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1
PDG
Candidate Name
(
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ANDERSON JUNIOR COLLEGE
2014 JC2 Preliminary Examination
PHYSICS
Higher 2
Paper 1 Multiple Choice
9646/01
Tuesday 2 September 2014
1 hour 15 minutes
Additional Materials: Multiple Choice Answer Sheet
READ THESE INSTRUCTIONS FIRST
Write in soft pencil.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Write your name, PDG and NRIC/FIN and shade the 7 digits of your NRIC/FIN in soft pencil on the Multiple
Choice Answer Sheet.
There are forty questions in this section.
Answer all questions.
For each question there are four possible answers A, B, C and D.
Choose the one you consider correct and record your choice in soft pencil on the separate Multiple
Choice Answer Sheet.
\
Read the instructions on the Answer Sheet very carefully.
Each correct answer will score one mark. A mark will not be deducted for a wrong answer.
Any rough working should be done in this question paper.
The use of an approved scientific calculator is expected, where appropriate.
This document consists of 19 printed pages and 1 blank page
9646/01/AJC2014/Prelim
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2
Data
speed of light in free space,
c = 3.00 x 108 m s-1
permeability of free space,
0 = 4 x 10-7 H m-1
permittivity of free space,
0 = 8.85 x 10-12 F m-1
(1/(36)) x 10-9 F m-1
elementary charge,
e = 1.60 x 10-19 C
the Planck constant,
h = 6.63 x 10-34Js
unified atomic mass constant,
u = 1.66 x 10-27 kg
rest mass of electron,
me = 9.11 x 10-31 kg
rest mass of proton,
mp = 1.67 x 10-27 kg
molar gas constant,
R = 8.31 J K-1 mol-1
the Avogadro constant,
NA = 6.02 x 1023 mol-1
the Boltzmann constant,
k = 1.38 x 10-23 J K-1
gravitational constant,
G = 6.67 x 10-11 N m2 kg-2
acceleration of free fall,
g = 9.81 m s-2
9646/01/AJC2014/Prelim
3
Formulae
s = ut + ½ at2
uniformly accelerated motion,
v2 = u2 + 2as
work done on/by a gas,
W = pV
hydrostatic pressure,
p = gh
gravitational potential,
 
- Grm
displacement of particle in s.h.m.,
x = x0 sin t
velocity of particle in s.h.m.,
v = v0 cos t
2
  x 0  x 2
mean kinetic energy of a molecule of an ideal gas
resistors in series,
E=
3
KT
2
R = R1 + R2 + …
resistors in parallel,
1/R = 1/R1 + 1/R2 + …
electric potential,
V =
Q
4  0 r
alternating current/voltage,
x = x0 sin t
transmission coefficient,
T  exp(-2kd)
where k =
8 2 mU  E 
h2
x = x0exp(- t)
radioactive decay,
decay constant,
=
0 .6 9 3
t1
2
9646/01/AJC2014/Prelim
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1
2
3
4
Which of the following pairs of physical quantities are both vector quantities?
A
work done, electric current
B
work done, electric field strength
C
force, electric current
D
force, electric field strength
The diameter D of a sphere is measured to be 5.0 cm with a fractional uncertainty of 0.02.
What is the absolute uncertainty and fractional uncertainty of the radius R of the sphere?
absolute
uncertainty of R
fractional uncertainty
of R
A
0.05 cm
0.01
B
0.1 cm
0.01
C
0.05 cm
0.02
D
0.1 cm
0.02
The precision and accuracy of a measurement is affected by:
precision
accuracy
A
Systematic error
Random error
B
Random error
Random error
C
Systematic error
Systematic error
D
Random error
Systematic error
A golf ball travels with an initial speed vo and it went only one-third of the way to the hole.
Assuming the force of resistance due to the grass remains the same, what new initial speed
should the ball travel for the ball to go into the hole?
A
vo
B
vo
C
vo
9646/01/AJC2014/Prelim
D 3 vo
5
5
The graph is a displacement-time (s-t) graph for a tennis ball during part of a game.
Which part of the graph shows the highest speed?
s
A
C
D
B
t
6
The diagram shows a trolley moving on a frictionless horizontal table at a speed of
0.5 m s─1. 500 g of sand is then released onto the trolley.
500 g of sand
0.5 m s─1
What is the change in the momentum of the trolley?
A
7
zero
0.15 N s
B
C
0.20 N s
D
1.80 N s
A pole of weight W is attached to a wall. It is held horizontal by a wire attached at point X of
the pole where T is the force of the wire. The pole experiences a reaction force R from the
wall.
wire
pole
wall
X
Which triangle of forces could correctly represent the three forces acting on the pole?
A
B
C
R
W
R
R
W
T
D
R
W
T
T
T
9646/01/AJC2014/Prelim
W
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8
A ball bearing was released from rest in a viscous liquid. Which of the following graphs
would represent the variation of the forces acting on the ball bearing with time?
B
A
force
weight
force
weight
viscous
force
viscous
force
upthrust
upthrust
0
0
0
0
time
C
force
time
D
weight
weight
force
viscous
force
viscous
force
upthrust
upthrust
0
0
9
10
time
A man lifts a 10 kg sack of rice from a ground to above his head using his both hands. The
sack of rice does not experience a change in kinetic energy between the two positions.
Which of the following statement is correct?
A
The work done on rice sack by gravity is less than the work done on rice sack by man.
B
The work done on rice sack by gravity is more than the work done on rice sack by man
C
The work done on rice sack by gravity is equal to the work done on rice sack by man.
D
There is no work done on the rice sack.
A motor driving a pump raises 0.10 m3 of water through a vertical height of 5.0 m in a time of
10 minutes.
If the efficiency of the pump is 60%, what is the power generated by the motor? (Take the
density of water to be 1000 kg m─3)
A
11
0
0
time
4.9 W
B
8.2 W
C
14 W
D
82 W
A body of mass m moves in a horizontal circle of radius r at constant speed v for one
complete revolution.
Which of the following statements is incorrect?
A
The angular velocity of body is directed perpendicular to the plane of circular motion.
B
The work done by the centripetal force is 2mv2.
C
The change in linear momentum of the body is zero.
D
The total energy of the body is constant.
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7
A straight length of tape unwinds from a roll rotating about a fixed axis with constant angular
velocity, the radius of the roll decreasing at a steady rate.
12
v
radius
Which of the following graphs correctly shows how the variation of speed v at which the tape
moves away from the roll with time t?
A
B
v
D
v
v
v
t
13
C
t
t
t
The gravitational potential at point X due to the Earth is 7.2 kJ kg1. At point Y, the
gravitational potential is 3.4 kJ kg1.
Earth
X
Y
7.2 kJ kg1
3.4 kJ kg1
The change in gravitational potential energy of a 4.0 kg mass when it is moved from X to Y
is
A 42.4 kJ
14
B 10.6 kJ
C +3.8 kJ
D +15.2 kJ
Two satellites, A and B, orbiting around Earth have the same kinetic energy. Satellite A has
a larger mass than satellite B. Which of the following statements is correct?
A
Satellite A has the same total energy as satellite B.
B
Satellite A has a smaller orbital radius than satellite B.
C
Satellite A has a smaller period than satellite B.
D
Satellite A has a larger angular velocity than satellite B.
9646/01/AJC2014/Prelim
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15
A body moves with simple harmonic motion about a point P. The graph shows the variation
with time of its displacement from P.
displacement / m
1.0
0.5
0
─0.5
0.2
0
0.4
0.8 time / ms
0.6
─1.0
What is its angular frequency?
A
16
1.6 × 10-3rad s─1
B
4.0 × 10-3rad s─1
C 4.0 × 103rad s─1
D
2.5 × 104rad s─1
Two horizontal springs are attached to a trolley. The trolley is then displaced along the axis
of the springs and then released. The trolley undergoes simple harmonic motion. The
variation of the kinetic energy EK of the trolley with horizontal displacement x from its
equilibrium position is shown in the figure below.
80
70
60
50
40
30
20
10
─3
─1
─2
0
1
2
x/cm
3
The trolley loses energy so that its maximum kinetic energy is reduced by 50 mJ. What is
the amplitude of the oscillations?
A
2.4 cm
B
1.9 cm
C
1.4 cm
9646/01/AJC2014/Prelim
D
0.95 cm
9
17
Two bulbs X and Y containing air at different pressures are connected by a tube P which
contains two mercury threads.
h1
h1 and h2 are
not to scale
h2
The density of mercury is 13 600 kg m─3.
Which pair of values of h1 and h2 is possible?
18
h1 / cm
h2 / cm
A
2.0
8.0
B
4.0
2.0
C
6.0
6.0
D
8.0
2.0
The intensity of a progressive wave, besides being dependent on the amplitude of the wave,
is also proportional to the square of the frequency.
The diagram shows two waves X and Y.
displacement
2a
a
wave X
wave Y
time
The intensity of wave X is I0.
What is the intensity of wave Y?
A
0.028 I0
B
0.11 I0
C
0.44 I0
9646/01/AJC2014/Prelim
D
2.25 I0
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19
The diagram shows an ice cube floating in water.
Both the ice cube and the water are at 0 °C.
Which statement correctly compares the molecular properties of the ice and those of the
water?
20
A
The mean kinetic energies are the same for both the ice molecules and the water
molecules.
B
The mean inter-molecular separations are the same for both the ice and the water.
C
The mean inter-molecular potential energies are the same for both the ice molecules
and the water molecules.
D
The mean total energies are the same for both the ice molecules and the water
molecules.
Polaroids P, Q and R are aligned such that the axes of polarisation of P and Q are aligned
to each other and the axis of polarisation R is perpendicular to that of P and Q as shown.
θ
light
Observer
P
Q
R
Polaroid Q is then rotated about its centre through an angle θ until its axis of polarisation is
aligned with that of R.
Which of the following describes the change in light intensity seen by the observer?
A
Light intensity remains zero all the time.
B
Light intensity was maximum initially and decreases to zero.
C
Light intensity decreases to zero and then increases back to maximum.
D
Light intensity increases to maximum and then decreases back to zero.
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11
21
A double-slit interference of green-coloured light was set up as shown and interference
fringes are formed on the screen.
green-coloured
light
interference
fringes
double-slit
screen
Which change would increase the distance between adjacent fringes?
22
A
Use orange-coloured light.
B
Reduce the width of each slit.
C
Use a double-slit where the slits are further apart.
D
Move the double-slit closer to the screen.
A speaker producing sound of frequency 2500 Hz is placed at the open end of a closed pipe
containing a gas.
piston
speaker
As the piston is moved along the pipe, a series of 6 loud sounds was heard. The first loud
sound was observed when the piston was 4.0 cm away from the open end, and the sixth
loud sound was observed when the piston was 37 cm away from the open end.
What is the speed of sound in the gas?
A
270 m s─1
B
330 m s─1
C
370 m s─1
9646/01/AJC2014/Prelim
D
400 m s─1
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23
Which one of the following statements about the electric potential at a point is correct?
A The electric potential at a point due to a system of joint charges is given by the sum of
the potentials at that point due to the individual charges of the system
B The electric potential is given by the rate of change of electric field intensity with distance
C The unit of electric potential is either the joule or the volt
D Two points in an electric field are at the same potential only when a unit positive charge
placed anywhere on the line joining them remains stationary.
24
Three identical electrical sources each with internal resistance r are used to operate a lamp
of resistance R as shown in figure below.
What fraction of the total power is lost due to the internal resistance of the sources?
R
A
25
3R + r
3R
B
3R ─ r
3R
C
r
3R + r
D
3R
3R + r
Four resistors of equal values are connected as shown.
W
Y
X
Z
How will the current through the resistors change when resistor W is removed?
A
The current through X will increase and the currents through Y and Z will decrease.
B
The current through X will decrease and the currents through Y and Z will increase.
C
The current through X will increase and the currents through Y and Z will remain
unaltered.
D
The currents through X, Y and Z will all decrease.
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13
26
The current I flowing through a component varies with the potential difference V across it as
shown.
I
V
Which graph best represents how the resistance R varies with V?
A
R
V
B
R
V
C
R
V
D
R
V
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27
A cell of emf E1 = 6.0 V and negligible internal resistance is connected to a uniform
resistance wire XY. Another cell of emf E2 = 3.0 V and negligible internal resistance is
connected as shown. When the movable contact C is placed at B, there is no current in the
branch CE.
E1 = 6.0 V
A
X
D
B
Y
C
G
E
E2 = 3.0 V
The contact C is then moved to point D.
What will be the direction of current in the branch CE and the change in current reading of
the ammeter A?
28
direction of current in
branch CE
change in current reading
in ammeter A
A
CE
decrease
B
CE
increase
C
EC
decrease
D
EC
increase
An -particle and a -particle both enter the same uniform magnetic field, which is
perpendicular to their direction of motion. If the -particle has a speed 15 times that of the
-particle, what is the value of the ratio
magnitude of force on β  particle
?
magnitude of force on α  particle
A
3.7
B
7.5
C
60
9646/01/AJC2014/Prelim
D
112.5
15
29
A coil PQRS mounted on an axle, has its plane parallel to the flux lines of a uniform
magnetic field B, as shown.
axle
P
Q
I
uniform
magnetic
field B
I
R
S
When a current I is switched on, and before the coil is allowed to move,
30
A
there are no forces due to B on the sides SP and QR.
B
there are no forces due to B on the sides PQ and RS.
C
sides SP and QR tend to attract each other.
D
sides PQ and RS tend to attract each other.
A conducting wire XY moves between the magnets as shown in the diagram.
Which of the following motions of XY will make the galvanometer deflect the most?
0
N
X
Y
S
A
It moves sideways along the magnetic field quickly.
B
It moves sideways along the magnetic field slowly.
C
It moves perpendicular to the magnetic field quickly.
D
It moves perpendicular to the magnetic field slowly.
9646/01/AJC2014/Prelim
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31
A flowmeter is used in many industries to measure the volume of liquid passing through a
pipe per unit time about the axis of the turbine. When the liquid flows through the turbine,
the 8 magnets attached on it rotate.
Which of the following voltage-time graphs shows the signal detected by the coil?
Voltage / V
A
Voltage / V
B
Time / s
C
Time / s
Voltage / V
Voltage / V
Time / s
32
D
Time / s
When a resistor is connected across an sinusoidal A.C. source of peak voltage 170 V, the
average power dissipated is 40 W. Two such identical resistors are now connected in series
to the electrical mains of 220 V r.m.s.
What would be the total power dissipated?
A
33.5 W
B
67.0 W
C
80.0 W
D
134 W
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17
33
In an experiment to investigate the photoelectric effect, monochromatic light is incident on a
metal surface. The photoelectric current and the maximum kinetic energy of the
photoelectrons are measured.
Which one of the following correctly shows the change, if any, in the photoelectric current
and in the maximum kinetic energy of the photoelectrons when light of the same intensity
but higher frequency is incident on the same metal surface?
34
photoelectric current
maximum kinetic energy
A
decreases
no change
B
decreases
increases
C
no change
decreases
D
no change
increases
The diagram below shows some possible electron transitions between three principal
energy levels in the hydrogen atom.
Which transition is associated with the absorption of a photon of the longest wavelength?
D
B
A
0 eV
─1.2 eV
C
─13.6 eV
35
According to the de Broglie hypothesis, matter waves are associated with
A
electrons only
B
charged particles only
C
neutral particles only
D
all particles
9646/01/AJC2014/Prelim
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36
A semiconductor X is made by doping germanium crystal with arsenic (donor). Another
semiconductor Y is made by doping germanium with indium (acceptor). The two are joined
end to end and connected to a battery as shown.
Which of the following statements is correct?
X
37
38
Y
A
X is p-type, Y is n-type and the junction is forward biased.
B
X is n-type, Y is p-type and the junction is forward biased.
C
X is p-type, Y is n-type and the junction is reverse biased
D
X is n-type, Y is p-type and the junction is reverse biased.
Which one of the following statements best describes stimulated emission in a laser?
A
Electrons collide with atoms in a metastable state and cause photons to be emitted.
B
Atoms in a metastable state de-excite and cause electrons to be emitted.
C
Photons interact with atoms in a metastable state and cause photons to be emitted.
D
Photons interact with atoms in a metastable state and cause electrons to be emitted
Radiation from a radioactive source enters an evacuated region in which there is a uniform
magnetic field perpendicular to the plane of the diagram. This region is divided into two by a
sheet of aluminum about 1 mm thick. The curved, horizontal path followed by the radiation is
shown in the diagram below.
A
B
Which of the following correctly describes the type of radiation and its point of entry?
type of radiation
point of entry
A
alpha
A
B
alpha
B
C
beta
A
D
beta
B
9646/01/AJC2014/Prelim
19
39
The diagram shows a graph of the binding energy per nucleon for a number of naturally
occurring nuclides plotted against their mass number.
binding energy per
nucleon / MeV 9
8
27
Al
13
7
23
11
6
238
92 U
Na
5
4
3
2
2
1H
1
0
20
40
60
80 100 120 140 160 180 200 220 240
mass number
Which of the following statements is a correct deduction from the graph?
A
Energy will be released if a nucleus with a mass number less than about 80
undergoes fission as a result of particle bombardment.
B
Energy must be supplied for a nucleus with a mass number greater than about 80 to
undergo fusion with any other nucleus.
C
D
40
238
92U is the stable end-point of a number of radioactive series.
27
23
13 Al will spontaneously emit an alpha particle to become 11 Na .
The probability of decay in one second of a radioactive nucleus is . During a particular one
second interval, a nucleus does not decay.
What is the probability of decay of this nucleus during the next one second interval?
A
1

