TERM 3 STPM TRIAL (COLLECTION) MODULE ANSWER

SMK LEDANG, TANGKAK JOHOR

SKEMA JAWAPAN TRIAL SEM 3 STPM 2021

BIL JAWAPAN MARKAH
1B

2C

3A

4C
5A

6C
7C

8D
9D
10 D

11 C

12 C

13 B

14 D
15 C

16 (a) The retarding force in critical damping is smaller than that in over
damping. Therefore, the time taken to return to the equilibrium position in
critical damping is shorter.
(b) Pendulum Q.
Pendulum Q and pendulum S oscillate at the same frequency because they have
the same length.
When the frequency of Q is equal to the frequency of S, resonance occurs and Q

oscillates with highest amplitude.
(c) When the boy's brother pushes the swing, the swing is forced to
oscillate with a forced frequency.
When the forced frequency is equal to the natural 'frequency of the swing,
resonance occurs.
17

18

19

20



PENGGAL 3

960/3 STPM 2022

JABATAN PENDIDIKAN NEGERI KELANTAN
SIJIL TINGGI PERSEKOLAHAN MALAYSIA

Mark Schemes: PHYSICS 3 (FIZIK 3)

Section A [15 Marks]

1. B In S.H.M displacement is measured from the equilibrium position. Amplitude of
oscillation is 1.0 cm.

2. C Increased damping: Lower peak and lower resonant frequency
2 2 (3.0)
3. D ℎ = = 5.0 = 3.8

4. A Frequency ⇒ = 2

⇒ =
2

⇒ = 200 = 31.8
2

Kinetic energy = 1 � � 2 − 2�2
5. D 2

= 1 (2) �200�52 − 32�2 = 640
2

6. D � 00�

′ = + = �333300 + 5100� 1000 = 1187.5
− −

1

7. A The brightness of the image depends on the quantity of light which is reflected by the
mirror. Because the upper half is closed, the image will be darker.

8. D Reason 1 :
1 1
+ =

1 1 1 0=+-21 1 0= 1
20

v = -20 and m = = -2


9. B Minimum distance between the object and the image u + v = 4f where u = v = 2f, hence

the graph’s minimum point is (2f, 4f)
1 1 + 1 +
also = =
=
(u+v)

4f2 = uv

f2 = 4 , [u = +, v = +]

f = √ , since f is positive, it is a convex lens
2

10. B Ans: B

= 2 2
0.3 = 0.5
= 39.2°

11. C ANS: C

=

= 0.04 ÷ 9
= 0.00444
(1.5)
0.00444 = 0.2 × 10−3
= 592

12. C ANS: C

From : Intensity ∝ 2; as a decreases, intensity I decreases

From : sin = ; as a decreases, diffraction angle increases

13. D use = ℎ = ℎ , the value of E must be bigger than work function

14. A radiation is emitted when an electron falls from higher energy level to a lower energy level

15. C 15 : 16 : 18 – because the mass, ∝

2

Section B [15 Marks]

16. (a)
-A body undergoing a simple harmonic motion has an acceleration directly proportional to
the displacement and the direction of the body is always towards to the center of the
oscillation/equilibrium position
- the frequency is a complete oscillation in a second/ amplitude is the maximum
displacement measured from the center of the oscillation

(b) = =
− 1= ma …………………………………………………………..1
− ( + ) = ……………..(ke = mg)………………………….2

= −

2 = �


= 2 � ………………………………………………………………3

(c) -when the loaded spring falls freely - the acceleration is zero/ the elongation of the
spring is zero/e=0………………………………………………………………1

when spring-mass displaced by x,

= −
ℎ , = 2 � ………………………………………………………2

-period of the oscillation was not change……………………………..…3

17. In the Bohr’s model, an electron of mass m moves in a circular orbit around its nucleus at a

speed of ℎ According to Bohr’s postulate,
2
= = ℎ ( = 1,2,3, … )
2

where n is an integer, h the Planck constant and r the radius of the orbit.

For the Hidrogen atom,

a) Show that the radius of the nth orbit is given by
2ℎ2 0
= 2

Where 0 is the permittivity of free space. [3 marks]

The ecelencttrriopsettaatlicfofrocrec,e , =2 == 4 4 2 2 0 20 2 2
The

-------- 1m

3

(2 ℎ )2 = 2 -------- 1m
4 0 2 -------- 1m

= 2ℎ2 0
2

b) Determine the radius of the smallest orbit. [2 marks]

Radius is the smallest when n=1
2ℎ2 0
= 2
1 = 12�6.63 10−34�2�8.85 10−12� -------- 1m
-------- 1m
(9.11 10−31)(1.60 10−19)2
= 5.31 10−11

c) The total energy level at the nth level is given by
�8 402 ℎ 2�,
= − 1
2

Calculate the ionisation energy of the atom. [2 marks]
Ionisation energy for a Hydrogen Atom = ∝ − 1
= 0 − 1
= �8 402 ℎ 2�
1
1100−−11924)2�9(6.1.613 1 100−−3314�)2� 2

= 1 �8(18..6805 -------- 1m
12
= 2.17 10−18
-------- 1m

4

Section C [30 Marks]

18 (a) Any 2 [2m]

Progressive wave Standing wave
No nodes or antinodes. Have antinodes and nodes where
displacement is 0.
The amplitude of each point is the same. The amplitude changes from zero at the node
to a maximum value at the antinode.
The distance between two consecutive point The distance between two consecutive nodes
vibrating in phase is equal to one or antinodes is half a wavelength.
wavelength.
Waveform moves in the direction of wave Waveform remains stationary.
propagation.

