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00 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
00’ AL Physics: Essay
Marking Scheme
1.
(a) (i) Momentum of the billiard ball is not conserved.
Its momentum perpendicular to the cushion is altered due to the normal reaction.
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(ii) Momentum of the rocket is not conserved as it is subjected to
external forces such as thrust due to ejecting gases, gravity, air resistance.
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(iii) When an -particle is emitted, the daughter nucleus recoils in opposite direction and
therefore its momentum is not conserved.
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(b) (i) In a perfectly elastic collision the kinetic energy of the system is conserved.
Let the velocity of m1 after collision be v1
By the conversation of linear momentum
m1u1 = m1v1 + m2v2 or m1(u1 - v1) = m2v2
…. 
As the collision is perfectly elastic,
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m1u12 = m1v12  m2 v22
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2
2
2
2
2
i.e.
m1 (u1  v1 ) = m2 v2 …. 

1
½
½
½
m12
(u1  v1 ) 2
m2
m2(u1 + v1) = m1(u1 - v1)
 m  m2 
u1
v1 =  1
 m1  m 2 
m1 (u12  v12 ) =
3
(ii) (I) m1 >> m2, therefore v1 ~ u1
For example the velocity of a bowling ball is hardly affected by collision with an
inflated beach ball of the same size.
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(II) m1 = m2, therefore v1 = 0
For example a billiard ball stops when it collides head-on with another stationary
billiard ball.
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(III) m1 << m2, therefore v1 ~ -u1
For example a ball drops vertically onto the ground (collision with the earth) and
rebounds with a reversed velocity and will reach the same height.
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To slow down the fast neutrons effectively, the stationary targets for collision should be of
nearly the same mass of neutrons. Therefore materials with numerous hydrogen centers
(such as water) are preferred.
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(c) (i) Use a spring-loaded gun (i.e. a compressed spring).
Use cotton wool or plasticine inside the hole to ensure the collision to be inelastic (i.e.
sticky).
(ii) The horizontal momentum of the system is conserved,
therefore mv = (m + M)V where V is the common velocity just after impact.
After the collision, the only forces acting on the system (block + bullet) are the weight and
the string tensions (which do not work), therefore the mechanical energy is conserved.
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00 AL Physics/Essay Marking Scheme/P.2
From energy conservation, ½(m + M)V2 = (m + M)gh.
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As h = L(1 - cos ), by eliminating V, we have
½
v= (
2.
mM
) 2 gL(1  cos ) .
m
During the impact the supporting strings may deviate a bit from vertical,
therefore some horizontal external force would act on the system
resulting in the experimental value lower than the theoretical value.
(or L should be measured to the center of gravity of the block instead of to the top of it)
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(a) (i) All points on a wavefront can be considered as point sources for the production of
spherical/circular secondary wavelets.
After a time t the new position of the wavefront will be the surface of tangency to these
secondary wavelets.
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(ii)
1½
Let v1 and v2 (v1 > v2) be the speeds of light in air (medium ) and in the denser medium
(medium ) respectively.
Let 1 and 2 be the wavelengths of light in air and in the denser medium respectively.
Consider the time t during which a Huygens wavelet from point B moves to include
point B’.
BB’ = v1t
½
Light from point A, traveling through the denser medium at a reduced speed will move
a shorter distance to A’ within this time.
AA’ = v2t
½
From the diagram,
v
sin 1
BB '
BB' / AB'
=
= 1 =
v2
sin  2
AA'
AA' / AB'
By Snell’s law of refraction n =
sin 1

v
(= 1 ) = 1
sin  2
2
v2
1
½
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00 AL Physics/Essay Marking Scheme/P.3
(b) (i)
screen
Filament
lamp
S1
S2
Light source should be strong (or black out the laboratory as much as
possible) and properly shielded so that no stray light falls on the screen.
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A monochromatic light source can be used in order to obtain sharper fringes.
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Both slits S1 and S2 should be as narrow as possible so that the light emerging
from the slits undergoes significant diffraction.
The slits should be separated by a very small distance (~ 0.5 mm) so that the
light from the two slits overlap somewhere in front of the screen.
