96 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
96’ AL Physics: Essay
Marking Scheme
1.
(a) For student A, when r increases, v also increases linearly with r as
2r
v2
period T =
is constant, therefore a =
actually increases.
v
r
For student B, when r increases, a = 2r increases provided that  =
1
2
is constant,
T
which is ensured by constant period T.
1
2
(b) (i)

T=W

mg
1
The string dips so that the vertical component of the tension balances
the weight of the rubber bung.
W cos  = m2r
 W cos  = m2 (L cos )
W = m2L
(ii)
-
ANY
TWO
(as T = W)
1
½
½
friction exists at the opening of the glass tube.
the rubber bung is not swirled with constant speed.
the string is not inextensible.
the rubber bung is not swirled in a horizontal circle.
(accept any other reasonable answers)
½
½
½
½
(iii)
ANY
TWO
- increase the weight W
- reduce the mass m of the rubber bung
- reduce the length L of the string
½
½
½
(iv) The rubber bung goes along tangential direction to A due to inertia and
falls as a projectile to the ground due to gravity.
½+½
½+½ 7
(c)
p2 p1
Tube + mixture
A B
p1 > p2
Consider the part of liquid between A and B inside the rotating tube,
the pressure at B is greater than that at A
so as to provide the necessary centripetal force acting inwards.
For that part of the liquid the force due to pressure difference supplies exactly
the centripetal force required.
If this part of the liquid is replaced by matter of smaller density/mass, the force is too large
and the matter moves inwards.
1
½
1
1
1
96 AL Physics/Essay Marking Scheme/P.2
Practical use:
ANY
ONE
- cream is separated from milk.
- separate solids in a suspension
- laundry drier spins to remove water from the clothes
½
½
½
or (Diagram.
The drum of a laundry drier has many tiny holes on its surface.
When the drum spins, the reaction from its circumference on the wet clothes
provides the necessary centripetal force acting inwards.
However there is no such reaction at those tiny holes and
the excessive water in the clothes can no longer keep a circular path and goes out
through the holes.)
(d) Similarities:
ANY
TWO
Differences:
ANY
TWO
2.
1
½
1
1
1
- the relatively massive body (earth/proton) remains practically
at rest at the centre.
- having negative potential energy (closed system)
- forces directed to centre obey inverse square law
5
½
½
½
- gravitational attraction versus electrostatic attraction.
- continuous range of orbital radii versus discrete orbital radii
- orbit of the satellite is definite versus orbit of the electron is
the probability of its distribution
½
½
½
2
(a) (i)
displacement
(right)
0
time
½
(left)
pressure
normal air
pressure when
wave absent
wave
travelling
to the left
1
time
(ii) The displacement leads/lags the air pressure by a phase angle of /2 (or ¼ period)
(b) (i) (I) Sound waves from the two loudspeakers reaching different points on XY
have different path differences, interference thus occurs.
For the path difference equals n, constructive interference occurs and
a maximum intensity results.
For the path difference equals (n + ½), destructive interference occurs and
a minimum intensity results.
(II) Waves from the loudspeakers take paths of different lengths to a minimum
position, by inversely square law, amplitudes of the waves are different,
or (When waves diffracted from a loudspeaker, the amplitude is maximum at the
central position but decreases when going sideways, so waves from the two
loudspeakers reaching a minimum point would have different amplitudes.)
½
1
½
½
½
½
½
½
½
2
96 AL Physics/Essay Marking Scheme/P.3
destructive interference of waves is not complete, hence a non-zero intensity
results.
(ii) Sound intensity is the energy per second crossing a unit area normally to the
direction of the sound (unit: Wm-2)
Sound intensity level of a source is its intensity relative to some agreed ‘zero’
intensity level (unit: dB)
When one loudspeaker is turned off, the amplitude of waves at O is halved,
intensity drops to ¼ of the original value, resulting in a change of intensity level
1
I
10 log10 4 = -6 dB.
I
(iii) If the frequency of the signal generator increases, the wavelength of the sound
waves produced would decreases, making the separation between positions of
maximum/minimum intensity decrease.
½
1
1
½
½
1
8
(c)
(or cork dust)
1
The loudspeaker produces progressive longitudinal waves travelling towards
the end of the cylinder where they are reflected to interfere/superpose the incident waves.
The frequency of the sound/signal generator is varied until resonance occurs.
The stationary wave formed is revealed by the lycopodium powder which swirls away
from the antinodes (where the air is vibrating strongly) and heaps are formed at the nodes.
By measuring the average separation between the heaps (or nodes), d
the speed of sound in air equals f(2d) where f = frequency of the sound waves
Precautions : 1. the tube should be dry
2. the layer of lycopodium powder should be thin.
3.
(a) (i) Pressure of a gas arises from the momentum change suffered by each gas molecule in
colliding elastically with the wall of the container.
1
1
1
½
½
½
½
6
1
Temperature of an ideal gas is a measure of the mean kinetic energy of the gas
molecules. (In fact mean k.e. of the gas molecules is proportional to the absolute
temperature of the gas.)
1
pV
= constant or pV = nRT)
T
or Boyle’s Law under all temperatures and pressures.
1
Real gas behaves like ideal gas under high temperatures AND low pressures.
1
(ii) Ideal gas is a gas that obeys the ideal gas equation (
4
96 AL Physics/Essay Marking Scheme/P.4
(b) Free electron gas model for metallic conduction.
Metal consists of a large number of free electrons which are in a state of rapid thermal
motion, moving randomly within the lattice at high speeds.
They make frequent collisions with the lattice ions, changing directions all the time
without net displacement.
