There Are Fifty Questions in This Paper. Answer All Questions. Marks Will Not Be Deducted For Wrong Answers
There Are Fifty Questions in This Paper. Answer All Questions. Marks Will Not Be Deducted For Wrong Answers
There Are Fifty Questions in This Paper. Answer All Questions. Marks Will Not Be Deducted For Wrong Answers
Time: 1h 45 min
There are fifty questions in this paper.
Answer all questions. Marks will not be deducted for wrong answers.
1. The dimension of angular velocity is
A T1
C LT2
1
B LT
D L2T1
1 mu2 1
Pt + mgh =
mv2
2
2
1
1
2
D Pt + mgh = mv mu2
2
2
A dW = F . d s
C dW = F d s
B dW = F . s
D dW = F s
6. The period of a particle which moves at speed
v in a horizontal circle of radius r is
2
2r
A
C
v
v
v
v
B
D
2
2r
4.00 m s1
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(a)
(b)
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100 N
106 m2
2 103 m
2m
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Booklet 2005 STPM Phy P1 & 2 (08).indd 4
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D
I
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D
I
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d 1
d+x 1
d
x m
B
m
D d+
1
d + 2x 1
2
P+
13
6
13
7
N+
1
0
N = 13.005739 u,
A
B
C
D
1
0
n = 1.008665 u.]
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Time: 2h 30 min
Section A [40 marks]
Answer all the questions in this section.
1. (a) Define a vector quantity.
(b) The diagram below shows a 5.0 kg street light suspended by two cables.
[1 mark]
[5 marks]
[2 marks]
[2 marks]
3. A lens of refractive index 1.52 is coated with a layer of transparent material of refractive index
1.38. The thickness of the coating material is such that the reflection of light at wavelength
400 nm can be eliminated.
(a) Determine the condition for interference to occur.
[3 marks]
(b) Calculate the minimum thickness of the coating material.
[2 marks]
4. (a) State the differences between gaseous, liquid and solid phases in terms of their atomic
arrangements and movements.
[2 marks]
(b) The variation of repulsive force Frep with distance r between two atoms can be represented
by the equation
a
Frep = ,
rn
where a and n are positive constants.
(i) Sketch a graph of Frep against r.
[1 mark]
(ii) Based on the graph in (b)(i), explain why a gas can be compressed but it is almost
impossible for a solid.
[2 marks]
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5. The diagram below shows two point charges 1.2 C and +1.0 C separated at a distance of
0.6 m. Point O is the midpoint of the two charges. Calculate the electric potential at point P which is
1.2 m vertically above O.
[5 marks]
[2 marks]
[1 mark]
[2 marks]
7. The work function for Cesium is 2.14 eV. Calculate the maximum kinetic energy of photoelectrons
emitted from Cesium surface when illuminated by light of wavelength 565 nm.
[4 marks]
8. A radioisotope tracer was injected into a human body. After 24 hours, the activity of the radioisotope
has reduced to 6% of its initial activity. Calculate the half-life of the radioisotope.
[4 marks]
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11. (a) (i) What is meant by work done in an isolated gas system?
[1 mark]
(ii) Differentiate between internal energy and thermal energy of a gas system.
[2 marks]
(iii) State the first law of thermodynamics and the meaning of each symbol used. [2 marks]
(b) An isolated system of 3.0 moles of an ideal gas is initially at pressure p1 and volume V1. It
is then allowed to expand at constant temperature T = 350 K to new pressure p2 and new
volume V2 which is twice the initial volume V1.
(i) Sketch the p-V diagram to show the expansion process and shade the region representing
the work done during the process.
[2 marks]
(ii) Calculate the work done during the process.
[3 marks]
(c) The diagram below shows an insulated window glass of a house which consists of two glass
panels separated by a layer of air.
The area of the window is 2.0 m 1.5 m. The thicknesses of the outer glass, inner glass
and air layer are 5.0 mm, 3.0 mm and 3.0 cm respectively. If the temperature outside the
house is 45 C and the temperature inside the house is kept at 20 C, calculate the energy
per hour that is prevented from entering the window.
[5 marks]
[Thermal conductivity of glass = 0.84 W m1K1,
thermal conductivity of air = 0.023 W m1K1.]