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ANY
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@1
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The screen should be placed at an appreciable distance (1 ~ 2 m) from the slits
so that the separation of fringes is observable while the intensity is not too slow.
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Make sure that the filament is parallel to the two slits S1 and S2. (or if a source
slit is used it should be parallel to the two slits S1 and S2.)
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(ii) The light from two slits is coherent while that from the two light bulbs is not and a
stable/observable interference pattern cannot be formed.
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(c) (i) Photoelectric effect involves the emission of electrons from metal surfaces when
electromagnetic radiation of high enough frequency falls on them.
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(ii) The maximum kinetic energy of the photoelectrons emitted does not increase
with the intensity of the incident radiation.
As the range of emission speeds (or kinetic energy) from zero to a maximum
is due to electrons having a range of possible kinetic energies inside the metal.
Those with the highest kinetic energy are emitted with maximum speeds.
According to the wave theory, we would expect a certain number of photoelectrons
to be ejected with greater speeds when the radiation intensity increases.
No time lag between the time when the metal surface being irradiated and the start
of emission of photoelectrons is observed, even when the radiation is weak.
According to the wave theory, radiation energy is spread over the wavefront and
since the amount incident on any one electron would be extremely small, some time
would elapse before an electron gathered enough energy to escape.
There exists a characteristic cutoff frequency 0 for a particular surface. For
frequencies less than this, the photoelectric effect disappears, no matter how
intense the illumination.
According to the wave theory, the photoelectric effect should occur for any
frequency of light, provided only that the light is intense enough.
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TWO
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00 AL Physics/Essay Marking Scheme/P.4
3.
(a) (i) Input to the primary is sinusoidal a.c. with small resistance in the primary
winding carrying small primary current.
The primary and the secondary windings must be wound on the same iron
core so that the flux leakage is negligible.
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The secondary is not on load (i.e. open circuit) or if on load the load
resistance is high so that the secondary current drawn is small.
(ii) Iron core is laminated to reduce eddy current loss.
Copper windings are thick to reduce joule heating loss.
(or soft iron is used for reducing hysteresis loss.)
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(iii) Secondary output voltage is reduced slightly and the secondary current becomes larger. ½ + ½
Back e.m.f. in primary would decrease and the primary current would increase.
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(b)
1½
Set up the current balance and place one pair of magnadur magnets around the current-carrying
arm.
With one rider placed on the arm, adjust the current by shifting the rheostat to restore the
½
balance.
With the current remains unchanged, place another pair of magnadur magnets next to the first
½
one.
Equilibrium can be restored by placing another rider on the arm.
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This shows that the magnetic force (i.e. no. of riders) is directly proportional to the length of
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current-carrying conductor in the magnetic field.
- To avoid overheating, the current should be switched off as soon as observations
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have been made and measurements taken.
- Make sure the direction of the magnetic field is perpendicular to the current-carrying
½
arm.
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- Shield the set-up from the disturbance of wind.
TWO ½
- Minimize the effect of the earth’s magnetic field by aligning the current-carrying arm
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along the N-S direction.
- The set-up should be far from any current-carrying conductors so as to avoid the
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effect of stray magnetic fields.
(c) (i) (I) The field is produced in the narrow air gap between the concave pole pieces of a
permanent magnet and a fixed soft iron cylinder and is radial, therefore it is always
parallel of the plane of the coil.
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00 AL Physics/Essay Marking Scheme/P.5
(II) When a current flows in the coil, the magnetic forces acting on the vertical sides
therefore form a deflecting couple which causes the coil to rotate.
Equilibrium is reached when the torque due to the deflecting couple is balanced by
that of the hair spring which is twisted a certain angle.
4.
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(ii) No. Air resistance is zero as soon as the pointer stops at the equilibrium position.
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(iii) When the a.c. mains is fed into a d.c. moving-coil meter, no deflection would be observed
as the moment of inertia of the coil and pointer is so large that the system was unable to
respond to the high frequency of the signal (50 Hz).
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(a) (i) Capacitance of 470 F means the capacitor holds 470 C of charge for every 1 V of
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voltage across it.
The maximum voltage across the capacitor should not exceed 50 V, otherwise (leakage or) ½
breakdown may occur.
