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STPM 2018 Term 3 Trial

Physics 960/3
1 hour 30 minutes

SEK.MEN.KEB. TINGGI MELAKA

(Malacca High School. Estd.1826)
Ke Arah Kecemerlangan Pendidikan

STPM 2018 TERM 3

TRIAL EXAMINATION

PHYSICS
PAPER 3
(1 hour and 30 minutes)

Instructions to candidates:

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO.

Answer all questions in Section A. Marks will not be deducted for wrong answers. For each question, four
suggested answers are given. Choose the correct answer and shade the corresponding letter on the objective
answer sheet provided.
Answer all questions in Section B. Write your answers in the spaces provided.

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

Values of constants are provided on last page in this question paper.

Prepared by: Checked by : Certified by:

____________________________ ________________ ________________

(Pn Ng Hwee Ing & Wee Choi Chiang ) (Mr Wee Choi Chiang) (En. Azmi B Sakmis)
Guru Physics Tkn 6 Ketua Panitia Physics Tkn 6 PK Tingkatan 6

This question paper consists of printed pages.

Section B: Answer both questions in this section in the space provided.

16. (a) A person is at a point 2.00 m from a point source of sound, the intensity at that point is
4.00 10-7 Wm-2.
Determine the intensity level heard by the person. [2]

(b) In the following figure, X and Y are two small loudspeakers having the same power as that found
in part (a). They vibrate in phase to emit sound of same frequency 170 Hz.

2.00m

2.00m

(i) Determine the intensity level heard by the person at Q. [3]

(ii) Explain what happens to the intensity level heard by the person at Q if the frequency of
loudspeaker Y is increased to 175 Hz but that of X remains unchanged.
[3]
17. (a) A thin plano-convex lens with radius of curvature 20.0 cm is made of glass of refractive index 1.50.
Determine the focal length of the thin lens in air. [2]

(b) The thin lens in part (a) above is then arranged co-axially with a concave mirror of radius of curvature
40.0 cm at distance 20.0 cm apart as shown in the diagram below.

80.0 cm 20.0 cm

If an object P is placed at distance 80.0 cm from the thin lens,

(i) determine the position of the final image formed by the system. [4]

(ii) State whether the final image is real or virtual. [1]

Section C: Answer one question only from this section in your answer sheets.

18. (a) The graph below shows the variation of acceleration, a with displacement x for motion of an object of
mass 2.00 kg.
a/ m s-2

8.00

- 0.40 0.40 x/m

-8.00

(i) Explain the motion of the object. [3]

(ii) Determine the period of the motion. [3]
(iii) Determine the total energy of the object. [2]

(b) The motion of a piston in a car engine can be considered as a simple harmonic motion. In a typical
engine, the top of a piston moves up and down between the points A and B, a distance of 0.10 m as shown
in the diagram below. The frequency of oscillation of the piston is 100 Hz.

(i) Deduce an equation for the displacement x of the piston with time t. [3]
(ii) If the mass of the piston is 480 g, determine the maximum force applied by the connecting rod
on the piston. [2]
(iii) Sketch the displacement-time graph and velocity-time graph on the same axis of time. [2]
 19. (a) Define Huygen’s Principle. [1]
(b) By using Huygen’s Principle, construct a suitable diagram and explain the phenomenon of
diffraction grating. [4]
(c) Light with a wavelength of 550 nm forms a diffraction pattern on a screen at distance 2.00 m
away after passing through a single slit of width 2.20x10-6m.
(i) Determine the width of the central bright fringe form on the screen. [3]
(ii) Sketch a graph to show the variation of the intensity with distance measured from the centre
of the maximum bright fringe. [2]
(d) Two small asteroids which emit yellow light of wavelength 550 nm are separated at a distance
of 5.00 km.
(i) Determine the maximum distance at which the Hubble Space Telescope which has an aperture
diameter of 2.40m can still resolve them. [3]
(ii) Explain the effect on the maximum distance to resolve them if blue light is emitted by the
asteroids. [2]

20. (a) By using quantum theory of light, explain the following observations in photoelectric effect experiment.
(i) The photoelectric current is directly proportional to the intensity of light. [2]
(ii) There is a threshold frequency for emission of photoelectrons. [3]

(b) The experimental arrangement in the figure below is used to study photoelectric effect. When yellow
light of wavelength 580 nm is incident on an alkaline metal C, the high resistance voltmeter will
achieve a final reading of stopping potential of 1.00 V.
yellow light
C

V high resistance
voltmeter

(i) Calculate the energy of each incident photon. [2]

(ii) Determine the maximum kinetic energy of the photoelectrons. [2]
(iii) Explain why the voltmeter reading remains unchanged when yellow light of greater intensity
falls on C. [2]
(iv) If violet light of wavelength 400 nm is used in the experiment, determine the final reading of
the voltmeter. [4]
 Values of Constants

