TERM 3 STPM TRIAL (COLLECTION) MODULE ANSWER

SMJK HUA LIAN, PERAK

Answer for 960/3/02/2022:

Section A

1. A

= = − 2

2 = − = − −12 = 66.7
(3)(0.06)

= 8.16 −1

2. D

3. D
1

5000 = (20 × 10−3)
= 5000 × 20 × 10−3
= 100

4. B
∝ 2

5. A

6. C

7. B
8. C
9. B
10. B
11. B

12. C
13. B
14. A
15. C

Section B:

16 a The speed of an electromagnetic wave is always equal to the speed of light 3 × 108 −1 in free space.

b = ; =

9 × 10−9
=
1 3 × 108
20 × 3 × 108

= 4.5 × 10−10

C frequency will always constant whereas its wavelength varies from one medium to another medium.
d 5.0 × 1014

17 a

b

c Reflection.

Section C:
18. gregre

19. c.
b. e.

d.

20. a. b.
c.

d.

Form : 6AS1 SMJK Yu Hua, Kajang, Selangor
Date :21st April 2022 (Thursday) STPM Trial Examination
Time :11:25 am – 12:55 am Physics Paper 3
April 2022
Number of printed pages :
Prepared by : Pn Rosmaya binti Mokhtar

Verified by : Pn. Fazida binti Saipan @ Saipol

Answer :
Section A.

1. D
2. A
3. D
4. D
5. B
6. C
7. C
8. A
9. B
10. A
11. D
12. B
13. A
14. C
15. B

Section B.

16. (a)

(b)

1

(c)

17. a)
b)
c) Less or no light is reflected ______________________1

Section C.
18. a) i)

2

ii)

c)

19. a) i) Light waves are transverse wave
ii) Direction of EM wave propagation is in the direction of the vector product,
E x B.
Direction of E-field +y direction, direction of B-field, -x direction or
Direction of E-field -y direction, direction of B-field, +x direction

3

b) i) Unpolarised light: Oscillations of the E-field (and B –Field) are in all possible
panes.
ii) Component of the electric field oscillations parallel to the polarization axis of
the polaroid are transmitted, but components perpendicular to the axis are
absorbed. / Therefore, only half of the energy is transmitted.
iii)

c)

20. a)

b) i)

b) ii)

4

5

JABATAN PENDIDIKAN NEGERI TERENGGANU

ANSWER SCHEME TRIAL TERM 3 2022

= 2
= (2 )2

1A 0.6
2D = √4 2 = √4 2(0.02) = 0.87
3B Compare = 3 sin(25 − 10 ) ℎ ℎ

= (2 − ), wave number, = 10 −1

For a point source, if r is fixed, intensity, = 2


Intensity=2I, when power = 2P

Amplitude, √ , = √2 ( )

4C

5C Intensity, ∝ 2 ∝ 1 ′ ∝ 12 ∝ 1 , ′ = 1 1 = 1
6B 2 (3 )2 9 3

7B 1 11
= (1.50 − 1) (10 + −20) ∴ = 40
8
B 1 11
= (1.33 − 1) (∞ + 20) ∴ = 60
9C
11 1 11 1
= + ∴ = 40 + 60 = 24.0

Optical path of PQ = (PQ – 1.2) + (1.5)(1.2) = PQ + 0.6

Optical path XY = PQ = -1.2 + 1.8 = PQ+0.6-PQ
Optical path difference = (PQ + 0.6) – PQ= 0.6 cm

x2 = 2 , 2 = 2


= 2 = 3.6 x 10-3 rad

= sin 1

= 1 ( 8.94 ) = = 590 m
2400×102 √8.942+62.52

10 B

11 1
+ =

11 A 1 + 1 = 1 but u’ = v = ∞ 11 1
′ ′ ′ −10 + = −10

1
= 0, = ∞

111
∞ + ′ =

v’ = f’ = -20 cm

Distance of emerging ray from the first lens = 25 – 20 = 5 cm

12 B E = E f − Ei
= hf
= hc


 = hc

E

13 A The K characteristic line is produced by electronic transition in the target atom. The intensity
is determined by the number of electrons hitting the target and the number of collisions
determines the number of electronic transition.