B

C
2
9646/01/AJC2014/Prelim
D
2
Suggested Solution to 2014 H2 Physics Prelim Paper 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
D
C
D
B
A
C
A
A
C
C
B
A
D
A
D
C
D
D
A
D
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
A
B
A
C
A
C
A
B
B
C
D
B
B
B
D
D
C
D
B
B
No
Answer & Solution
1
Ans: D
Even though work done can be negative, it is a scalar since the negative does not
represent its direction.
Electric current flows along the wire and therefore has no fixed direction in space.
2
Ans: C
The absolute uncertainty of the diameter is 0.02 x 5.0 = 0.1 cm.
The absolute uncertainty of the radius will be 0.1 cm / 2 = 0.05 cm
The fractional uncertainty of the radius will be 0.05 cm / 2.5 cm = 0.02.
3
Ans: D
Random error causes a scatter of points about an average which affects the precision.
Systematic error causes a shift in the value away from the true value which affects the
accuracy.
4
Ans: B
1
1
1
0 = vo2 – 2a (3s) => vo2 = 2a(3s) ------------------- (1) OR using 2mv2 = Fx
0 = u2 – 2as => u2 = 2as --------------------(2)
vo2 1
(1)/(2) : u2 = 3
u = 3vo
5
Ans: A
A has the steepest gradient.
1 6
Ans: C
Applying conservation of momentum to trolley.
mtrolleyvtrolley = (mtrolley+msand)vtrolley&sand
vtrolley&sand = (2.0)(0.5)/(2.5) = 0.4 m s─1
ΔPtrolley = |(2.0)(0.4 ─ 0.5)| = 0.2 N s
7
Ans: A
The three forces must through a common point. Hence direction of R must act along
the dotted line from the wall.
wire
pole
wall
X
8
Ans: A
Since object is released from rest, no viscous force would act on it at the start.
Upthrust is dependent on the pressure difference acting on the object. As pressure
difference remain the same as object falls, upthrust will remain unchanged.
9
Ans: C
Based on Energy Work Theorem and isolating rice sack as system,
Work done on rice sack by man + Work done on rice sack by gravity = Change in KE
Work done on rice sack by man = - Work done on rice sack by gravity
|Work done on rice sack by man| = |Work done on rice sack by gravity|
10
Ans: C
Pout = 0.6 x Pin
Vρgh
t = 0.6 x Pin
Pin = 13.6 W = 14 W
2 11
Ans: B
A Correct. Using “right hand grip”, the direction of angular velocity is perpendicular to
the plane of circular motion.
B Incorrect. The work done by the centripetal force is zero since centripetal force (ie
resultant force) acting on the mass is always perpendicular to the displacement
travelled by the mass.
C Correct. When the object is back to the same position in one complete revolution,
the velocity is the same. Hence, change in linear momentum of the body is zero.
D
12
Correct. The total energy (ie KE & GPE) of the body remains constant.
Ans: A
Using v = r, since angular velocity  remains constant, hence speed v varies
proportionally with r, which decreases at a constant rate.
13
Ans: D
There is work done on mass to move it from X to Y. Hence, a gain in its GPE.
Change in GPE = [─3.4 ─ (─7.2)]x4 = +15.2 kJ
14
Ans: A
Since mAvA2/rA = GMmA/rA2, mAvA2 = GMmA/rA
Similarly,
mBvB2 = GMmB/rB
Since ½ mAvA2 = ½ mBvB2, GMmA/rA = GMmB/rB,
mA/rA = mB/rB,
Since mA > mB , rA > rB , and vA2 < vB2, vA < vB,
Since v = rω, rA ωA < rB ωB , ωA < ωB , as T = 2π/ω, TA > TB.
Total energy E = -1/2[GMm/r],
Since GMmA/rA = GMmB/rB ,
15
EA = EB
Ans: D
3


