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HOH FUK TONG COLLEGE
Mock Examination, 2004 - 2005
Physics
Paper I
Secondary: Seven
Date: 28/02/2005
Time allowed: 3 hours (8:30 - 11:30 a.m.)
Marks: 122
Name : ______________________________
Number : ______
1.
Answer ALL questions.
2.
Write your answers in the spaces provided in the question paper.
you should show all the main steps in your working.
3.
Assume:
1.
(a)
In calculations
velocity of light in air = 3 108 m s-1
acceleration due to gravity = 10 m s-2
Figure 1(a) shows a skier descending the ramp of a ski jump. Figure 1(b)
shows a graph of the distance travelled along the ramp against time, from
the time the decent starts until the skier leaves the end of the ramp.
Figure 1 (a)
110
100
90
80
70
60
50
40
30
20
10
0
distance / m
Figure 1 (b)
0
2
4
6
8
time /s
10
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The skier of mass 80 kg (including equipment) skies down the ramp and
leaves it horizontally. The skier gains 55% of the available gravitational
potential energy as kinetic energy when descending the ramp.
(i)
Use Figure 1(b) to find the speed at which the skier leaves the
ramp. Show your working.
(2)
Draw tangent to graph at the end of run.
slope = speed = 23 m s-1
(ii)
Assume uniform acceleration when descending the ramp, estimate
its value
(2)
By v = u + at
23 = 0 + a(8.8)
a = 2.61 m s-2
(iii)
OR
By s = ut + ½ at2
100 = 0 + ½ a(8.8)2
a = 2.58 m s-2
Determine the height of the ramp.
(2)
P.E. lost  55% = K.E. gained
m  10  h  0.55 = ½  m  (23)2
h = 48 m
(b)
Figure 1(a) shows the path taken by the skier after leaving the ramp.
Assuming that there was no lift or drag due to the air during this jump,
calculate
(i)
the time for which the skier was in flight;
(2)
By s = ut + ½at2
80 = 0(t) + ½(10) t2
t=4s
(ii)
the horizontal distance jumped by the skier before landing. (2)
By s = ut
= 23 (4) = 92 m
(c)
The landing surface is suggested to be inclined at an angle to the
horizontal. Explain briefly.
(2)
Skier takes longer time to reduce (vertical) momentum
compared with time when surface is flat.
mv
By F = t , force experienced by the skier is reduced.
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2.
Figure 2(a) shows the displacement-time graphs for two mass-spring systems X
and Y that are performing simple harmonic motion.
0.3
displacement / m
0.2
0.1
X
Y
0
1.0
0.5
1.5
time / s
-0.1
-0.2
-0.3
(a)
Figure 2(a)
Deduce an equation for the motion of system X.
(2)
Let x = A sint
2
2
 = T = 1.2 = 5.24 rad s-1
Hence x = 0.2 sin(5.24 t) m
(b)
The springs used in X and Y have the same spring constant.
ratio of the
(i)
the mass used in Y to that in X
m
Since T 
(given: same spring constant)
k
Find the
(2)
From graph, they have the same period.
Hence, the have the same mass, i.e. mY : mX = 1 : 1
(ii)
the maximum energy stored in Y to that in X.
(2)
1
1
EY : EX = 2 kY (AY)2 : 2 kX (AX)2 (same k)
= (AY)2 : (AX)2 = (0.1)2 : (0.2)2 = 1 : 4
(c)
Figure 2(b) shows how the potential energy of X varies with displacment.
(i)
Determine the spring constant of the spring used and the mass of X.
(3)
1
By PE = 2 kX (AX)2
1
0.16 = 2 kX (0.2)2
kX = 8 N m-1
By T = 2
m
m
 1.2 = 2
k
8
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m = 0.29 kg
(ii)
(iii)
Draw on it how the kinetic energy of X varies with its diplacement.
Label this graph A.
(1)
Draw on it how the kinetic energy of Y varies with its diplacement.
Label this graph B.
(2)
0.2
0.15
0.1
0.05
0
-0.2
3.
(a)
-0.1
0
0.1
0.2
A small loudspeaker placed in open space emits sound energy uniformly
in all directions with a total power of 5 10-5 W.
(i)
What is the sound intensity at a distance of 2 m from the
loudspeaker?
(2)
P
510-5
I=
=
= 9.95  10-7 W m-2
4  r2
4  (2)2
(ii)
