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04 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
04’ AL Physics: Essay
Marking Scheme
1.
(a) (i)
By subtracting vA from vB,
vB - vA = vB + (-vA) = v
Since vA is perpendicular to OA and vB is perpendicular to OB, if t is small,  will also be small,
so v = XZ = v 
Arc AB = vt = r, for  being small
v2
t , direction of v towards the centre
r
v2
Thus the acceleration is given by
, its direction is perpendicular to vA and towards the center i.e.
r
centripetal.

v = XZ =
(ii) Suppose the string makes an angle  with the vertical when the bob's angular speed is  and the tension in the
string is T (or using a diagram).
and

T sin  = m2 L sin  (or mr2)
T cos  = mg
mg
= m2L
cos 
g
2 =
L cos 
2
L cos 
Period =
= 2
g

(iii)
There is frictional force F at the ground to provide the centripetal force (mv2/r) towards the centre. The force
F has a moment about the centre of gravity G which tends to turn the rider outwards.
When the rider leans inwards as shown, this is counterbalanced by the moment due to the normal reaction R
from the ground.
04 AL Physics/Essay Marking Scheme/P.2
(b) (i)
RE
E
RE
earth
Spring tension T indicates the apparent weight
At the poles, there is no rotation.
 mg0 – T = 0
At the equator, there is a resultant force m2E RE ,
2
E =
rads-1 and RE is the radius of the earth
24  60  60
 mg0 – T ’ = m2E RE
T ’ = mg0  m2E RE < T
which is smaller than mg0 at the poles by m2E RE
(ii) Simple pendulum: T = 2
l
where l is the length of the pendulum
g
m
where m is the mass and k is the force constant of the spring
k
The period of the simple pendulum will decrease as the apparent weight and hence the apparent acceleration
due to gravity at the poles is greater than that at the equator.
The period of the mass-spring system will remain unchanged as m and k are constants (independent of g)
Mass-spring system: T = 2
2.
(a) visible spectrum: approx. 4  10-7 m to 7  10-7 m
A filament lamp (continuous spectrum light source) is used as the source and a lens is used for producing a pure
spectrum. The various wavelengths of electromagnetic radiation suffered different degree of deviation after
passing through the glass prism.
A phototransistor, connected to a milliammeter and a battery, used as an IR detector is moved slowly through the
spectrum produced, the milliammeter gives a reading beyond the visible red, region, i.e, in the infra-red.
This indicates the presence of infra-red radiation.
A photoelectric cell, connected to a galvanometer and a battery, used as a UV detector is moved slowly through
the spectrum produced, the galvanometer gives a reading beyond the visible violet region, i.e. in the ultra-violet.
This indicates the presence of ultra-violet radiation.
or allowing the light from an overrun filament lamp to fall on strips of fluorescent paper and non- fluorescent
white paper as shown.
04 AL Physics/Essay Marking Scheme/P.3
The fluorescent paper also glows outside the violet (shown on the white paper) of the visible spectrum. This
indicates the presence of ultra-violet radiation.
(b) With the presence of an obstacle, light spreads round it into region that would be in shadow if the light travelled in
straight lines. Behind the obstacles or apertures at which diffraction occurs, a diffraction pattern of dark and
bright fringes can be observed.
When the aperture is wide (compared with the wavelength of light), incident waves emerge from aperture almost
unchanged with little diffraction (a fairly sharp shadow of the opening is obtained). If the size of the obstacle or
aperture becomes smaller compared to the wavelength of light used, the central maximum of the diffiacted light
spreads more to the geometrical shadow.
Since the diffraction of light can be minimized using shorter wavelengths, blue light is preferred when
photographing tiny objects through microscope as its wavelength is shorter than that of red light.
(c) (i) Diffraction occurs at the narrow slits that act as sources of secondary wavelets which superpose beyond the
slits.
Consider wavelets coming from corresponding points A and B on two successive slits and traveling at an angle
 to the direction of the incident beam. The path difference AC between the wavelets is dsin, as it is for all
pairs of wavelets from other corresponding points in these two slits and in all pairs of slits in the grating.
Hence if d sin  = n, where n is an integer giving the order of the spectrum then reinforcement of the
diffracted wavelets occurs in direction .
(ii) Advantages: sharper
large angular separation within the same order.
(Accept any other reasonable answers.)
3.
(a) An insulated conductor has a certain capacitance (i.e. charge-storing ability) in a way that when it is given a charge
Q and consequently acquires a potential V, the capacitance of the conductor is the charge required to cause unit
Q
change in its potential (i.e. C  ).
V
Q
(b) (i) V =
for r  R
4 0 r
V
Q
4 0 R
0
R
r
04 AL Physics/Essay Marking Scheme/P.4
The potential inside the charged sphere is uniform (E = 0 inside a conductor in a steady state) and equals that
Q
at its surface (
). The potential of the charged sphere is inversely proportional to the distance r from its
4 0 R
centre for all points outside the sphere (r > R).
Q
Q
at the sphere’s surface and C =
4 0 R
V
 C = 40R
(ii) V =
When the charged metal sphere is earthed/connected to the earth, charges flow from the metal sphere to the
earth until their potentials are the same.
Since the earth has such a large capacitance, the metal sphere loses most of its charges to the earth but the
change in the earth's potential due to the gain (or loss) of charges is negligible.
The earth thus provides a practical zero of potential.
(c) (i)
p.d.
V
Q
0
Q
charge
Consider a capacitor of capacitance C and when the p.d. across it is V it stores charges Q.
If the capacitor starts to discharge such that a very small charge Q passes from the negative to the positive
plate, then the resulting energy loss is VQ, assuming that the decrease in V is negligible as Q is small.
From the graph,
energy loss = area of shaded strip
If the capacitor discharges completely,
Energy stored = total energy loss
= area of all strips or triangle below graph
1
= QV
(Q = CV)
2
1
= CV 2
2
04 AL Physics/Essay Marking Scheme/P.5
With a piece of metal is inserted in between the plates, the p.d. between the plates decreases.
1
The energy stored QV also decreases.
2
p.d. before
p.d. after
+Q
-Q
metal
(ii) - smoothing/rectification
- tuning
- blocking d.c.
(Accept any other reasonable answers.)
wire
4.
(a) For a wire carrying current I lies at right angles to
a magnetic field B (or using a diagram), the
magnetic force F is directly proportional to I and
the length l of the wire in the field.
The magnetic flux density B is given by the force
F
acting per unit current length, i.e. B =
.
Il
F is at right angles to both I and B.
I
magnetic field B
(into paper)
F
normal to area A
(b) (i) The magnetic flux  through a small plane
surface of area A is given by
A

