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93 AL Physics/Essay Marking Scheme/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
93’ AL Physics: Essay
Marking Scheme
1.
(a) Steady flow
- liquid elements which start at a given point always follow the same path
and have the same velocity at each point on the path.
1
½
Turbulent flow
- liquid elements which start at a given point take random paths
and their velocities vary in magnitude and direction
1
½
3
(b) Consider a cross-section of the pipe,
the liquid layer touching the pipe wall is always stationary
due to adhesive force between the liquid molecules and pipe wall.
The velocity of liquid is greatest at the centre.
Internal friction exists between liquid layers with different velocities
because of intermolecular forces
so velocity falls off gradually as the pipe wall is approached
½
½
½
½
½
½
3
(c) Liquid in contact with the bottom surface of the block moves at the block’s velocity v
Liquid in contact with the floor is stationary.
v
Thus a velocity gradient
is set up in the liquid of thickness t.
t
So, force required = liquid frictional force
=   Area  Velocity gradient
Av
=
t
Assumptions :
- uniform thickness of liquid
- the liquid is Newtonian
ANY
- constant velocity gradient across the thickness of liquid
THREE
- only the liquid lying below the block moves (ignore edge effect)
(d)
Marker A
falling ball-bearing
long glass tube
containing liquid
Marker B
½
½
½
½
½
½
½
½
½
4
93 AL Physics/Essay Marking Scheme/P.2
- The diameter of the tube is large compared with the diameters of ball-bearings
so that streamline conditions are satisfied.
- Marker A is far enough below the liquid surface
for the ball-bearing to have its terminal velocity at A
- Dip the ball-bearing in the liquid and thereby coated, before dropping so as
to reduce the chance of air bubbles adhering to the falling ball-bearing
- Avoid using ball-bearings of large radii as
their terminal velocities are high and vortices may form
- Release the ball-bearing at the centre of the tube
to reduce the effect of the wall of the tube on the streamlines
- Marker B is at a considerable distance from the bottom of the tube so as
to reduce the effect of the bottom of the tube on the streamlines
ANY
FOUR
2.
(a) progressive wave
- waveform advances as time goes on
- energy is transmitted along the direction
of travel of the wave
- particles within one wavelength have
ANY
different phases
FOUR
- all particles are vibrating
- all vibrating particles have the same
amplitude
stationary waves
- waveform does not advance
- energy is confined within the region of the
stationary wave
- all particles between two adjacent nodes
are in phase
- some particles (at nodes) have no vibration
- different particles have different amplitudes,
in particular, amplitude is maximum at
anti-nodes
A stationary wave is formed when there is superposition of
two waves of nearly equal amplitude
and equal frequency
travelling in opposite directions
1
½
1
½
1
½
1
½
1
½
1
½
½+½
½+½
½+½
½+½
½+½
½
½
½
½
(b) (i) At T1, the waves arrive in phase to produce a loud sound
Phase difference then increases between the waves due to different frequencies.
At T2, the waves are completely out of phase, little or no sound is heard.
Later at T3, the waves are in phase again and a loud note is heard.
½
½
½
½
y1
T2
T1
T3
time
½
y2
T1
T2
T3
time
Variation of
amplitude
y3
resultant
T2
1
T1
T3
T
Beats (not to scale)
time
6
6
93 AL Physics/Essay Marking Scheme/P.3
(ii) Suppose the beat period = T, then in time T
number of cycles of f1 = f1T
number of cycles of f2 = f2T
Assume f1 greater than f2, then
f1T - f2T = 1
1
f1 - f2 =
T
1
 beat frequency =
= f1 - f2
T
3.
½
½
½
1
½
½
3½
(c) Radar installed near the road sends microwaves of frequency f1 to a travelling car, then
the microwaves are reflected back to the radar.
Due to Doppler effect,
the observed frequency f2 of the reflected microwaves is slightly different from f1.
Hence by comparing the transmitted and reflected microwaves, beats are formed.
As the beat frequency (= f1 - f2) depends on the car speed, the car speed can be checked.
½
½
½
½
½
½
3
(a) (i) average drift velocity: ~10-4 m/s
(ii) speed of electrical signal: ~ 108 m/s