 24800 rads-1
3 
 0.76  10 
angular frequency,  = 2f = 2 
3 16
Ans: C
First method
Second method
Shift the graph down by 50 mJ.
1
2
m 2 x0
2
2
 x0
KEmax 
Read off the x-intercept value.
KEmax
x0 = 1.40 cm
 2.4 
75

 
75  50  x0 
x0  1.39 cm
2
1.40
17
Ans: D
For pressure in P, 16000 - ρgh1 = 8000 - ρgh2,
h1 – h2 = [16000 – 8000]/[13600 x 9.81] = 0.05996 m = 6.0 cm
since pressure in X > pressure in Y,
18
h1 > h2 hence D is the answer.
Ans: D
Amplitude of X = 2 x Amplitude of Y
Period of X = 3 x Period of Y  Frequency of X = 1/3 Frequency of Y
Hence, IX / IY = (2)2(1/3)2
IY = 9/4 IX = 2.25 I0
4 19
Ans: A
As the ice and the water are at the same temperature, they have the same mean
kinetic energy.
C is wrong as distance between molecules in ice is less than water. Hence
inter-molecular potential energy of ice is less than water.
B is wrong as explained in A.
D is wrong as the ice and the water have different inter-molecular potential
energy, they do not have the same total energy.
20
Ans: D
At the original position, the light from Q is polarized vertically. Hence no light will pass
through R as the incoming light from Q is in the plane that is perpendicular to the axis
of polarisation of R.
At the final position where the axes of polarisation of Q and R are aligned, the light
that passes through P is polarized vertically. Hence no light will pass through Q as the
incoming light from P is in the plane that is perpendicular to the axis of polarisation of
Q. Therefore the intensity of light through R is zero.
When Q is rotated, the light intensity will increase until it reaches a maximum at an
angle of 45°, and it reduces again as the angle is increased.
21
Ans: A
Fringe separation, x = λD/d, where λ is the wavelength of the light, D is the distance
between slits and screen and d is the slit separation.
Options C and D reduces the fringe separation while option A increases fringe
separation (as the wavelength of orange light is larger).
Option B has no effect on fringe separation.
22
Ans: B
Since 6 nodes form 5 segments,
Distance between adjacent nodes = (37 – 4.0) / 5 = 6.6 cm
Wavelength of sound = 2 x 6.6 = 13.2 cm
Speed of sound = 2500 x 0.132 = 330 m s─1
Be careful: 4 cm 
λ
because there is an end correction.
4
5 End
correction
(37 – 4.0 ) cm
loud loud loud loud loud loud
23
Ans: A
The electrical potential at a point due to a system of joints charges is given by the sum
of the potentials at that point due to the individual charges of the system.
(C)
(A) Unit of potential cannot be Joule,
dE
(B) V  dr
(D) +ve charge may not remain stationary if there is presence of lower potential in
the field.
24
Ans: C
Fraction of total power lost
Power lost by internal resistance
= Power lost by internal resistance+ Power of external resistor R
i2r+ i2r + i2r
r
= i2r+ i2r + i2r + (3i)2R = 3R + r
25
Ans: A
Original distribution of resistance across the two loops in the circuit is R/2 to R/2 i.e.
1:1. The new distribution is R to R/2 i.e. 2:1.
Due to the new distribution of resistance, the voltage across X will increase (2/3 V)
and the voltage across Y and Z will decrease (1/3 V). Since current is V / R, and R is
6 constant, the increase in voltage across X will cause an increase in current through X
while the decrease in voltage across Y and Z will cause a decrease in current through
Y & Z.
26
Ans: C
Resistance is the inverse of I/V ratio.
27
Ans: A
E2 = 6.0 V
A
I1
I3
X
I2
D B I1
I2
G
E2 = 3.0 V
When C is connected at D, the pd across XD will take on the pd of 3.0 V. The pd
across DY will then also be 3.0 V as the pd across XY is 6.0 V. The current through
the ammeter is VCY/RCY. When C is shifted from B to D, R of CY will increase, while
the voltage across CY remain the same, the current through the ammeter will
decrease.
VXD = VDY = 3 V. Since RXD < RDY => I3 > I1.
At node X, I1 and I2 combine to give a higher current I3.
Direction of current in branch CE is C
E.
Extension question:
What if C is shifted to the right of B?
28
Ans: B
F=Bqv
F
2
B( 2e)v


 7.5
F  B(e)(15v ) 15
29
Ans: B
F = BIlsin 
For sides PQ and RS,  = 0o and 180o, hence F = 0
For sides SP and QR,  = 90o, F = BIl.
Direction of the forces on SP and QR can be found by Fleming’s left-hand rule. The
force on QR points out of the plane of the diagram and the force on SP points into the
plane of the diagram.
7 30
Ans: C
For the galvanometer to deflect the most, the rate of change of magnetic flux linkage
in the wire is to be the largest. Hence, the wire moving perpendicular to the magnetic
field quickly will give a larger induced emf than moving perpendicular slowly. For A &
B, wire is moving sideways along magnetic field, this will not induce any emf.
31
Ans: D
When each magnet approaches the coil, there is an increase in the flux linkage in the
coil. The maximum flux linkage in the coil occurs when the magnet is closest to the
coil. The increase in flux linkage with respect to time results in an induced emf. When
that magnet moves away from the coil, there is a decrease in the flux linkage in the
coil. The minimum flux linkage in the coil occurs when the coil is at equidistant from
the 2 magnets that are closest to the coil. Thus, the decrease in flux linkage with
respect to time results in having an induced emf in the opposite direction.
Recall the Demo Lab EMI experiment “Activity B: EMI - Swinging bar magnet above
solenoid”
32
Ans: B
Let peak voltage of 1 resistor be Vo.
1 Vo2
<P> = 2 R
Given <P> = 40 W,
40 = ½ (170)2/R
R = (170)2/80
Ptotal = 2202 / [2x(170)2/80] = 67 W
33
Ans: B
From photoelectric effect equation: KE max of electrons  hf   , using a light of higher
frequency, maximum energy of ejected electrons increases.
As intensity of light = n p 
hf
, with a higher frequency, the rate of number of
Area
incoming photons decreases. Hence, the photoelectric current decreases.
Ans: B
For absorption of a photon to take place, the energy transition is from a lower energy
level to a higher energy level. For absorption of a photon with highest wavelength, the
34
difference between the 2 energy levels is the smallest. i.e. E 
hc

8 35
Ans: D
The de Broglie hypothesis is a fundamental principle exhibited by both radiation and
matter. The de Broglie hypothesis can ‘predict’ wavelengths for all particles.
36
Ans: D
X being doped by Gp V arsenic, would have valence electrons as mobile charge
carriers, hence n-typed. Y being doped by Gp III indium, would have holes as mobile
charge carriers, hence p-typed. The connection of XY with battery increases the
potential barrier, hence reverse biased.
37
Ans: C
A – Photons interact with electrons/atoms in metastable state, resulting in stimulated
emission of photons.
B- Electrons in a metastable state de-excited releasing photons.
D- Photons and not electrons are emitted.
38
Ans: D
It is beta as the particles are able to penetrate through the thin film of aluminum where
alpha particles will be stopped by the aluminum film.
Entry is at B because the beta particles lose energy after passing through the
aluminum. Hence, they have a lower speed and hence, a smaller radius, greater
mv2
curvature of path under the influence of magnetic field. (Bqv = r => r = mv / Bq)
39
Ans: B
A Incorrect. When a nucleus with a mass number less than about 80 splits into
smaller nuclei, there is a decrease in the binding energy per nucleon, hence, energy is
required to trigger the fission process i.e. energy is absorbed..
B Correct. When a nucleus with a mass number greater than 80 fuses with another
nucleus, there is a decrease in the binding energy per nucleon, hence, energy is
required to trigger this fusion process.
238
C Incorrect. For 92U to be the daughter nucleus of a radioactive decay, the parent
nucleus has to have a greater mass number than 238. However, from the graph,
238
92U is the nucleus with greatest mass number. Hence, it is not the stable end-point of
a number of radioactive series.
27
D Incorrect. 13 Al is more stable than
23
11 Na . Hence, the former will not
spontaneously emit an alpha particle to become the latter.
Ans: B
Radioactive decay is a random process where the probability of decay per unit time is
constant.
40
9 1
PDG
(
Candidate Name
)
ANDERSON JUNIOR COLLEGE
2014 JC2 Preliminary Examination
9646/02
PHYSICS
Higher 2
Paper 2 Structured Questions
Thursday 18 September 2014
1 hour 45 minutes
Candidates answer on the Question Paper.
No Additional Materials are required.
READ THESE INSTRUCTIONS FIRST
Write your name, index number and PDG on the spaces provided at the top of this page.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough working.
For Examiner’s Use
Do not use staples, paper clips, highlighters, glue or correction fluid.
1
The use of an approved scientific calculator is expected where
2
appropriate.
3
Answer all questions.
4
At the end of the examination, fasten all your work securely together.
5
The number of marks is given in brackets [ ] at the end of each
question or part question.
6
7
8
9
Deduction
Total
This document consists of 19 printed pages and 1 blank page
9646/02/AJC/2014Prelim
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Data
speed of light in free space,
c  3.00 x 108 m s-1
permeability of free space
0  4 x 10-7 H m-1
permittivity of free space
0  8.85 x 10-12 F m-1
(1/(36)) x 10-9 F m-1
elementary charge,
e  1.60 x 10-19 C
the Planck constant
h  6.63 x 10-34 J s
unified atomic mass constant
u  1.66 x 10-27 kg
rest mass of electron,
me  9.11 x 10-31 kg
rest mass of proton,
mp  1.67 x 10-27 kg
molar gas constant,
R  8.31 J K-1 mol-1
the Avogadro constant,
NA  6.02 x 1023 mol-1
the Boltzmann constant,
k  1.38 x 10-23 J K-1
gravitational constant,
G  6.67 x 10-11 N m2 kg-2
acceleration of free fall,
g  9.81 m s-2
9646/02/AJC/2014Prelim
3
Formulae
uniformly accelerated motion,
s  ut  21 at 2
v 2  u 2  2as
work done on/by a gas,
W  pV
hydrostatic pressure,
p  gh
gravitational potential,
 
