Download H2 Physics - Yearly Solutions - 2008 - 29+1pp - v3.07

2008  1
2008
nov
advanced
physics
complete yearly solutions
2008 Nov
2008  2
3.
2008 Nov Paper 1
[Measurement]
Solution
MCQs
Manufacturer’s specification of a digital voltmeter is:
“accuracy ± 1 % with an additional uncertainty of ±
10 mV”
Answer all questions.
1.
The meter reads 4.072 V.
[Measurement]
 Recorded reading  (4.072 ± 1% ± 10 mV)
Solution
 (4.072 ± 1% ± 0.010)
Velocity is a vector. Time is a scalar.
 (4.072 ± 0.04072 ± 0.010)
 To define velocity with respect to time, the
measurement of distance travelled must be of vectorform.
 (4.07 ± 0.05) V (error has only 1sf and the main
reading follows the decimal place of the error)
Recorded reading, together with its uncertainty:
 Distance travelled in unit time
 (4.07±0.01) V
 Distance travelled per unit time
 (4.07±0.04) V
 Speed in a particular direction
 (4.072±0.052) V
 Displacement per unit time
 (4.07±0.05) V
(A) (ans)
 themis 
2.
[Measurement]
(D) (ans)
 themis 
4. [Kinematics]
Solution
Solution
Reasonable estimate:
Air resistance is negligible.
 The average kinetic energy of a bus (say, m 
3000kg) travelling on an expressway (say, v 
100km/h) is 12 mv 2  1 MJ.
A metal ball is at rest and dropped over a bed of sand
for 1.0 s before impact making a depth of 8.0 mm.
 Just before hitting the sand, the metal ball (mass m)
from rest (initial speed u = 0) has a final speed
the kinetic energy of a bus travelling on an
expressway is 30,000 J
(v)  u + gt  (0) + (10)(1)  10 ms-1.
 The average power of a domestic light is 12 –
100W.
Kinetic energy before impact is
 12 m 10   50m
2
the power of a domestic light is 300 W
 The average temperature of a hot oven is 300C.

Work done against sand
 Ffrictions  mas
the temperature of a hot oven is 300 K
1 mv 2
2

where a is the deceleration of ball in sand and
s is its depth of displacement.
 The average volume of air in car tyre is say,
1.0m (tube)  0.2m (height)  0.2m (wide)
:
 0.04m3 (close)
the volume of air in a car tyre is 0.03 m3
50m = mas  a 
(D) (ans)
 themis 
50
0.0080
 a  6.3  103 ms-2 (2sf) (close)
The average deceleration of the ball:
 6.0  102 m s-2
 1.2  103 m s-2
Advanced Physics solutions - yearly
 themis
2008  3
 1.2  104 m s-2
3.0 ms-1
5.0 ms-1
 6.0  10 m s
3
-2
6.0 kg
10 kg
(C) (ans)
 themis 
5.
[Kinematics]
pin
cork
The trolleys had a head-on collision that lasted 0.20 s.
After the collision, the trolleys stick together.
Solution
 Taking   to be positive,
A cyclist accelerates down a hill, then constant speed,
before decelerating back up another hill.
Total momentum  6.05.0  10 3.0
 To deduce the graph of distance s moved by the
cyclist against time taken, t:
Since objects stick together,

When the cyclist goes down slope, for every t ,
there would be an increasing s , giving an
upwardly increasing graph.

When the cyclist is travelling at a constant speed,
for every t , there would be a constant s ,
giving a constant increasing graph.

When the cyclist goes up another slope, for every
t , there would be a decreasing s , giving an
downwardly decreasing graph.
= 0 kgms-1
Total kinetic energy = 0 J
Average force  change in momentum per unit time of
impact
 6.0 5.0  0.20  150 N
Since the cyclist is moving further away from his
original position, the graph of s is ever
increasing.
The shape of graph:

s
0
t
0
0
The average force
acting on each
trolley during the
collision
(N)

0
300

60
150 

60
300

0
150 
(C) (ans)
t
0

s

s
Total momentum of
the two trolleys
before the collision
(kgms-1)
 themis 

s
7.
[Forces]
Solution
0
t
0
0
t
A ladder of weight W rests against a vertical wall.
0
(D) (ans)
 themis 
6.
Friction exists at each contact point prevents the
ladder from slipping.
[Dynamics]
Solution
Trolley 1:
Mass = 6.0 kg, Velocity = 5.0 ms-1
Direction: to the right
Trolley 2:
W
 According the Lami’s theorem, the lines of action of
the 3 co-planar forces must meet at a common point.
Mass = 10 kg, Velocity = 3.0 ms-1
Direction: to the left
2008 Nov
2008  4

The diagram that shows the directions of the forces on 9.
the ladder:
[Work, Energy & Power]
Solution


A car, of mass m, has a driving force F.
W
In a time t, it travels a distance s and its speed
increases from u to v.
W
 Useful work done by the car engine

= final energy  initial energy

=
W
=

m v 2  u2

2

Fs
, it has a unit equivalent to power
t

mv - u  , it has a unit equivalent to momentum

Ft , it has a unit equivalent to impulse
(C) (ans)
 themis 
8. [Forces]
Solution

l
rubber
handle

m v 2  u2
2
 , it has a unit equivalent to energy
(D) (ans)
 themis 
wooden
section
2.10 l
10. [Electromagnetic Induction]
A uniform rod with a rubber section and wooden
section is made to balance as shown.


1 mu2
2
The useful work done by the car engine:
W
4.00 l
1 mv 2
2
4.00 l
l
Ww
2.10 l
2.00 l
Solution
A square loop of wire is placed in a region of uniform
magnetic field.
Wr
Direction of field  perpendicular to plane
magnetic field
Taking moment about the pivot:
Ww  4.00  1.00  2.00  2.10  l  Wr  2.10  0.5  l
v
0.90 Ww  1.60 Wr  
:
loop of
wire
W  V  g  AL g
0.90  ALw wg  1.60  ALr r g
The loop is pulled out of the field at a uniform speed v.
 