In a current-carrying conductor, electrons tend to accelerate along the opposite
direction of the electric field inside.
Due to collisions between electrons and atoms (ions) of the conductor,
electrons move in zig-zag paths
and drift with small displacement in unit time.
½
½
½
½
½
½
electric field direction
with electric field
e-
without electric field
small
displacement
1
Electric field travels at an extremely high speed in a circuit
so electrons at every point of the circuit are influenced by the electric field
nearly simultaneously as the switch is closed,
electrical signal results at once.
(b) v = average drift velocity
e = electronic charge
A = area of cross-section of the wire
n = no. of conduction electrons per unit volume
½ mark for ANY TWO of them
1
½
½
5
1
Suppose electrons drift a distance l along a wire in time t, the charge q flows through a cross-section
of the wire is
q = nlAe
½
Drift velocity
v = l/t
½
current in the wire
i = q/t
= (nlAe)/(l/v)
½
= nAve
3
93 AL Physics/Essay Marking Scheme/P.4
(c)
When a current flows through the meter, the coil experiences a couple,
the coil then turns until it is stopped by the increasing tension in the springs.
Thus the larger the current through the meter, the greater the forces on the coils,
and a greater angular deflection results.
To achieve a linear scale:
Set up a radial magnetic field
then flux density B of the field is nearly constant at the coil
The coil stops rotating when
 coil =  spring
NBAI = k ( = angular deflection)
hence I  
(N, B, A & k are constants)
2½
2½
½
½
½
½
½
1
½
½
½
½
8
93 AL Physics/Essay Marking Scheme/P.5
4.
1 2
mvm : maximum kinetic energy of photoelectrons
2
h : energy of incident photon
 : work function of metal
- the work required to remove an electron from the metal surface
(a)
½
½
½
½
2
½
+½
(b) (i)
Current i
- Vs
0
p.d. V
Photoelectrons emerge with different speeds (or K.E.).
When V is positive, current is constant
because all photoelectrons can reach electrode D
Current falls when V is negative
because the less energetic photoelectrons cannot overcome the potential barrier.
When V reaches the stopping potential (V = -Vs)
even the most energetic photoelectrons are repelled back so no current flows.
½
½
½
½
½
3½
(ii)
Current i
increased intensity (I)
½+½
original curve
½+½
increased frequency (II)
- Vs
0
p.d. V
(I) increased light intensity
constant frequency  same maximum K.E.
 Vs remains unchanged
intensity increased  no. of photons increases
 no. of photoelectrons ejected increases
 current increases
(II) increased light frequency
frequency increased  energy of each photon increases
 maximum K.E. of photoelectrons increases
 magnitude of Vs increases
For the same intensity, if energy of each photon increases
 no. of photons decreases
 no. of photoelectrons ejected decreases
 current decreases
½
½
½
3½
½
½
½
½
2
93 AL Physics/Essay Marking Scheme/P.6
(c)
Light is focused on the ‘soundtrack’ of a moving film.
The ‘soundtrack’ varies the intensity of light falling on a photocell.
The photocell creates a varying current,
thus produces a voltage which is amplified for driving a loudspeaker.
½
½
½
½
2
(d)
screen
diffraction
pattern
carbon film
+ 4000 V
1
electron
diffraction
A beam of electrons strikes a thin film of graphite just beyond the anode.
A diffraction pattern,
consisting of concentric rings, is observed on the fluorescent screen
showing the wave-like behaviour of electrons.
5.
½
1
½
3
(a) (i)
a.c. supply
L, R
C
a.c. supply
mA
OR
L
R
C
mA
LRC in series
a.c. supply & milliammeter
Set up the above circuit,
set the signal generator output to a value, say 3 V, with a measurable current,
and increase the frequency stepwise from a low value, say 10 Hz,
check whether the output is constant at the previous setting, 3 V,
then record the corresponding current readings on the a.c. milliammeter,
when frequency increases, the current reading rises and then drops.
½
½
½
½
½
½
93 AL Physics/Essay Marking Scheme/P.7
(ii) (iii)
smaller
resistance (I)
I
EACH curve 1/2 mark
larger
resistance (II)
f
f0
f0 = resonant frequency
current I and voltage across resistor VR are in phase,
I = VR/R
½ mark
for V0 lags behind VR0
½ mark
VL0