displacement of particle in s.h.m.,
x = x0 sin  t
velocity of particle in s.h.m.,
v = v0 cos  t
Gm
r
2
v    xo  x 2
mean kinetic energy of a
molecule of an ideal gas,
E
resistors in series,
R  R1 + R2 + …
resistors in parallel,
3
2
kT
1/R  1/R1 + 1/R2 + …
Q
4  0 r
electric potential,
V 
alternating current/voltage,
x  x0 sin  t
transmission coefficient,
T  exp(-2kd)
where k 
8 2 mU  E 
h2
x  x0 exp(- t)
decay constant,
 
3
1
9
6t
.
0
radioactive decay,
2
9646/02/AJC/2014Prelim
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1
(a) A body has an initial velocity u and an acceleration a. After a time t, the body moved a
distance s and has a final velocity v. One of its equation of motion is
1
s = ut + 2at2.
State the conditions that must be satisfied for the above equation to be valid.
………………………………………………………………………………………………………
……………………………………………….……………………………………….………... [2]
(b) A hot air balloon was rising steadily at a speed of 10.0 m s-1 when weather conditions
turned windy. A constant breeze of 3.0 m s-1 blew horizontally across the sky, which
caused the hot air balloon to travel with a resultant velocity of vR at an angle  to the
horizontal, as shown in Fig. 1.1 below.
vR

Fig. 1.1
(i)
Calculate the magnitude and direction of the resultant velocity vR.
 = …………………….. o
vR = …………………….. m s-1 [2]
(ii)
A sandbag was dropped from the balloon.
1. In Fig. 1.2, sketch the path of trajectory of the sandbag as it drops from the
balloon.
vR

sandbag
[1]
Fig. 1.2
9646/02/AJC/2014Prelim
5
2. Determine how far below the balloon would the sandbag be after 4.0 s. You may
assume that the sandbag had not landed on the ground, the dropping of
sandbag did not affect the velocity of the hot air balloon and that air resistance
on the sandbag is negligible.
distance = ……………………………. m [3]
2
Fig. 2.1 shows a signboard suspended by two elastic ropes of tension T1 and T2. The tension
in T1 is 300 N and the tension in T2 is 252 N.
T1
40
50
T2
signboard
Fig. 2.1
(a)
(i)
State the conditions for equilibrium.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[2]
(ii)
On Fig. 2.1, mark the centre of gravity of the signboard with a dot and label the
point as G. Show clearly your construction to determine the centre of gravity on
Fig. 2.1.
[1]
(iii)
Determine the weight of the signboard.
weight = …………………………….N [1]
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(b) The signboard is pulled vertically downwards with a force of 20 N so that the ropes are
stretched. Determine the acceleration of the signboard immediately after it is released.
acceleration = …………………………….m s-2 [2]
3
The curve in Fig.3.1 shows the way in which the gravitational potential energy of a body of
mass m in the field of the Earth depends on r, the distance from the centre of the Earth, for
values of r greater than the Earth’s radius RE.
A
Energy
0
RE
r
R
B
Fig.3.1
9646/02/AJC/2014Prelim
7
The body referred in Fig. 3.1 is a rock which is projected vertically upwards from the Earth.
Assume air resistance can be neglected.
At a certain distance R from the centre of the Earth, the total energy of the rock (i.e. its
gravitational potential energy plus its kinetic energy) may be represented by a point on the
line AB.
(a) For a rock that is momentarily at rest at the top of its trajectory, which is at the
distance R from the centre of the Earth,
(i)
mark its total energy on the line AB in Fig.3.1 with a cross and label it P.
[1]
(ii)
sketch the variation of its kinetic energy with distance from the centre of the Earth
on Fig.3.1.
[2]
(b) The escape speed of the rock is the minimum speed that the rock must possess when it
is on the Earth’s surface so that it can escape to infinity.
(i)
On Fig.3.1, mark on the line AB with a cross and label it Q to represent the total
energy of the rock if it were projected from the Earth with the escape speed.
[1]
(ii)
Explain why the minimum speed for the rock to reach the moon from the surface of
the Earth is less than the escape speed.
…………………………………………………………………………………....................
…………………………………………………………………………………....................
…………………………………………………………………………………............... [1]
4
(a) Outline and explain experimental observation(s) which provided evidence for the wave
nature of light.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
………………………………………………………………………………………………………[3]
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(b) The photoelectric effect provides evidence for the particulate nature of electromagnetic
radiation. State three experimental observations that support this conclusion.
1. …………………………………………………………………………………………………....
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
2. …………………………………………………………………………………………………....
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3. …………………………………………………………………………………………………....
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………………………………………………………………………………………………………[3]
5
In Fig 5.1. below, an electrical device (load) is connected in series with a cell of e.m.f. 2.5 V
and internal resistance r. The current I in the circuit is 0.10 A. The power dissipated in the
load is 0.23 W.
e.m.f. = 2.5 V
I = 0.10 A
load
Fig 5.1
(a)
S.I. unit for resistance is defined as the Ohm (Ω).
Define the Ohm.
……………………………………………………………..……………………………….……
………………………………………………………………….……………………………..[1]
9646/02/AJC/2014Prelim
9
(b)
Show that the internal resistance r of the cell is 2.0 Ω.
[1]
(c)
A second identical cell is connected into the circuit in (a) as shown below in Fig.5.2.
I = 0.15 A
load
Fig. 5.2
The current in this circuit is 0.15 A. Deduce that the load is a non-ohmic device.
……………………………………………………………………………………………….……
………………………………………………………………………………………………... [3]
6
(a) Define the tesla.
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….………..
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….......... [2]
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(b) A large horseshoe magnet produces a uniform magnetic field of flux density B between
its poles. Outside the region of the poles, the flux density is zero.
The magnet is placed on a top-pan balance and the wire XY is situated between its
poles, as shown in Fig. 6.1.
Y
pole P
magnet
X
top-pan
balance
Fig. 6.1
The wire XY is horizontal and normal to the magnetic field. The length of wire between
the poles is 4.4 cm.
A direct current of magnitude 2.6 A is passed through the wire in the direction from X to
Y. The reading on the top-pan balance increases by 2.3 g.
(i)
State and explain the polarity of the pole P of the magnet.
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….…………........
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….………….... [3]
(ii)
Calculate the flux density between the poles.
flux density = ……………………………...T [2]
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(c) The direct current in (b) is now replaced by a very low frequency sinusoidal current of
r.m.s. value 2.6 A.
Calculate the variation in the reading of the top-pan balance.
variation in reading = …………………………….g [1]
7
(a) X-rays are produced when electrons are accelerated through a potential difference
towards a metal target such as tungsten. Fig. 7.1 shows a typical X-ray intensity
spectrum that can be produced from an X-ray tube.
Intensity, I
K
K
L
L
Wavelength, 
Fig. 7.1
Using conservation of energy, explain why there is a minimum wavelength for the
emitted X-rays.
……………………………………………………………………………………………………….
……………………………………………………………………………………………………….
…………………………………………………………………………………………………... [1]
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(b) In a chest X-ray, a photographic film receives photons which have travelled through flesh
and bone from a source.
(i)
Estimate the area of a film which covers the chest of an adult.
area = ……………………. m2 [1]
(ii)
Assume that on average, 10 x-ray photons fall on each grain of the photographic
film and the grains are about 1.0 µm apart.
Use your estimate in (b)(i) to calculate the total x-ray energy falling on the film.
Each x-ray photon has a quantum energy of 10-15 J.
total energy = …………………..…. J [3]
(c) When all possible photon paths are summed, the amplitude of the wave function for
paths travelling through flesh is four times the size of the amplitude for paths travelling
through bone.
Calculate the ratio
probabilit y of photon arrival though flesh
.
probabilit y of photon arrival through bone
ratio = …………………. [1]
9646/02/AJC/2014Prelim
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8
A student is investigating the stopping distance for a motorcycle with high-performance
brakes. A motorcyclist riding and stopping a motorcycle on a test track is recorded on film.
The stopping distance d is measured for different speeds v.
The variation with speed v of the stopping distance d is shown in Fig. 8.1
110.0
d/m
100.0
90.0
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
v / m s-1
Fig. 8.1
It is suggested that v and d are related by the equation
v2
d = 2A + Bv
where A and B are constants.
9646/02/AJC/2014Prelim
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d
Data from Fig. 8.1 are used to obtain values for v and v. These are plotted on the graph of
Fig. 8.2.
3.40
d
v /s
3.00
•
•
2.60
2.20
•
1.80
•
1.40
•
1.00
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
v / m s-1
Fig. 8.2
(a)
(i)
d
Determine the value of v for v = 25.0 m s-1.
d
v = …………………………….s [1]
(ii)
On Fig. 8.2,
1.
plot the point corresponding to v = 25.0 m s-1
2.
draw the line of best fit for all the points.
[2]
9646/02/AJC/2014Prelim
15
(b)
(i)
Use Fig. 8.2 to determine the gradient of the line drawn in (a)(ii).
gradient = …………………………… [2]
(ii)
Hence find the value of A with appropriate units.
A = …………………………….. [2]
(iii)
Use Fig. 8.2 to determine the magnitude of B.
B =………………………s [1]
(iv)
The stopping distance d consists of the thinking distance and the braking distance.
Suggest which part of the equation given represents the thinking distance and
which part represents the braking distance. Explain your answers clearly.
…………………………………………………………………………………....................
…………………………………………………………………………………....................
…………………………………………………………………………………....................
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……………………………………………………………………………….………….... [4]
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(c) The motorcyclist riding at a speed of 50.0 m s-1 towards an obstacle which is 200 m
away along the test track applies the brake. Determine the distance he will be from the
obstacle when the motorcycle stops.
distance =…………………….m [2]
(d) Explain why the stopping distance has to be increased when it is raining.
…………………………………………………………………………………………….………..
…………………………………………………………………………………………….………..
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….………..
…………………………………………………………………………………………….......... [2]
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9
Two flat circular coils carrying current can be used as Helmholtz coils to produce uniform
magnetic field between the coils. Fig. 9.1 illustrates the magnetic flux pattern due to the
Helmholtz coils.
Fig. 9.1
The magnetic flux density B between the Helmholtz coils is thought to depend on the current
I in the coils and may be written in the form, B = kIn where k and n are constants.
You are provided with two flat circular coils. You may also use any of the other equipment
usually found in a Physics laboratory.
Design an experiment to determine the value of n.
You should draw a labeled diagram to show the arrangement of your apparatus. In your
account you should pay particular attention to
the identification and control of variables,
the equipment you would use,
the procedure to be followed,
how the magnetic flux density between the Helmholtz coils would be measured,
(e) any precautions that would be taken to improve the accuracy of the experiment.
(a)
(b)
(c)
(d)
Diagram
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BLANK PAGE
9646/02/AJC/2014Prelim
2014 Prelim H2 Physics Paper 2 Solutions
1a
Conditions for equations to be valid:
 Constant acceleration
 Motion in a straight line
1b(i)
vR 
v
2
x
 vy
2