If the loudspeaker were immersed in water, what change would be
produced on the sound intensity at the same distance of 2 m from it?
Explain briefly.
(2)
The sound intensity would be smaller (1)
due to absorption of wave energy by water. (1)
(b)
In the figure below, the curves AB and EF represent the threshold of
hearing and threshold of pain for the average human ear respectively.
CD represents curve of equal loudness.
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threshold of pain
equal loudness
threshold of hearing
(i)
Deduce the lowest intensity and lowest intensity level of sound that
can be heard at a frequency of 0.25 kHz.
(2)
From graph AB,
log10(I) = -9
I = 10-9 W m-2 (1)
10-9
h = 10 log ( 10-12 ) = 30 dB (1)
(ii)
What is the difference in intensity level between two sounds of
frequencies 1 kHz and 0.25 kHz which would give the same
loudness level corresponding to curve CD?
(3)
At 1 kHz, log10(I1) = -10
I1 = 10-10
At 0.25 kHz, log10(I2) = -7
I2 = 10-7
10-7
10-10
h2 – h1
= 10 log10( 10-12 ) - 10 log10( 10-12 )
= 10 log10(105) - 10 log10(102)
= 50 – 20 = 30 dB
(iii)
A person stands in front of a small sound source of frequency 0.25
kHz so that the loudness level is the same as that corresponding to
curve CD, i.e. log10(I) = -7. He cannot hear the sound if he walks
a further distance of 13.5 m away from the source. Estimate the
initial distance of the person from the source.
(3)
D
d
source
13.5 m
I1(10-7)
1
10-7
D2
Since I  r2  10-9 = d2
I2(10-9)
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D = 10d
d + 13.5 = 10d
9d = 13.5
d = 1.5 m
4.
A source emits red and blue light, parallel beam of lights is produced when the
lights pass through a collimator. It is then incident normally on a diffraction
grating placed on a spectrometer turntable.
(a)
In the space below, draw a ray diagram showing the path of the rays
passing through the spectrometer, when the spectrum of the source is
observed. The diagram should include the collimator and the telescope of
the spectrometer.
(2)
diffraction grating
telescope
source
(b)
cross wire
collimator
Explain why the spectrometer telescope must be adjusted to receive
parallel lights in order to view a pure spectrum of the source.
(1)
It is adjusted so that all images will be brought into focus on the
cross-wire.
(c)
The source, collimator and telescope are set initially in a straight line and
the telescope is turned in one direction only. For the first order, what
colour of the spectral lines will be seen first? Explain briefly.
(2)
Apply a sin = m
For the same order, as the wavelength of blue light is smaller, it
will be seen with a smaller diffraction angle, i.e. blue light will
be seen first.
(d)
At a diffraction angle of 54.8o, a blue line and a red line coincide. If the
grating has 600 rulings per mm and the wavelength of the blue light is 454
nm,
(i)
Find the order of blue light at that diffraction angle.
(2)
By a sin = m
1
sin54.8o = m (45410-9)
600103
m=3
(ii)
Hence, estimate the order number and wavelength of red light.
State the reason of your estimation.
(3)
At the same diffraction angle, the order number of red
light must be smaller than that of blue as its wavelength
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is longer, i.e. order of red light must be 1 or 2.
When two lights coincide,
mb b = mr r
3  454 nm = mr r
If mr = 1,
If mr = 2,
(e)
r = 1364 nm which is not reasonable.
r = 681 nm which is reasonable.
The same diffraction grating is used to view a monochromatic light source
placed 50 cm away from the grating. The observer holds the grating close
to his eye, and observes an image on either side of the source. Calculate
the wavelength of the light if the images are 37.8 cm apart.
(2)
18.9 cm