for VC0 > VL0
VR0
½
½
½
For frequency < f0,
V lags behind VR,
so VR0 < V0
V
V
 current I0 = R0  0 = Imax
R
R
½
V0
VC0
VL0
½ mark
for V0 , VR0 in phase
½ mark
VR0
for VC0 = VL0
V0
For frequency = f0
V and VR in phase,
so VR0 = V0
½
V
 current I0 = Imax = 0
R
½
For frequency > f0,
V leads VR,
so VR0 < V0
V
V
 current I0 = R0  0 = Imax
R
R
½
VC0
VL0
½ mark
for V0 leads VR0
½ mark
V0

for VC0 < VL0
VC0
VR0
11
93 AL Physics/Essay Marking Scheme/P.8
(b) (i) Radio signals from different transmitting stations induce e.m.f.s
of various frequencies in the aerial,
which cause currents flowing in the aerial coil.
Then currents of the same frequencies are induced in coil L by mutual induction.
If C is adjusted so that
the resonant frequency of the LCR circuit equals the frequency of the wanted station,
a large current and p.d. at that frequency only will develop across C
This selected and amplified p.d. is then applied to the next stage of the receiver.
(ii) Use an inductor of high L/R ratio
so increases the resonant current and hence the voltage across C.
6.
½
½
½
½
½
½
½
1
½
5
(a) (i) (ii)
Umax
P.E.
K.E.
P.E. Curve
- shape correct
- max. at xc - A and xc + A
½
½
K.E. Curve
- shape correct
- zero at xc - A and xc + A
½
½
xc marked
½
x
xc - A
xc
xc + A
In S.H.M., the total mechanical energy E is conserved
At maximum displacement, K.E. = 0, P.E. = Umax
Total mechanical energy of the system
E = K.E. + P.E. = Umax (constant)
so K.E. = Umax - P.E. for xc - A  x  xc + A
(b) (i) mass of each spring element, dM =
=
1 dl
l2
  M  v2  2
2 L
L
Mv 2
 l 2 dl
2L3
L Mv 2
Total kinetic energy of the spring =
 l 2 dl
0 2L3
Mv 2 1 3
=
 L
2L3 3
=

=
l
L
1
½
½
½
½
Mv 2
6
(ii) Extension of spring, y = x - xc
Total energy = K.E. of spring + K.E. of block + P.E. of spring
Mv 2 mv 2 ky 2


6
2
2
dE M dv
dv
dy

v  mv  ky
dt
3 dt
dt
dt
4
1
 kinetic energy of this spring element = ½ (dM) (V1)2
E=
½
dl
M
L
speed of spring element at a distance l from the fixed end, V1 = v 

½
½
4
½
1
½
93 AL Physics/Essay Marking Scheme/P.9
dx dy
dv d 2 x d 2 y

;


dt dt
dt dt 2 dt 2
dE
and
=0
dt
M
d2y
 (  m) 2  ky  0
3
dt
 v
or
d2y
dt
2
For S.H.M.
 
and
k

(
M
 m)
3
d2y
dt 2
k
½
½
1
y
  2 y
M
(  m)
3
2
T =

M
(  m)
= 2 3
k
(c) Damped oscillation takes place,
energy of the system decreases,
amplitude of oscillation gradually decreases to zero,
the frequency is smaller and the period is longer
½
½
½
½
6
½
½
½
½
2