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92 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
92’ AL Physics: Essay
Marking Scheme
1.
(a) (i) 2nd law - The rate of change of momentum of a body
is proportional
to the resultant force acting on the body and this occurs
in the direction of this force.
(ii) 3rd law - if a body A exerts a force on a body B,
then body B exerts an equal but opposite force on body A.
(b)
A
B
Before collision m1, u1 and m2, u2
After collision m1, v1 and m2, v2
If time of impact is t and u1 > u2
Body A will exert a force F on body B for a time t and by Newton's
3rd law body B will exert a force -F on body A (opposite direction).
By Newton's 2nd Law, force = rate of change of momentum (constant
of proportionality = 1)
m u  m1 v1
m u  m2 v2
 [ 2 2
],
Hence F  1 1
t
t
Since forces on bodies are equal and opposite
so m1u1 + m2u2 = m1v1 + m2v2,
and linear momentum is conserved.
No - K.E. may not be conserved
since some energy may be converted
to another form such as heat or sound.
Total energy is, however, conserved.
½
½
½
½
2
½
½
1
1
½
½+½
½
½+½
½
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½
½
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(c) (i) Franck Hertz experiment
4.9 V
current
D
C
B
0
5
10
15
V in volts
The voltage +V is varied between the cathode and anode of a
vacuum tube containing Hg vapor.
A small retarding potential exists between an intermediate
grid to prevent electrons reaching anode.
From O  B electrons accelerate and current increases.
At B, C, D inelastic collision
take place with Hg atoms raising their internal energy to a
quantised value.
Current I drops due to a loss of K.E. of electrons.
½
½
½
½
½
½
½
3½
92 AL Physics/Essay Marking Scheme/P.2
(ii)
No of beta
particles per
unit energy
range
Energy
½
Emax
In -decay, XZA  YZA1  e
But e- can have a large range of K.E.’s. Since all electrons
result from same internal process
- total energy should be constant
Hence postulation of another ‘mass-less’ particle - the neutrino
to carry rest of energy.
2.
(a) Kinetic theory assumptions are :
(1) Intermolecular forces are negligible,
except in collisions.
(2) Volume occupied by molecules is negligible,
compared with volume of gas.
(3) All collision between molecules
and with the wall of container
are elastic (no energy loss).
(4) Time of contact during collisions
is negligible compared with time between collision.
½
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2½
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4
(b) (i)
B
½
A’
A
Molecule A travels to A’ rapidly with high average speed.
However to travel from A to B it has to suffer many collisions
and changes of direction
resulting in a slow diffusion rate.
½
½
½
2
92 AL Physics/Essay Marking Scheme/P.3
(ii) MFP is the average distance moved by a molecule between
2 successive collisions with other molecules
Molecules have little probability of colliding with their
neighbouring molecule
since they are all moving rapidly and randomly in direction.
Hence the M.F.P. >> average separation of molecules.
½
½
½
½
2
(c) (i)
A
T
R = 0.1m
in
t = 500s
1
The glass tube T containing bromine capsule connected via
rubber tubing.
Capsule broken by pliers
and bromine liquid forms in bottom of tube A.
Using stop watch, time taken for ‘half-brown’ level to travel
halfway up tube.
From kinetic theory,
Given, P = Nmc2/3V, density  = Nm/V
hence Cr.m.s. = (3p/), p - pressure (S.T.P.) (no need to derive)
If no collisions, distance travelled by molecule, s = ct
If collisions are N, M.E.P.  = s/N
From random walk theory, range of molecule R =
Hence  = R2/ct
(ii) Since PV = RT =
1
3
N
Nmc2,
at same temperature T molecules of lower mass will have
greater r.m.s. speed.
Would expect in earth’s atmosphere that lower mass molecules
such as hydrogen would diffuse to greater heights
compared with the higher mass molecules such as oxygen,
which would predominate at lower levels.
½
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1
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6
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2
92 AL Physics/Essay Marking Scheme/P.4
3.
(a)
Progressive transverse wave
Progressive longitudinal wave
8
(b)
scale
2
Microphone M is moved along bench away from loudspeaker L.
Distance is measured on scale.
Y2
T
in phase
Y1
Y2
M moved by /4
 = /2
Y1
1
Y2
M moved by /2
=
Y1
1
If voltage waveform in phase at a particular location of M
then on moving through a distance  (wavelength) they would
again be in phase.
Period T is measured using time-base of C.R.O.
½
½
92 AL Physics/Essay Marking Scheme/P.5
Then propagation velocity =

