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97 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
97’ AL Physics: Essay
Marking Scheme
1.
(a)
F'
F
1
Fr
The pushing force acting on the rock is balanced by the friction/resistance from the
ground,
but not by the reaction force which acts on the man.
½
½
(b) (i) The earth is not a perfect sphere (actually an oblate ellipsoid),
i.e. the north pole is closer to the centre of the earth than at the equator.
1
The gravitational field strength (g  2 ) is greater at the north pole, consequently
r
weight (mg) is larger.
The rotation of the earth ‘reduces’ the apparent gravitational field strength g’ at
the equator,
apparent gravitational field strength at equator = g0 - centripetal acc. = g0 - 2RE
where g0 = gravitational field strength at the equator if the earth is at rest
 = angular speed of rotation of the earth
RE = radius of the earth
(ii) For the same object, the beam balance readings would be the same at the north
pole and at the equator (i.e. the same mass is needed to balance the weight of the
object). This is because beam balance measures the ‘mass’ of the object, which is
unchanged, by comparing the weights at both arms.
(c) Similarities : the man is said to be ‘weightless’ in both cases (i.e. no reaction)
both are accelerating motions
the respective weights are exactly used for acceleration (i.e. mg = ma)
total mechanical energy is conserved
(d) (i) n
R
T
N
m
c
- number of moles
- universal gas constant
- absolute temperature
- number of molecules
- mass of a molecule
2
- mean square speed
2
½
½
½
½
½
½
1
1
ANY
THREE
@1
1
1
5
3
97 AL Physics/Essay Marking Scheme/P.2
(ii) According to Avogadro’s law, the gases have the same number of molecules. (As
p, V and T are the same, n is the same according to pV = nRT.)
1
As
pV = Nmc 2 = nRT
3
3 pV
1
3nRT
=
mc 2 =
2N
2
2N
therefore the average molecular kinetic energy is the same in both cases since T is
the same (or p, V and N are the same).
1
1
2
As
=
mH c H
m0 c02
2
2
c H2 > c 02
(  mH < m0)
As the molecular mass of hydrogen is smaller than that of oxygen, the mean
square speed of the hydrogen molecules is higher than that of the oxygen
molecules since their average molecular kinetic energy is the same.
2.
1
½
1
1
½
6
(a) (i) Consider the diffracted rays from the slits in a certain direction
1

d

When all these rays are in phase, they will superpose to give a local maximum, i.e.
a principal maximum.
The path difference between two consecutive rays must be n, where n is an
integer.
In the diagram, the path difference between two consecutive rays = d sin.
Therefore, the angular position  of the principal maxima is given by d sin = n.
(ii) Gratings give sharper and brighter patterns.
Also the angular separation of the fringes is larger by using gratings as the slit
separation can be made many times smaller.
1
1
½
1
½
5
97 AL Physics/Essay Marking Scheme/P.3
(b)
Straight
filament lamp
iodine
vapour
1
grating
Vaporize some iodine crystals in a test tube by warming.
View a straight filament lamp through iodine vapour with a diffraction grating.
The grating should be set with its lines parallel to the filament of the lamp.
½
½
½
A continuous spectrum consisting of some dark lines is observed.
The light from the lamp is a continuous spectrum consisting of photons of a range
of energies.
When the light is incident on an iodine molecule, it can only absorb energy from a
photon whose energy is just enough for exciting it to a higher energy state.
When the excited molecule returns to ground state, it re-emits light of the same
wavelength of the photon but equally in all directions.
So the intensity of the radiation in the direction of the incident photon is reduced.
However, photons of other wavelengths will pass straight through.
Thus a dark line, whose wavelength is that of the absorbed photon, is seen on the
background of a continuous spectrum.
½+½
(c) The sun emits a continuous spectrum of photons.
Vaporized elements in the (relatively cooler) outer part of the sun’s atmosphere absorb
those photons whose energy is just sufficient to excite them to a higher energy state.
The sun’s spectrum is now darker at wavelengths characteristic of the elements in the
sun’s atmosphere.
Elements can then be identified by studying the wavelengths of those dark lines.
3.
(a) Potential difference between two points is the work done (or change in p.e.) per unit
positive charge moving from one point to another.
By selecting a reference point (usually the earth or a point at an infinite distance from
any electric charges) in the electric field to be the zero potential, the potential of a point
is taken as the potential difference between that point and the reference point.
1
1
1
½
½
½
8
½
1
1
½
1½
½+½
½
(b) (i)
Field lines
radial field
arrows
circular equipotential lines
correct spacing
½
½
½
½
Equipotentials
(ii) (I) The lines of force are drawn such that the density of the lines represents the
strength of the field.
3
½
3
97 AL Physics/Essay Marking Scheme/P.4
At the region nearer to the conductor, the lines are closer together and thus
the field strength is stronger.
½
The tangent at a point on a line gives the direction along which an electric
force will act on a positive charge if it is placed at that point and the arrow
gives the direction of the force (radially outward in this case).
1
(II) Electric field strength is equal to the (negative of) potential gradient
dV
(E = 
).
dr
In the diagram, the equipotential lines are closer together at the region
nearer to the conductor and thus the field strength there is stronger.
1
1
6
(c)
1½
A flame probe can be used to investigate the potential in an electric field.
The probe, in the form of a small gas flame at the point of a needle, is connected to a
calibrated electroscope which has the same potential as the probe, therefore its deflection
indicates the potential at the point where the probe is situated.
½
½
The conducting sphere is charged to about 1 kV by an EHT power supply.
Negative charges are induced on the probe, and positive charges go to the electroscope.
The flame ionizes the surrounding air to produce ions to neutralise the charges on the
probe. The potential is therefore unaffected by the presence of the probe.
½
½
½
½
Move the probe round the sphere keeping a fixed distance from the centre of the sphere
shows that the potential does not vary. This shows that the equipotential line is circular.
½
½
Move the probe outward from a point quite near the surface of the sphere along a radial
direction shows that the potential is inversely proportional to the distance from the centre
of the sphere (V  1/r).
½
½
+½
7
97 AL Physics/Essay Marking Scheme/P.5
4.
(a)
N
S
½
galvanometer
Wind the copper wire into a coil of a few turns and connect it with the light beam
galvanometer.
½
Move the bar magnet towards and away from the coil along a direction normal to the
coil, the galvanometer would deflect to opposite directions showing that the induced
e.m.f. / current depends / opposes to the change in magnetic flux.
1
Repeat the experiment by moving the magnet with a higher speed, the deflection of the
galvanometer would increase showing that the induced e.m.f. / current increases with the
rate of flux change.
1
Repeat the experiment by using two magnets with the same pole facing the coil and move
them at the same speed as before, the deflection of the galvanometer would increase
showing that the induced e.m.f. / current increases with the strength of the magnetic field
responsible for the change.
1
Wind more turns for the coil and repeat the experiment, the deflection of the galvanometer
would increase showing that the induced e.m.f. / current increases with the number of
turns of coil experiencing the flux.
1
(b) (i) A plane coil rotates with constant angular speed  about an axis perpendicular to
a uniform magnetic field of flux density B.