0.90
 4.00l  l  r 
1.60
 w 
The ratio,
The ratio,
 Initially when the loop is pulling, the loop
experienced no change in magnetic flux linkage, i.e., no
work done.
density of rubber  r 
= 2.25
density of wood   w 
When the loop is partially out of the magnetic field,
the loop experienced a change in magnetic flux
linkage, i.e., work done needs to be done to oppose
that change. Since the change of flux is constant, the
work needs to be done is also constant.
density of rubber

density of wood
 1.71  2.50  3.27  2.25
(B) (ans)
 themis 
Advanced Physics solutions - yearly
 themis
2008  5
The graph shows work done W versus the speed v:  Tension in string  W  Fc — in its usual notations


0
 mg  m
W
W
v
0
0
Tension in the string:


 11.8 N
W
W

0.5752 
 1.27  9.81 
  13.2 N
0.600 

v
0

0.5752 
v2
 1.27  9.81 

0.600 
r

 12.5 N
 13.7 N
0
v
0
0
v
0
 13.2 N
(C) (ans)
(D) (ans)
 themis 
 themis 
13. [Gravitational Field]
11. [Motion in a Circle]
Solution
Solution
A satellite has:
The rotor of an engine is rotating at 3000 rpm
(revolutions per min); radius  8.0 cm.
Mass  m
Orbit  Circular, initially at radius r1 around the
Earth, then moved to a new circular orbit of
radius r2 , as shown.
 Centripetal acceleration
 r 2 — in its usual notations
2
 3000  2 
-2
 8.0  102 
  7900 m s
60




Mass of the Earth  M
Gravitational constant  G
The centripetal acceleration:
 25 ms-2
 3.1104 ms-2
 7.2107 ms-2
C
Earth
mass M
satellite
mass m
r1
 7900 ms-2
r2
(B) (ans)
 themis 
 Change in potential energy of the satellite, 
12. [Motion in a Circle]
 m  m2  1 
Solution

1 
 1  
 m  GM     GM   

 r2  
 r1  
A pendulum has:
Bob mass  1.27 kg
Radius of oscillation  0.600 m
1 1 
 GMm    (note: ve)
 r1 r2 
Velocity (horizontally)  0.575 ms , at the centre
of its motion when the string is vertical.
-1
Increase in the potential energy of the satellite:
T
a
W

1 1
GM   
 r2 r1 

1 1
GM   
 r1 r2 
2008 Nov
2008  6

1 1
GMm   
 r2 r1 

1 1
GMm   
 r1 r2 
a / m s-2
10
(D) (ans)
5
 themis 
14. [Gravitational Field]
x / cm
0
-5
-10
Solution
 Deduce from graph, amplitude of motion  5.0 cm
Earth is orbiting round the Sun.
Radius of the Earth's orbit  1.50  1011 m
gradient 
Period  365 days
GMSmE
—
r2
— in its usual notations
 Centripetal force, mEr2 
: r2 
GMS
r2
 Mass of sun, MS 
r 32
G
 2 
r3 

T 
 
G
2
1.50  10   365  242 60  60  
11 3


6.67  10
2

10
 2 
 2  

2
5.0  10
 T 
2
 Period of motion, T  0.44 s
Amplitude (cm)
Period (s)

5.0 
14

10
0.44 

10
14

5.0 
0.44 
(A) (ans)
 themis 
11
16. [Oscillations]
 2.01  1030 kg
Solution
The mass of the sun:
 6.41029 kg
 1.161033 kg
 3.311033 kg
 2.011030 kg
An object is said to undergo lightly-damped
oscillations.
(B) (ans)
 themis 
15. [Oscillations]
Solution
Defining equation for a particle moving in simple
harmonic motion is
Although the object is undergoing lightly-damped
oscillations, i.e., its amplitude will reduce slowly, but
its period remains unchanged.
Diagram shows the displacement, y of that object
against time, t:


y
y
a  2 x
t
where, acceleration of the particle  a ,
displacement  x , and
angular frequency  
How a varies with x for a particle moving in simple
harmonic motion, is as shown.
t


y
y
t
t
(D) (ans)
 themis 
Advanced Physics solutions - yearly
 themis
2008  7
 Original Ideal gas equation:
17. [Thermal Physics]
pV  nRT
Solution
 1 
 1 p
V
 nRT 
—
A container contains a hot vapour and is allowed to cool New Ideal gas equation:
and loses heat to its surroundings.
 1 
pV  2n R 2T   1 p  14 
V — 
The graph shows how its temperature changes with
 nRT 
time.
The new gradient would be 14 that of the original
temperature
graph.
P
X
Line represents the new gas:
Q
1/p
R
S
Y
0
0

0.4
T

position of
original graph
0.3
time
 The length of the line Q is greater than the length of
the line S, implies it needs to lose more heat in
liquefaction than freezing.
0.2

0.1

V
0
The part of the graph shows that the specific latent
0
heat of vaporization of the substance is greater than its
(D) (ans)
specific latent heat of fusion:
 The gradient of the graph at P is greater than the
gradient at R.
1
2
3
4
 themis 
19. [Thermal Physics]
 The gradient of the graph at T is greater than the
gradient at R.
Solution
 The value of X is greater than the value of Y.
An electric kettle contains:
500 g of water at 15°C
Power  2.2 kW
 The length of the line Q is greater than the length
of the line S.
Specific heat capacity of water  4.2  103 J kg-1 K-1
(C) (ans)
 themis 
The temperature of the water is raised to 100°C.
18. [Thermal Physics]
 Recall,
Solution
A fixed amount of an
ideal gas:
Pressure  p
Volume  V
The graph shows the
variation of 1 p with
V at a constant
temperature.
Pt  mc — in its usual notations
Time taken, t 
1/p

0.4
mc
P
0.500  4200  100  15 
 81 s (2sf)
2200
0.3
Minimum time taken to raise the temperature of the
water to 100°C:
0.2
 22 s
 8.1104 s
 81 s
(B) (ans)
0.1
V
0
0
 95 s
1
2
3
 themis 
4
The amount of gas and the thermodynamic
temperature are both doubled.
2008 Nov
2008  8
20. [Wave Motion]
c
E  hf  h   — in its usual notations
 
Recall,
Solution
If measured in the other units, then

c
E ' 1.6  1019  h 
  ' 109

A point source produces waves is made to pass
through an area that is 2.0 cm wide, as shown.

16 cm
point
source of
waves



hc
E'  
 109 1.6  10 19




hc
K 
 109 1.6  1019





Within this area, the intensity of the waves is I and
their amplitude is A. The waves reach a second area of
width 16 cm.




6.63  10 3.00  10 
10 1.6  10 
8
9
19
 1240 (3sf)
Numerical value of K:
I
:  Area  64    I  64  
64
 3.1810-53
 3.1810-35
A
:  I  64    A  8  
8
 1.2410-15
 1.24103
(D) (ans)
 themis 
Intensity at 2nd area
Amplitude at 2nd area
I
8
A

4
I

64
A
8

I
256
A
16

I

64
A

4


1

'

34

 Recall, I 






2.0 cm
1
—  and I  A2 —
Area
— in its usual notations

22. [Superposition]
Solution
Diagrams show air particles moving in a column.
A stationary wave is formed.
L
node
antinode
node
antinode
(B) (ans)
 themis 
21. [Quantum Physics]
Solution
Energy E of a photon and its wavelength  is
E
K

, where K is a constant.
The first diagram shows the displacement of some
particles at one instant and the second diagram shows
the displacement of some particles half a cycle later.
 Deduce from diagrams that
3
4
L .
Maximum varying pressures occur at points with
maximum particle movements (anti-nodes).
 If E is measured in electronvolts E ' and  in
nanometres   ' , then:
Advanced Physics solutions - yearly
 themis
2008  9
Length L in terms of
wavelength A

3
4
Position within the
column that pressure
change by the largest
amount

(D) (ans)
 themis 
25. [Electric Fields]
Solution
node
In an electric field, the electric potentials at points P1
and P2 in free space are V1 and V2 respectively.