(b) (i) Compare y = 6.0 sin π(4.0t + 0.020x) with y = A sin (2πft + 2π xx), [1m]
A= 6.0 cm , f = 2.0 Hz and λ = 100 cm [1m]

V=fλ [1m]
=(2.0) (100) cm s-1

= 2.00 m s-1

(ii) Maximum speed , vmax [1m]
[1m]
Vmax = ω A [1m]
= (2π) (2.0) (6.0) cm s-1
= 75 cm s-1

(iii) Period , T = 1 = 1 = 0.50 s [1m]
2.0
1 1
T = 0.125 s = 4 T, the wave travels to the left (- x direction) a distance of = 4 λ

(c) (i) y = yP + yQ [1m]
= 0.015 sin(2t - 8x) + 0.015 sin (2t + 8x) [1m]
= 0.030 sin (2.0t) cos (-8x) [1m]
=0.030 cos (8x) sin (2.0t) [1m]

(ii) Amplitude, A = 0.030 cos 8x [1m]
= 0.030 cos 8(0.12) [1m]
= 0.0172 m
5

19 a) Huygen’s principle state that the points on a wave front act as a source to produce wavelets

which spread out in the direction of the wave. -1m

The wavelets superpose to form a new wave front which is tangent of all these spherical

wavelets. -1m

Two sources of the wave must be coherent sources, they produces waves with constant phase

difference. -1m

Two sources must have the same amplitude. -1m

b) For = 1, 1 = � 1 �, 2 = � 2 �

separation between the two first order bright fringes,

2 − 1 = � 2 � − � 1 � -1m

= 1.50 (450 − 400) 10−9 = 3.75 10−3 -1m
0.02 10−3

At the location the bright fringes of the two wavelengths overlap,

� 1 1 � = � 2 2 � -1m

� 1 2 �= � 2 1 � = 450 = 9 -1m
400 8

At 9th bright fringe of 400 nm overlaps with the 8th bright fringe of 450 nm. -1m

The distance 9th bright fringe of 400 nm from the centre bright fringes,
10−9
8 400 10−3 1.50 = 0.24 -1m
0.02

c) = = 3.00 108 = 3.75 10−7
8.00 1014

= 1 = 400 1 103 = 2.50 10−6


sin ∅ = -1m

2.50 10−6 sin ∅ = 2 3.75 10−7
∅ = 26.74
-1m

sin ∅ =
≤
1 -1m

≤ 6,

n=6 -1m

The number of maximum produced= 6+6+1 = 13 -1m

6

20 (a)(i)The decay constant λ of a radioactive isotope can be defined as the probability that a
radioactive nucleus of the isotope in the sample would decay in one second. [1]

a(ii) = − = − 1


integrating from t = 0 when N = No , to t = t when N = N
∫ − ∫0
=

[ln ] = − [ ]0 1
1
= − = − = −


b) 2400 + −10 1
(i) 4109

(ii) Mass loss = 39.964000 - 39.962591 = 0.001409u 1
Energy released = 0.001409 x 1.66 x 10-27 x ( 3 x 108 )2 1
= 2.105 x l0-13 J

(Twiihhia)etrBeeeaxcstartauhseeelaettchotemroaintcommmaaiscsssmionafsths41e09ofa t o24im00nc i cl u mdienascslstuhodefems42ta00hs es om fiasosnsclooynfs12i9d0eoorrerbbdiittaiinsnggtheeelleeccttrroonnss., 1
mass of the beta particle. 1
1
c) (i) In N = -λt + In No 1
λ = 0.02 min-1 1
Gradient = - λ 1
Gradient = - λ = − 630= -20 X 10-3 min-1;
ln 2 ln 2
1 = = 0.02 = 34.7
2

(ii) ln No = 60 1
No = 1.14 x 1026 1

7

KOLEJ T6 PETALING JAYA, SELANGOR 1

SKEMA PEMARKAHAN FIZIK
PEPERIKSAAN PERCUBAAN SEMESTER 3/2021

Section A: ANSWER MARK

QUESTION B 1

1  Acceleration of the metal bob in simple harmonic motion is directly proportional to its
displacement. Thus, acceleration is maximum at the maximum displacement, which is the
amplitude of the oscillation. The minimum time taken by the metal bob to reach another
amplitude from the initial position is half of the period. If the period of the oscillation is
3.6 s, the minimum time taken by the metal bob to reach another amplitude is 1.8 s. Thus,
in 1.2 s, the acceleration of the metal bob is not maximum.

 The metal bob reaches the amplitude in 1.2 s. The minimum time taken by the metal bob

to reach the amplitude is half of the period, T of oscillation. This 1.2 s possibly be , ,

, , …….. . If 1.2 s equals ,

 At the amplitude, the speed of metal bob is zero. Thus, the kinetic energy of the metal bob
is also zero.

 The point which is half of the amplitude from the initial position of the metal bob is the
equilibrium point. At the equilibrium point, the acceleration of the metal bob is zero.

2D

 When an oscillatory system is forced by another system. The oscillatory oscillates at the 1
forced frequency.

3B . Thus,

 Assuming that particle P is 1.0 cm from the source of disturbance,
.

 Take the time, . Thus, the displacement of P after 1.0 s,

 The displacement of Q after 1.0 s, 1
 The difference between the displacement of P and Q,

4 C 1

 Electromagnetic waves can cause resonance in other system, but the electromagnetic wave

2

itself does not undergo resonance. When electromagnetic wave is absorbed by an object,
and the natural frequency of the object is same as the frequency of the electromagnetic
wave, resonance will occur in the object.

5A

-------------------------- (i) 1
------------------------- (ii)

(ii)  (i),

6D

 When observer moves towards the source of sound with a velocity of , there is sound
wave that propagates towards the observer with the velocity of .

