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01 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
01’ AL Physics: Essay
Marking Scheme
1.
(a) (i) The inertia of a body is its reluctance to change the velocity.
A force is required to change the velocity of the body.
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(b) (i) A body (e.g. a satellite revolving round the earth) moving with uniform circular motion.
Although the speed is unchanged, the body continuously changes its direction (velocity).
The momentum changes. This is brought about by the (centripetal) force.
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½
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(ii) Set up a friction-compensated runway.
To investigate the relation between force and acceleration, a trolley is pulled by one, two
and three identical elastic strings which are stretched by the same amount.
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½
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(ii) A trolley moves (with uniform velocity) on a friction-compensated runway. No net force is
acting on the trolley, however it is moving but is not at rest. (Accept any example that is
correct in principle, including motion under terminal velocity or motion on a frictionless
surface)
The corresponding accelerations are recorded and a graph of the force (number of elastic
strings) is plotted against the acceleration, which shows a straight line passing through the
origin (linear relationship).
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½
To investigate the relation between mass and acceleration, use one elastic string to pull one,
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two and three trolleys. The corresponding accelerations are recorded and a graph of
mass
is plotted against the acceleration, which shows a straight line passing through the origin
(linear relationship).
force
Thus, acceleration 
.
mass
For a body of mass 1 kg and moves with acceleration 1 ms-2, the force acting on it is 1 N.
½
(c) Consider the head-on collision between two billiard balls A and B moving with velocities u1 and u2
respectively (u1 > u2) in the same direction.
u1
FA
½
½
½
1
½
u2
FB
A
B
½
Let FA and FB be the average forces acting on A and B respectively during collision and t be the
time during which the two balls are in contact.
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By Newton’s second law, the impulse FAt will change the momentum of ball A, i.e.
½
FAt = m1v1 - m1u1
½
Similarly, the momentum of ball B will change by FBt,
FBt = m2v2 - m2u2
½
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01 AL Physics/Essay Marking Scheme/P.2
By Newton’s third law, FA = - FB (equal in magnitude but are opposite in direction), therefore
m1v1 - m1u1 = -(m2v2 - m2u2)
m1u1 + m2u2 = m1v1 + m2v2
2.
½+½
½
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(a) Mechanical waves are vibrations of particles, which require a medium for propagation.
Electromagnetic waves are oscillating electric and magnetic fields, which do not necessarily
require a medium for propagation.
The speed of mechanical waves depends on the elasticity and the density of the medium for
propagation.
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(b) (i) Set up the apparatus as shown.
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The filament lamp acts as a single slit, and the diffracted light beams from the two narrow
slits overlap in the region beyond the slits (accept idea presented using diagram).
Use a filter to obtain monochromatic light so that alternate bright and dark, equally spaced
interference fringes are observed (at the cross-wires of the traveling eyepiece). This shows
the wave nature of light.
The average fringes spacing y is found by measuring across as many fringes as possible with
the traveling eyepiece.
A metre rule is used to measure the separation d between the double slit and the traveling
eyepiece. The slit separation a is measured directly with a traveling microscope.
The wavelength  of the monochromatic light is approximated by  =
½
½
1
½
1+½
ay
d
(ii) Light waves can propagate through vacuum to the earth, which is a property of
electromagnetic waves.
Light can be polarized, which implies that it is transverse.
(c) When two waves travel through a medium, their resultant displacement at any point is the sum of
the separate displacements due to the two waves.
(i) At an instant such as A, the waves from the sources arrive in phase and reinforce to
produce a loud sound.
The phase difference then increases until a compression (or rarefraction) from one source
arrive at the same time as a rarefraction (or compression) from the other.
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½
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01 AL Physics/Essay Marking Scheme/P.3
The observer hears little or nothing, point B. Later the waves are in phase again (point C)
and a loud note is heard.
Beats are thus due to interference but because the sources are of slightly different
frequencies there is sometimes reinforcement at a given point and at other times
cancellation.
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(ii) Stationary waves result from the superposition of two trains of progressive waves of equal
amplitude and frequency traveling with the same speed in opposite directions.
At the nodes (N), the interference of the two waves is always destructive and the resultant
amplitude is zero.
Or At the antinodes, the interference of the two waves is constructive and the resultant
amplitude is largest.
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01 AL Physics/Essay Marking Scheme/P.4
3.
(a) The electric field intensity E at any point is the force per unit positive charge which it exerts at
that point.
1
The electrical potential V at any point is the work done per coulomb of positive charge from infinity
(or any zero reference) to that point.
O
A
x
E
B
1
E
x
Suppose A and B are two neighbouring points on a line of force, so close together that the electric
field intensity between them is constant and equal to E.
Electric potentials at A and B are respectively V and V + V.
½
VBA = potential difference between A and B
= VA - VB
= - V
The work done in taking a charge Q from B to A
Force  Distance = charge  potential difference VBA
EQ  x = Q (-V)
V
E= 
x
dV
or, in the limit
E= 
dx
½+½
½
(b) (i) The capacitance (C) or charge-storing ability, of an isolated conductor is the charge (Q)
required to cause unit charge in the potential (V) of a conductor.
Examples: tuning circuit of a radio
ANY
smoothing circuit of a power supply
TWO
flash unit of a camera
@1
(any other relevant examples)
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(ii)
Potential
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Distance
After conductor AB is brought near, the end A closer to the sphere is negatively charged
while the far end B is positively charged (accept answer shown in diagram).
As a result the electric potential of the sphere is lowered.
Thus the capacitance of the sphere increases as it stores the same amount of charges for a
smaller potential.
