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87 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
87’ AL Physics: Essay
Marking Scheme
1.
(a) (i) The rate of change of momentum produced in a body
is proportional to the resultant force acting on it
and occurs in the direction of the force.
½
½
(ii) The body will move with an acceleration a in the
direction of the resultant force F and a = F/m,
where m is the body mass.
½
½
(b) (i) The mass m of a body is constant and is a measure
of its inertia to any enforced change of state,
either stationary or moving. For a force F the
body experiences an acceleration  1/mass.
½
(ii) The weight W of a body can be variable and is the
force which would act if the body were allowed to
fall freely under the influence of gravity.
W = mg, where g is the local 'free-fall'
acceleration due to gravity.
½
(iii) In a lift moving downwards with an acceleration
(or for a lift moving up, suddenly stops), there
will be less reaction between the weight (force)
of the passenger and the floor of the lift, giving
the impression of weightlessness.
2
½
½
½
½
1
1
5
(c)
P
F
v
C
A constant (centripetal) force F acts towards the center
O of the circle. This provides an acceleration in this
direction for the rotating body. Only the speed v of
body is constant, its velocity (and momentum) are
continuously changing - due to the accelerating
force F.
½
½
1
2
87 AL Physics/Essay Marking Scheme/P.2
(d)
l
glass
tube
paper
marker
rubber
bung
screw nuts
1
The glass tube is held vertically, the bung is whirled
around above his head by one student and the speed of
bung is increased until the marker is just below tube.
Another student times, say, 50 revolutions of the bung.
By moving marker the length l of the string can be varied
and the relation between l and the angular velocity ()
obtained. Since T (tension provided by screwnuts*) and
mg are constant from T cos = mg it is clear that  is
constant. Also T sin = mr²,   1/r.
1
½
1
½
1
1
l