 Since the direction of velocity of the sound wave is opposite to the velocity of the
observer, the relative velocity between the observer and the sound,

 Therefore, the relative velocity between the observer and the sound becomes higher. 1

 The source of the sound is stationary. Thus, the wavelength of the sound wave is
unchanged

 Based on the equation , the frequency heard by the observer, .

 Hence, the higher the relative velocity between the observer and the sound, the higher
the frequency heard by the observer. (the higher frequency heard by the observer is an
apparent frequency)

7D

(negative sign shows that the mirror is convex) 1

8A

1

3

9A Screen
S1 P

3.0 cm

S2

4.0 cm

1

Optical path, S2P

Optical path, S1P

Optical path difference, S1P  S2P

 Since the path difference and the sources are coherence and in-phase, maximum

constructive interference occurs at P.

10 C 1

 When the light from air is reflected by optical material, there is a phase change of ,

which is equivalent to a path difference of

 If , the reflected light has a phase change of , which is equivalent to a path
difference of .

 When the thickness, , the additional optical path in the optical material
 Thus, the path difference between lights X and Y
 Hence, constructive interference occurs

11 B

 When the slit is wider, more light rays could pass through the slit. Thus, the brightness

of the central maximum increases. 1

 Based on the equation , when increases, increases. Thus, the position of the

first minimum becomes further, while the width of the central maximum increases.

12 A

 Let intensity of light before passing through P1 is I1, intensity of light transmitted by P1
is I2, and intensity of light transmitted by P2 is I3.

1

Thus,

4

13 C

 Photoelectric current is the rate of photoelectron emitted by the metal.

 The intensity of radiation illuminated onto the metal refers to the numbers of photon of

the radiation.

 The more the numbers of photon of the radiation, the higher the intensity of the 1
radiation.

 Each photon ejects one photoelectron from the metal. Thus, the higher the intensity of

the radiation, the more the numbers of photon reach the metal, the more photoelectrons

emitted at an instant.

 Hence, the higher the intensity of radiation, the larger the photoelectric current.

14 D

1

15 B

 The energy required by the electron at the lowest energy level to excite to the first higher

energy level,

—

 While the energy required by the electron at the lowest energy level to excite to the
second higher energy level,

— 1

 Thus, the energy provided by the electron (5.4 eV) that collides the atom is only enough
for the electron at the lowest energy level to excite to the first higher energy level.

 The excited electron then drops from the first higher energy level back to the lowest
energy level. When the electron drops to the lowest energy level, a photon whose energy
equals the difference in energy between the first higher energy level and the lowest
energy level (4.9 eV) is emitted from the atom.

 Thus,

5

TOTAL 15

Section B: ANSWER MARK
2/0
QUESTION  Simple harmonic motion is defined as a periodic motion where the acceleration of the body
is directly proportional to its displacement, and is always direct towards its equilibrium
16(a) point/position

------------------------------------------------------------------------ 1M
--------------------------------------- 1M

16(b)(i) 4

----------------------------------------------------------- 1M

 The equation for displacement of the load is --------- 1M

16(b)(ii)  Add electrical vibrator to oscillate the loaded spring. [1M] 2
17(a)  The loaded spring system oscillates with the frequency of the vibrator. [1M] 8
17(b)(i)  As the vibrator vibrates, the loaded psring system will keep oscillating without damping 1

TOTAL 2
 Focal point of a lens means a point where all the parallel light rays passing through a

converging lens refract/converge towards it. OR

 Focal point of a lens means a point where all the parallel light rays passing through a
diverging lens refract away/diverge from it.

------------------------------------------ 1M

--------------------------------------------------- 1M

----------------------------------------------------------- 1M

17(b)(ii) 3

6

17(c) ----------------------------------------------------- 1M 1
7
--------------------------------------------------- 1M
 Focal length of the lens becomes longer, while the lens remains as a diverging lens.

TOTAL

Section C: ANSWER MARK
2/0
QUESTION  Doppler effect is a phenomenon where an apparent change in the frequency of sound heard
by an observer when there is relative motion between the source of sound and the observer.
18(a)

………………………………………………… 1M

………………………………………………… 1M

18(b) 4

18(c)(i) …………………………………………………….. 1M 4
……………………………….. 1M
18(c)(ii) 2
18(c)(iii)  The source of sound leaving the detector/approaching the wall, 1
18(c)(iv) ………………………………………………….. 1M 2
………………………………… 1M

 The sound wave reflected by the wall approaching the detector,
Frequency of reflected sound = Frequency incident of the wall

……………………………. 1M
…………... 1M

……………………………………... 1M
……………………………………... 1M

………………………………….. 1M

7

19(a) ………………………………………….. 1M 15
19(b)(i) 2/0
TOTAL
 Polarisation of light is a phenomenon where the vibration of a non-polarised light is limited

only to one plane.

---------------------------------------------------- 1M 3
The rotational angle between Q and P is 2 ----------------------- 1M

----------------- 1M

19(b)(ii)  Malus’s law 1

---------------- 1M

19(b)(iii) 2

19(c)(i) ---------------------------------------------------- 1M 4

 When non-polarised light incidents onto a transparent medium, the light is partially
reflected. [1M]

 The reflected light is partially (horizontally) plane-polarised. [1M]
 If the angle of incidence of the non-polarised light equals Brewster’s angle, [1M]
 The reflected light will be fully plane-polarised. [1M]

--------------------------------------------------- 1M

19(c)(ii) ------------------------------------------------------- 1M 3
20(a) --------------------------------------------------- 1M
TOTAL 15
 Photon is a quantum of electromagnetic radiation. 1