3
2
node

3
2
antinode 
A point electric charge q is brought from P1 to P2 by
an external agent.
antinode 
 Work done on the charge  q V2  V1 

3
4

 q V1  V2 
(B) (ans)
 themis 
Work done on the charge:
23. [Superposition]
Solution
A beam of incident monochromatic light is shined
normally to grating:
For maximum images, set sin  1 :

1 3.0  105
600  109
q  V2  V1 

q V2  V1 

q V1  V2 
26. [Current of Electricity]
Solution
 Recall, d sin  n — 


 themis 
Diffraction grating  3.0  105 lines per metre
d
q V2  V1 
(D) (ans)
Wavelength  600 nm
: nmax 

A wire has:
Resistivity  1.3  108 m
Diameter  0.50 mm
Length  30 times round an insulating rod
Insulating rod’s diameter  1.5 cm
 5.5
 5 images, maximum
Total number of images produced by light transmitted
through this grating:
 8
 9
 11
wire wound
round rod 30
times
 5
insulating rod
diameter 1.5 cm
(A) (ans)
 themis 
 Recall, resistance, R  
—
A
— in its usual notations
24. [Electric Fields]
Solution
Electric field strength is defined:
: R  1.3  108 
“force per unit positive charge on a small test charge”
 The key definition of electric field strength came
from Coulomb. Hence, it is necessary for the test
charge to be small,
 so that the test charge does not distort the
electric field

so that the force on the test charge is small
 so that the test charge does not create any forces
on nearby charges

30  2  1.5  10 2  2

 0.5  103  2


2
 9.4  102 
Resistance of the wire:

1.1  101 

4.7 105 


7.0 104 
9.4 102 
(B) (ans)
 themis 
 so that Coulomb's law for point charges is obeyed
2008 Nov
2008  10
Thermistor is connected to the potential divider circuit
as shown.
27. [Current of Electricity]
Solution
+9.0 V
1.5 k
A
A battery, connected to an external resistor, has:
e.m.f.  6.0 V  E 
Internal resistance  0.40   r 
Rv
External resistor  2.90   R 
 Recall, current i 
B
output
1.5 k
E
6.0

 R  r   2.90  0.40 
0V
 1.82 A
At 0730, an output of 6.O V is required.
Power dissipated P  i 2R  1.82  2.90
 If the output is to be set at 6.0 V, then across the
other section (A), the voltage drop needed to be 3.0 V.
2
 9.6 W
At 0730, the resistance of the thermistor  1.5 k
Power supplied to the external resistor:
 1.3 W
 5.3 W
 12.4 W
 9.6 W
The current flowing through section (A):
i
(C) (ans)
 themis 
3.0 V
 0.001 A
1.5 k  1.5 k 
Similarly, the same current flows through the section
(B):
6.0 V   RV  1.5k0.001 A 
28. [D. C. Circuits]

Solution
The resistance R of a thermistor varies during part of a
day is shown.
R/ k
4.0
Rv  4.5k
Value of the variable resistor be set:
 1.5 k
 3.0 k
 6.0 k
 4.5 k
(C) (ans)
 themis 
29. [D. C. Circuits]
3.0
Solution
2.0
The circuit consists:
1.0
0.0
0500
0600
0700
0800
time
0900 of day
A battery of e.m.f. 6.0 V, of negligible internal
resistance;
Three resistors, each of resistance R, and a variable
resistor T, as shown.
T
X
R
6.0 V
+6.0 V
+0.0 V
V
R
R
Y
The resistance of T changes from R to 5R.
Advanced Physics solutions - yearly
 themis
2008  11
 When variable resistor, T is set to R:
Potential X  +3.0 V
Potential Y  +3.0 V
Potential difference, V  0.0 V
must exactly opposed in order for the beams to
emerge undeflected.
The arrangement that would allow the beam to pass
through undeflected:

When variable resistor, T is set to 5R:
Potential X  +1.0 V
Potential Y  +3.0 V
Potential difference, V  2.0 V
Change in the reading of the high-resistance voltmeter
 2.0 V  0.0 V  2.0 V
electrons
protons


 4V
 5V

magnetic field
Change in the reading of the high-resistance
voltmeter:
 zero



 2V



(B) (ans)
 themis 
electrons
30. [Electric Fields]
protons

Solution

(B) (ans)
 themis 
A charged particle, in vacuum, travels in a straight line,
enters a uniform field. The particle then travels in a
curved path that is not the arc of a circle.
32. [Electromagnetic Induction]
Solution
 Only electric field influences a charged particle in a
non-circular way.
A coil has area A and n turns.
For the charged particle to go on a subsequent nonparallel path, the field direction has to be non-parallel.
A uniform magnetic field of flux density B acts at an
angle  to the plane of the coil, as shown.
Type of field
Initial direction of the
particle compared to
the field

Electric 
parallel

magnetic
parallel

magnetic
perpendicular 

electric 
perpendicular 
pivot
coil area A
(B) (ans)
 themis 
31. [Electromagnetism]
Solution

magnetic flux
density B
Different particle beams are made to enter a region
between two metal plates in which there are uniform
electric and magnetic fields.
 By applying Fleming’s left hand rule and electric
field direction, the forces acting on the particle beams
2008 Nov
2008  12
 Initial flux linkage i  BAn sin
34. [Alternating Currents]
The coil is rotated by an angle of  .
Solution
Final flux linkage  f  0
The graph of an alternating current with a square
waveform is shown.
Change in flux linkage   BAn sin
current
Change in magnetic flux linkage when the coil rotates
so that the angle  is reduced to zero:
 BAn cos
 2BAn cos

2BAn sin

square wave-form
Io
BAn sin
time
(B) (ans)
- Io
 themis 
Its peak value for the current is lo .
33. [Electromagnetic Induction]
 Consider a complete period, T.
Solution
current
The magnetic flux linking a coil, with respect to time is
as shown.
Io2
Io
flux
0
time
- Io
t
0
2t
time
Squaring the wave to give the  Io2  waveform.
 The induced e.m.f. of a coil, varies with the rate of
change of flux linkage, i.e., the gradient of the fluxtime graph.
Further increasing or decreasing flux linkage induces
differently signed e.m.f..