r
T
mr2
mg
*However there is friction between string and glass rod
which may vary throughout experiment.
2.
(a) (i) In a gas the molecules move randomly with high
speeds throughout all available space and are
comparatively far apart (~ 10  molecular
diameter). Thus normally the forces between
molecules are very small - except that on
collision with one another large repulsive
forces will occur. Molecules will give rise to
a pressure (force) on the sides of a container
due to impact of the moving molecules. When there
is an increase in temperature the average kinetic
energy of the molecules will increase and (1) in
free space gas will expand with molecules moving
into cooler gas or (2) if contained there will be
an increase in pressure on the sides of the
container.
1
7
1
1
½
½
½
½
4
87 AL Physics/Essay Marking Scheme/P.3
(ii)
Resultant force
Repulsion
Force
Separation
of atoms
r0
Attraction
1
Most solids are crystalline with molecules formed
of atoms at ‘fixed’ separations in a lattice.
The separations r0 are very close (< 1 molecular
diameter) and if atom nearer a repulsive force
predominates while farther away force is
predominantly attractive. The atoms at any
particular temperature vibrate about this mean
separation r0. At higher temperatures there is
asymmetrical vibration with the displacement
greater on the extension side - hence solid
expands.
1
1
1
4
(b) (i) Ionic Bonding
Cl
ions
Na
½
e.g. NaCl consists of +ve Na ions and -ve Cl ions formed because an electron is easily transferred
from the Na to the Cl atom. The ions are held
together in the lattice by attractive electrostatic
forces.
½
½
1½
(ii) Metallic Binding
Electron sea
Positive metal ion
nucleus plus inner
electrons
(Optional)
Electron
Metal atoms have ‘outer’ electrons which are
loosely held and can become detached, drifting
around randomly. All atoms share the ‘free’
electrons. The electrostatic attraction
½
½
1
87 AL Physics/Essay Marking Scheme/P.4
constitutes the metallic bond.
(iii) Covalent Binding
C
C
C
Carbon atom
Outer electron
Covalent
bonds
C
C
(optional)
½
C
Electron sharing occurs between two or more atoms.
e.g. 4 ‘outer’ electrons of C can be shared with
4 other C atoms to make 4 bonds - through
interlocking electron ‘clouds’. Bonds are very
strong.
½
½
1½
(c)
(Optional)
The glass fibres are strongly bound by covalent binding
forces and are difficult to break. They are imbedded in
a weaker yielding plastic material (matrix). The matrix
(1) binds and holds the fibres together so that any applied
load is transmitted to them, (2) it protects the surface of
the fibres from scratches/and possibility of cracks/breaks
occurring) and (3) if cracks do appear it should prevent
them spreading from fibre to fibre - crack deflected along
interface. Such a material combines great strength with
low weight.
3.
1
1
1
1
4
(a) (i) Light waves
E
v
B
E
v
1
B
These are a particular electromagnetic wave of high (1014 Hz)
frequency. They consist of an electric field E and
magnetic field B component which undergo time
variations in planes at right angles to each other
and the direction of propagation. The changing
magnetic field produces the changing electric field
(and vice-versa). Waves do not require any medium
in which to propagate with the relatively high speed
of 3  108 m/s.
½
½
1
1
½
½
5
87 AL Physics/Essay Marking Scheme/P.5
(ii) Sound waves
Displacement
(right)
C
R
C
R
Distance
0
(b)
(left)
C
R
C
Normal air
pressure when
wave absent
R
Pressure
Distance
(a)
2
These are of much lower frequency (50 - 2  104 Hz)
and are due to momentum transfer of energy from
pockets of air vibrating with S.H.M. about their
mean locations. There are associated pressure
changes of rarefactions R and compressions C.
Waves propagate outwards from vibrating source and
need a medium for propagation. Speed is much lower
~ 330 m/s in air.
½
½
½
½
½
½
5
(b) (i) Refraction
NIGHT
DAY
v
warmer air 
T
cool air 
OR
cool air 
warmer air 
v2 > v1
ground
v1 > v2
1
ground
Refraction is a bending of the propagation path of
the waves due to change of speed experienced when
either moving into another medium - or in the example
given, due to a change of temperature. Snell’s Law
of refraction holds good sin i / sin r = 1n2 = v1/v2.
½
½
2
(ii) Diffraction
P
obstacle
1
87 AL Physics/Essay Marking Scheme/P.6
Diffraction is a bending of the waves around the
edge of obstacles. there are centres of disturbance
on the wave front in plane of obstacle, such as P,
and these send out waves (spherical) as shown
propagating behind obstacle.
(c) For observable diffraction the wavelength of the waves has
to be comparable with the dimension of the opening. Briefly
for a ‘slit’ maxima occur due to re-inforcement of waves
from secondary sources at an angle  to normal of slit when
sin  = /a. ‘a’ being width of slit. For sound  ~ 1 m
while for light  ~ 6  10-7 m. Thus very narrow slits
needed.
4.
½
½
2
½
½
½
½
2
(a) Electromagnetic Induction takes place when
(i) there is a relative motion between a magnetic field
and an object which cuts the magnetic field and
(ii) there is a magnetic flux change e.g. through a coil
in a circuit, due to the changing current.
electric field, or e.m.f. E is induced, E = -d/dt
d/dt is the rate of change of magnetic flux.
Direction of E is so as to oppose the magnetic field
change (drive current/produce magnetic field for a
conductor).
(b) (i)
½
½
½+½
½
½
3
L/R
L
R
VL
VR
E
S
On closing switch S a back e.m.f. VL is produced
across L due to changing I and magnetic field
through coil. At first VL  E (slightly less
since a current has to flow) and there is a large
rate of change of current dI/dt, though I is small.
dI/dt reduces with time and also VL - leading to an
increase in I. Current I is a maximum when
dI/dt and the back e.m.f. VL = 0.
½
1
1
1
I
E/R
½
t
4
87 AL Physics/Essay Marking Scheme/P.7
(ii)
C/R
C
R
VC
VR
E
S
On closing switch S a back e.m.f. VC is set up
across C due to the flow of charge Q to C from the
battery. At first Q  0 and dQ/dt is large (no back
e.m.f.). The back e.m.f. VC increases with time as
the charge builds up on C and dQ/dt reduces. Charge
Q becomes constant (maximum) when VC = E and
I = dQ/dt = 0
½
1
1
Theory
VC = Q/C and I = dQ/dt = (E - VC)/R
Q
1
dQ
, where final charge Q0 = EC
dt 
0 ( Q0  Q )
CR
½
Integrating we obtain Q = Q0 (1  e t / RC )
2


Q
EC
t
(c)
Charge max
C
L
Charge max
(i)

(ii)
Current max
Charge max
(iii)
(iv)
Current max
Q
(v)
(not expected; for
reference only)
charge
I
time