8

20(b)(i)  Photoelectric current increases with the increase of intensity of incidence radiation. 2
 Electrons are emitted immediately after a metal is illuminated by electromagnetic radiation. /
20(b)(ii) 4
There is no delay of emission of photoelectrons.
20(c)(i)  The maximum kinetic energy of photoelectrons increases with the increase in frequency of 2
20(c)(ii) 3
incidence radiation.
 There is a minimum frequency of incidence radiation / threshold frequency to eject electrons

from a metal target.
** Any two observations

Point 1:
 Each incidence radiation/photon ejects one electron from the target metal. [1M]
 The intensity of radiation depends on the number of photon. The more the number of

incidence photon, the higher the intensity of the radiation. Thus, the higher the intensity of the
incidence radiation, the more electrons ejected. Hence, the higher the photoelectric current.
[1M]
Point 2:
 According to the principle of quantization of electromagnetic radiation, energy of the
radiation/photon is in quantum/packet / is quantized. [1M]
 If the energy of incidence radiation/photon is larger than the work function of the target metal,
electrons will be ejected immediately / no delay in emission of photoelectrons. [1M]
Point 3:
 According to the principle of quantization of electromagnetic radiation, energy of the
radiation/photon depends on its frequency. The higher the frequency of the radiation, the
higher the energy of the radiation/photon. [1M]
 Since the work function of the target metal is constant, the remains energy of the incidence
radiation/photon will be converted into kinetic energy of the photoelectrons.
 Thus, the higher the energy of incidence radiation/photon, the higher the maximum kinetic
energy of the photoelectrons. [1M]
Point 4:
 According to the principle of quantization of electromagnetic radiation, the energy of the
incidence radiation/photon is discreet / fixed. [1M]
 Thus, if the energy of the incidence radiation/photon is lower than the work function, no
electron will be ejected. [1M]
** Must be the same points/observations stated in (b)(i).

-------------------------------------------------------------------- 1M

--------------------------------------------------- 1M

---------------------------------------------------- 1M

9

20(c)(iii) --------- 1M 3

--------------------------------------------- 1M
------------------------------------------------------ 1M

------------------------ 1M

---------------------------------------------------- 1M

TOTAL 15

VICTORIA INSTITUITION, KL

skema 2022 P3 Physics

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

16(a) Explain what is meant by Doppler effect ? [1 mark]

The apparent change in the frequency (or pitch) of a sound when there is relative motion
between the source and the observer

Explain what an observer would experience in terms of wavelength and speed of wave

when a source of sound approaches him. [2 marks]

The apparent wavelength would decrease,

The speed of sound remains unchanged

(b). An object is moving in a straight line with speed 40 ms-1, towards a stationary sound
source which is emitting sound of frequency 6000 Hz. If the speed of sound in air is

330 ms-1, determine

(i) the frequency of the sound received by the object. [2 marks]
f '= v + u0 f

v
= 330 + 40 (6000) = 673Hz

330

(ii) the frequency of the sound reflected from the object [2 marks]

13

f '' = v (673) = 766Hz
v − us

17 (a) (a)(i)State two characteristics of the reflected light when unpolarised light is

incident with Brewster’s angle onto a glass surface. [2]

The reflected light is fully polarised
The intensity of the reflected light is half the intensity of the incident light

(ii)Polarisation occurs when sunlight falls onto a glass panel at the side of a
building. Determine Brewster’s angle of the polarised light.

[The refractive index of glass is 1.52.] [2]

tan i =n

tani=1.52
i= 56.60

(b)(i)Two polarising sheets are placed together with their axes of polarisation
making an angle of 35  . Unpolarised light passes through them.
Determine the percentage of the light intensity transmitted through both

polarisers. [2]

I1= 2 0
I2=I1 2Ө
= 2 0 235
=0.335 0
Percentage of I2 transmited=33.3%

14

(ii)State two advantage of wearing a pair of Polaroid sunglasses during an afternoon

stroll when the sun is bright hot. [2]

-polaroid removes the glare
-intensity of light is reduced by half

18. (a) With the help of a diagram, explain the meaning of the following terms when
applied to wave motion.

(i) Diffraction

(ii) Coherence

(iii) Superposition [6 marks]

(i) or similar diagram that shows spreading of wave

Diffraction is the spreading of waves when they pass through an opening or
around an obstacle into the geometrical shadow regions

(ii)
or similar diagram that shows constant phase difference
Two waves are coherent if they have a constant phase difference

15

(iii)
or similar diagram that shows the vector sum.
Superposition is the phenomenon whereby two or more waves of the same
kind meet at a point. The resultant displacement is the vector sum of the
individual displacement.

(b) Two microwave sources A and B are in phase with one another. They emit waves of
equal amplitude and of wavelength 30.0 mm. They are placed 140 mm apart and at a
distance of 810 mm from a line OP along which a detector is moved as shown in the
Figure below.

(v) Given the source emits the microwave at a frequency of 10 x 109 Hz,

calculate the wavelength of the microwave. [2 marks]

v=f
3 × 108

= 1 × 109 = 0.3

16

(vi) Determine the path difference between AP and BP, in terms of number of

wavelengths. [2 marks]

AP = √(514 − 70)2 + 8102 = 923.7

BP = √(514 + 70)2 + 8102 = 998.6

| − | = 998.6 − 923.7 = 74.9

74.9 × 10−3
| − | = 3 × 10−2 = 2.5

(vii) Describe and explain what is observed as the detector is moved from point P

to point O on the central axis. [3 marks]

Since the path difference at P is 5/2 , the minimum intensity is at the 3rd order.

The detector will detect 2 maximum intensity and 2 minimum intensity before
reaching O.

At O, the detector will receive a maximum intensity.

(viii) If source A and B are now set to be at anti-phase, state and explain whether a
maximum or minimum intensity will be received by the detector when it is at
P. [2 marks]

When source A and B are anti-phase,

The path difference of 5/2  will cause the 2 waves to arrive in phase.