Acceleration of free fall g = 9.81 ms-2

Avogadro constant NA = 6.02 x 1023 mol-1

Boltzmann constant k, kB = 1.38 x 10-23 J K-1

Gravitational constant G = 6.67 x 10-11 N m2 kg-2

Magnitude of electronic charge e = 1.60 x 10-19 C

Mass of the Earth ME = 5.97 x 1024 kg

Mass of the Sun MS = 1.99 1030 kg

Molar gas constant R = 8.31 J K-1 mol-1

Permeability of free space μo = 4π x 10-7 H m-1

Permittivity of free space εo = 8.85 x 10-12 F m-1

1
= (36𝜋) × 10− 9 Fm-1

Planck’s constant h = 6.63 x 10-34 J s-1

Radius of the Earth RE = 6.38 x 106 m

Radius of the Sun RS = 6.96 x 108 m

Rest mass of electron me = 9.11 x 10-31 kg

Rest mass of proton mp = 1.67 x 10-27 kg

Speed of light in free space c = 3.00 x 108 ms-1

Stefan-Boltzmann constant σ = 5.67 x 10-8 W m-2 K-4

Unified atomic mass unit u = 1.66 x 10-27 kg

Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

16. a)

I 
  10 log 
 0
I
 56.02dB
b) (i) constructive interference occurs at Q
 A  A0  A0
 2 A0
I  A2
I  (2 A0 ) 2
I 0  A0
2

 I  4I0

 4  4  10 7 
  10 log  12 

 1 10 
 62.04dB
(ii) Beat occurs
The intensity level will vary periodically between maximums and minimums at beat
frequency of 5 Hz

17. a)

1 1 1
 (n  1)  
f  r1 r2 
 1 1
 (1.5  1)  
 20  
f  40.0cm

1
Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

b) (i) For the thin lens

1 1 1
 
u v f
1 1 1
 
u v f
1 1 1
 
80 v 40
v  80.0cm
For mirror
r
f   20.0cm
2
U 2  60cm
1 1 1
 
 60 v2 20
v2  15cm

For lens
u3  5cm
1 1 1
 
5 v3 40
v3  5.714cm
At 5.714cm right of thin lens

(ii) Virtual

18. a) (i) The acceleration is directly proportional displacement and

always acts towards the equilibrium position
Therefore it undergoes simple harmonic motion

(ii)
a  w 2 x
w 2  gradient  20
2
w
t
2
 20
t
T  1.405s

2
Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

(iii)
1
E mw2 A2
2
1
  2  20  0.4 2
2
 3.20 J

b) (i)
x  A cos wt
0.1
A  0.05m
2
w  2f
 2  100
 200
x  0.050 cos(200t )

or x = 0.050sin(200πt)

(ii)
F = ma
F= mw2A ( a = w2x)
= 0.480 × (200π)2 × 0.05
= 9475N or 9470N

(iii)

x, v

0 t
v

3
Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

19. (a) Huggen’s Principle states that every point on a wave front can be considered as a
secondary point source. The surface tangential to the secondary wavelets forms the new
wavefront

b)

S1

S2
3

2
S3 S1, S2, S3 are secondary point
sources Tangent to the secondary
1 wavelets 1 , 2 , 3
are the new wavefronts

c) (i) asinθn = nλ
2.2× 10-6sinθ1 = 550×10-9
sinθ1 = 0.25
θ1 = 14.480
y
tan 1 
D
y  0.5164m
 width  2 y  1.03m

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 Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

(ii)

d) (i) Resolving power

1.22 
  2.796 107 rad   2.292 107 rad
D or D

y y
 
D D
y y
D D
 
or
5  10 3
5  103
 
2.796 7 2.292 7
 1.79  1010 m  2.18  1010 m

(ii) The max distance will be greater because blue light has shorter wavelength, thus the
resolving power increases

20.a) (i) The intensity of light is directly proportional to the rate of arrival of photons
Thus rate of emission of photoelectrons increases

ii) To emit a photoelectron, the energy of the photon must be greater or equal to the work
function
Since the energy of a photon is directly proportional to frequency, therefore the frequency
of the light must be greater than the threshold frequency

5
Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

b) (i)
hc
E

6.63 10 34  3 108

580  10 9
 3.429  10 19 J

or = 2.143eV

(ii)
E max  eV s
 1.6  10 19  1
 1.60  10 19 J

(iii) Increasing the intensity will only increase the rate of incident of photons but the
energy of each photon is unchanged
Therefore them maximum kinetic energy of photoelectrons remains unchanged
(iv) Einstein’s Equation
hc
 W0  eVs

6.63 10 34  3 108
9
 W0  1.6 10 19 1
580 10
W0  1.829 10 19 J

For violet light

hc
 W0  Emax

hc
 1.829 10 19  Emax
400 10 9
Emax  1.96eV

Therefore stopping potential = 1.96V

6
Marking Scheme Physics Paper 3 STPM 2018 Trial Exam

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