.
T
½
(c) The propagation of light waves differs from that of sound waves
in air since :
(1) no medium is necessary.
(2) wave is transverse.
(3) wave energy is carried by the electric and magnetic field
of the electromagnetic wave.
6
½
½
½
E
direction of propagation
½
2
B
4.
(a) (1) Electric field intensity, E = F/Q,
where F is the force acting upon a small
charge +Q placed at the particular location. It is in the
same direction as F
(2) Potential, V = W/Q,
where W is the work done in moving a small
charge +Q from infinity to that location.
dV
Relationship is E = 
ds
½
½
½
½
½
½
E
ds
V
1
V + dV
4
(b)
Q
a
E
r
P
At P, if sphere is very small (a  0), and small charge +dq
Qdq
Q

is placed there, then Er 
2
4  0 r dq 4  0 r 2
Sphere Potential V 
Clearly

a


Q
4 0 r
2
dr 
Q
= a constant  capacitance
V
Q
4 0 a
½
½
1
1
1
4
92 AL Physics/Essay Marking Scheme/P.6
(c)
Reed switch
(1)
(2)
25 cm
coil
25 cm
f ~ 400 Hz from low
impedance output of
signal generator
V
12 V
Diode (to rectify a.c.)
Capacitor
R = 100 k
A
Metal
plates
A
Sockets for
4-mm plugs
C
2 Marks
Polythene
spacer d
1 Marks
Reed switch, switches alternately between contacts (1) and (2), charging
C and then discharging C with a frequency f.
The generated current pulses are so rapid that the micro-ammeter
deflection remains steady,
indicating an average current I, where I = Qf
- Q being charge stored in C. Hence capacitance of capacitor
C = Q/V, V being measured by a voltmeter.
Clearly the area of overlap of plates A
and the separation d (using various spacer thicknesses)
can be varied and effects on C determined.
Sources of error include (any 2)
(1) R should prevent excessive current pulses but not be too
large otherwise C does not completely discharge
(2) Finding effect of electric field at edges of plates affects
dependence of C on areas A.
(3) Stray capacitances to earth could affect the effective
capacitance of C.
(4) leakage of charge from capacitor
5.
(a)
Semiconductor
½
½
½
½
½
½
2
8
+
t
E
I
VH
d
1
v
Force on a free electron of charge -e is -Bev
v being velocity in opposite direction to I.
Electrons move to bottom surface
which becomes negatively charged w.r.t. top.
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½
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92 AL Physics/Essay Marking Scheme/P.7
-Ee
v
_ electron
½
-Bev
Charge stops flowing to bottom surface
when the electric field force due to the negatively charged
surface -Ee = -Bev.
electric field E = Bv
½
½
½
5
(b)
B
Y
v
E
l
X
1
There will be a force -Bev on
the free electrons which will move these to end X of conductor.
This end will become negatively charged
and set up an electric field E’.
Flow of electrons will stop
when -eE’ = -Bev, electric field E’ = Bv
½
½
½
½
½
½
(c)
V
O
vacuum
1
Crossed fields are
(1) electric field E across plates Y1 and Y2
(2) magnetic field B due to Helmholtz coils X1 and X2
These are adjusted so there is no deflection of electron beam,
then forces are equal and opposite and -Ee = -Bev.
If electrons emitted from cathode with zero speed
½mv2 = eV.
If d separation of Y1, and Y2, E = V/d
Hence e/m = V/(2B2d2).
B could be determined using e.g. a Hall probe
Main difficulties are (1) magnetic/electric fields uniform and
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92 AL Physics/Essay Marking Scheme/P.8
(2) extend over same length of electron path.
6.
½
(a) Definition of s.h.m. - at all times the acceleration of the moving
body is proportional to its displacement from a fixed point
and is directed towards this point.
Examples Swinging suspended ceiling lamp, boat bobbing up and down
in the sea, child on swing, etc. (one only)
7
½
½
1
2
(b) (i)
Unstretched
position
Equilibrium
position
F = k (l + x)
l
P
x
mg
½
Suppose mass is moving downward below P.
Mass is decelerating and m
d2x
= k(l + x) - mg
dt 2
= kx, since mg = kl
1
½
2
k
d x
=  x is x = a cos 0t,
m
dt 2
k
where 02 =
m
Solution of
½
½
(ii) Modified equation of motion is :
d2x
dx
 b  kx  0
dt
dt 2
Substitute solution x = Aet cos t .
dx
Then
= Ae t cos t  Ae t sin t
dt
m
d2x
=  2 Aet cos t  Aet sin t
2
dt
Ae t sin t  2 Ae t cos t
Substituting, and taking t = 0, we obtain :
m[ 2 Aet  2 Ae t ]  bvAet  kAet  0
i.e. m(2 - 2) - b + k = 0
.......... (1)
½
1
2
1
3
92 AL Physics/Essay Marking Scheme/P.9

, we obtain :
2
........... (2)
m[Ae t  Ae t ]  bAe t  0
i.e.  = b/2m
Substituting in (1)
we derive 2 = 02 - 2
Substituting, and taking t =
(c)
L
I
Total p.d. across circuit is given by :
Q
dI
 L  IR  0
C
dt
d 2Q
dQ Q
R
 0
2
dt C
dt
By analogy with (b)(ii) we can write :
  R/2L,
2 = 02 - R2/4L2,
02 = 1/LC
or L
R
1
½
½
½
7
C
Q
½
1
1
½
½
½
4