Therefore a maximum intensity will be detected at P.

19(a) The potential difference across an X-ray tube with tungsten target is 30 kV.

(i) Calculate the minimum wavelength of X-ray produced.
ℎ

= 
17

(30 ) = (6.63 10−34)(3 108
)

= 4.14x10-11m

(ii) Give two properties of tungsten as a target metal.

High melting point

High density

(iii) Sketch a graph to represent the X-ray spectrum produced.

Label y axis as intensity and x axis as wavelength

Draw and label continuous spectrum

draw and label characteristic lines

label minimum wavelength

(iv) Explain the production of the continuous background spectrum and the

characteristics line spectrum.

continuous background spectrum

When electrons hit the target they will be decelerated. The deceleration will
be over a range of values and hence the energy lost by the electron has a
range of values. This energy appears as X-rays which will have a range of
wavelengths as different energies are involved…………………….1
If all the energy of an electron is lost in a single collision, then the radiation
emitted will have the maximum possible energy or smallest possible
wavelength……………………………1

Characteristics line spectrum

The bombarding electrons may be able to penetrate and displace electrons
very close to the nucleus (inner shells). When an electron from the outer shells
fall into the vacancy, it will results in emission of radiation whose energy
corresponds to the energy difference between the levels. …….1
Thus the emitted wavelengths have particular wavelengths which are
characteristic of the element used as a target. …….1
The probability of this happening is very high giving rise to high
intensities………………………1

18

(b)X-rays are diffracted by a crystal, but not by an optical diffraction grating.
(i) Deduce one characteristic of X-ray

-High frequency wave//em wave//does not require medium to travel

Deduce one characteristic of the crystal.
- Atoms arranged in order//can diffract x ray//well defined planes//fixed

shaperigid//incompressible
(ii) Write down the Bragg’s equation for X-rays diffraction. Identity symbols

in the equation.

dsinӨ=n
d= distance between atomic planes
Ө=glancing angle
n=order
= wavelength of incident x-ray beam

(iii) Why X-rays are not diffracted by an optical diffraction grating?

Slit width of the diffraction grating is too large compared to the
wavelength of x-ray

20. a. [2]
i. State what is meant by nuclear binding energy

energy required to remove completely all the nucleons in a nucleus

ii. Explain why the mass of an alpha particle is less then the total mass of the two

individual protons and the two individual neutrons in its nucleus [2]

when 2 protons and 2 neutron fuse to form an alpha particle, energy is released. This
energy is the binding energy of the alpha particle. The mass equivalent to the binding
energy is the mass defect between the alpha particle and its constituent nucleons

19

iii. the variation with nucleon number A of the binding energy pe nucleon BE is shown in

figure 8.1 . Explain how stability of elements in the periodic table, can be described by

the graph [3]

for light elements,as the nucleon number increases , the binding energy per nucleon
number increases.hence the stability for lighter elements increases as the nucleon
number increases

The curve reaches a peak around Fe-56. Hence it can be deduced that elements around
iron are most stable, with iron being the most stable element

Beyond iron, the curve gradually decreases, as the nucleon number increases. Hence
beyond iron ,as elements get heavier , the stability of these elements decreases

b. An equation for one possible nucler reaction is shown below
Data for the masses of the nuclei are given in the table below.

i. Calculate the mass change ,in u with this reaction [2]
Mass defect=(16.99913+1.00728) –(4.00260+14.00307) [3]
= 0.00074 u

ii. Calculate the energy in J, associated with the mass change in (i)
= 0.0097 x 1.66 x 10-27 kg

20

Mass defect convert to energy , E= mc2
=0.0097 x 1.66 x 10-27x (3.0 x 103)2 ...1 (convert to J method)

= 1.11X10-13 J

iii. Explain why for this reaction to occur , the helium-4 nucleus must have a

minimum speed [3]

Mass of the reactants is less then the mass of the products. The mass defect is the
equivalent energy for the reaction to occur

This energy is provided in the form of the kinetic energy of the alpha particles

Hence the kinetic energy of the alpha particle enables it to overcome the repulsive
force of the oxygen nucleus and fuse together to form the nitrogen nucleus

21

Name: (Marking Scheme) 1
Class: (Pn Kang Seow Hung)

960/3 Trial STPM 2021 (*Exam in 2022)
SMK King George V
PHYSICS
PAPER 3
One and a half hours

SIJIL TINGGI PERSEKOLAHAN MALAYSIA
(MALAYSIA HIGHER SCHOOL CERTIFICATE)

Instructions to candidates: For examiner’s use
DO NOT OPEN THIS QUESTION PAPER UNTIL YOU
Question Total Marks
ARE TOLD TO DO SO. marks obtained

Answer all questions in Section A. Indicate the correct answer on Section A
the Multiple-choice Answer Sheet provided.
1-15 15
Answer all questions in Section B. Write the answers in the spaces
provided. Section B

Answer two questions only in Section C. You may answer all the 16 8
questions but, only the first two answers will be marked. Write the
answers in the answer sheets provided. All working should be shown. 17 7
For numerical answers, units should be quoted wherever appropriate.
Begin each answer on a fresh answer sheet. Section C

18 15

19 15

20 15

Total 60

___________________________________________________________________________
This question paper consists of 9 printed pages

2

Section A [15 marks]
Answer all questions in this section.

1. Which of the following statements concerning simple harmonic motion is true?

A The velocity is maximum when the displacement is zero.
B The velocity is directly proportional to the displacement.
C The acceleration is zero when the displacement is maximum.
D The acceleration is positive when the displacement is positive.

Answer: A a α x
When x=0, a=0, v is maximum.

2. The variation with time t of the displacement x of a particle in simple harmonic motion from
point O is given as

= 5.0 sin 20

Where x is in cm and t in seconds. What is the period of the simple harmonic motion?