50 cm
18.9
tan = 50   = 20.7o
a sin = 
1
sin20.7o =    = 589 nm
600103
5.
Two conducting spheres A and B as shown below, are of equal mass and equal
charge Q. They are suspended by two fine threads of equal length l.
1
Take
= 9  109 N m2 C-2 .
4 π εo


l
l
A
B
x
In equilibrium, the threads connecting the spheres make equal angle  with the
vertical and the centers of the spheres are x apart.
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(a)
Draw and label the forces acting on sphere A.
electrostatic force F
(2)
tension T

weight mg
(b)
In terms of  and tension T, write down an expression for
(i)
the electrostatic force acting on sphere A
(4)
F = T sin (½)
(ii)
the weight of sphere A
mg = T cos (½)
Hence, deduce that the separation x can be expressed as
 Q2 l 

x= 
 2 π εo m g 


What is the value of n?
F=
n
(Hint: tan  sin)
1 Q2
4  o x2
From (b)(i) and b(ii), F = mg tan  mg sin = mg
Hence,
(x/2)
l
1 Q2
(x/2)
2 = mg
l
4  o x
1
 Q2 l  3
1
 and n =
Solving gives x = 
3
 2 π εo m g 


(c)
(3)
If m = 2.0  10-2 kg, l = 0.1 m and x = 0.02 m, find the value of the charge
Q.
(2)
 2 π εo
From (b), Q = 