0
2t
time
 0
0
t
0
2t
time



Io
2

Io
2

Io
(D) (ans)
 themis 
35. [Quantum Physics]
e.m.f.
e.m.f.
Io2  Io2  Io
Root-mean-square value for the current:

t
Io2  T
 Io2 .
T
Irms 
e.m.f.
0
Io2 
Take the square root of this average value,
The corresponding e.m.f. induced in the coil, with
respect to time:
e.m.f.
Take its average over its period,
Solution
0
0
t
2t
time
0
0
t
2t
time
A beam of electrons is made to pass through a thin
carbon film, velocity, v, as shown.
(C) (ans)
 themis 
alternating
regions of bright
and dark zones
electron beam
graphite
plate
Advanced Physics solutions - yearly
fluorescent
screen
 themis
2008  13
The electrons produce a pattern of concentric circles
on the fluorescent screen.
Graph shows the probability of finding an electron at
each position:

 According to de Broglie’s principle of wave-particle
duality, the electrons exhibited wave-like property, i.e.,
diffraction.
From mv 
h
probability
— in its usual notations

When  mv   
0
From d sin  n — in its usual notations
When     

distance
  

probability
  
More fringes and diameters of circles decrease.
0
Cause of the pattern




Changes to pattern
when the velocity is
increased
diffraction 
diameters of circles
increase
refraction
diameters of circles
increase
refraction
diameters of circles
decrease 
diffraction 
distance

probability
0
distance

probability
diameters of circles
decrease 
0
(B) (ans)
 themis 
distance
36. [Quantum Physics]
(B) (ans)
 themis 
Solution
Sketch shows how the wave function  of an electron
varies with position.

37. [Lasers and Semiconductors]
Solution
The symbols    and    represent the majority
carriers in the p-type and n-type sides of a p-n
junction.
0
distance
 As interpreted in standard text.
 We usually interpret the absolute square of the
wavefunction   r ,t  as the probability density for the
particle to be found at each point in space. In other
words,   r ,t  d 3r is the probability, at time t, of
2
finding the particle in the infinitesimal region of
volume d 3r surrounding the position, r .
2008 Nov
2008  14
Pair of diagrams illustrates how a p-n junction acts as
a rectifier:



voltage
source
p
n
conventional
current



  

  
(D) (ans)
 themis 
39. [Nuclear Physics]


p
n

no
current
Solution
  
A detector has:



voltage
source
p
n



  

  
conventional
current
p


n


no
current
  

A radioactive source has:




voltage
source
p
n



  

  
conventional
current


p
n
Ionising radiation
Background count rate  24 counts per minute
Reading  532 counts per minute
 True initial reading  532  24  508

no
current
  
After 2 half-lives,
the final true reading 

 voltage
source

p
n



  

conventional
current


p
n

no
current
508
 127
4
We now have to add back the background count,
the final meter reading  127  24  151
  
Reading after two half-lives of the source:
(D) (ans)
 themis 
 127
 133
 157
 151
(C) (ans)
38. [Lasers and Semiconductors]
 themis 
Solution
40. [Nuclear Physics]
Solution
In a helium-neon laser:

Helium atoms collide with neon atoms and excite
them

It produces a population inversion which allows
stimulated emission.
A nucleus of bohrium yx Bh decays to mendelevium
255
101Md
by a sequence of three -particle emissions.
bohrium yx Bh  dubnium  
 The spontaneous infra-red emission is low energy.
 lawrencium  
The stimulated emission of red light is high energy.
The total excitation energy should equal its total
emissions.
 Rewriting,
Neon energy level diagram that correctly shows the
excitation of the neon atoms by the helium atoms, the
spontaneous infra-red emission from the neon, and
the stimulated emission of red light:


spontaneous emission
stimulated
emission
excitation
stimulated emission
spontan.
emission
excitation
 mendelevium
x
y Bh

255
101Md
255
101Md
 3

—
: x  255  3 4   267
y  101  32  107

no. of neutrons  267  107  160
No. of neutrons in a nucleus of yx Bh :
 267
 261
 154
 160
(C) (ans)
 themis 


spontaneous
emission
excitation
spontaneous emission
stimulated
emission
Advanced Physics solutions - yearly
excitation
stimulated
emission
 themis
2008  15
[2008N P1 MCQ Key]
Q.
Key
Q.
Key
Q.
Key
Q.
Key
1.
A
11.
B
21.
D
31.
B
2.
D
12.
C
22.
B
32.
B
3.
D
13.
D
23.
A
33.
C
4.
C
14.
B
24.
D
34.
D
5.
D
15.
A
25.
D
35.
B
6.
C
16.
D
26.
B
36.
B
7.
C
17.
C
27.
C
37.
D
8.
B
18.
D
28.
C
38.
D
9.
D
19.
B
29.
B
39.
C
10.
D
20.
B
30.
B
40.
C
 themis 
2008 Nov
2008  16
(b) Upon impact, the speed of the stone is reduced
from 34 ms-1 to 2.0 ms-1 in a time of 0.95 s.
2008 Nov Paper 2
 Using momentum considerations,
Questions
Impulse, Faverage t  mv  u — in its notations
Answer all questions.
1.

[Kinematics]
Average force acted on the stone during this
time,
Faverage 
Solution
 4.38N (ans) [2]
At the top of a cliff of height 32m, a stone of mass 130
g is thrown horizontally as shown.
(c) If the stone causes a big splash on hitting the sea,
using energy considerations, the high kinetic
energy from the stone is not only transferred to its
immediate vicinity but over a wide area. Hence,
the whole water surface and column is made to
resist the impact. Together with a louder sound,
the stone will be slowed down in a shorter time
(shorter distances) than when its impact is
restricted to a small water column, where no
splash is produced. (ans) [2]
path of
stone
32 m

0.130  2.0  34 
0.95 s
A
sea
 themis 
Air resistance is negligible.
2.
The stone hits the sea at speed of 34 ms-1 .
Solution
(a) At point of impact,
(i)
Let the velocity of impact be
v.
[Oscillations]
vy
v

A
vx
The diagram showed a flat horizontal plate vibrating in a
vertical plane.
Consider its vertical descent,
plate
v y  uy  2gs — in its usual notations
2
2

vy 2  02  2  9.81  32

vy  25.1 ms-1 (the vertical component of the
vertical
oscillations
The plate’s the variation with displacement x of the
acceleration a is as shown.
velocity of the stone) (ans) [2]
a / m s-2
20
(ii) Deduce from the final velocity,
v 2  v y 2  v x 2  342  25.12  vx 2

10
vx  23.0 ms-1
The angle  to the horizontal of the stone’s path
 vy 
 25.1 
 tan1    tan1 

v
 23.0 
 x
-8
-6
-4
-2
0
0
2
4
6
8
x / mm
-10
-20
 47.5 (3sf) (ans) [2]
Advanced Physics solutions - yearly
 themis
2008  17
(a) From the straight-line graph, the negative gradient
gives a relationship i.e., a   x (oppositely
directed).
It is also given that it is a periodic oscillation, i.e.,
the graph does not extend itself indefinitely.
balanced by the satellite’s kinetic urge to leave the
orbit.