A 0.050 s B 0.20 s C 0.22 s D 0.31 s

Answer: D

= 20 −1
2

= 20
= 0.31

3. The wave intensity at a distance of 5.0 m from a point source is 4.0 W m-2. What is the
intensity of the wave at a distance of 2.0 m from the point source?

A 0.64 W m-2 B 1.6 W m-2 C 10 W m-2 D 25 W m-2

Answer: D
1

∝ 2
2 22 = 1 12
2(2.0)2 = 4.0(5.0)2
2 = 25 −2

4. The equation for a standing wave is = 4.0 cos 4 sin 5 , where x and y are in cm.
What is the displacement of the point at x=2.0 cm when t=0.50 s?

A -0.58 cm B 0 cm C 0.54 cm D 4.0 cm

Answer: A
= 4.0 cos 4(2.0) sin 5 (0.50)

= −0.58

3

5. A string of length L is stretched between two fixed points. Two successive resonant
frequencies emitted by the string are 315 Hz and 420 Hz. What is the fundamental resonant
frequency of the string?

A 105 Hz B 140 Hz C 210 Hz D 280 Hz

Answer: A

Resonant frequencies of a stretched string are: fo, 2fo, 3fo, 4fo, 5fo, …
Differences between successive resonant frequencies: fo
Therefore, fundamental frequency, fo =420 – 315 =105 Hz

6. A 30.0 cm long guitar string with a mass per unit length of 9.00 x 10-3 kg m-1 is stretched to
a tension of 28.0 N.

What is the fundamental frequency of the guitar?

A 27.9 Hz B 84.0 Hz C 93.0 Hz D 185 Hz

Answer: C

1
= 2 √

= 1 √9.002 8 1.00−3
2(0.300)

= 93.0

7. A sound wave has an intensity of 4.00 μW m-2. What is the intensity level of sound?
[Threshold of hearing is 10-12 W m-2]

A 25.0 dB B 40.0 dB C 66.0 dB D 76.0 dB

Answer: C
Intensity level of sound, = 10 ( )



4.00 10−6
= 10 ( 10−12 )

= 66.0

4

8. An audio source emitting sound of frequency 700 Hz is attached to an end of a 0.50 m rod.
The rod is rotated in a horizontal circle with an angular velocity of 50 rad s-1. What is the

maximum frequency of the sound received by a stationary observer?
[Speed of sound in air = 330 m s-1]

A 360 Hz B 640 Hz C 648 Hz D 757 Hz

Answer: D

= = 0.50(50) = 25 −1

′ = ( − )

′ = 330 700
(330 − 25)
′ = 757

9. An object is placed 15.0 cm from a concave spherical mirror with focal length of magnitude
10.0 cm. The characteristics of image formed are

A real, inverted and magnified.
B real, upright and diminished.
C virtual, upright and magnified.
D virtual, inverted and diminished.

Answer: A

11 1
+ =

11 1
15.0 + = 10.0
= +30.0 > u

5

10. A point of object O is at a distance of 20.0 cm from the concave end of a glass block of
refractive index 1.50 as shown in the figure. The radius of curvature of the concave surface is
50.0 cm.

Where is the image formed due to refraction of light at the concave surface?

A 25.0 cm to the left of the concave surface
B 37.5 cm to the left of the concave surface
C 25.0 cm to the right of the concave surface
D 37.5 cm to the right of the concave surface

Answer: A

1 + 2 = 2 − 1

1.00 1.50 1.50 − 1.00
20.0 + = −50.0
= −25.0

11. An astronaut is in a space craft orbiting 250 km above the Earth. Assume that the
wavelength is 550 nm and pupil diameter = 5.0 mm.

What is the approximate size of the smallest object on the Earth that can be resolved by the
astronaut's eye?

A 25 m B 34 m C 58 m D 65 m

Answer: B
= 1.22 , When angle is small, = =

(550 10−9)
250 103 = 1.22 5.00 10−3

= 33.55

6

12. Electrons are accelerated from rest through a potential difference . The de Broglie
wavelength λ of the electrons

A is directly proportional to
B is inversely proportional to .

C is directly proportional to √ .
D is inversely proportional to √ .

Answer: D

= 1 2
2
2 = 2 2

ℎ2
2 = 2
ℎ
=
√2
= ℎ
Memorize this formula
√2
= ℎ ➔ 1
√2 √

13. The angular momentum of electron in the nth orbit of a hydrogen atom is

A 2ℎ B ℎ C 2ℎ D ℎ
4 2
4 2

Answer: D

Bohr’s First Postulate

The angular momentum of the electrons in an atom is quantised and have discrete values that
are multiple of 2ℎ .
Angular momentum, L = mvr = n ( ℎ ), n = 1, 2, 3, 4, …

2

(n corresponds to successive orbits beginning

with the one nearest to the nucleus.)

14. Which of the following experiment illustrates the wave nature of a particle?

A Photoelectric effect
B Diffraction of a beam of electrons by a crystal
C Line spectrum emitted by a hydrogen discharge tube
D Production of X-rays when electrons collide into a metal target

Answer: B

Diffraction is a wave property.

7

15. A nuclear reactor uses U-235 as fuel to generate 100 MeV of energy for the fission of each
U-235 nucleus. If the output power of the nuclear reactor is 1.0 MW, how many U-235 nuclei
has undergone fission in one second?

A 3.1 106 B 4.0 106 C 6.3 1016 D 7. 1 1016

Answer: D

Energy produced by one nucleus = 100 x 106 x 1.6 x 10-19 J

= 1.0 106 = 6.25 1016
100 1.6 10−13

8

Section B [15 marks] [2 marks]
Answer all questions in this section.

