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88 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
88’ AL Physics: Essay
Marking Scheme
1.
(a) (i)
Force
repulsion
Molecules in surface layer more
widely spaced and in state of tension
separation
attraction
Inner molecules
in liquid
(optional)
At the surface of the water molecules escape and in
the vapour state are more widely separated (on average).
Also at surface the molecular concentration decreases
and molecular forces are strongly attractive producing
a surface layer under tension.
A needle is supported by this ‘elastic skin’,
with repulsive forces from the water molecules
underneath needle preventing sinking.
½
½
½
½
½
½
3
(ii)
solid
B
A
liquid
R
C
At point B in liquid there is an attractive force
(I) BC due to the liquid molecules
(II) BA due to the neighbouring molecules in the
pin.
If forces BA > BC (as for water) the resultant BR
will be as shown in diagram.
The liquid surface has to be at 90° to this and
so it curves up sides of pin.
The resultant BR will be opposed by repulsive
(liquid) molecular forces.
(iii) Work has to be done in separating further the
molecules, at a liquid surface,
against the molecular attraction forces.
This leads to an increase in molecular P.E.
(or free surface energy of liquid).
In nature this will be a minimum and so any created
liquid film has a minimum area.
½
½
½
½
½
½
3
½
½
½
½
2
88 AL Physics/Essay Marking Scheme/P.2
(b)


(optional)
(a)
(b)
Liquid (a) with an acute angle of contact spreads while
liquid (b) with an obtuse angle of contact does not.
½
½
(i) Good spreading, e.g. flux in soldering, wetting
agents in paints, lubricating oils.
(ii) Poor spreading, e.g. silicones for waterproofing
fabrics.
1
1
3
(c)
Capillary tube
Travelling
microscope
h
Pin
Liquid
2.
1
Since contact angle ~ 0
2rT = r2hg
T = ½hrg
1
(i) Travelling microscope focused on bottom of meniscus
of water in tube and then moved to focus on pin
which just touched surface of water in beaker
(beaker removed) - hence measurement h.
1
(ii) Tube broken at meniscus level (in tube) and the
diameter of bore measured in two directions at
right angle - hence measurement r.
1
(iii) Cleanliness of apparatus - should be cleaned by
immersing in caustic soda, dilute nitric acid and
distilled water, in turn.
1
(a)
R
long distance
moon
L
dollar
O2
O1

a
b

image
1
5
88 AL Physics/Essay Marking Scheme/P.3
The size of the image formed on the retina of the eye only
depends upon the angle subtended by the object at the eye.
This angle is similar for the larger object (moon) at a
long distance and the smaller object (dollar) at a nearer
distance.
½
½
2
(b)