14 D Energy released, E
15 B
= [(1.007825+13.003355)(934 MeV) + 2 MeV]
– (13.005739+1.008665)(934 MeV) = –1 MeV

Reaction cannot occur because total energy before

< total energy after

SECTION B

From the principle of superposition, the standing wave equation is
= 1 + 2
=2(4)sin(3x)cos(2t)
Amplitude of standing wave varies with x according to the expression, 8 3

At the nodes the amplitude of the standing wave is zero

when amplitude = 8 sin 3 = 0 1

16(a) 3 = , n=0,1,2,3…….. 1

= = 0, 6 ,23 , , … … ….= 0, 1.05 cm, 2.10 cm, 3.15 cm, …….. 1
3 1

The antinodes are located at the midpoint between two nodes, where 1
= 6 , 2 ,56 , …….cm 1
= 0.52cm, 1.57 cm, 2.62 cm, …………… 1

(b) At an antinode, 1
amplitude of simple harmonic motion = (4.0 +4.0 ) cm = 8.0 cm

At, = 1.5 ,

(c) = 8 sin(3.0) (1.5)
== 8 sin(4.5 ) = 7.82

17(a)

1

1
1

(b)(i)

1

(b) (ii) 1
1

SECTION C

18(a) Propagation of sound waves in a steel bar 1
-When one end of a steel bar is knocked, the molecules along the bar will 1
experience compression and rarefaction.
-Compression points are points of high pressure, rarefactions are points of 1
low pressure. 1
-Sound waves are being propagated in the steel bar by compression and 1
Rarefaction.
-Hence, longitudinal sound waves travels very fast along the bar because 1
molecules in a steel bar are so closely packed. 1
1
-Intensity of sound
It is the wave energy per unit time per unit area passing perpendicularly through
a surface.

(b) -Intensity level of sound,
(c)(i)

= 10
ℎ = ,

= ℎ ℎ ℎ = 1.0 × 10−12 −2
ℎ



Applying = 10 log 0


90 = 10 log 1.0 × 10−12

109 = 1.0 × 10−12 ∴ = 1.0 × 10−3 −2


=

1.0 × 10−3 = 80
4 2
= 79.8

(c)(ii) Amplitude or frequency of the waves

-Doppler effect is apparent change in the frequency of sound when there is 1
1
(d) relative motion between the source and the observer 1
-Apparent frequency is higher and pitch is higher. 1

′ = ( ) 1
′ = ( − )
1
+ 1
1
630 = ( − )

(e)(i) 580 = ( + )

630 = 300 + )
580 (300 −

= 12.4 −1

300 − 12.4
= 630 ( 300 )
= 604 Hz
(e)(ii)

Due to Doppler effect, as the police car accelerates, frequency of siren heard
(f) keeps increasing, pitch keeps increasing

19a) Transverse wave

b)i) Unpolarised light is a light wave in which the electric field of the light 2
wave oscillates in all direction, perpendicular to the direction of propagation
of the wave.

ii) - Every electric vector that oscillates in the light wave can be resolved into two 1

mutually perpendicular components. The electric component parallel to the

axis of transmission of the polaroid is transmitted, and the electric component

perpendicular to the axis of transmission is blocked

- Intensity is directly proportional to the square of the amplitude of oscillations 1

- Therefore, half of the light intensity will be blocked, half will be transmitted

1

iv) I = 0.5 x Io cos2 θ, 1
0.25Io = 0.5Io cos2 θ, 1
θ = 45o

c) Brewster’s law : 1
1
n = tan i
1.52 = tan i

i = 56.7o

d)
2

core

cladding

The core and cladding are made of glass or plastic 1
The core has a higher refractive index than the cladding 1
Light wave travels through the core by total internal reflection between the core- 1
cladding boundary

20 (a) The process is said to be random because every atom has an equal probability to 1
decay 1

The process is said to be spontaneous because radioactive decay cannot be predicted.
It happens at any time and it is not triggered by any factor

b 4109 → 2400 + 10 1

Mass loss = 39.964000u - 39.962591u = 0.001409u 1

Energy realease = 2 1

= 0.001409u × 1.66 × 10−27 × (3 × 108)2 1
= 2.105 × 10−13 1

c(i) is time taken by half of the number of nuclei in a radioactive sample to decay 1
1
c( ii) = 0 −
c(iii) 1 = 0.06 − (86400) 1
1
= 3.2562 × 10−5 1
1
ln 2 ln 2
1 = = 3.2562 × 10−5

2

1 == 2.1886 × 104

2

∆ = 7 + 8 −

∆ = 7(1.007276u) + 8(1.008665u) − 15.000108u = 0.120144u 1
Binding energy per nucleon = 0.120144×931.5 = 7.46 1

15