16. (a) State Huygen’s principle.

All points of a wave front of a primary wave serve as point sources of secondary waves.

Subsequent wave front of the primary wave is the surface that is tangent to all these secondary

waves when these waves overlap. [2/0]

(b) State two conditions to obtain interference pattern. [2 marks]

The two sources of waves must
- be coherent (constant phase difference) [1]
- same wavelength/frequency [1]

(c) A beam of a laser of wavelength 635 nm is directed to a single slit of width 0.050 mm. The
pattern formed on a screen placed 1.50 m from the slit as shown in the diagram below.

o

(i) Name the property of light wave that causes the pattern on the screen. [1 mark]

Diffraction [1]

(ii) Calculate the distance x of the first minimum of the pattern from the centre of the screen O.

[3 marks]

= [1]

0.050 10−3 = (1)635 10−9

When angle is small, = =

0.050 10−3 = (1)635 10−9 [1]
0.050 10−3 ( ) = (1)635 10−9 [1]

1.50

= 0.0191

= 1.91

9

17. (a) Define the half-life and decay constant of a radioactive substance. [2 marks]

The half-life T1/2 of a radioactive nuclide is the time taken for a sample of the nuclide to decay
to half of the original number of nuclei. [1]

Decay constant, = − ⁄ = Number of Rate of decay in the sample [1]
radioactive nuclei

(b) A doctor uses 1.49 μg of iodine-131 to treat thyroid disorder. Iodine-131 has half-life of

8.0 days. Calculate

(i) the number of iodine-131 initially present. [2 marks]

= 1.49 10−6 6.02 1023 [1]
131 [1]

= 6.85 1015

(ii) the initial activity of iodine-131. [3 marks]

2 [1]
= 1

2

= 2

8.0 24 3600


= −


− =
=
[1]

= 2 (6.85 1015)
8.0 24 3600

= 6.87 109Bq [1]

10

Section C [30 marks]
Answer any two questions in this section.

18. (a) Define simple harmonic motion. [2 marks]

A simple harmonic motion is a motion where the acceleration is directly proportional to the
distance from the equilibrium position, and is always directed towards the equilibrium position.

Or

A simple harmonic motion (SHM) is a periodic motion acted by a restoring force which is
always directed towards the equilibrium position. The restoring force is directly proportional
to the displacement. The mathematical definition is:

= − 2 where is the acceleration, the displacement and 2 is the

proportionality constant. [2/0]

(b) Derive the expression for the period of oscillations for vertical spring-mass system.
[5 marks]

At equilibrium, = [1]
[1]
Restoring force, = − ( + ) [1]
= − −
= − − [1]
[1]
= −
= −

= − ( )



It is a SHM where = − 2
2 =



(2 )2 =



Period, = 2 √

11

(c) The maximum speed of a 1.2 kg mass attached to a vertical spring with a force constant of

35 Nm-1, is 0.95 ms-1.

(i) Determine the frequency of oscillation, and [2 marks]

= 1 √ [1]

2

= 1 √35

2 1.2

= 1 √ 35
2 1.2

= 0.86 [1]

(ii) Determine the total energy of the spring-mass system. [2 marks]

Total energy of the system, = 1 2
2
= 1 (1.2)(0.95)2

2

= 0.54

(d) A simple harmonic motion has amplitude = 5.00 cm and angular frequency = 5.20
rad s-1.

At time = 0, = 2.00 cm. The equation for the SHM is given by:

= sin( + ) where is measured in centimeters and in seconds

(i) Determine the phase angle . [2 marks]

= 0, = 2.00 , = 5.00 : = sin( + )
2.00 = 5.00 sin
= 0.41 rad

(ii) Determine the first time it reaches = −2.5 cm. [2 marks]

= −2.5, =?: −2.5 = 5.00 sin(5.20 + 0.41)

−0.5 = sin(5.20 + 0.41)
5.20 + 0.41 = +

6

= 0.626 s

12

19. (a) (i) Write the thin lens formula. [2 marks]
[2/0]
1 +1 = 1 where u = object distance

v = image distance
f =focal length

(ii) The distance of an object from a converging lens is 5 times the focal length of 5.0 cm.

Determine the distance of the image from the lens. [2 marks]

1 +1 = 1 [1]
[1]


1 +1= 1

25.0 5.0

= 6.25

(b) A bulb, a spherical mirror and biconvex lens are placed on the same axis as shown below.

The mirror and the surface of the lens have the same radius of curvature of 8.0 cm.
The refractive index of the lens is 1.5. All the rays from the bulb striking the mirror are to be
directed towards the lens. The rays emerging from the lens are parallel to one another.

(i) Sketch the related ray diagram. [2 marks]

rf

(ii) Determine the distance between the mirror and the lens. [3 marks]

Given = 8.0

1 = ( − 1) ( 1 − 1 ) [1]
1 2 [1]
1 (81.0 −18.0)
= (1.5 − 1) −

= 8.0

∴Distance between the mirror and the lens = +
= 8.0 + 8.0

= 16.0 [1]

13

(c) A lens maker wants to make a converging glass lens (n = 1.50) with the same curvature on
both sides and a focal length of 20.0 cm.

(i) Determine the radius of curvature for both surfaces. [2 marks]

1 = ( − 1) ( 1 − 1 ) [1]
1 2
1 1 1
20.0 = (1.50 − 1) ( − − )

2
= 0.1
= 20.0
[1]

(ii) If the lens is placed in carbon disulfide (n = 1.63), the converging lens becomes a diverging

lens. Explain qualitatively how this can occur. [4 marks]

1 = ( 2 − 1) ( 1 − 1 )
1 1 2
1 = (1.50 − 1) ( 1 − 1 )
1.63 20.0 −20.0 [1]

= −125 [1]

The focal length is negative. Hence the lens is now a diverging lens. [1]
This happens because the lens is surrounded by a medium with a higher refractive index. [1]
Therefore, the converging rays at the surface are reversed.

