuit 1
1
To limit base current to the transistor 1
1
(ii) refer diagram in (b) 1
1
(d) (i) LDR 1
9M
To detect light 1
1
(ii) interchange at Q
1
To produced greater base voltage and allow base current to flow
1
TOTAL 4M
13 (a) Npn 1
(b) 2 = ( +1 500) 6 1
2R + 3000 = 6R 1
R = 750 Ω
1
(c) Current amplifier
TOTAL
CHAPTER 12 – NUCLEAR PHYSICS
1 (a) Radioactive is an unstable nucleus
(b) (i) Strength of electric field is the same
(ii) diagram 1.1 the emission is positively charge and Diagram 1.2 is
negatively charge.
(iii) deflection in Diagram 1.2 is greater
(iv) mass of radioactive in 1.2 is lower 1
(v) the lower the mass the greater the deflection 1
(c) (i) increase the voltage (EHT) 1
(ii) to increase the strength of electric field 1
TOTAL 8M
2 (a) GM tube 1
(b) Beta particle 1
(c) Fast moving electron 1
(d) (i) gamma ray 1
Can kill all bacteria // can penetrate box 1
(ii) long half-life 1
Long lasting // can use for a longer time 1
(iii) solid 1
Does not spill // easy to be handled 1
TOTAL 9M
3 (a) Half-life is time taken for a sample of a radioactive nuclei to decay to half 1
of its initial number.
(b) N = (1/2)n N0 1
1200 – 600 – 300 – 150 150 = (1/2)n 1200 1
3 t ½ = 3 x 14 day
= 42 days log 0.125 = n log 0.5 1
n=3
T = 3 x 14 = 42 days
(c) (i) show on the graph time when activity is 400 count/s 1
t ½ = 1.2 minutes 1/8 = (1/2)n 1 1
(ii) 1 --- ½ --- ¼ --- 1/8 1
T = 3 x 1.2 = 3.6 minutes log 0.125 = n log 0.5 1
n=3
t = 3 x 1.2 = 3.6 min 1
TOTAL 9M
4 (a) Helium nucleus 1
(b) Radiation ionised air molecule into ion positive and negative 1
The ion flows to plate negative and positive. 1
Thus current flow and circuit is complete 1
(c) Beta particle has medium ionizing power 1
Produced low number of ion positive and negative. 1
(d) 24915 --- 23937 + 42 1
1
TOTAL 8M
5 (a) Nuclear fission 1
(b) (i) X = 92 1
Y = 36 1
(ii) m = (236.03290 – 235.86653) x 1.66 x 10-27 1
1
= 2.762 x 10-28 kg
(c) (i) solid 1
Not spill // easy to handle 1
(ii) gamma ray 1
Can penetrate metal 1
(iii) long half-life 1
Long lasting TOTAL 1
(d) Californium-252 1
12M
6 (a) Chain reaction 1
(b) 23925 + 01 ------ 3948 + 14540 + 2 01 1
(c) (i) m = (235.04395 + 1.00867) – (139.9216 + 93.9154) – 2(1.00867) 1
1
= 0.19828 x 1.66 x 10-27 1
= 3.29 x 10-28 kg 1
1
(ii) E = mc2 1
= 3.29 x 10-28 x ( 3 x 108 )2 1
= 2.962 x 10-11 J 9M
1
(d) Control rod / Boron – absorb access neutron 1
Moderator / graphite – slow down the neutron 1
TOTAL 1
1
7 (a) Nuclear fission 5M
1
(b) Access neutron released will bombard other uranium nucleus
1
Heat energy released will heat water and produced steam to rotate the 2
1
turbine 1
1
(c) E = mc2 1
= 0.1876 x 1.66 x 10-27 x (3 x 108 )2 1
= 2.803 x 10-11 J 9M
1
TOTAL 1
1
8 (a) Time taken for a sample of radioactive nuclei to decay half of its initial
1
number. 1
5M
(b) (i) 15 minutes
1
(ii) 2000 ---- 1000 --- 500 ---- 250 activity. 1
(c) (i) uranium – 235 1
1
Produced nuclear fission // released greater heat energy
4M
(ii) Boron 1
Absorbed access neutron
(iii) R
TOTAL
9 (a) Uranium-235
(b) Mass defect
(c) M = (236.0529 – 235.8653) x 1.66 x 10-27
E = mc2
= (3.114 x 10-28 ) x (3 x 108 )2
= 2.803 x 10-11 J
TOTAL
CHAPTER 13 – QUANTUM PHYSICS
1 (a) Object that emits thermal radiation determine by its temperature
(b) Max Planck
(c) Wave-particle duality
(d) E = hf // E = h
TOTAL
2 (a) Absorber and radiators
(b) Visible light 1
(c) (i) light intensity on the left side of the peak does not continue to increase 1
with the increase of wave frequency
(ii) energy is directly proportional to wave frequency // light is a form of 1
energy packet
TOTAL 4M
3 (a) (i) line spectrum 1
(ii) coloured line with unique wavelength and frequency 1
(b) Discrete packet of energy 1
(c) The greater the frequency the greater the energy 1
1
(d) E = h = 6.63 x 10-34 x 3 108 1
6.5 10−7
= 3.06 x 10-19 J
TOTAL 6M
4 (a) Quantum energy // discrete energy packets 1
(b) Planck constant 1
(c) J s 1
(d) Energy decreases 1
TOTAL 4M
5 (a) Minimum frequency of light needed for a metal to emit electron 1
(b) (i) 4 x 1014 Hz 1
(ii) W = h fo 1
= 6.63 x 10-34 x 4 x 1014 1
= 2.652 x 10-19 J
(c) (i) E = W + Kmax
3.15 x 10-19 = 2.652 x 10-19 + Kmax 1
1
Kmax = 4.98 x 10-20 J
1
(ii) E = 1 m v2 1
2
1
4.98 x 10-20 = 2 x 9.11 x 10-31 x v2
v = 3.3065 x 105 m s-1
TOTAL 8M
6 (a) Emission of photoelectron from surface of metal when illuminate by a 1
beam of light at certain frequency
(b) Light exceed threshold frequency for the metal 1
(c) (i) E = hf 1
1
= 6.63 x 10-34 x 7 x 1014
= 4.641 x 10-19 J 1
(ii) Kmax = E – W 1
= ( 4.64 x 10-19 – 2.32 x 10-19 )
= 2.321 x 10-19 J
TOTAL 6M
7 (a) Photoelectric effect 1
(b) The minimum energy required for a photoelectron to be emitted from a 1
metal surface.
(c) E = hf
= h = ( 6.63 x 10-34 ) x ( 3 108 ) 1
10−9 1
59
= 3.371 x 10-18 J
(d) E = W + Kmax 1
(3.371 x 10-18) – (9.92 x 10 -20) = W 1
TOTAL 6M
W = 3.272 x 10-18 J