B
 = BA cos 
(ii) Consider a magnet approaching a coil of N turns such that its N-pole faces the plane of the coil (or using
diagram), the magnetic flux  through the coil increases and an e.m.f.  would be induced in the coil ( =
d
d
) which opposes the change ( N
) producing it.
N
dt
dt
04 AL Physics/Essay Marking Scheme/P.6
(iii)
When the bar magnet is moved towards the coil, the total flux linkage through the coil increases, an induced
current should flow in a direction such that the coil behaves like a magnet with a N-pole towards the bar
magnet. The motion of the magnet towards the left will be opposed.
Therefore, work has to be done by an external agent in moving the magnet against the opposing force. This
work done then becomes the electrical energy in the circuit.
(c) When a coil (armature coil in a motor) rotates and cuts the magnetic flux of the field magnet, an e.m.f.  called the
back e.m.f. is induced in it which opposes the applied p.d. V causing current I in the coil.
(i) The armature coil resistance r of a d.c. motor is small (< 1 ). When the motor is switched on, the armature
V
is at rest ( = 0) and the back e.m.f.  is zero. The armature current I =
is very large at first and it
r
gradually decreases to its working value as the motor speeds up and the back e.m.f.  increases due to the
increasing rate of change of flux through the coil. The back e.m.f.  (slightly less than the applied p.d. V) then
limits the current V -  = Ir.
.
(ii) When a mechanical load is to be raised by a running motor, the motor's speed decreases as well as the back
e.m.f. in its coil will decrease as the rate of change of flux through the coil decreases, the current drawn from
V 
the supply will then increase since I =
.
r
5.
(a) (i) The pressure exerted by a gas on its container is due to the force exerted by molecules rebounding from the
container's surface, and it is measured by the average rate of change of momentum of the molecules per unit
surface area.
(ii) Internal energy of a gas consists of two components:
- kinetic energy due to the random (translational, rotational and vibrational) motion of the molecules, which
depend on the temperature
- potential energy due to the intermolecular forces, which depends on intermolecular separation
(b) (i) For ideal gases,
- the intermolecular forces are negligible except during a collision (or the internal energy only consists of k.e.
because there is no intermolecular forces and thus no p.e.)
- the volume of the molecules themselves can be neglected compared with the volume occupied by the gas
(Accept any other reasonable answers)
(ii)
No. of molecules N in each
unit range of speed
higher
temperature
molecular
speed v
04 AL Physics/Essay Marking Scheme/P.7
When temperature increases, the curve spreads toward the high speed end of the distribution. Therefore the
proportion of gas molecules with speeds higher than a certain value would increase.
(iii) From the kinetic theory equation and the ideal gas equation of state, for a mole of an ideal gas
1
pV = N A mc 2 and pV = RT
3
where NA = Avogadro constant and R = universal gas constant
1 2 3 R
we have
T
mc =
2
2 NA
since
c2 =
cH2
cO2
=
3RT
NA
mO
=
mH
1
m
16
=4
1
(c) (i) Some forms of energy, say, electrical energy are more suitable for doing work or changing into other forms
compared to internal energy, which is not easily transferred to anything else. Eventually, all energy changes
result in the surroundings being heated, say, as a result of work done against friction. This process is called
'degradation of energy' and that is why there is a need ( crisis) for new sources of such energy (as chemical
energy from fossil fuels is limited).
(ii) Nuclear energy - large initial capital required (dis)
- (relatively) low operation cost (adv)
- potential environmental hazards (dis)
- no 100% safe way to handle the nuclear waste (dis)
- practically unlimited supply (adv)
ONE advantage &
ONE disadvantage
Hydroelectric power - large initial capital required (dis)
- (relatively) low operation cost (adv)
- only economical for large scale production (dis)
- may cause some environmental drawbacks (dis)
- relatively clean compared to coal-fired power plant (adv)
(Accept any other reasonable answers.)
ONE advantage &
ONE disadvantage