= (10.02 + 3.02)
= 10.44
 10 m s-1
10
 = tan-1 ( 3 )
= 73.3
 73
1b(ii)
1.
Initial velocity of sandbag is vR with angle  to horizontal.
Path is symmetrical and parabolic
vR

vR

1bii
2. Taking downwards as positive,
1
Sbag = -10.0 (4.0) + 2(9.81)(4.0)2
= 38.48 m
Sballoon = - 10.0 (4.0) = - 40.0m
Distance apart = 40.0 + 38.48 = 78.48  78 m
1
2ai
There is zero net force acting on the signboard.
There is zero net torque about any axis acting on the signboard.
ii
T1
40
50
T2
signboard
G
G is at the intersection of lines of action of T1 and T2
iii
Weight = T1 cos 40 + T2 cos 50
= 300 cos 40 + 252 cos 50
= 391.8
= 390 N
b
Net force acting on the board immediately after released = 20 N
Initial net acceleration = F/m
= 20/(391.8/9.81) = 0.501  0.50 m s-2 [allow ecf from (aiii)]
2
3ai
aii
Correct point P
A
Energy
Correct curve with:
 KE = 0 at R and KE + PE = ET
Roughly correct energy at RE
 KE + PE = ET
Roughly correct shape between RE
and R
Kinetic
energy
3bi
Correct point Q
0
RE
XQ
R
r
XP
total energy for
rock in (a)
B
When the rock is momentarily at rest, KE is zero and total energy = GPE
(ai) Hence, total energy of rock is the intersection of line AB and GPE graph
(aii) At distance R, KE is zero. At distance RE, KE = total energy – GPE
(bi) For a rock to escape to infinity with minimum speed, total energy at infinity is zero.
3bii
Possible explanation includes
 Moon would attract the rock
 potential at Moon(’s surface) is
o not zero
o less than 0
o lower than at infinity
 moon is nearer; less gain in GPE from Earth to moon
Thus, escape speed would be lower.
3
4a
Diffraction
Diagram (if drawn) must show light source (e.g. light, laser), object causing diffraction
(e.g. single slit) and means of observation (e.g. screen)
For observable diffraction, the size of the slit should be of the same order as that of
the wavelength.
OR Description of setup must mention the above items.
It was expected that the light on the screen would be the size of the slit but the width
of the band of light seen on the screen is greater than the slit width. This demonstrates
the spreading effect on a wave as it passes through an aperature.
OR
2-Source Interference
Diagram (if drawn) must show light source (e.g. light / laser), object causing
interference (i.e. double slit) and means of observation (e.g. screen)
Light source must be monochromatic and coherent
OR Description of setup must mention the above items.
Instead of two bright fringes of the screen, a fringe pattern containing light and dark
fringes is observed.
OR
Difraction grating
Diagram (if drawn) must show light source (e.g. light / laser), diffraction grating and
means of observation (e.g. screen)
Light source must be monochromatic and coherent
OR Description of setup must mention the above items.
A pattern of bright narrow lines/spots with dark regions between them are observed on
the screen. The spacing between the lines/spots are not equal.
Note: drawing of wavefronts / ripples is not accepted as there are multiple planes for
light waves.
4b
1. For a given metal, there is a threshold frequency below which no emission of
photoelectrons occurs regardless of the intensity of radiation.
2. Emission of photoelectrons occurs with no observable time lag at frequencies of
greater than threshold frequency, even at very low intensity.
3. Increasing the intensity of the radiation has no effect on the maximum energy of
the electrons
4. The rate at which electrons are emitted increases proportionally with the intensity
of the radiation.
5. The maximum kinetic energy (not kinetic energy) of the emitted photoelectron is
dependent on / increases linearly (not proportional) with the frequency of the
incident radiation (above threshold frequency), even with low intensity.
Any 3 from the above.
4
5
(a)
One ohm is the resistance of a conductor in which the current is 1 ampere when a
potential difference of 1 volt is applied across it.
OR
One ohm is the resistance of a conductor in which the ratio of potential difference
across it to the current flowing through it is 1 volt per ampere.
(b) Power dissipated in r = Total Power – Power dissipated in Load
I2 r = ε I – PL
(0.10)2 r = (2.5)(0.10) – 0.23
r = 2.0 Ω
(c)
Single cell
Resistance of Load, R1
Two cells
5.0 = 0.15R + 0.15 x 4.0
Resistance of Load, R2 = 29 Ω
As an ohmic device has constant resistance, since the load’s resistance has
changed/ increased, the load can be deduced as a non-ohmic device.
5
6a
One tesla is the magnetic flux density which causes a force per unit length of one
newton per metre on a straight wire carrying a current of one ampere and is at
right angles to the direction of the magnetic field.
6bi
As seen from the increased balance reading, there is a downward force on
magnet due to wire carrying current.
By Newton’s third law, there is an upward force on wire by magnet.
By Fleming’s left hand rule, pole P is a north pole
6bii
By Newton’s 2nd law, W – BIL =0  W= BIL
2.3 × 10–3 × 9.81 = B × 2.6 × 4.4 × 10–2
B = 0.20 T (g = 10, loses this mark)
6c
Reading for maximum current = 2.6 x 2 = 3.68 A
F = BIL =(0.20)(3.68)(4.4x10-2) =0.032 N
mg = 0.032 N, m = 0.032/9.81 = 3.3 g
total variation of mass = 2 x 3.3 = 6.6 g
OR
m