By definition, the oscillations of the plate are
simple harmonic (SHM) as it is a periodic motion
in which acceleration (a) is proportional to, but
oppositely directed to, the displacement (x) from
its equilibrium position. (ans) [3]
GMm
v2
— in its usual notations

m
R
R2
GMm
 kinetic energy of the satellite
2R
(ans) [2]
1 mv 2
2

(ii) Gravitational potential energy  
GMm
— in its
R
usual notations
(b) Some sand is introduced onto the plate.
Amplitude of vibration of the plate is gradually
increased from zero.
gravitational potential energy of satellite
kinetic energy of satellite
At one particular amplitude, the sand is seen to
lose contact with the plate.
GMm
R

GMm
2R

= 2 (QED) [1]
acceleration
(i) When the sand particle is resting
reaction, R
on the vibrating plate, the
sand grain,
external forces acting on the
mass m
particle are as shown in the free
body diagram (upward positive).
ma  R  W — 
weight, W
When the sand particle loses contact, the reaction
force, R become zero.
: ma  0  W  ma  mg
 a  g  9.81 m s-2 (ans)
10.0
energy
9
/ 10 J
8.0
6.0
4.0
At a  g (upward ), the plate is travelling
downward from its top displacement position. At
this point, the acceleration due to gravity  g  and
the acceleration due to the vibrating plate is the
same, hence, if the plate were to accelerate faster
than g (downwards), the sand particle will not be
able to “catch-up” and stay in contact, thereby
losing contact with the plate. (ans) [3]
(ii) From graph, when a  g , the amplitude of
vibration of the plate at which the sand first
loses contact is at x  3.8 mm. (ans) [1]
 themis 
3.
(b) The variation with orbital radius R of the
gravitational potential energy of the satellite is
shown.
[Gravitational Field]
Solution
A planet, with a mass M and radius Rp , has an
orbiting satellite of mass m and the orbit is R .
2.0
0
Rp
2Rp
3Rp
4Rp
R
-2.0
-4.0
-6.0
-8.0
-10.0
(i) The variation with orbital radius of the kinetic
energy of the satellite (The line should extend from
R  1.5Rp to R  4Rp ):
 Relating the ratio in (a)(ii) with the actual given
K.E. values in the graph. We get,
K.E. of satellite   12 G.P.E. of satellite
(a) In terms of M, m and R:
(i) For the satellite to be in a stable orbit, it implies
that the attractive gravitational force that
provides the centripetal acceleration is perfectly
2008 Nov
2008  18
battery
e.m.f. E
10.0
energy
/ 109 J
8.0

A ammeter
6.0
kinetic
energy of
satellite
4.0
current i
resistor R
V
2.0
0

internal
resistance r
switch
voltmeter
p.d. V
Rp
2Rp
3Rp
4Rp
R
The resistance of resistor R can be varied.
Take the potential difference across R to be denoted
as V and the power dissipated in R as P .
-2.0
-4.0
The graph thus show the variation with V of P .
-6.0
5.8
P/W
-8.0
5.6
-10.0
(ans) [2]
5.4
(ii) The radius of the orbit of the satellite is changed
from R  4Rp to R  2Rp .
5.2
Given the mass m of the satellite is 1600kg.
5.0
3.0
From K.E.  12 mv 2 ,
v
4.0
4.5
(i)
5.5
V/V
P is maximum at 5.62 W.
Its corresponding potential difference across the
resistor is 4.50 V.
2  1.25  109
 1250 m s-1
: v 
1600
The current in the circuit
 current in the resistor
When R  2Rp , K.E.  5.00  109 J
i
: v 
5.0
(a) For the maximum value of P,
2  K.E.
—
m
When R  4Rp , K.E.  1.25  109 J
2  5.00  109
 2500 m s-1
1600
5.62W
P

 1.25 A (3sf) (ans) [2]
4.50V
V
(ii) Correspondingly,
Resistance in R 
Change in orbital speed of the satellite
 2500  1250  1250 m s-1 (ans) [5]
[Current of Electricity]
Solution
4.50 V
V

1.25 A
i
 3.6 (QED) [1]
 themis 
4.
3.5
(b) Set the resistor R to 2.03 , the current in the
circuit is 1.60 A.
 When R  2.03  and i  1.60 A ,
An electrical circuit (as shown) has:


A variable resistor R connected
A battery of e.m.f. E with internal resistance r .
Advanced Physics solutions - yearly
Using the data and answers to part (a),
Let E be the e.m.f. of the battery,

E  i R  r 
 themis
2008  19
From (a),
E  1.253.6  r  — 
From (b),
E  1.60 2.03  r 
6.
—
Solution
Substituting E of  into ,
Strontium-90 is a radioactive nuclide.
1.253.6  r   1.60 2.03  r 
3.6  r   1.28 2.03  r 
3.6  r  2.5984  1.28r
1.0016  0.28r  r  3.58
Internal resistance r of the battery  3.58 (this is
a proof of maximum power theorem) (ans) [3]
 themis 
5.
[Nuclear Physics]
[Lasers and Semiconductors]
(a) A substance is said to be radioactive if it
comprised unstable nuclei that will disintegrate
into more stable configurations by the emission of
alpha-particles (helium nuclei), beta-particles
(electrons or positrons) and/or gamma radiation
(electromagnetic waves of short-wavelengths).
(ans) [2]
(b) A sample of Strontium-90 has a mass of
2.40  10 8 g. The average activity of this sample
during a period of 1 hour is found to be 1.26  105
Bq.
Solution
The electrical resistance of an intrinsic semiconductor
material is found to decrease as its temperature rises.
The explanation using the band theory of conduction,
with a diagram, is shown below.
(i) The decay constant () of a radioactive nuclide is
defined as the constant of proportionality relating
its activity to the number of undecayed nuclei.
(ans) [2]
A
— in its usual notations — 
N
: The decay constant of Strontium-90,
energy of
electrons,
E
(ii) From (i),  
conduction band

small energy gap
valence band
In semiconductors, the conduction and valence bands
are spaced closely enough together (1 eV) that, at
room temperature, a nontrivial number of electrons is
found in the conduction band. These materials have
significant conductivity that is highly temperaturesensitive.



2.40  10  103 kg 90  1.66  1027 kg
 7.84  1010 s-1 (3sf) (ans) [3]
energy bands in
semiconductors

1.26  105 Bq
8
(c) For nuclides that have relatively small decay
constants, it will have a relatively large half-life. It
is thus suitable to accurately measuring its
changes in mass and activity to determine its
decay constant; whereas if the half-life is short, its
mass and activity measurements would not be too
accurate. (ans) [1]
 themis 
At low temperatures, semiconductors behave like
insulators. The valence electrons have no adjacent
energy levels to transit into, and do not have
sufficient energy to the energy gap.
7.
At high temperatures, semiconductors behave like
conductors. The valence electrons have sufficient
energy to cross over the small energy gap to assist
in the conduction. (ans) [4]
A serious hazard for fire-fighters is the explosion of
containers of `liquefied gas' (butane) that have been
heated in a fire. When the butane suddenly burns in
an explosion, the fire spreads very rapidly in the form
of a spherical fireball of increasing radius that is at very
high temperature.
 themis 
[Thermal Physics] [Data Response]
Solution
In order to study such fireballs, a series of experiments
is carried out. Some butane of volume 12.5  103 m3
is put in a sealed container and is then heated until it
2008 Nov
2008  20
20
1.4
lg (R / m)
explodes. The variation with time t of the radius R of
the fireball is determined. The results are shown in the
diagram.
1.3

1.1
16


1.2

R/m
1.0
12

0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
8
1.7
lg (t / ms)
(i) On the diagram,
(1) At t  40 ms, R  14.6 m
4
 lg R  1.16 and lg t  1.60
0
0
10
20
30
40
50
The point corresponding to time t  40ms is
plotted at (1.60,1.16). (ans) [1]
t / ms
(a) Use the diagram,
(2) The best-fit line for all the plotted points is
drawn. (ans) [1]
(i) Without any calculation,
From t  0  2 ms, initially the rate at which
the radius of the fireball increases is very
high.