1
3 2
mgx
l



1
 2 π (8.85 10 -12 ) (2 .0  10 - 2 )(10)(0.02 ) 3  2

= 

0.1


= 2.9  10-8 C
(d)
(i)
(2)
If the spheres carry the same sign but with different amount of
charges, will the thread be inclined at an equal angle to the vertical
when the system is in equilibrium?
(2)
If spheres carry different amount of charges, the
electrostatic forces acting on both spheres are still the
same. Therefore the threads will incline at equal angle
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to the vertical as before. (2)
(ii)
If the spheres carry the same amount of charges but different
masses, will the thread be inclined at an equal angle to the vertical
when the system is in equilibrium? Explain
(2)
From (b) F = mg tan
For same F, mg tan = constant
Different m implies different angle  (2)
6.
The primary coil of a transformer is connected to a 3.0 V d.c. supply and a switch
S1. The secondary coil is connected to a resistor and a switch S2 .
S1
S2
3.0 V
(a)
Figure below is a graph showing the growth of current in the primary coil
when switch S1 is closed and switch S2 is left open.
current / A
time / ms
(i)
Use the graph to determine the initial rate of growth of current in
the primary coil.
(2)
dI
dt (at t = 0)
0.3
=
150  10-3
= 2.0 A s-1 (2)
(1.8 – 2.2 A s-1)
(ii)
Calculate the inductance of the primary coil.
dI
E = L dt
(2)
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3=L2
L = 1.5 H (2)
(1.3 – 1.7 H)
(b)
Now consider the situation when switch S2 is closed before switch S1 is
closed. Draw the variation of the current in the circuit connected to the
secondary coil when switch S1 in the primary circuit is closed. Assume
that the primary current varies as shown above.
(2)
current
finite Imax at t = 0
becoming asymptotic at ~ 700 ms
(2)
0
(c)
100
200
300 400 500
600 700
time / ms
A transformer is not perfectly efficient when used with alternating current.
State and explain TWO sources of energy loss and explain how they can
be eliminated.
(2)
eddy current losses: use laminated core
hysteresis losses: use soft iron as core (any 2)
heat loses: use copper as wires
(d)
(i)
Complete the circuit diagram below to show how you would use a
capacitor and diodes to rectify and smooth the output from a
transformer.
(2)
a.c. supply
(ii)
output
Figure below shows the rectified output voltage when a current is
drawn from the secondary of the transformer. Draw the output
voltage you would get if a larger capacitor were used.
(2)
voltage
Shallower drop in voltage
Same frequency and same peak voltage
time
Page 11 of 17
7.
In Figure 7, a uniform magnetic field B pointing into the paper and a uniform
electric field E pointing upwards are applied on the left-hand side and right-hand
side of y-axis respectively. Suppose that a charged particle of mass m and charge
q enters the magnetic field with an initial velocity Vo at point P. After leaving the
magnetic field, it will move into the electric field and leave it at point S. Neglect
the gravitational effect.
y
E
B
Vo
S
P
(a)
What is the sign of the charges carried by the charged particle?
paths of the charged particle in the two different fields.
x
State the
(2)
By Fleming’s left hand rule, it can be found that the charged
particle carries negative charges.(1)
In the uniform magnetic field, the path is part of a circle. (½)
In the electric field, the path is a parabola. (½)
(b)
Deduce an expression, in terms of m, q, E and B, for the time taken for the
charged particle to travel through.
(i)
the magnetic field and
(3)
In the magnetic field , the charged particle undergoes a
circular motion with radius R. The centripetal force
acting on the charged particle is the magnetic force.
F=
m Vo2
= q Vo B
R
qBR
(1)
m
Since the charged particle describes a quarter of circle in
the magnetic field only, the time taken for the charged
particle to move in the magnetic field is given by
2πR
1
Tm= V
 4 (1)
o
πm
Tm = 2 q B (1)
∴Vo=
(ii)
the electric field
(3)
When the charged particle enters the electric field, its
horizontal component of velocity remains uncharged,
whereas its vertical component of velocity is accelerated.
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By Newton’s second law of motion, the acceleration of
the charged particle is given by
F= ma = q E
qE
∴a = m (1)
For the vertical displacement, we have
1
R = 2 a te2 (te is the time taken for the charged particle to move
in the electric field) (1)
te2
te =
(c)
2R
2mR
= a = qE =
m
q
2 Vo
BE
m Vo
qB
qE
2m
2 m2 Vo
= q2 E B
(1)
Suppose that the velocity of the charged particle at point S is Vs, deduce
down an expression for the total work done on the charged particle by the
fields in terms of m, Vo and Vs.
(2)
Since the motion of the charged particle is perpendicular to the
magnetic force, there is no work done on the particle by the
magnetic field. (1)
The work done on the charged particle comes from the electric
field such that it is equal to the change of kinetic energy.