20. (a) State two properties of X-rays. [2 marks]

It has high penetrating power.
It cannot be deflected by electric or magnetic field.
It can undergo diffraction by a crystal surface.
It is an electromagnetic wave with very short wavelength.

Any two [1,1]
(b) Explain the formation of characteristics lines spectrum in the X-ray spectrum. [3 marks]

When the incident electrons have enough energy to knock out electrons from the innermost
orbit of the anode atoms, the atoms become excited and unstable. [1]

Transition of electrons from the higher orbit to the innermost orbit occurs and the difference
between the energy levels will be radiated in the form of photons/ electromagnetic wave. [1]

Since atoms has discrete energy levels, thus the photons emitted has discrete wavelengths. [1]

(c) The graph shows the X-ray spectrum from an X-ray tube.

14

(i) Copy the graph above, and on the same axes, draw a new X-ray spectrum if the potential

difference across the X-ray tube is increased. [1 mark]

(ii) Explain how the X-ray with minimum wavelength is produced, and hence, derive the

equation for minimum wavelength. [4 marks]

Electrons are accelerated by high potential difference. When the high-speed electrons
bombarding the anode metal target, the electrons undergo deceleration. [1]

During deceleration, some kinetic energy of the electron will be changed into photons of
various wavelength. Thus, a continuous spectrum is produced. [1]

When all the kinetic energy of the electron is radiated as a photon in a single collision, the
photon emitted has the maximum energy and thus it has minimum wavelength. [1]

= ℎ

ℎ
=

= ℎ [1]


(iii) Estimate the potential difference across the X-ray tube. [2 marks]

= ℎ

6.63 10−34(3.00 108)
= 1.60 10−19 (2.00 10−11) [1]

= 62.1 [1]

15

(iv) It is known that the target is made up of one of the following elements, copper,

molybdenum, argentum or tungsten which respectively have the line of frequencies
1.9 1018 , 4.3 1018 , 5.4 1018 , and 1.4 1019 .

Identify the target element of the X-ray tube. [3 marks]

= [1]

3.00 108
= 7.00 10−11
= 4.29 1018Hz [1]

The frequency corresponds to that of molybdenum. [1]

End of question paper
Prepared by Pn. Kang Seow Hung

0

SMJK CHUNG HWA PENANG

STPM 2022 Trial Exam
Physics Paper 3

Answer scheme

Section A (15 marks) 6D 11 A
7A 12 C
1B 8D 13 D
2B 9C 14 C
3B 10 C 15 A
4A
5B

Section B (15 marks)

16. (a) SHM is a motion in which the acceleration is always directly proportional to the displacement from

and directed towards a fixed point.

∝ − [1]

(b) (i) - Maximum velocity and angular velocity [1]
= − ' cos(2 ) - Negative cosine [1]

(ii) [1]
= 1

= − ' sin(2 ) [1]
2

(iii) K - Correct shape [1]
0 - Label T [1]

1 t

=

17. (a)

n1 M n2

O PC I
u

Spherical boundary - Correct direction [1]
- Label I [1]

(b) 8 ; ; − 8

+ = [1]

1.00 + 1.50 = 1.50 − 1.00 [1]
15.0 15.0

= −45 cm [1]

The image is virtual(, upright) and magnified. [1]

(c) → ∞ [1] is always negative/ = − EF . [1]
EG

Section C (30 marks)

18. (a)

= = 0.220 m [1]


Fundamental mode: = K = 5.5 cm [1]
L

First overtone: = MK = 16.5 cm [1]
L

2 resonance modes. [1]

(b)
[1]

[1]

(c) The end correction is negligible. [1]

(d) i) OPQ = 413 Hz. [1] OUV = 825 Hz. [1]

ii) = K [1] = 825 Hz [1]
;

[1]

iii) When temperature decreases, speed of sound decreases.[1] Wavelength remains.[1] Therefore,
frequency decreases.[1]

19. a) The waves must have same frequency/wavelength/constant phase difference. [1]
To produce constant optical path difference(at every point). [1]

b) = XY [1] = KZ = [['×8']^∙8.` = 9.9 mm [1]
Z X '.8'×8']a

c) = cY [1] = KZ = [['×8']^∙8.` = 39.6 mm [1]
Z c '.';[×8']a

OR sin = [1] = 1.8 tan gsinh8 cKi = 39.6 mm [1]

width = 2x = 79 mm [1]

d) I
sin ; = 2 [1]
; = 1.8 tan gsinh8 ;cKi = 79 mm [1] - Correct pattern [1]
- Correct number of interference fringes

in each diffraction maxima [1]
- Labels [1]

e) the refractive index of the water is higher than the air, wavelength in the water is shorter. [1]
As decreases, sin /x decreases. [1] The fringes are closer to each other. [1]

20. a) X-rays are produced through conversion of kinetic energy from fast moving electrons into photons. [1]
Photoelectric effect is a conversion of photon energy into kinetic energy of electrons. [1]

b) i) = lm F = . [1] q = r [1]
;nm lm

q = r [1]
s;qnmt

ii) ru = [1] OPQ = ru = ;nmru [1] = ;nmu q ; [1]
Kvwx qt lm F r

c) i) OPQ = ru [1] = 6.9 × 10-11 m. [1]
qt

ii) = = 2.0 × 10hM ∙ 18 × 10M = 36 W [1]

rate of heat energy produced = 0.995 × 36 W [1] = 35.8 W [1]

iii) I

- Correct shape [1]
- Correct labelling. [1]

0 OPQ