From t  2  25 ms, the rate reduces.

From t  25 ms onwards, the rate is reduced
further to an approximate constant (still
positive). (ans) [2]
(ii) In a room of length 12 m, width 5 m and height 3
m, such an explosion would be very hazardous.
From the graph, it will take less than t  25 ms for
the explosion to reach every part of the room. This
is a very short reaction time.
The space is relatively small and enclosed, the
blast and sound will add sudden pressure onto the
ears and may confuse the victim subsequently.
The debris created by the exploding gas will
behave like bullets from a shotgun thus harming
the victim quickly. (ans) [3]
(b) It is thought that, for a fixed volume of butane, the
radius R of the fireball varies with time t
according to the expression
Rn  kt m ,
where n and m are integers and k is a constant.
Some corresponding values of lg t and lg R for
the data in the diagram are plotted on the graph
as shown following.
Advanced Physics solutions - yearly
1.4
lg (R / m)

1.3


1.2


1.1
(b)(i)1.

1.0

(b)(i)2.
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
lg (t / ms)
(ii) From the best-fit line drawn in (i) part (2),
The gradient  
1.24  0.935
 0.386 (ans)
1.79  1.00
[2]
(iii) Recall,


Rn  kt m
n lg R  m lgt  lg k
 m
1
lg R    lg t    lg k
n
n
m
 
:    0.386  0.4
n
Given that n and m are integers, making m as
small as possible, implies n  5 and m  2 (ans)
[3]
(c) The experiment is repeated using similar
containers but with different volumes of butane.
The results are shown in the next diagram.
 themis
R/m
2008  21
20
volume of
container / m3
16
12.5  10-3
10.0  10-3
7.5  10-3
5.0  10-3
12
2.5  10-3
8
4
0
10
0
20
30
40
50
60
t / ms
Without drawing a further graph,
Proof:
R5  cV , where c is a constant
 5lg R  lgV  lg c

lg R  15 lgV  15 lg c — 
From the graph at t  40ms,
R / m V / 10-3 m3 R5 / 105
R5 V / 107
14.6
12.5
6.63
5.30
14.0
10.0
5.38
5.38
13.1
7.5
4.01
5.35
12.2
5.0
2.70
5.41
10.6
2.5
1.34
5.35
average
5.36
Within limits of 1.1% error, it is found that
R5
is a
V
constant; hence, the relationship
R5  cV is true. (QED) [3]
(d)
(i)
The equation in (c) may also be applied to other
exploding gases.
 One suggested physical quantity on which the
constant c will depend is its surrounding pressure.
(ans) [1]
(ii) The data were collected for butane in a container
in a room.
One other situation where the theory developed
predicts a high level of hazard for fire-fighters is
fire-fighting onboard ships where flammable gases
build up in enclosed small cavities onboard the
ship. (ans) [1]
 themis 
2008 Nov
2008  22
(ii) Within limits of errors, inverse proportion is
demonstrated by the plotted graph, since the
plot of 1 p against V is approximated to a
straight line passing through the origin, p is
shown to be inversely proportional to V . (QED)
[1]
2008 Nov Paper 3
Questions
Section A
(b) F and d are in the same direction.
Answer all questions in this section.
1.
The variation with displacement d of the force F
applied to an object is as shown below (C).
[Work, Energy and Power]
30
F /N
Solution
(a) At constant temperature, for a given fixed mass of
gas, the variation of volume V of the pressure p
is as shown (A).
20
(C)
10
p / 105 Pa
3
0
2
0
1.0
2.0
3.0
(A)
0
100
200
300
400
V / cm3
(i) Using values from (A), using the axes drawn, plot
in diagram below (B) to show that p is inversely
proportional to V .
work / J
Draw a graph showing the variation with d of the
work done in (D).
1
0
4.0
d/m
(D)
0
(B)
1.0
2.0
3.0
4.0
d/m
Work done is area under the force  displacement
graph.
0
work / J
60
0
1/p / 10-5 Pa-1
3.0
40
(D)

(B)
0

1.0

0.0
20

2.0
0

0
1.0
2.0
3.0
4.0
d/m
(ans) [4]
100
200
300
400
V / cm3
(ans) [2]
Advanced Physics solutions - yearly
 themis
2008  23
(c) Measurements are made of the Earth's
gravitational field strength g for different
distances r from the centre of the Earth. Diagram
(E) shows the variation of lg r against lg g .
lg (g / ms-2)
1.0
Hence, when the coil rotates, its flux linkages
changes with the magnetic field resulting in an
e.m.f. being generated between the ends of the
coil. (ans) [2]
0.5
0.0
6.7
6.8
6.9
7.0
7.2
7.1
7.3
-0.5
7.4
lg (r / m)
(E)
The gradient of the graph (E) 
 0.73   0.85
7.5  6.7
 1.98 (3sf) (ans) [3]
(ii) Within limits of errors, since the plot of lg g
against lg r is approximated to a straight line, lg g
is shown to have a linear relationship with lg r
From (i), the gradient  2 or lg g relates to
2lg r


g relates to
(ii) Two factors that affect the magnitude of the
maximum e.m.f.:



-1.0
(i)
(i) Faraday's law of electromagnetic induction states
that the magnitude of the induced e.m.f. in a
conductor is directly proportional to the rate at
which magnetic flux is cut by the conductor or flux
linkage is changed in a coil.
the number of turns in the coil
the field magnetic field
(the coil is rotated at a relatively faster speed)
(ans) [2]
(iii) When the coil is at its horizontal position, its
direction of motion is perpendicular to the
magnetic field. Its induced e.m.f. is therefore at its
maximum.
When the rotation of the coil reaches its vertical
position, its direction of motion is parallel to the
magnetic field. No e.m.f. would be induced.
e.m.f / V
1
r2
+Eo
The value of gradient relates g to vary
inversely with the square of the distance
1
 2  from the centre of the earth. (ans) [2]
r 
A
A
B
B
B
0
T
T/2
one revolution
[Electromagnetism]
Solution
B
A
time/ t
-Eo
 themis 
2.
A
A
B
Voltage output-time
graph
Since the motion of the coil is circular, its induced
e.m.f. follows a sinusoidal curve. (ans) [2]
(a) A simple generator has:
A coil with a large number of turns
Rotates at a constant rate in a uniform
magnetic field, as shown.
Q
C
permanent
magnet
B
D
S
(b) The output from a similar generator is connected
to the input of an ideal transformer:




primary coil  30 turns
secondary coil  600 turns
input e.m.f.  72 V r.m.s..
output is connected to a resistor of resistance
160 , as shown in diagram.
transformer
input
A
30
turns
600
turns
160 Ω