BL
I,mI
g
mmax 2.6 2
mmax I max



2.3
2.6
mdc
I dc
mmax = 2.3 2 = 3.3 g
total variation of mass = 3.3 x 2 = 6.6 g
6
7(a) The cut-off wavelength corresponds to the most energetic photon that can be
produced. That happens when all the kinetic energy of an accelerated electron is
lost in a single collision/interaction with the target atom in producing one photon.
(b)
(i)
area = 0.200 m x 0.300 m = 0.0600 m2
Accept a range of 0.0300 m2 ≤ area ≤ 0.120 m2
(ii)
10 m
-6
Area of a graph:
no of grains =
-6
10 m
0.0600
= 6.00×1010
6
6
10 10
no of photons = 6.00×1010 ×10 = 6.00×1011
energy = 6.00×1011 × 10-15 = 6.00 x 10-4 J
(c)
implied probability of arrival proportional to square of amplitude
ratio = 42 = 16
7
8ai
From Fig. 8.1, d = 57.5 m, d/v = 57.5/25.0 = 2.30 s
Accept 57.0 m < d < 58.0 m, 2.28 s < d/v < 2.32 s
aii
Plot the point corresponding to v = 25.0 m s
Draw the line of best fit for all the points (fair scatter of points about the best fit line)
-1
3.40
d/v / s
3.00
•
(36.0,3.00)
•
2.60
•
2.20
•
1.80
•
1.40
•
1.00
5.0
ˡ
10.0
(11.5,1.40)
ˡ
15.0
ˡ
20.0
ˡ
25.0
ˡ
30.0
ˡ
35.0
ˡ
40.0
v / m s-1
bi
Gradient = (3.00-1.40)/(36.0-11.5)
= 0.065312 = 0.0653
Accept range : 0.062 < gradient < 0.069
bii
v
From d = 2A + Bv,
biii
d
v
Sub. (11.5, 1.40) into the equation v = 2A + B
1.40 = 0.0653(11.5) + B ,
B = 0.6489=0.649 s
biv
2
d
v
gradient = 1/(2A)
v = 2A + B,
-2
A = 1/(2x0.0653) = 7.652 = 7.66 m s
range : 7.3 < A < 8.1
range : 0.61< B < 0.69
2
v
Since A has the unit of acceleration, 2A is most likely representing the braking distance or
the stopping distance where A is the deceleration of the motion. The braking distance
depends on the speed and the deceleration of the braking force applied.
B has the unit of time. Bv will represent the thinking distance where B is the thinking time or
reaction time of the rider. Before the braking force is applied, the thinking distance depends
on the speed and the thinking time.
8
c
2
v
Using d = 2A + Bv,
2
d = 50.0 /(2x7.656) + 0.6489x50 = 195.7= 196 m
Distance from obstacle = 200 – 195.7 = 4.28 m
d
Friction between the tyres and track surface will be reduced, reducing the magnitude of the
motorcycle’s deceleration. This increases the required braking distance.
Visibility of motorist’s is reduced, increasing the reaction time and hence the thinking distance.
Thus stopping distance has to be increased.
9
Suggested Solution
Aim
To investigate how the magnetic flux density B between the 2 circular coils varies with the current I
in the coils
Procedure
datalogger
Hall
probe
Power source
A
1) Set up the apparatus as shown above.
2) Switch on the power source to pass a current through both coils to produce a magnetic
field.
3) Record the value of I from the ammeter.
4) Place the hall probe between the coils. Record the value of the magnetic flux density B
from the datalogger.
5) Repeat steps 2 to 4 to obtain further values of B for different I by changing the resistance of
variable resistor.
n
6) From B = kI  lg B = n lg I + lg k . Plot a graph of lg B against lg I. The value of n can be
obtained from the gradient of the graph.
Control of variables
1) The separation of the 2 coils should be kept constant by using the meter rule to check and
adjust the separation before every reading.
2) The position of the hall probe between the 2 coils should be kept constant by clamping the
hall probe to a fixed position between the coils.
Safety and Accuracy
1) The plane of the hall probe should be perpendicular to the magnetic field within the 2 coils.
2) The axes of the 2 coils must coincide to ensure that the magnetic field in the region
between the 2 coils is uniform.
3) The coils are to be close to one another to ensure that the magnetic field between the coils
is uniform.
4) The position of the hall probe is along the centre axis of the 2 coils and it is equidistant from
the coils.
5) The range of the current used should be kept small to reduce heating effect in coils by
using the variable resistor.
10
Name: _________________________________ (
)
PDG: ______/13
ANDERSON JUNIOR COLLEGE
2014 JC2 Prelim Examination
PHYSICS Higher 2
9646/03
Paper 3 Longer Structured Questions
Monday 15 September 2014
2 hours
Candidates answer on the Question Paper.
No Additional Materials are required.
READ THESE INSTRUCTIONS FIRST
Write your name and PDG in the spaces at the top of this page.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough working.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Section A
Answer all questions.
For Examiner’s Use
Section B
Answer any two questions.
Section A
You are advised to spend about one hour on each section.
1
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each
question or part question.
2
3
4
5
6
Section B
7
8
9
Deduction
Total
This document consists of 22 printed pages and 0 blank page.
9646/03/AJC2014Prelim
2
Data
speed of light in free space,
c = 3.00 x 108 m s−1
permeability of free space,
 0 = 4 x 10−7 H m−1
permittivity of free space,
 0 = 8.85 x 10−12 F m−1
(1/(36)) x 10−9 F m−1
elementary charge,
e = 1.60 x 10−19 C
the Planck constant,
h = 6.63 x 10−34 J s
unified atomic mass constant,
u = 1.66 x 10−27 kg
rest mass of electron,
me = 9.11 x 10−31 kg
rest mass of proton,
mp = 1.67 x 10−27 kg
molar gas constant,
R = 8.31 J K−1 mol−1
the Avogadro constant,
NA = 6.02 x 1023 mol−1
the Boltzmann constant,
k = 1.38 x 10−23 J K−1
gravitational constant,
G = 6.67 x 10−11 N m2 kg−2
acceleration of free fall,
g = 9.81 m s−2
9646/03/AJC2014Prelim
3
Formulae
uniformly accelerated motion,
1
s = ut + 2 at2
v2 = u2 + 2as
work done on/by a gas,
W = pV
hydrostatic pressure,
p =  gh
gravitational potential,
m
ϕ = −G r
displacement of particle in s.h.m.,
x = x0 sin t
velocity of particle in s.h.m.,
v = v0 cos t
v = ± ω x0  x2
2
mean kinetic energy of a
3
E = 2 kT
molecule of an ideal gas,
resistors in series,
R = R1 + R2 + …
resistors in parallel,
1/R = 1/R1 + 1/R2 + …
electric potential,
V=
alternating current/voltage,
x = x0 sin t
transmission coefficient,
Q
4  ε0 r
T  exp(−2kd)
where k =
8 2 mU  E 
h2
radioactive decay,
x = x0exp(− t)
decay constant,
=
0.693
t1
2
9646/03/AJC2014Prelim
4
Section A
Answer all the questions in this Section
1
Fig. 1.1 shows the cross-section of a symmetrical object that is partially submerged in water
and displaced from its equilibrium position. The thickness of the object is uniform throughout.
½h
½h
h
water
h
Fig. 1.1
The density of the object is uniform and is less than the density of water.
(a) On Fig. 1.1, draw arrows to show the forces acting on the object.
[1]
(b) Describe the subsequent motion of the object.
…...………………………………………………………………………….……………….......[1]
(c) On Fig. 1.2, draw the final equilibrium position of the object and indicate the forces
acting on the object with arrows.
Fig. 1.2
[2]
(d) If the amount of water displaced by the object at equilibrium is the same as that shown in
Fig. 1.1, determine the density of the object in terms of the density of water, w.
density of object = ……………………………. [2]
9646/03/AJC2014Prelim
5
2
(a)
(i)
State the first law of thermodynamics.
…………………………………………………………………………………....................
……………………………………………………………………………….………….... [1]
(ii)
Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy
of the molecules of the gas.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
(b) A cylinder contains 1.0 mol of an ideal gas.
(i)
The volume of the cylinder is constant. Calculate the thermal energy required to
raise the temperature of the gas by 1.0 K.
energy = …………………………….J [2]
(ii)
The volume of the cylinder is now allowed to increase so that the gas remains at
constant pressure when it is heated. Explain whether the thermal energy required
to raise the temperature of the gas by 1.0 K is now different from your answer in
(b)(i).
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[2]
(c) In order for an atom to escape completely from the Earth’s gravitational field, it must
have a speed of approximately 1.1 x 104 m s-1 at the top of the Earth’s atmosphere.
(i)
Estimate the temperature at the top of the atmosphere such that helium, assumed
to be an ideal gas, could escape from the Earth. The mass of a helium atom is
6.6 x 10-27 kg.
temperature = …………………………….K [2]
9646/03/AJC2014Prelim
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(ii)
Suggest why some helium atoms will escape at temperatures below that
calculated in (c)(i).
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
3
(a) Explain what is meant by the principle of superposition.
………………………………………………………………………………………………………
………………………………………………………………………………………………………
……………………………………………………………………………………………………[1]
(b) Light from a mercury vapour lamp is incident normally on a diffraction grating and the
diffraction pattern is observed on a screen placed 30 cm away from the diffraction
grating. The positions of the first order orange spectrum of wavelength 580 nm are
20 cm apart, as shown in Fig. 3.1 below.
screen
first order orange
light from
mercury
vapour lamp
diffraction
grating
20 cm
first order orange
30 cm
Fig. 3.1
(i)
Show that the diffraction grating has 545 number of lines per mm.
[2]
9646/03/AJC2014Prelim
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(ii)
Fig. 3.2 shows some of the energy levels in a mercury atom. Another spectrum that
was also emitted from the mercury lamp is shown by the transition A.
energy / eV
- 1.60
- 2.71
A
- 5.77
- 10.44
Fig. 3.2
Calculate the wavelength of the spectrum associated with the transition A.
wavelength = ……………………………… m [2]
(iii)
Hence, determine the highest order of diffracted beam that can be observed on the
screen in Fig. 3.1 for the spectrum calculated in (b)(ii).
highest order = ……………………………… [2]
9646/03/AJC2014Prelim
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4
A short magnet is released and falls straight through a long solenoid as shown in Fig. 4.1.
A voltage sensor and datalogger are used to measure the e.m.f induced in the solenoid.
short magnet
long
solenoid
voltage sensor
datalogger
Fig. 4.1
Fig. 4.2 shows how the induced e.m.f  in the solenoid changes with time, t.