L
I’
h’
O’
I

h
F
’
O
C
D
1
Following from (a) larger objects at a fixed distance will
appear larger since the angle subtended at eye is greater.
Clearly, from ray diagram, the viewed image I formed at the
distance of nearest clear vision D appears greater than
the image I’ formed at infinity
since  > ’,
and hence magnification is greater for the former arrangement.
Magnification is limited by the focal length of the lens.
½
½
½
½
1
4
(c)
D
objective
I2
Fo
h
O
Fe I1
”
eyepiece
h1
Fo
L1
f1
h2
Clearly, from ray diagram, a magnified image I1 is produced
for object O by lens L1.
In this case the final image I2 of this ‘object’ I1
produced by lens L2 subtends an angle ”.
Since h1 > h the angle ” > ’
and the magnification has been increased.
(d) (i) The intensity of light collected by the camera 
aperture area, i.e. d2 and this determines the
intensity of the image.
(ii) The recorded brightness (and darkening of chemicals
on film) depends upon the time integration of the
light energy received, i.e. t.
L2 f2
1
½
½
½
½
1
1
3
88 AL Physics/Essay Marking Scheme/P.4
(iii)
F1
d
F2
f1
1
f2
For distant objects it can be seen, from ray diagram,
that the size of the image is  f
and the area of image  f 2
 Brightness of image  1/f2
½
½
½
(iv) The f-stop is defined as f-n, where the aperture
diameter d = f/n, f being the focal length of
camera lens.
In a camera the effective aperture diameter can be
varied by a mechanical ‘iris’
e.g. by decreasing the f-stop the time exposure can
be reduced which could produce a shaper image of a
moving object. (or any other reasoning)
3.
1
½
1
7
(a)
(i)
(ii)
(iii)
(iv)
2
(v)
(i)
(ii)
(i) Both ends of the spring are momentarily moved
sideways at the same time.
(ii) One end of the spring is attached to a rigid support
and the other end given a momentarily ‘flick’ as
in (i). After reflection amplitude of return pulse
is reversed and a further pulse is produced from
the free end as in (i).
2
(b) When the pulses pass through each other :
(i) they remain unaffected (in amplitude).
1
4
88 AL Physics/Essay Marking Scheme/P.5
(ii) Where they cross the total displacement is the
vector sum of the individual displacements.
1
(c) Following from the principle of superposition, in a region
where 2 wave-trains meet the amplitudes either reinforce
or cancel, depending upon whether they are in-phase or
anti-phase - giving rise to an interference pattern
(or fringes).
2
1
(i) Sound
Displacement
f2 f1
0
Time
(a)
A
B
Resultant
Displacement
C
Variation of
amplitude
0
Time
(b)
Beat period = T
Hence, Beat frequency = 1/T = f1 – f2
e.g. beats produced between two sources of slightly
different frequencies - as in music if one
instrument is slightly out of tune.
OR any other example
1
1
(ii) Light
eye
N
(i)
A
i
M
(ii)
D
O
C
E
r
B
t
r t
1
e.g. different colours observed as eye is moved at
different angles, viewing a thin oil film on water.
Different path distances for white light (i) and (ii),
2t cos r = n for re-inforcement.  varies for
different coloured light.
OR any other example
1
5
88 AL Physics/Essay Marking Scheme/P.6
(d) (i) At observing location the interfering waves must have
a constant (or slowly observable changing) phase
difference.
1
(ii) Interference between any 2 sound sources is possible
if condition (i) holds good.
1
(iii) Interference for light waves only possible from
single source via 2 different path distances.
1
(iv) Light is not coherent (normally) since waves are
emitted in random quanta or wave packets. To maintain
phase relation path distance has to be < wave packet
length.
1
1
5
[Bonus mark 1 - if laser mentioned,
then path distances longer.]
4.
(a)
+


½
E
A
R
B
(i) Electric field, E, is the force per coulomb of charge
acting on the free electrons which move in a net
direction along the connecting wires  to .
Direction of electric field is  to .
1
½
(ii) Potential difference, VA - VB, is the energy converted
to other forms, from the electrical P.E., per coulomb
of charge passing from B to A.
1
(iii) E.m.f., , is the energy imparted by the battery per
coulomb of charge passing internally between its plates.
1
(b) The moving free electrons, under the influence of the electric
field
suffer collisions with the atoms in the resistor.
They loss part of their K.E.
which is transformed into making the atoms vibrate,
generating heat.
(c) (i) According to (a) (ii) for a charge Q the converted
energy in R is (VA - VB)Q.
dQ
Energy rate = (VA - VB)
or (VA - VB)I
dt
V  VB
= I2R, since by Ohm's law R = A
I
If all the energy is converted to heat,
rate of heat dissipation = I2R.
4
½
½
½
½
½
½
½
2
88 AL Physics/Essay Marking Scheme/P.7
(ii)
I
V
½
IR
In an a.c. circuit containing reactance  :
V
I=
, where V is the voltage across the
2
  R2
circuit [V, I are r.m.s. values]
Rate of heat dissipation, still, = I2R
½
½
(d) The power is transmitted as an a.c. high voltage (132 kV)
along transmission lines over long distances from the power
stations to sub-stations
where it is transformed down to 6 kV, - to 220 V for
domestic consumption.
(i) Reason for high voltage
Transmitted power, P = IV
Power loss in transmission line = I2R = (P/V)2R
- by increasing V a significant reduction in power
loss
(ii) Reason for use of a.c.
It is relatively easy to transform the voltages up
or down using transformers.
(e)
RL
Vm
3
½
½
½
½
½
½
1
4
I
R
V
½
There is a voltage line drop due to the resistance of
mains cables RL and current I taken by appliance
V = Vm - IRL.
½
(i) Length (and condition) of cable leads from transformer
station will vary RL for any particular location.
1
(ii) Even at a particular location variation of current
taken will vary mains voltage across appliance.
1
3
88 AL Physics/Essay Marking Scheme/P.8
5.
(a)
v