12. (a) State two functions of the dielectric in a capacitor.
[2 marks]
(b) Two pure capacitors of capacitance C1 = 5.0 F and C2 = 10.0 F are connected in series
to a 10.0 V battery.
(i) Derive an expression for the effective capacitance CT of the circuit and calculate the
value of CT .
[4 marks]
(ii) Calculate the total energy stored in the circuit.
[2 marks]
(c) The battery in (b) is replaced by an alternating current supply of maximum voltage 10 V
and frequency 120 Hz.
(i) Calculate the reactance of the combined capacitor.
[2 marks]
(ii) Calculate the maximum current in the circuit. Comment on the phase of the current
with reference to the voltage supplied.
[3 marks]
(iii) Describe briefly the behaviour of the energy stored in the combined capacitor during
a cycle of the alternating current supply.
[2 marks]
13. (a) State de Broglies hypothesis and give the relationship between momentum p and wavelength
of a particle.
[2 marks]
(b) In an electron diffraction experiment, an electron beam which is accelerated on a potential
difference is incident normally on a very thin gold film. Several circular diffraction rings
are seen on a photographic film.
(i) If the voltage at the anode is increased, what happens to the circular rings? [1 mark]
(ii) If a particular ring of radius R is chosen and different values of accelerating voltage V
1 Deduce that the experiment is in agreement
are recorded, sketch a graph of R against .
V
with de Broglies hypothesis.
[6 marks]
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(c)
(i) A 60 kg marathon runner runs at a speed of 5.1 m s1. Calculate the de Broglie
wavelength of the marathon runner.
[2 marks]
(ii) Explain briefly the production of continuous and characteristic X-rays.
[4 marks]
[1 mark]
(i) Express v in terms of E and B1.
(ii) Derive an expression for the mass m of an ion in terms of E, B1, B2, r and q. [2 marks]
(iii) If B1 = B2 = 0.01 T and r = 20.0 cm, calculate two different electric fields to enable
63
65
the mass spectrometer to differentiate singly charged copper isotopes Cu and Cu.
29
29
[3 marks]
63
29
65
29
Cu = 62.929601 u,
Cu = 64.927794 u.]
Cu +
14
[1 mark]
N
[3 marks]
13
1
1
14
7
Cu = 13.003355 u,
H = 1.007825 u,
N = 14.003074 u,
1u 931 MeV.]
(d) Explain briefly the controlled fission reaction in a nuclear reactor.
11
[3 marks]
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PAPER 1
1. A Unit is rad s1
2. C R is maximum when = 45
3. B Vertical motion: v = u sin 45, v = 0, a = g,
s = 1.0 m. Use v 2 = u 2 + 2as
1
1
4. A Work done = Pt = mv 2 mu 2 + mgh
2
2
5. A Work done is the dot () product of force and
displacement
distance travelled in a complete circle
6. C Period =
speed
7. C After collision, distance of centre of mass from
1.0 kg mass
(0.5 + 0.5)(4.00) + (1.00)(0)
x = = 2.00 m
(0.5 + 0.5) + 1.00
Angular momentum is conserved,
[(0.5+0.5)(2.00)2 + (1.00)(2.00)2] = (0.5)(4.00)(2.00)
= 0.50 rad s1
1
mv 2
2
m(r )2
5
8. B = =
1 2
2 2 2
2
mr
I
2
5
[Note: The moment of inertia of the ball about the
2
axis through the centre is mr 2.]
5
9. B E = constant for the whole region
10. C W must act through the centre of gravity (C.G.).
Taking moments about C.G.,
clockwise moments of T and F = anticlockwise
moment of N.
In D, there is no anticlockwise moment about the C.G.