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½
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01 AL Physics/Essay Marking Scheme/P.5
(c)
1½
During the half-cycle when the reed is in contact with X, the capacitor is charged by the
battery. The charge Q = CV stored in capacitor is then discharged through the 100-k
protective resistor and the light beam galvanometer when the reed is in contact with Y in the
following half-cycle.
½
The number of charge-discharge actions per second equals the frequency f of the a.c. supply
to the coil of the reed switch. If this is high enough current pulses follow one another so
rapidly that the galvanometer deflection is steady and represents the average current I
through it.
I = Qf = CVf
½
In order to ensure complete discharge, a CRO is connected across the resistor to check
whether the discharging current drops to zero within a half-cycle.
Factors affecting the accuracy:
- stray capacitance arises from any nearby conductors
- as very small currents are concerned, apparatus should be made of high quality
insulating material, otherwise leakage of currents will lead to serious error.
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½+½
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01 AL Physics/Essay Marking Scheme/P.6
4.
(a) (i) One tesla is the flux density in a magnetic field when the force on a conductor 1 m long,
placed perpendicular to the field and carrying a current of 1 A, is 1 N. (F = IBl)
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(ii) Consider two long, straight, parallel conductors, distance a apart in air, carrying currents I1
and I2 respectively.
0
B1
F
½
I1
I2
a
Long straight
parallel conductors
The magnetic field at the right-hand conductor due to the current I1 in the left-hand one is
directed into the paper.
½
A magnetic force F therefore acts on length l of the right-hand conductor carrying current I2
½
The left-hand conductor experiences an equal and opposite force due to being in the field of
the right-hand conductor.
½
(b) A long solenoid carrying constant current will give a uniform magnetic field inside the solenoid.
Set up the apparatus as shown.
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Adjust the rheostat so that there is a current of about 1 A through the solenoid.
Insert the Hall probe well inside the solenoid and adjust for zero deflection of the galvanometer
before switching on the current.
Switch on the current and set the galvanometer to give a (large) deflection.
Move the probe about inside the solenoid over a cross-section and along the length of the solenoid.
The deflection of the galvanometer remains unchanged, which indicates the magnetic field due to
the solenoid is uniform.
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01 AL Physics/Essay Marking Scheme/P.7
4.
(c) (i)
uniform magnetic
field (into paper)
P
v
l
Q
The electron in the moving rod will experience a magnetic force towards Q.
Therefore electrons will be forced to end Q, making Q negative and P positive.
As a result of the charge separation and electron accumulation, an electric field is created
inside the rod which causes a repulsive electric force to be exerted on other electrons being
urged towards Q by the magnetic force.
These magnetic force Bev and electric force Ee act oppositely and when they become equal
there is no further charge accumulation, i.e. when
Bev = Ee
½
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½
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½
where E is the equilibrium electric field and e the charge on an electron.
and E = potential gradient = potential difference across PQ/length
potential difference across PQ = Blv
(ii) According to Faraday’s law, the induced e.m.f. across the rod is
d
dA
= 
= B
= Blv
which is consistent with the result.
dt
dt
(iii) The potential difference across PQ will increase and the accumulation of electrons will
continue as the magnetic force (Bev) will keep pushing electrons towards end Q.
5.
(a) Rutherford scattering - the majority of the -particles were scattered through small angles and
some particles were scattered through very large angles. It meant that some -particles had
come into the repulsive field of a highly concentrated positive charge at the center of the atom.
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½+½
½
½
Frank-Hertz experiment (or hydrogen emission spectrum) - energy of an atom can only be raised
½
by specific amounts as inelastic collisions occur at certain accelerating potentials for the
½+½
colliding electrons. As the atom can only be excited to an energy level with a fixed gap, the
energy levels in an atom are therefore quantized/discrete.
½
(b) (i) Ground state - the lowest energy level of an atom (An atom is most stable, i.e. in its
ground state, when it has minimum energy)
If an atom is in its ground state and absorbs an amount of energy eV which just removes
an electron completely from the atom, then V is said to be the ionization potential of the
atom.
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01 AL Physics/Essay Marking Scheme/P.8
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(ii) Bombard the atom by accelerated electrons of high enough energy, in the order of eV.
Irradiate the atom by electromagnetic radiation of energy h which matches with the
energy difference between two energy levels.
An excited hydrogen atom is unstable due to the extra energy gained with respect to the
ground state.
This energy is eventually radiated in the form of electromagnetic radiation (photon) when
the excited atom returns to the ground state directly (or through an intermediate energy
level). Thus a number of frequencies of electromagnetic radiation may be produced,
which give rise to the emission line spectrum of hydrogen.
(c) For photoelectric effect, electrons are emitted from metal surfaces when electromagnetic
radiation of high enough frequency falls on them.
Thermionic emission occurs when a metal is heated to a temperature that the loosely held
electrons in the metal acquire kinetic energy high enough to escape from the metal against the
inward attractive force.
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½
½
½
½
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For radioactivity, -particles or high speed electrons are emitted spontaneously from nuclides of
relatively high proportion of neutrons (unstable with respect to its daughter nuclide).
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Electrons emitted from radioactive nuclides have the highest maximum kinetic energy (ranging
from 0.025 to 3.2 MeV) depending on the characteristics of the nuclides. The maximum kinetic
energy of the photoelectric emission electrons is of the order of eV depending on the difference
between the energy of the incident photon and the work function. Electrons emitted in
thermionic emission have the lowest maximum kinetic energy (less than 1 eV) as they are just
able to escape from the metal.
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