N
P
(i) The peak input e.m.f.
 Vrms  2  72  2  102 V (3sf) (ans) [1]
2008 Nov
2008  24
(ii) The r.m.s. value of the p.d. across the resistor 
Noutput
Ninput
 Vinput 
(d)
600
 72
30
light
 1440 V (3sf) (ans) [1]
unpolarised
wave
(iii) The r.m.s. value of the current in the resistor

 themis 
4.
P  Irms R   9.00  160
Solution
 13.0 kW (3sf) (ans) [1]
(a) Briefly define:
(v) The r.m.s. value of the current from the
generator
Vrms, input

(i) Every minute particle (such as ions, electrons,
atoms or molecules) in a body has random
potential energy  Ep  , due to their state and
12960
72
position, and, random kinetic energy  Ek  , due to
 180 A (3sf) (ans) [2]
 themis 
3.
[2008N P3] [Thermal Physics]
2
2
P
planepolarized
Polarization, also called wave polarization, is an
expression of the orientation of the lines of
electric flux in an electromagnetic field. (ans) [1]
Vrms
1440

 9.00 A (3sf) (ans) [1]
R
160
(iv) The mean power dissipated in the resistor,

Polaroid
lens
[Superposition]
their motion. Collectively, the sum of these
randomly distributed energies is termed the
internal energy U  of the body, i.e.,
U  Ep  Ek . (ans) [2]
Solution
Explanations of the meaning of each of the following
terms as applied to waves.
(ii) The first law of thermodynamics is defined as the
heat  q  absorbed by a system either raises the
internal energy  U  of the system and/or does
work by the system on the environment  wby  ,
(a)
i.e., qin  U  wby . —  (ans) [1]
A standing wave, also known as a stationary
wave, as opposed to progressive wave, is a wave
that remains in a constant position, i.e., their
waveforms do not move. (ans) [2]
(b) Diffraction is the bending and
spreading of waves when they
meet an obstruction or gap
into regions where a shadow
might be expected. (ans) [2]
(c) Sources are said to be coherent, if the waves
leaving them bear the same phase relationship to
each other at all times. The same phase
relationship implies the same frequency, and
therefore the same wavelength. (ans) [2]
(b) An ideal gas undergoes a cycle of changes
A  B  C  A , as shown in the diagrams
below.
pA
cylinder
area A
x
pressure
p / 105 Pa
B
7
1
A
C
5
Advanced Physics solutions - yearly
frictionless
piston
gas
20
volume
V / cm3
 themis
2008  25
(i) The work done by a gas during the change C  A
expanding against an external pressure p is given
by area under p  V graph or pave  V


wby  1  105 Pa  20 - 5  106 m3 
work done
on gas / J
A B
zero
4.2
BC
8.5
(a)
(i) The electric field strength, E , or electric field
intensity or electric field is defined by the ratio of
the electric force on a charge at a point in free
space to the magnitude of the charge placed
there, i.e.,
increase in
internal
energy / J
electric field strength, E 
CA
section of
cycle
[Electric Fields]
Solution
(ii) The figure below is a table of energy changes
during one cycle.
heating
supplied to
gas / J
Answer two questions in this section.
5.
 1.5 J (2sf) (ans) [2]
section of
cycle
Section B
electric force
charge
(ans) [1]
heating
supplied to
gas / J
work done
on gas / J
increase in
internal
energy / J
A B
zero
4.2
4.2
BC
8.5
zero
8.5
CA
5.8
1.5
4.3
Rearranging ,
(ii) The diagram below shows a uniform electric field
of electric field strength E where a charge q is
placed at point.
d
q
uniform field
E
U  qin  won — 
Process A  B :
Y
X
: U  0  4.2  U  4.2 (ans)
The charge at X is moved to Y through a distance
d.
Process B  C :
Using the definition in (i), Since the charge is
positive, positive work done has to be
won  p 0.0   0.0 (ans)
W.D.  FE  d  qE  d (ans) [1]
: U  8.5  0  U  8.5 (ans)
(iii) The potential difference between X and Y is V.
Process C  A :
Using the answer from (ii),
Ucycle  UAB  UBC  UCA  0

The electric potential (V) at a point in free space in
an electric field is defined as the work done in
bringing a unit positive charge from infinity to the
point, i.e.,
0   4.2   8.5  UCA
 UCA  4.3 (ans)
: 4.3  qin  1.5  qin  5.8 (ans) [4]
 themis 
electric potential, V 
: VXY 
work done
—
charge
work doneXY
qEd
V 
charge
q
 Ed (ans) [2]
(b) An X-ray tube has:


a vacuum
anode and cathode
2008 Nov
2008  26


electrons are accelerated from rest through a
potential difference of 60 kV between the
cathode and the anode.
The current in the tube is 8.6 mA.
 5.38  1016 (n) (3sf) (ans) [2]
(ii) Work done on the electron  qV
This work done is used to accelerate the electron
to a high speed – converted to pure kinetic energy,
qV  12 mv 2 — in its usual notations
 the speed of electrons arriving at the anode,
v
2  1.6  10  60  10
2qV

m
9.11  1031
3
(i) By referring to the above diagram, If the
conducting sphere is replaced by a point charge of
the same value, beyond the boundaries of the
original conducting sphere, it will produce exactly
the same field pattern and strength as that of the
conducting sphere at points A, B and C. Hence, to
points A, B and C, it appears as if the charge is
concentrated at the centre of the sphere. (ans)
[1]
(ii) Electric field strength at the surface of the
sphere,
E
 1.45  108 m s-1 (3sf) (ans) [4]
 Power supplied by the electrons hitting the
anode
 nqV
 5.38  1016  1.6  1019  60  103
 516 W (3sf) (ans) [2]
(c) X-ray production:


uses a negligible fraction of the power
reaching the anode in (b)
has to be cooled by passing a coolant through
the anode.
The coolant has specific heat capacity of 3500
J kg-1 K-1 and the temperature rise is 30 K.
 The heat that needed to be removed  (iii)
Rate at which the coolant must be pumped
work done
through the anode 
c
516
 4.91  103 kg s-1 (ans) [3]

3500  30
1 Q
— in its usual notations
4 o r 2

(iii) Work done by the electrons per second

field of
isolated
point charge
A
B
8.6  10-3 A
current

chargeelectron
1.6  10 19 C
19
0.060 C
C
(i) The number of electrons passing through the
tube in one second

The electric field around the sphere is as shown in
the diagram below.
1 0.060  106
4 o
0.102
 5.39  104 N C-1 (ans) [2]
(iii) Point A and B are 0.40 m and 0.50 m from the
centre of the sphere respectively.
Potential difference (V) for AB  V for BC.
 VAB 
Q 1 1
  