uniform magnetic field
 = t
½
axis of rotation
coil
By Faraday’s law of electromagnetic induction, the e.m.f. induced in the coil
 = - d / dt
= - d(NAB cos ) / dt
= -d(NAB cos t) / dt
= NAB sin t
where A is the area of the coil, B is the magnetic field strength and t is time.
½
½
½
½
5
97 AL Physics/Essay Marking Scheme/P.6
(ii)
Coil wound on
an armature
_
+
1½
The coil wound on an armature is connected to a commutator which consists of
two half-rings insulated from one another.
Brushes press against the commutator and are connected to the light bulb.
½+½
½
When the coil rotates, by Fleming’s right hand rule, the current flow are as
shown:
motion
B
P
P
E
E
Q
P
Q
E
Q
Q
E
P
P
current
Q
1½
(iii) By Fleming’s left hand rule, when the generator produces a current, that current
flowing in the coil would produce a torque which opposes its rotation.
FB
I
I
1

B
FB
A larger current is drawn from the generator when connecting two light bulbs in
parallel.
When the current drawn increases, the opposing torque will increase.
The rotation of the coil will then slow down, thus the driving torque has to be
increased in order to maintain the speed of rotation of the coil.
When the current drawn increases, more electrical energy is consumed in the light
bulbs.
½
½
1
½
97 AL Physics/Essay Marking Scheme/P.7
Therefore more work has to done by the driving torque to provide the electrical
energy.
5.
½
11
(a) (i)
(soaked with
alcohol)
1½
The upper compartment in the chamber contains air at room temperature at the
top and at about -78 ºC at the bottom due to the layer of ‘dry ice’ in the lower
compartment.
The felt ring at the top of the chamber is soaked with alcohol, which vaporizes in
the warm upper region, diffuses downwards and is cooled. Thus a layer of
saturated alcohol vapour is formed.
The ions produced along the track of, say, an -particle attract the molecules of
the saturated vapour, condensation of these molecules produces liquid droplets
which constitute the visible track.
Range of radiations (or penetrating power) is shown by the length of the track.
Relative ionizing power of radiations is indicated by the thickness of the track.
½
1
½
½
1
1
(ii)

v1

u

½
He

v2

Let u = velocity of -particle before collision

v1 = velocity of -particle after collision

v2 = velocity of the atom after collision
By conservation of momentum,



m u + 0 = m v1 + mHe v2
As the track is right-angled,
(m u)2 = (m v1)2 + (mHe v2)2
m
u2 = v12 + ( He v2 )2
m
½

m v1

mHe v 2
½
½
½

m u
(1)
½
97 AL Physics/Essay Marking Scheme/P.8
As the collision is elastic, k.e. is conserved
1
1
1
m u2 = m v12 + mHe v22
2
2
2
m
2
He
u2 = v12 +
v2
m
½
(2)
½
Comparing (1) and (2), m = mHe
10
(b) We cannot tell which particular atoms are going to decay in unit time, just as we cannot
tell which particular dice turn up a certain value in each throw.
We cannot tell when a particular atom is going to decay, only the probability of decay in
unit time () is known, just as we cannot tell in which throw a particular dice turn up a
certain value, only the probability of this is known.
The number of atoms disintegrate per second,
[ln N ]NN 0 = -kt

N = N 0e kt
1
dN
, is directly proportional to the number
dt
of undecayed atoms, N, present
dN
= -kN
dt
where k is a constant characteristic of the atom concerned called the decay constant,
which is the probability of decay per unit time.
N dN
t
=  k dt
N0 N
0

1
1½
½
½
1
½
6