t
Fig. 4.2
With reference to Fig. 4.2, explain
a) why the second peak values of the induced e.m.f. is greater than the first peak value.
………………………………………………………………………………….…….....................
………………………………………………………………………………….…….....................
………………………………………………………………………………………......................
……………………………………………………………………………………......................[2]
b)
the difference in the signs of the induced e.m.f.
…………………………………………………………………………………………….........….
……………………………………………………………………………………….........……….
………………………………………………………………………………………….........…….
……………………………………………………………………………………….........…….[2]
9646/03/AJC2014Prelim
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5
Free neutrons (neutrons not contained within a nucleus) are unstable and decay by
-emission with a half-life of 770 s.
(a) Explain what is meant by
(i) the neutrons are unstable.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
(ii)
half-life.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
(b) Write down a possible nuclear equation for the decay of a free neutron.
………………………………………………………………………..…………….…………....[1]
(c) Using your relationship in (b) and the following data, calculate the energy released in the
decay of a free neutron.
rest mass of neutron = 1.008665 u
rest mass of proton = 1.007276 u
rest mass of electron = 0.000549 u
[2]
(d) Sketch, on the axes of Fig. 5.1, the following graphs to show the variation with time of
(i)
(ii)
the number of undecayed neutrons and label as N,
the number of -particles and label as B.
number of
particles
time / s
Fig. 5.1
(e) Label on the axis of Fig. 5.1, the half-life of neutron.
9646/03/AJC2014Prelim
[2]
[1]
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6
The energy bands in an intrinsic semiconductor are illustrated in Fig. 6.1
conduction band
valence band
Fig. 6.1
Using band theory,
(a) describe how the electrical resistance of an intrinsic semiconductor material decreases
with increase of temperature.
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….………..
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….………..
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….......... [3]
(b) suggest and explain a difference in the resistance of an intrinsic semiconductor
compared to a p-type semiconductor.
…………………………………………………………………………………………….………..
…………………………………………………………………………………………….………..
…………………………………………………………………………………………….………..
…………………………………………………………………………………………….………..
………………………………………………………………………………………….…………..
…………………………………………………………………………………………….......... [3]
9646/03/AJC2014Prelim
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Section B
Answer two questions from this section
7
(a)
(i)
State Newton’s first law of motion.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
(ii)
State Newton’s second law of motion.
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[1]
(iii)
With a suitable definition of the unit of force, Newton’s second law can be written in
the following relationship
force = mass x acceleration
for a body of constant mass.
Hence, together with Newton’s third law, derive the principle of conservation of
momentum.
…………………………………………………………………………………....................
…………………………………………………………………………………....................
…………………………………………………………………………………....................
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[2]
(b) An alpha particle collides head-on with a stationary nitrogen-14 atom. The nitrogen atom
moves off in the same direction as the approaching alpha particle with a speed of
0.005 c where c is the speed of light.
(i)
Calculate the change in momentum of the alpha particle.
change in momentum = …………………………….N s [2]
9646/03/AJC2014Prelim
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(ii)
Show quantitatively whether the interaction described in (b) is elastic in nature if
the initial speed of the alpha particle is 0.02 c.
[3]
(c) Alpha particles of speed 4.41 x 106 m s-1 enter a region of uniform electric field at an
angle 25  with the horizontal. The region of field is set up by two plates P and Q in a
horizontal plane as shown in Fig. 7.1.
plate P
region of uniform
electric field
alpha
particles
25 
plate Q
Fig. 7.1
(i)
State and explain how the horizontal component of the velocity of the alpha
particles would vary along the length of the plates.
…………………………………………………………………………………....................
…………………………………………………………………………………....................
……………………………………………………………………………….…………....[2]
9646/03/AJC2014Prelim
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(ii)
Fig. 7.2 shows how the vertical displacement of an alpha particle varies with time
as it passed through the plates.
vertical displacement from plate Q / m
0.75
0.50
0.25
0
0
1.
Fig. 7.2
time
With reference to the features of the graph in Fig. 7.2, state and explain the
direction of the resultant force acting on the alpha particle. Ignoring the effect
of gravity.
.…………………………………………………………………………………............
………………………………………………………………………………….............
………………………………………………………………………………….............
.……………………………………………………………………………….……… [2]
2.
Sketch on Fig. 7.3 how the vertical displacement of the alpha particles would
vary with position along the length of plate Q as they travel between the
plates. You may assume that the vertical displacement above plate Q to be
positive.
vertical displacement from plate Q / m
0.75
0.50
0.25
0
left edge
of plate Q
Fig. 7.3
9646/03/AJC2014Prelim
right edge position
of plate Q
[2]
[Turn Over
14
(iii)
Using Fig. 7.2, determine the resultant force experienced by an alpha particle.
resultant force = …………………………….N [3]
(iv)
The stream of alpha particles entering the region between PQ is replaced with a
stream of protons with the same entry speed and angle of projection.
On Fig. 7.2, sketch the variation of vertical displacement with time for a proton.
[2]
9646/03/AJC2014Prelim
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8
A toy car’s wheel is set up as a pendulum by hanging it vertically from a fixed support. The
wheel oscillates about the fixed support as illustrated in Fig. 8.1.
Fig. 8.1
The variation with displacement x of the acceleration a of the wheel is shown in Fig. 8.2.
a
0
x
Fig. 8.2
(a)
(i)
Use Fig. 8.2 to explain why the motion of the wheel is simple harmonic.
…………………………………………………………………………………………….…
…………………………………………………………………………………………….…
…………………………………………………………………………………………….…
…………………………………………………………………………………………….…
……………………………………………………………………………………….………
……………………………………………………………………………........................[3]
(ii)
State a condition that must be satisfied for the oscillation of the wheel to be simple
harmonic.
.……………………………………………………………………………………….………
………………………………….………………………………………….......................[1]
(b) The wheel in Fig. 8.1 is displaced and released at time t = 0. The oscillations of the
wheel have amplitude 14.7 cm and angular frequency 2.40 rad s-1.
(i)
Define angular frequency.
.……………………………………………………………………………………………….
……………………………………………………………………………………………..[1]
9646/03/AJC2014Prelim
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(ii)
State an expression for the displacement x of the wheel in terms of time t.
……………………………………………………………………………………………..[1]
(iii)
Use your expression in (b)(ii) to sketch the variation of the kinetic energy of the
[2]
wheel EK with time t for one complete oscillation in Fig 8.3.
Ek/J
0
t/s
Fig. 8.3
(c) The angular frequency value stated in (b) is calculated from the period of the simple
harmonic motion. An accurate value for the period is found by timing a large number of
oscillations.
(i)
Explain why a large number of oscillations would help to achieve a more accurate
value for the period.
………………………………………………………………………………………………..
….………………………………………………………………………………………….[1]
(ii)
Calculate the number of oscillations that are measured in a total time of 83.8 s.
number of oscillations = ………………………… [2]
(iii)
A stopwatch with display provides timing to 1/100th of a second over a range of
9 hours, 59 minutes and 59.99 seconds was used to record the duration of the
oscillations in (c)(ii).
Explain why the duration of the oscillations in (c)(ii) is recorded to 0.1 seconds.
………………………………………………………………………………………………..
……………………………………………………………………………………………..[1]
9646/03/AJC2014Prelim
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17
(d) In order to time the oscillations of the wheel in Fig. 8.1, a stopwatch is started when the
centre of the wheel passes a marker. This marker is 2.7 cm from the equilibrium position
as shown in Fig.8.4.
14.7 cm
equilibrium
position
marker
2.7 cm
Fig. 8.4
(i)
The wheel has a mass of 0.165 kg. Calculate the restoring force acting on the
wheel as it passes the marker.
restoring force = ………………………… N [2]
(ii)
Calculate the speed of the wheel at the instant it passes the marker.
speed …………………………m s-1 [2]
9646/03/AJC2014Prelim
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18
(e) In an experiment to demonstrate resonance, the wheel in Fig. 8.1 is made to oscillate by
an external periodic driving force of frequency f. Fig. 8.5 shows the variation with
frequency f of the amplitude of the forced oscillations of the wheel.
amplitude
f
Fig. 8.5
(i)
Describe how the amplitude of oscillation depends on the forcing frequency.
……………………………………………………………………………………………..…
…………………………………………………………………………………………….….
……………………………………………………………………………………………..…
……………………………………………………………………………………………..[2]
(ii)
Sketch in Fig. 8.5 the shape of the graph if the forced oscillations of the wheel were
repeated in a vacuum. Label your sketch as A.
[2]
9646/03/AJC2014Prelim
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19
9
(a) Two small charged metal spheres A and B are situated in a vacuum. The distance
between the centres of the spheres is 12.0 cm, as shown in Fig. 9.1.
12.0 cm
P
sphere A
sphere B
Fig 9.1 (not to scale)
The charge on each sphere may be assumed to be a point charge at the centre of the
sphere.
Point P is a movable point that lies on the line joining the centres of the spheres and is
distance x from the centre of sphere A.
The variation with distance x of the electric field strength E at point P is shown in
Fig. 9.2.
Fig 9.2
9646/03/AJC2014Prelim
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20
(i)
Define electric field strength.
………………………………………………………………………………….....................
……………………………………………………………………………….………….....[1]
(ii) State the evidence provided by Fig. 9.2 for the statements that
1. the spheres are conductors,
…………………………………………………………………………………..............
……………………………………………………………………………….……….[1]
2.
the charges on the spheres are both positive.
…………………………………………………………………………………..............
…………………………………………………………………………………..............
……………………………………………………………………………….……….[2]
(iii)
An -particle is placed at x = 4.0 cm on the line joining the centres of the spheres.
Describe what will happen to the -particle?
…….………………………………………………………………………………................
……….……………………………………………………………………………................
…………………………………………………………………………………….……….[2]
(b) Fig. 9.3 shows two charged parallel plates which are 12.0 cm apart. A flame probe
connected to an electroscope or electrostatic voltmeter can be used to measure the
electric potential in the space between the plates. The probe is a tiny gas flame which
makes the air near the probe conduct. By moving the probe about, between the plates,
the equipotential surfaces can be mapped out. The dashed lines A, B, C, D and E
represent some equipotential surfaces. (Not to scale)
- 40 V
A
B
X
- 30 V
C
D
- 35 V
- 25 V
- 20 V
Y
E
- 10 V
- 15 V
Fig. 9.3
9646/03/AJC2014Prelim
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21
(i)
Define electric potential.
………………………………………………………………………………….....................
………………….…………………………………………………………….………….....[1]
(ii) Suggest how the flame probe enables air between the plates to conduct electricity.
………………………………………………………………………………….....................
………………………………………………………………………………….....................
………………………………………………………………….…………….………….....[2]
(iii) On Fig. 9.3, draw lines to show the electric field between the plates.
(iv)
[2]
On Fig. 9.4, sketch the variation of the electric potential, V, at different points along
the line XY where X and Y are midpoints of the parallel plates.
Label the positions of equipotential lines A, B, C, D and E clearly on your graph.
V/ V
0
d / cm
Fig. 9.4
[2]
9646/03/AJC2014Prelim
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22
(v) 1.
A proton, which is initially at rest at P as shown in Fig 9.5, is moved within the
electric field. Complete the table below by stating the work done in moving the
proton along each of the following path.
A
B
C
D
- 40 V
X
Q
8.0 cm
R
- 35 V
- 30 V
14.0 cm
30o
- 25 V
P
- 20 V
E
Y
- 10 V
- 15 V
Fig 9.5
Path taken
Work done / eV
P→Q
Q→R
P→Q → R→P
[3]
2.
The proton, back at position P, is released from rest.
Describe the motion of the proton.
…………………………………………………………………………………..............
…………………………………………………………………………………..............
.……………………………………………………………………………….…….…[2]
3.
Give a quantitative explanation as to why gravitational force due to Earth is not
considered when predicting the motion of the proton between the plates.
…………………………………………………………………………………..............
…………………………………………………………………………………..............
…………………………………………………………………………………..............
.……………………………………………………………………………….……….[2]
9646/03/AJC2014Prelim
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2014 AJC H2 Phy Prelim Solutions
Paper 3 (80 marks)
1a
upthrust
weight
b
Object rotates anticlockwise as weight of object and upthrust form a couple which
causes an anticlockwise moment.
OR
Object will rock back and forth.
c
-
Weight and upthrust will lie along the same vertical axis along the geometric centre
of the object when in equilibrium.
Figure is drawn symmetrical as seen below and the 3 points A, B and C lie at the
water surface.
upthrust
C
A
B
weight
d
By principle of floatation and Archimedes’ principle,
Upthrust = weight of object = weight of fluid displaced
obj gVobj = w gVw
where Vw is the volume of object submerged in water
obj = w Vw / Vobj
= w(0.5h2t)/ (0.75h2t)
where t is the thickness of object
= 0.67w
2ai
The increase in internal energy of a system is equal to the sum of heat supplied to the
system and the work done on the system.
aii
Internal energy is the sum of total kinetic energy and potential energy of the molecules
of gas. For ideal gas, no intermolecular forces so no potential energy (only kinetic).
Therefore, internal energy of ideal gas = total kinetic energy of molecules.
bi
∆U = Q + W, since W = 0, Q = ∆U = ∆Ek = (3/2)Nk ∆T
( or (3/2)nR∆T)
Thermal energy required Q = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023
= 12.5 J
bii
Additional energy is needed to push back piston /do work against atmosphere besides
the increase in internal energy of gas. So thermal energy required is greater.
cii
Ek = (3/2) kT
½ x 6.6 x 10-27x(1.1x104)2 = (3/2)x1.38x10-23xT
T = 1.93 x 104 K
At a given temperature, helium atoms have a range of kinetic energy around the mean
value. Some helium atoms have kinetic energy higher than the mean kinetic energy of
a temperature below that in (c)(i). These atoms have a velocity more than 1.1 x104 m
s-1 and hence they have sufficient energy to escape.
3a
The Principle of Superposition states that when two waves meet at a point, the
resultant displacement is equal to the vector sum of the individual displacements.
ci
Note:
- paraphrasing the question statement is not accepted (i.e. “when two waves
superpose” is not acceptable)
- use of “amplitude” instead of “displacement” is not accepted.
3bi
Angle of diffraction of the first order = tan-1 (10/30) = 18.43°
Using d sinθ = nλ, we have
d = (1)(580 x 10-9) / sin(18.43°)
= 1.834 x 10-6 m
Hence, number of lines per mm = 1 mm / 1.834 x 10-6
= 545 lines [shown]
3bii
Using E = hf, we have
[-2.71 – (-5.77)] x 1.6 x 10-19 = 6.63 x 10-34 x 3.0 x 108 / λ
λ = 4.0625 x 10-7
= 4.06 x 10-7 m
3biii
To determine the highest order, sinθ ≤ 1.
Hence, using d sinθ = nλ,
nλ / d ≤ 1
n ≤ 1.834 x 10-6 / 4.0625 x 10-7 = 4.51
Therefore, highest order = 4
e.c.f. for wrong values of λ in b(i) and b(ii) that are of the same order as light
(i.e.10-7 m).
4a
Due to the acceleration of free fall, the speed of magnet leaving the solenoid is higher
than when it is entering the solenoid.
Hence, the rate of change of magnetic flux linkage with solenoid is greater when the
magnet is leaving the coil, leading to a larger second peak value of the induced e.m.f.
according to Faraday’s Law.
b
When the magnet is entering the solenoid, the magnetic flux linkage with solenoid is
increasing but as it is leaving the magnetic flux linkage is decreasing.
By Lenz’s law, the induced e.m.f acts in a direction such that it tends to produce
effects to oppose the change causing it.
Hence the induced e.m.f have opposite signs when entering and leaving the solenoid.
9646/03/AJC2014Prelim
It is the average time taken for the activity of a particular radioactive nuclide to fall to
half its initial value.
b
1
0
c
Mass defect = (mass of initial – mass of product)
= (1.008665 u - 1.007276 u - 0.000549 u) = 8.4 x 10-4 u
 01 11
H
ii
e
It means neutrons will spontaneously and randomly disintegrate/decay into smaller
particles.
n
5ai
Energy released
= (mass of initial – mass of product)c2
= (1.008665 u - 1.007276 u - 0.000549 u) c2
= 1.25 x 10-13 J
di, ii
e
number of
particles
N0
B
½N0
N
770 (half-life)
time /s
6a
An intrinsic semiconductor has a small energy gap
(Describe the
(about 1eV) which decreases as temperature
band structure) Band
increases.
theory
As temperature increases, more electrons from the
(Describe how
valence band are excited to become free electrons in
free electrons &
the conduction band while leaving more holes in the
holes are formed)
valence band.
Although there is also an increase lattice vibrations of the ions but this effect is
outweighed by the large increase in the number of charge carriers (free electrons and
holes).
Hence the electrical resistance of the intrinsic semiconductor will decrease.
b
A p-type semiconductor has an acceptor energy level
very close to the valence band.
Hence it is easier for the electrons from the valence
band to be thermally excited to the acceptor level,
leaving behind more positive holes in the valence band.
(Describe the
band structure)
(Describe how
more holes are
formed)
Due to the increase in the positive charge carriers (holes) for the p-type
semiconductor, p-type semiconductor has a lower resistance than intrinsic
semiconductor.
9646/03/AJC2014Prelim
Band
theory
7ai
ii
iii
Newton’s first law states that a body stays at rest or continues to move at constant
velocity unless a resultant force acts on it.
Newton’s second law states that the rate of change of momentum of a body is
proportional to the resultant force acting on the body and takes place in the direction of
the resultant force.
From Newton’s third law, if body A exerts a force on body B, then body B will exert a
force of the same type that is equal in magnitude and opposite in direction on body A.
Hence, (ma)A = -(ma)B
where m represents mass and a represents acceleration
By Newton’s second law, mA(vA-uA)/t = - mB(vB-uB)/t
For the same duration of interaction, the gain in momentum by body A is equal to the
lost in momentum by body B.
This shows principle of conservation of momentum.
bi
By COM,
lost in momentum of alpha
= gain in momentum nitrogen
= 14 u (0.005 c)
= 3.486 x 10-20 = 3.49 x 10-20 N s
bii
By COM,
4u(0.02c) + 0 = 4uva + 14u(0.005c)
final speed of alpha, va = 0.0025 c (after collision, alpha continues in its original dirn)
Relative speed of approach = ua – un = 0.02 c
Relative speed of separation = vn – va = 0.005 c – (0.0025 c) = 0.0025 c
Since relative speed of approach is not equal to relative speed of separation, the
interaction is not elastic.
ci
As there is no net horizontal force acting on the alpha particles,
Hence, horizontal component of velocity will remain unchanged along the length of the
plates.
cii1.
From the gradient of the graph, velocity decreases till zero and then increases in the
opposite direction.
OR the vertical distance covered per second decreases and then increases.
Hence, there is a resultant force acting vertically downwards towards plate Q.
9646/03/AJC2014Prelim
cii2.
vertical displacement from plate Q / m
0.75
0.50
0.25
0
position
0
ciii
v y2  u y2  2as y
0  ( 4.41 x 10 6 sin 25) 2  2a(0.70)
a = 2.481 x 1012 m s-2
F = ma
= 4(1.66 x 10-27) (2.481 x 1012)
= 1.647 x 10-14 = 1.6 x 10-14 N
civ
vertical displacement from Q/ m
0.75
0.50
0.25
time
Since charge of proton is half that of alpha, net force on proton is half of alpha.
Since mass of proton is a quarter of alpha, net acceleration of proton is 2 times that of
alpha. Proton would have half the maximum vertical displacement of alpha.
With the same initial speed and doubled acceleration, proton takes half the time of
alpha, to reach maximum height. Hence range of proton is half of the range of alpha.
Path of proton should be symmetrical as same net force acts on it throughout the
region of field.
9646/03/AJC2014Prelim
8ai
SHM is the motion of a body such that its acceleration is proportional to its
displacement from a fixed point and is always directed towards that point.
The oscillations are simple harmonic because:
1. The straight line passes through the origin indicating that the acceleration a of the
plate is directly proportional to the displacement x from its equilibrium position
2. The negative gradient of the graph indicates that its acceleration a is opposite in
direction to its displacement x, and is directed towards the equilibrium position.
Note: acceleration and displacement not inversely proportional
ii
The angle of oscillation must be small.
bi
Angular frequency  of a body undergoing simple harmonic motion is a constant of a
given oscillator and is related to its natural frequency f by   2f
t
 0.147 cos( 2.40 )
 14.7 cos( 2.40 )
x
 14.7 cos(2.40 ) OR
OR
t
x
t
x  0.147 cos( 2.40t ) OR
x
ii
iii
T
Correct shape of cos2 function
2 cycles within one period with Ek max at T/4, 3T/4 and Ek = 0 at 0, T/2, T.
ci
To reduce fractional error of time measurement.
ii
T
2