½
½
When a p.d. is applied, the electric field would cause the free electrons to accelerate (in a
direction opposite to the electric field) and hence gain velocity/k.e.
1
On colliding with the lattice ions, the electrons give up their additional k.e. to the lattice ions. ½
The electrons slow down and then accelerate again. (This process is repeated many times.) ½
The net effect is that the vast number of free electrons drift slowly, giving a net flow
of electrons in one direction, which constitute an electric current macroscopically.
1
When collided by the accelerating electrons, the lattice ions gain vibrational energy (both k.e.
and p.e. as they vibrate more vigorously and with greater amplitude).
1
Macroscopically the internal energy of the metal increases resulting in a temperature rise.
Consequently thermal energy is released.
1
(c) (i) Rutherford’s model (nuclear model)
Most of the mass of an atom and all the positive charges are concentrated in a tiny
nucleus at the centre of the atom.
Outside the nucleus, most of the space in an atom is ‘empty’, in which the negativelycharged electrons (of negligible mass) are orbiting around the nucleus.
1
(ii) In -particle scattering experiment, most of the -particles pass through the thin gold
foil with no or small deflection,
since most of an atom is ‘empty’ and the very light orbiting electrons
outside the nucleus have negligible effect on the energetic and massive -particles.
½
Some -particles (about 1/8000) were scattered by angles greater than 90º and very
few even rebounded back along original paths.
Only for a few -particles which come very close to the tiny nucleus experience a
strong electrostatic repulsive force (Coulomb force) and hence suffer large angle
deflection.
(iii) Rutherford’s model could not explain the existence of the characteristic atomic line
spectra from atoms of a particular element.
4.
(a) (i) One major source is coal (or oil).
The fuel (coal/oil) is burnt in the furnace for heating water in a boiler.
This produces high pressure stream for
driving the turbine of the generator (alternator, rotor) to produce electric power.
(ii)
1
1
½
1
1
½
½
1
½
Coil
N
S
Carbon Brush
Slip Ring
Load
Magnet
Coil
Slip rings
Brushes
6
½
½
½
½
6
96 AL Physics/Essay Marking Scheme/P.5
The output is sinusoidal if the coil is rotating with uniform angular speed (50 rps) in
a uniform magnetic field between two pole-pieces.
(b)
Power loss = I2R
=
½
P2 R
as P = IV
V2
If output power P and resistance R of the cables are constant, then power loss  1/V2
where V = output voltage.
(c) (i)
1
½
Maintenance is easy as overhead cables are bare.
Also no insulation is required, so it is many times cheaper.
Less installation compared with underground cables, therefore much cheaper
Compared with underground cables, overhead cables can use higher voltages
for reducing power loss.
1
1
1
Potential danger of electric shock due to high voltage.
Overhead transmission is suspected to cause environmental pollution (e.g. leukemia a blood cancer) due to electromagnetic radiation consisting of oscillating electric
and magnetic fields.
(Accept any other reasonable answer)
1
ANY
TWO
(ii)
1½ 6
Aluminium is lighter, but copper is a better electric conductor.
(or Copper is easily oxidized when bare but aluminium is not.)
(or Compared with copper, aluminium is less brittle.)
2
1
1
½+½
Steel has a high breaking stress, so it can increase the strength of the overhead cable. 1
6
(d)
Transformer
Rectifier
Smoothing
Battery
(polarity correct)
5.
½
½
½
½
2
(a)
C
H
G
Y2
A1
A2
X2
Y1
Screen
X1
Heater H &
Cathode C
½
Grid G
½
Anodes A1, A2
½
Deflecting
plates X, Y
e.h.t.
Graphite
coating
½+½
Tube, screen
½
Potentiometer
1
4
96 AL Physics/Essay Marking Scheme/P.6
(b) (i) The filament (heater H) is heated by passing a current from a low voltage supply.
The cathode C is indirectly heated by radiation from the hot filament, electrons
are emitted from the cathode surface. (Thermionic emission)
(ii) The grid potential is more negative to the cathode so that less energetic electrons
are repelled.
The grid potential can be adjusted to control the number of electrons passing
and hence the intensity of the beam (i.e. the brightness of the spot).
The focusing anode A1 and the accelerating anode A2 together form an electron
lens system so as to focus and accelerate the beam.
The potential of anode A1 is adjustable and in turn controls the sharpness.
(iii) The graphite coating provides a return path for the electrons reaching the
screen back to the cathode.
- The anode A2, one of the X-plates and one of the Y-plates are also earthed so
that they together with the graphite coating form a field-free region between A2
and the screen. (i.e. shield the beam from external electric fields)
- As a result the electron beam leaving anode A2 is not deflected if no external
p.d. is applied to the deflecting plates.
- This avoids the disturbance by nearby earthed objects (such as an observer’s
hand) outside the CRT.
ANY
ONE
(c) (i) Time base circuit is an electronic circuit which generates a saw-toothed voltage
applied to the X-plates internally.
½
½
1
1
½
½
1
1
1
1
6
1
V
sweep
0
t
½
flyback
The time base circuit repeatedly moves the spot from left to right across the
screen at a steady speed (i.e. sweep) and then brings it back nearly instantaneously
to the left side (i.e. flyback).
Thus it allows a waveform to be displayed on the screen.
(ii) A capacitor for blocking the unwanted d.c. component.
(d)
ANY
TWO
ANY
TWO
1½
½+½ 4
Advantages:
Nearly infinite internal resistance.
Quicker response than pointer.
Not easily damaged by overloading.
½
½
½
Disadvantages:
No direct reading/need calibration.
It is bulky.
It needs electric supply to operate.
½
½
½
2