1
W.D.=△K.E. = 2 m(Vs2 - Vo2) (1)
(d)
From the result in part (c), express Vs in terms of Vo, q, E and B.
Electric work done = ∫Fe dy = Fe ∫dy = Fe R = q E R
1
m Vo
2
2
m(V
s - Vo ) = q E R = q E
2
qB
1
Vo
2
2
(V
s - Vo ) = E
2
B
Vo
Vs2 = 2E B + Vo2
Vs =
Vo2 
2 E Vo
B
(3)
(3)
Page 13 of 17
8.
Figures below show two characteristics of an n-p-n transistor.
after saturation.
IB / A
Take VCE = 0.2 V
IC / mA
3.6
0
IB / A
VBE / V
0.8
0
30
The following circuit is connected to investigate the input/output voltage
characteristics of the transistor.
6V
IC
RL=2.2 k
Vout
IB
Vin
RB=15 k
0V
(a)
Find the values of IC, IB and Vin when the transistor is just saturated. (3)
6 - 0.2
= 2.64 mA (1)
2.2103
3010-6
-3
IB = (2.6410 ) 
= 22 A (1)
3.610-3
Vin = IBRB + VBE = (2210-6)  (15103)+ 0.8 =1.13V (1)
IC =
(b)
Draw the input/output voltage characteristics of the transistor on the
following diagram, inserting the results in part (a) and their corresponding
values of Vout.
(2)
6
Vout / V
(½)
5
4
(1)
3
2
1
(½)
(0.2)
Vin / V
0
0
(0.8)
1
(1.13)
2
3
4
5
6
Page 14 of 17
(c)
Find the voltage amplification from the graph.
G=
(d)
(2)
Vout
6 - 0.2
= 0.8 - 1.13 = -17.6
Vin
The voltage amplification of the above transistor circuit depends on the
current amplification factor . Unfortunately,  varies greatly even among
the same model of transistors. The following circuit design overcomes this
drawback.
6V
IC
RC
Vout
Vin
IE
RE
0V
(i)
Express IC in terms of Vout and RC.
(1)
Ic = (6 – Vout)/RC
(ii)
Express IE in terms of Vin and VBE.
(1)
IE = (Vin –VBE)/RE
(iii)
Deduce the voltage amplification of the circuit in terms of RC and
RE. State the assumption you need.
(3)
IC  IE before saturation (1)
(6 – Vout)/RC = (Vin –VBE)/RE
Vout
RC
= - R (2)
Vin
E
Page 15 of 17
9.
A mass of dry air rises above a patch of warm ground. As it rises its volume and
pressure change adiabatically according the graph shown below.
presure / kPa
100
90
80
70
60
50
40
30
20
10
volume / 103 m3
0
0
(a)
(i)
1
2
3
4
5
6
7
8
9
Explain why the temperature of the air falls as the air rises. (2)
Air does work as it expands.
Energy comes from the gas’s internal energy.
(ii)
Use the graph to determine the change in internal energy of the air
for the adiabatic change shown.
(3)
U = Q + W = W =  P dV = area under the P-V graph
1
= 2 (100103 + 75103)  [(8.5 – 7) 103]
= 1.3 108 J
(iii)
The mean relative molecular mass of the air is 28 g and the density
of the air at ground level is 1.2 kg m-3. How many moles of air
are there?
(2)
At ground level, volume = 7000 m3
mass = 7000  1.2 = 8400 kg
8400
5
number of moles =
-3 = 3 10 moles
2810
(iv)
Calculate the temperature change of the air as it rises. (molar gas
constant = 8.31 J K-1 mol-1)
(2)
P1 V1 = n R T1 and P2 V2 = n R T2
P2 V2 - P1 V1 = n R (T2 - T1)
(75 103)(8.5103) - (100103)(7103) = (3105)8.31 T
T = -25 K
Page 16 of 17
(b)
Another mass of air containing water vapour rises by the same amount.
Some of the vapour condenses into water droplets. State and explain
how the temperature change differs from that of the dry air.
(3)
On condensation, latent heat of vaporisation is released.
The energy goes into the air.
The air becomes higher in temperature or less temperature drop.
10.
One nuclear power station produces an average useful output power of 800 MW.
The mass of uranium 235 in the fuel rods is reduced to 70% of its original value
in a four-year period. The average yield of each fission reaction is 200 MeV and
35% of this is converted into useful power.
Charge of an electron, e
Avogadro constant, NA
(a)
= 1.610-19 C
= 6.01023 mol-1
Calculate the average power generated by fission.
(1)
100
power generated = 800 MW  35 = 2.286  109 W
(b)
Calculate the average change in the mass for each fission.
(2)
Energy per fission = (200106)  (1.610-19) = 3.210-11 J
E 3.210-11
-28
m = c2 =
kg
8 2 = 3.5610
(310 )
(c)
Calculate the total number of fission reaction that occurs in the four-year
period.
(3)
Energy per fission = (200106)  (1.610-19) = 3.210-11 J
2.286  109
Number of fission in 1 second =
= 7.1251019
3.210-11
Number of fission in 4 yr = 7.1251019  (4365243600)
= 8.991027
(d)
Hence, deduce the mass of uranium that was originally present in the
reactor core.
(3)
Number of molecules under fission in 4 yr = 8.991027
8.991027
Number of moles =
= 1.5 104
6.01023
Mass = (1.5104)  (235 10-3) = 3521 kg
100
Original mass = 3521 kg  30 =11740 kg
Page 17 of 17
(e)
In such a reactor about 5% of the reactions produce the strontium isotope
90
38Sr . This nuclide is radioactive with a half-life of 28 y.
(i)
Calculate the initial rate at which strontium is being formed in the
reactor.
(1)
initial rate = 7.1251019  0.05 = 3.5631018 s-1
(ii)
Estimate the number of strontium atoms after one day and its
activity.
(3)
Number of strontium atoms after one day
= 3.5631018  (2460 60) = 3.0781023
Decay constant k =
Activity
ln 2
= 7.8510-10 s-1
28365246060
=kN
= (7.8510-10)  (3.0781023) = 2.41014 Bq