11. D Distance travelled by the wave in 0.40 s
= (0.10 0) = 0.10 m
0.10
Speed v = = 0.25 m s1, = 0.02 m
0.40
v
0.25
f = = = 12.5 Hz
0.02
12. A amax = 2x0 = (2f)2(0.02) = 0.6
f = 0.87 Hz
13. A F = m( 2x), varies with displacement x
v
320
14. C = 2(0.2) m, f = = = 800 Hz
0.40
15. D
16. B Area under F-r curve = work done (or energy required)
1
1
17. D Energy stored = Fe = (100)(2 103) = 0.10 J
2
2
18. A f depends on whether the molecule is monatomic,
diatomic or polyatomic, and f is greater at high
temperatures
19. D pV = nRT, c T . (2p)(2V ) = nRT1
T1 = 4T, c1 4T , c1 = 2c
20. B pV = nRT, T pV. At P, pV = 100 J.
At Q,
pV = 400 J. At R, pV = 500 J
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
1
1
1 2
(1 2)
2
2
dQ
A = kA = kA
dt
x
0.5 x
RA
1
B = , =
=
RA
A
1
= and R =
C Volume = C A = A(3A),
3
A
c
RA A
1
= =
3
RC C
Power in resistor IV
R
8.0
C = = = = 0.80
Power from cell
IE
R + r 8.0 + 2.0
A Charge conserved, Q = (5 + 10)V1 = (5)(12)
V1 = 4.0 V
D Time constant of discharge circuit = CR 3
5
A I = (2.0) = 0.50 A
5 + 15
BNA
A BINA = k, = BAN
I
k
A When v is parallel to B, force on the charge, F = 0.
Path is a straight line
D Currents in the same direction, wires attract; currents
in opposite directions, wires repel
12
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T1
T2
W
=
=
sin 30
sin 50
sin 100
(5.0 9.81)(sin 30)
T1 = = 24.9 N
sin 100
(5.0 9.81)(sin 50)
T2 = = 38 N
sin 100
2. (a) Conditions of equilibrium of a rigid body:
Resultant force = 0
Resultant torque about any point = 0
(b) (i)
PAPER 2 Section A
1. (a) A vector quantity is a physical quantity which has
magnitude and direction
(b)
3. (a)
1
Path difference between AP and BQ = m
2
1
2n1t = m for m = 1, 2, 3,
2
1
2(1.38)t = m (400 nm)
2
[t = thickness of coating material]
(b) Thickness t is minimum when m = 1.
1
1 (400nm)
2
Minimum thickness =
2(1.38)
= 72.5 nm
4. (a)
Gaseous
Atomic
No
arrangement arrangement
at all,
constantly
changing
13
Liquid
Solid
Constantly
random
nonuniform
pattern
over short
range
Regularly
repeated
pattern
over long
range
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Movement
(b)
(i)
Random
Free
motion
random
motion with
high speeds
Section B
Atoms
vibrate
about
mean
positions
Frep
Large
force
Solid
Very small
force
r
Gas
vy
6.27
= tan1
at an angle tan1
vx
4.38
= 55.1 below the
horizontal
(iv) Average power of the marble,
mv 2 (30 103)(7.65)2
P = =
2t
2(0.639)
= 1.373 W
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2f = 315
2
Frequency, f = 50.1 Hz. Also, = 1.05
= 5.98 m
Velocity, v = f = (50.1)(5.98) = 300 m s1
(iii) = 5.98 m phase difference 2 rad.
For a distance of 2.0 m, phase difference
2.0
= (2)
5.89
= 2.10 rad
(ii)
(c)
(i)
(ii)
(iii)
V2
(ii) Work done = nRT ln
V1
= (3.0)(8.31)(350)
2V1
= 6.05 103 J
ln
V1
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14. (a)
29
(64.927794)(1.66 1027)
= 29.7 V m1
(c) (i) Hydrogen nuclei require a large amount
of energy to overcome the repulsive force
between each other.
Kinetic energy of the hydrogen nuclei is raised
by increasing its temperature to a very high
value
(ii) Mass difference, m
= (13.003355 + 1.007825)u 14.003074 u
= 8.106 103 u
Energy released
[1 u 934 MeV]
= (8.106 103)(931 MeV) = 7.55 MeV
(d) Water is used as a moderator to slow down the
neutrons.
Contol rods of cadmium are used to absorb two out
of the three secondary neutrons so that the fission
reaction occurs at a constant rate
2eV
1
mv 2 = eV, v =
m
2
[m = electron mass]
h
de Broglies wavelength, =
mv
h
h
= =
2eV
2meV
m
m
(c)
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