EM
energy:
electrical
(E) only
ME
magnetic
(M) only

E
only
current
ME
EM
M
only
The p.d. across the capacitor C. VC = Q/C and so voltage
observed in C.R.O. will be of same wave form as that of Q.
E
only
½
87 AL Physics/Essay Marking Scheme/P.8
(i) C, initially, has max. charge Q - ENERGY IN
ELECTRICAL FIELD and I = 0. Current I starts to
flow due to p.d. across C and back e.m.f. set up
across L (greatest at first, and also dI/dt).
1
(ii) Q = 0, all charge transferred from one plate of C
to the other. I is a maximum (no back e.m.f.
VC = Q/C) - ENERGY IN MAGNETIC FIELD.
1
(iii) Q increases so that polarity of C (plates) is
reversed I = 0 when back e.m.f. VL = VC ENERGY IN ELECTRICAL FIELD, Cycle continues.
1
(Deduct 1 mark if energy drawn not correct.)
N.B. Due to dissipation of power in the resistance
of the circuit oscillations will decay in
amplitude with time.
½
4
(Answer need not be in above form - an explanation
of change of charge on C only required.)
5.
(a) Disintegrations follow the statistical law of chance as
to which particular nucleus at a particular time will
disintegrate. At any time the rate of disintegrations
is proportional to the remaining number of nuclei which
have not yet undergone disintegration (N).
dN
 N ,
Hence
 constant
dt
N dN
t
N
  dt  ln N  N
0
N0 N
0

i.e.
1
1

N  N0 et
1
3
(b) Completed reactions are :

 234
 234
238

234

 91 Pa 

 92 U
92 U 
90Th 
1
’s mass is 4 (a.m.u.) and charge is +2 e
’s mass is negligible and charge is -e
All reaction steps must show agreement in regard to
conservation of mass and charge.
1
1
(c)
+V (EHT)
R
source
*
GM tube
The Geiger tube has to have a very thin mica window (or a
spark counter could be used - better).
C
amplifier/scales
or ratemeter
1
½
3
87 AL Physics/Essay Marking Scheme/P.9
(i) -particles. With source and Geiger tube almost
touching the introduction of a sheet of paper will
stop all -particle with a resulting slight drop
in count rate.
(ii) -particles. The insertion of increasing thickness
of Al sheets will reduce the count rate, stopping
for a thickness greater or ~ 5 mm - when all ’s will have
been absorbed.
(iii) -rays. These are very penetrating and a Pb sheet
of thickness ~ 2.5 cm will be needed to reduce the
count rate to ~ zero.
(d) (i) Factors affecting dangerousness of a radioactive
source are :
(I) the activity strength (disintegrations per
sec.)
(II) -ray emitter has more penetrating power.
(III) the greater the half-life, the longer it will
be dangerous.
(ii) Precautions are :
(I) screening off of source by lead shielding
(absorber).
(II) monitor total radiation dose using photobadge or clip-on electrometer - check if
reasonable.
6.
dI
= LI0 cos t
dt
or VL = LI0 sin (t + /2),
i.e. VL has phase advance of /2 on current I
Reactance = (VL)rms/(I)rms = L
(a) (i) I = I0 sin t, VL = L
(ii) I = I0 sint =
I
0
dV
dQ
= C C
dt
dt
½
1
½
½
½
½
5
1
1
1
3
1
1
2
½
½
½
½
2
½

sin t dt = C dVC
1
1

cos t =
sin(t  )
C
C
2
i.e. VC has a phase retardation of /2 on current I
1
Reactance = (VC)rms/(I)rms =
C
VC = 
½
½
½
2
87 AL Physics/Essay Marking Scheme/P.10
(b)
VL - VC
= (L - 1/C)I
VL = LI
OR
VR = IR
V
I
I
VR = IR
1
VC = I/C
Hence impedance, Z  V / I  R2  (L 
(c) At resonance 0L =
R 2  (L 
I/I0 =
[ 0   ]L 
1 2
)
C
and
(e) (i) At resonance
2,
=
2
½+½
1
½
½
1
 1
[1 
] =R
 0C
0
[ 0   ]L 
simplifying,
1
1
and I0 = V/R
 0C
R
since Power = I2R
1
hence, L =R
C
(d)
1 2
)
C
1

[1 
] =R
 0C
0
[L 
1
]=R
 02C
 =
3
½
since  << 0
1
1
L
 02C
1
since
R
2L
½
(VC)rms = Irms /(0C)
(VC)rms = Vrms /(0RC)
= QVrms
½
½
½
(ii) This is possible since an equal anti-phase voltage exists
‘across’ the inductance - these voltage contributions
cancel - giving a much smaller voltage in complete
circuit, Vrms = IrmsR.
1
½
½+½
3
4