2
 2.6180 s
2.40
number of oscillations =
iii
83.8
 32 (round to lower integer)
2.6180
As the average human reaction time is between 0.2 s and 0.4 s, the timing for the
oscillations is recorded to the same precision as the human reaction time.
9646/03/AJC2014Prelim
2

x
m

a
m
F
di
 0.165 2.40  0.0270 
2
 0.0257 N (3sf)
ii
1st method
v   x 0  x 2  2.40 0.147  0.0270
= 0.347 m s-1 (3sf)
2nd method
x  x o cos t
2
2
2
2.7  14.7 cos(2.40t )
t  0.57753 s
v  x 0 sin t  2.40(14.7  10 2 ) sin(2.40t )  0.347 ms -1
ei
The amplitude of oscillation increases when f increases to natural frequency f0. When f
is larger than f0, the amplitude decreases.
Amplitude is maximum when the frequency is equalled to f0.
(Learning point: f0 = 1/T or /2 = 2.40/2 = 0.382 Hz)
ii
amplitude
A
f
f0 = 0.382 Hz
- greater values of amplitude for all values of f
- the peak is shaper and at the right of f0
9ai
Electric field strength at a point is defined as the force per unit positive charge acting
on a stationary charge placed at that point.
ii1.
Electric field strength inside spheres is zero
2.
There is a point between the spheres,
where electric field strength is zero.
iii
The particle will move towards sphere B and
oscillate about x = 8.0 cm
bi
The electric potential at a point is defined as the work done per unit positive charge in
moving a point charge from infinity to the point.
9646/03/AJC2014Prelim
ii
Heat from the flame creates ions in the air.
These ions then act as charge carriers.
Iii
-within plates vertical parallel lines with equal spacing pointing towards upper plate
-near edge of plates, field perpendicular to potential lines.
iv
Shape of graph is correct
Points A, B, C, D & E are labeled at the correct positions
v1.
Path taken
P→Q
Q→R
P→Q→R→P
Work done / eV
-5
15
0
2.
Proton moves upward in a vertical straight path towards -40 V plate with increasing
speed at constant acceleration.
3.
Correct values of FG = 1.6 x 10-26 N and FE = 4.0 x 10-17 N
Since FE is about 109 times larger than FG, hence gravitation force is not taken into
account.
9646/03/AJC2014Prelim