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(ii) When the current in the coil decreases, a back e.m.f. would be induced in the coil in the
same direction as the current so as the oppose the ‘change’ according to Lenz law.
This phenomenon is called self induction.
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(b) (i)
energy
Electric (in capacitor)
2
Magnetic (in inductor)
0
T/2
T
time
(ii) Both involve two kinds of energy which are transferred between the two components
within the system, i.e. potential energy of the compressed or stretched spring and the
kinetic energy of the mass versus the electric energy in the capacitor and the magnetic
energy in the inductor.
Mechanical
spring
1 2
kx
2
mass
1 2
mv
2
dx
v
dt
2
d x k
k
( 2 
x)

m
m
dt
Electromagnetic
capacitor
1 Q2
2 C
inductor
1 2
Li
2
dQ
i
dt
d 2Q  1
1
( 2 
Q)

LC
LC
dt
2
1
1
1
1
00 AL Physics/Essay Marking Scheme/P.6
Certain electromagnetic quantities “correspond” to certain mechanical ones
Q  x
i  v
1
C 
k
L  m
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(iii) There is inevitable resistance somewhere in the circuit such as resistance in connecting
½
wires, contact resistance etc. (or Energy loss through the emission of EM waves due to the
oscillation of electrons.)
The current oscillates with frequency close to f0 (with small resistance) but with
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decreasing amplitude.
Current
½
Time
2
(c) Resistance of resistor is independent of the applied frequency.
½
Impedance of inductor is directly proportional to the applied frequency.
½
Impedance of capacitor is inversely proportional to the applied frequency.
½
Any ONE circuit:-
multifrequency
signal
Low f
Low f
Low f
When frequency is low, VC is large while VL is small and we have VC >> VR >> VL.
5.
(a) (i) (I) An appropriate example given.
Energy transfer can be achieved through work or heat. When work (W) is done on
the object, say, by rubbing a metal bar with a cloth, or heat (Q) is transferred to
the object, say, by heating the metal bar with a burner, its temperature would
increase as a result of internal energy increase (U > 0 and k.e.  T).
Hence the first law of thermodynamics U = Q + W, which relates the change in
internal energy and the total energy transfer, is consistent with the law of
conservation of energy.
(II) An external force is required to pull a metal frame from a region with a magnetic
field perpendicular to it. The is due to the induced current in the frame which
produces a magnetic force to ‘oppose’ the change as suggested by Lenz Law.
Therefore the work done (W) by the external force eventually becomes the joule
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00 AL Physics/Essay Marking Scheme/P.7
heating (i2R) within the frame, and this is consistent with the law of conservation
of energy.
(ii) Any appropriate example.
(b) (i) - Radioactive decay is a spontaneous process while nuclear fission is not.
- Radioactive decay of a particular kind of atoms only has one definite
way to decay while nuclear fission can have various modes and
may give different products.
- Radioactive decay involves a characteristic half-life while nuclear fission
can be initiated whenever it is desirable.
- Radioactive decay is not affected by any change in the environment while
nuclear fission can be controlled by physical means such as the number of
neutrons for initiating fission.
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THREE
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(ii)
1½
The binding energies per nucleon of the heavy nuclei are smaller than those in the middle
part of the graph. Therefore when the heavy nuclei are split into two daughter nuclei of
smaller mass numbers and with larger binding energies per nucleon (i.e. more stable), a
large amount of energy would be released.
½
(c) (i) The fission of a uranium-235 nucleus produces two or more neutrons which cause further
fission of uranium-235 nuclei and the reaction multiplies and goes on as a chain reaction.
Natural uranium consists of very little uranium-235 (U-235 : U-238 = 1 : 140) by weight
which is not enough for sustaining a chain reaction.
This is because only uranium-235 nuclei undergo fission by capturing a slow neutron
while uranium-238 nuclei capture neutrons without fission.
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(ii) Control rods are used for controlling the reaction rate by absorbing the neutrons
produced in fission such that chain reaction is steady and controlled so that on average
only one neutron from each fission produces another fission.
Moderator is used for very quickly slowing down the fast neutrons produced in fission so
that they can cause fission of uranium-235 instead of being captured by uranium-238.
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