m1
u
q  m1v
OR
s  m2w

m2

w

p  m1u
½
(i) Electric collision
(I) Total kinetic energy conserved,
1
1
1
m1u2  m1v2  m2w2
2
2
2
(II) Total momentum (vector) is conserved,
p  q  s or m1u  m1v  m2w
½
½
(ii) Non-elastic collision
(I) Total kinetic energy is not conserved
some of energy appearing in another form due
to work done against an internal force,
increasing the internal energy such as
e.g. heat.
(II) Total momentum (vector) is conserved.
½
½
½
3
(b) (i) -particle scattering
Alpha
particles
+Ze
+2e
P
Nucleus
½
-particle lose their K.E. on approaching the +ve
charged nucleus, being repelled by an electrostatic
force.
At P, distance of closest approach K.E. lost = P.E.
due to -particle location in electric field of
nucleus.
-particle is then ‘reflected’ away from nucleus
and finally acquires the same K.E. as it had
initially.
Collision is elastic.
(ii)
½
½
½
½
1 235
0 n/ 92U collisions
Neutrons penetrate the 235
92 U nuclei
and cause fission,
148
85
1
1
( 235
92 U + 0 n  57 La + 35 Br + 3 0 n )
(equation optional)
There is an apparent loss of mass
½
½
½
2½
88 AL Physics/Essay Marking Scheme/P.9
resulting a release of energy, according to relation
E = mc2.
Collision is inelastic.
½
½
(iii) electron/Xe collisions
On collision some of the K.E. of the electron will
be ‘absorbed’ by atom,
exciting the atoms to a higher internal (quanta)
energy level
and the electron will continue with a reduced K.E.
Thus K.E. is not conserved and collision is
non-elastic.
2½
½
½
½
½
2
(c) Franck-Hertz experiment
C
Gas
I
G
A
Electrons
I
xenon-filled
thyratron
V1
0 - 25 V
Q
G
V2
V2 ~ 1.5 V
P
At low voltages V1 electrons repelled by -V2 do not reach
anode.
As V1 increases K.E. of electrons increase and more electrons
are collected by A, increasing current I.
Collisions with xenon atoms are elastic.
At VC some electrons have sufficient Kinetic energy to
excite the atoms to their first higher internal energy level,
electrons lose energy and current decreases
(velocity reduced, less no. of electrons reaching A per
sec.)
On further increasing V1 electrons (which have lost energy
on collisions)
now have sufficient energy to reach A and current increases.
Importance of experiment is the demonstration of internal
discrete (quanta) energy for atoms.
6.
S
Galvo
reading
1+½
R
VC
V1
½
½
½
½
½
½
½
½
½
(a)
B
r
O
x

N

 = 2r
1
v = r
P
A
Consider the projection of point P2 on the diameter AB point N. Acceleration of P2 =  = 2r along PO.
½
6
88 AL Physics/Essay Marking Scheme/P.10
Acceleration of N is
d2x
(or 
x ) = 2r cos , directed
dt 2
towards O.
Displacement of N from O is x = r cos .
Hence the relation, 
x = - 2x
½
½
½
(b) Simplest method is to attach spring to a rigid support at one
end and at the other to the object resting on a smooth table then pull object, extending spring.
Object will oscillate about unstretched length of spring with
S.H.M. since extension of spring  force applied. Restoring
force = -kx, where k is force constant of spring [force/
extension].
[Note: for suspended object, S.H.M. is about ‘normal’
extended end of spring]
(c) (i) The force between atoms, F(x) = 
so, F  
dU
dx
3
1
1
1
3
½
a
b
 3
2
x
x
½
F
0
(general shape)
b/2a
+
x0 = b/a
xm
x
P
-
Q
For equilibrium location F = 0
b
so x0 
at P
a
At small value of ‘x’ the 2nd (+ve) term predominates
while
for large values of ‘x’ the 1st (-ve) term
predominates.
At Q the force F is a minimum corresponding to
dF
 0.
condition
dx
3b
2a 3b
i.e. 3  4  0 or x m 
2a
xm xm
(ii) Consider a small displacement dx from equilibrium
position P,
a
b

F =
2
(x0  dx )
(x0  dx )3
=
a
dx
b
dx
(1  )2  3 (1  )3
x0
x0
x02
x0
a
dx
b
dx
(1  2 )  3 (1  3 )
2
x0
x0
x0
x0
using Binomial expansion
= 
½
½
½
½
½
½
½
½
½
½
4½
88 AL Physics/Essay Marking Scheme/P.11
1
1
3b
a
a4
mx  3 [2a  ]dx   3 dx   3 dx ,
2
x0
x0
x0
b
since x0 = b/a
i.e. F =
2
3½
(d) Clearly this represents S.H.M. and
a4
b3
(ii) the oscillation period = T
(i) the force constant, k 
= 2
m/2
mb3

2

a 4 / b3
2a4
1
1
2