4 o  rB rA 
Similarly, VBC 
— in its usual notations
Q 1 1
  .
4 o  rC rB 
Since VAB  VBC , rearranging
1 1 1 1
    
 rB rA   rC rB 
1  1
1 
 1
 


 
0.50
0.40
r
0.50

  C

Distance from the centre of the sphere to C,
rC  0.67 m (ans)
 themis 
(d) A conducting sphere has:


radius  0.10 m
charge  0.060 C
Advanced Physics solutions - yearly
 themis
2008  27
6.
[Oscillations]
At its equilibrium position,
kx  mg
Solution
(a) Frequency  f  is defined as the number of
complete oscillations produced per unit time, is
distinct from angular frequency   which is
defined as the magnitude of the vector quantity
angular velocity, that is defined as the rate of
change of angular displacement with respect to
time. (ans) [2]
(b) A spring has:




mg
0.400  9.81
 19.6 N m-1

e
0.200
At its lowest point of its movement, the resultant
force on the load
 T ' W  k  x  e   mg
 19.6  0.400  0.400  9.81
 3.92 N (ans) [2]
(ii) Angular frequency of the oscillation,
An unstretched length  0.650 m
Attached to a fixed point
A mass  0.400 kg is attached to the spring
and gently lowered until equilibrium is
reached.
The spring has then stretched elastically by a
distance of 0.200 m.
For the stretching of the spring,
(i)
 k

 7.00 rad s-1 (ans) [2]
(iii) Maximum speed of the mass,
v   x02  x2  7.00 0.2002
 1.40 m s-1 (ans) [1]
The loss in gravitational potential energy of the
mass
 mgh   0.400 9.81 0.200
(e) Complete the table below which is a list of
energies of the simple harmonic motion.
 0.785 J (ans) [1]
gravitational
potential
energy / J
elastic
potential
energy / J
kinetic
energy / J
total
energy / J
gravitational
potential
energy / J
elastic
potential
energy / J
kinetic
energy / J
total
energy / J
lowest point
0
1.57
0
1.57
equilibrium
position
0.785
0.392
0.392
1.57
highest point
1.57
0
0
1.57
(ii) The elastic potential energy gained by the spring
 12 Fx  12  0.400 9.81 0.200
 0.392 J (ans) [2]
k
19.6

m
0.400
lowest point
0
equilibrium
position
highest point
(c) The two answers to (b) are different because the
springmass system is a rather inefficient
mechanical energy storage system. It only manages
to store half of the energy supplied, while the rest
are lost through heat, sound and other forms of
energy. (ans) [2]
(d) The load on the spring is now set into simple
harmonic motion of amplitude 0.200 m.
(i)
simple
mass-spring
system
spring
By observation,
(ans) [5]
T
e
equilibrium
position
T’
e+x
W
W
2008 Nov
2008  28
(f) Based on the axes below, sketch four graphs to
show the shape of the variation with position of
the four energies. Label each graph.

Most of the incident alpha particles pass
straight through the foil, to hit the fluorescent
screen.

There are few particles, however, which suffer
deviations in the forward and backward
directions.

A very few particles even retrace their original
path, backwards.
energy
Conclusions:
lowest point
equilibrium position

The observation can be explained by proposing
that the atom is made up of a very small,
positively charged nucleus surrounded by a
cloud of electrons. The atom is mainly empty
space, so that most alpha particles pass
through the foil with practically no deviation.

If the alpha particles, however, come too close
to the nucleus, the strong Coulomb repulsion
between the nucleus and the positively
charged alpha particle will cause the alpha
particle to deviate from its original direction.
The repulsive force indicates that the nucleus
must be of the same nature as the alphaparticle.

The retracing of a few alpha-particles resolved
that the atomic nucleus must be real, massive,
hard and physical. (ans) [4]
highest point
By conservation of energies:
energy
TE
EPE
GPE
KE
–xo
0
xo
x
(ans) [3]
 themis 
7.
[Nuclear Physics]
Solution
(a) In the past, experiments were performed to
provide evidence for a small charged nucleus in
the atom.
Rutherford and his students conducted an
experiment on alpha particles scattering. Alphaparticles from a radium source are made to
impinge on a very thin sheet of gold foil,
approximately 10-6 m thick.
(b) Uranium-235 nucleus:


at rest, and
absorbs a slow neutron and undergoes fission.

Sometimes the fission don’t emit any
neutrons.
One such fission:
235
1
92U  0 n
235.0439 u
zirconium
235
92U
x
52Te
98
y Zr
neutron
1
0n
1.0087 u.
uranium
tellurium
fluorescent
screen
(i) Nucleons:

gold atom
x
98
52Te  y Zr
The masses of these particles are
Observations:
 -particles

137.9603 u
97.9197 u
235  1  x  98
x  138 (ans)
92  0  52  y
Protons:
 y  40 (ans) [2]
radium
source
gold foil
Advanced Physics solutions - yearly
alpha nuclei
 themis
2008  29
 Speed of the zirconium nucleus,
(ii) Assume all particles are initially at rest.
Energy released in the fission  mass excess
equivalent
 (235.0439 u  1.0087 u  137.9603 u 
97.9197 u)  c 2  (0.1777 u)  c 2

 0.1777  1.66  1027 3.00  108

2
 2.65  1011 J (ans) [4]
vZr  1.29  107 m s-1 (ans) [3]
(vi) Two assumptions made in the calculation in (iv)
part 1.:
1.
The fission takes place in isolation.
2.
Newtonian laws of motion are obeyed at
atomic level. (ans) [2]
 themis 
(iii) Part of the energy released, 2.3  1011 J were
used to become kinetic energy of the tellurium
and zirconium nuclei.
 The remaining energy released may be in a
form of photon(s). (ans) [1]
(iv)
1. Initial momentum is zero, hence final momentum
must also be zero

mZrvZr  mTevTe (oppositely directed)
 the ratio

speed of zirconium nucleus
speed of tellurium nucleus
m
137.9603 u
vZr
 Te 
v Te
mZr
97.9197 u
 1.41 (3sf) (ans) [2]
2.
The ratio
kinetic energy of zirconium nucleus
kinetic energy of tellurium nucleus
1m v 2
 m  v 
 K.E.Zr 
2 Zr Zr
  Zr  Zr 

 1
2
m v
 K.E.Te 
 mTe  v Te 
2 Te Te
1
m  v 
  Te   Zr 
 mZr   v Te 
2
2
v 
  Zr   1.41 (ans) —  [2]
 v Te 
(v) K.E.Zr  K.E.Te  2.3 1011 — 
Combining  and :
K.E.Zr 
K.E.Zr
 2.3  1011 
1.41
K.E.Zr  1.346 1011 
1m v 2
2 Zr Zr
1
2
 1.346  1011 
97.9197 1.66 10 v
27
2
Zr
 1.346  1011
2008 Nov
2008  30
Notes:
Advanced Physics solutions - yearly
 themis