2010 h2 physics ans - ri
DESCRIPTION
RI PHYSICS ANSTRANSCRIPT
Question Answer
1 D
2 B
3 D
4 C
5 C
6 B
7 C
8 B
9 B
10 C
11 B
12 C
13 C
14 D
15 C
16 B
17 B
18 D
19 C
20 D
21 B
22 A
23 B
24 B
25 A
26 C
27 B
28 D
29 A
30 C
31 D
32 D
33 C
34 B
35 D
36 A
37 B
38 A
39 D
40 C
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1
1 (a) vernier callipers
zero error (do not accept parallax)
(b)
dV h
V d h
V d h
V
V
2
3
3
3
=π4
= 964.665 cm
∆ 2∆ ∆= +
2×0.01 0.1∆ = + ×964.665
8.50 17.0
= 8 cm
= (965 ±8) cm
2 (a) U W T
ρVg
V
U
ρ
1 1
1
-3 3
2
2 -3
-3
= -
= 50 - 40
=10 N
10 =
10=
1000×9.81
=1.02×10 m
= 50 -34
=16 N
16=
1.02×10 ×9.81
=1600 kg m
(b) (i)
(ii)
s
s
Taking moments about O,
5.00×500+(70×9.81× )=1000sin53.0°×10.0
=7.99 m
(iii)
R T
R T
R
x
y
2 2
At equilibrium, the net vertical and horizontal forces must be zero.
= cos53.0°= 602 N
= 500+(70×9.81) - sin53.0°= 388 N
= 602 +388
= 716 N
2010 JC2 Prelim H2 Physics Paper 2 Suggested Solutions
weight of
beam force by worker
on beam
tension reaction
force
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2
3 (a) Gravitational field strength acts in the direction of decreasing potential / As distance
from the centre of a spherical mass increases, gravitational field strength decreases
but the gravitational potential increases.
(b) Total energy on surface = Total energy at infinity where PE=0 & KE=0 (since
spacecraft launched with minimum kinetic energy)
−
E s
E
GM m
Rmin
KE + = 0
∴
-11 24
min
11
6.67×10 ×5.98×10 ×2250KE =
6370000
=1.41×10 J
(c) Gravitational potential energy heat (air resistance) + light (fire over its body) +
sound energy (due to rapid vibrations of the body of the spacecraft)
(d) True weightlessness occurs in a situation where the astronaut is remote from the
gravitational field of celestial bodies i.e. gravitational force is zero / net gravitational
field strength is zero / experiences no net acceleration due to zero gravitational
force on him.
The astronaut and the spacecraft he is in are both accelerating with the same
acceleration, hence the normal contact force on him is zero / the gravitational force
on the astronaut entirely provides for the centripetal force for him to be able to
move in circular orbit, hence the normal contact force on him is zero. Here,
weightlessness is just a sensation as the gravitational force on him still exists.
4 (a) It is the sum of the kinetic energies and potential energies of the particles in the
gas.
(b) (i)
=pV nRTUsing ,
−× × ×
= =×
=
pVT
nR
5 31.0 10 5.0 10
0.20 8.31
300 K
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3
(ii)
5 (a) (i) 0.25
sin 14.48 14.5R
Rθ θ= ∴ = = °
Sine rule and cosine rule are also accepted.
(ii)
(iii)
( )( )
( ) ( ) ( )
29
3
2
0
tan
15 1010.5 10 9.81 tan 14.5
4 0.5
0.0799 m
8.0 cm
EF mg
R
R
θ
πε
−
−
=
×= ×
=
=
(b) θ will be smaller.
R R
0.50 R
hemispherical bowl
2θ
Normal reaction
force due to the
bowl, N
Electrostatic
Force due to
other ball, FE
Weight of ball, W
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4
6 (a)
Magnetic flux
through small
coil
time
Induced
e.m.f in
small coil
.
time
(b) Magnetic flux through the small coil is proportional to the magnetic flux density
which is proportional to the alternating current in the large coil. Hence the magnetic
flux-time graph through the small coil has the same shape as that of the current-
time graph for the large coil.
Induced e.m.f. in the small coil is proportional to the rate of change of magnetic flux
linkage which is obtained from the negative of the slope of the magnetic flux- time
graph.
(c) When the planes of the two coils are at 900 to each other, the magnetic field due to
the current in the large coil is parallel to the plane of the small coil. Hence there is
no magnetic flux linkage with the small coil and no e.m.f. will be induced in the
small coil.
The trace on the c.r.o. will show the amplitude of the induced e.m.f. reduced to
zero.
7 (a) (i)
4
6
4
5.6 24 14 1881.6
1881.6 3000 60 60 2.0321 10 MJ
1kWh 1000 60 60 3.60 10 J
2.0321 10 MJ 5644.7 kWh
P UA T
E Pt
∆= = × × =
= = × × × = ×
= × × = ×
∴ × =
(ii) 1 2
9
( ) (5.6 3.2) 24 14 806.4
806.4 3000 60 60 8.7091 10 J 2419.2 kWh
Savings 2419.2 0.25 $604.80
P U U A T
E Pt
∆ ∆
∆ ∆
= − = − × × =
= = × × × = × =
= × =
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5
(b) (i)
1 2 3
1
1 1 1 1 1 1 1
1.4 1.9 1.4
1 1 10.51154
1.4 1.9 1.4
0.51154 60 14 430
C
C
C
U U U U
U
P U A T W∆
−
= + + = + +
= + + =
= = × × =
(ii) t / mm P / W
50 430
100 250
150 170
200 130
(iii)
Note: Heat transfer via conduction dominates and the relationship between rate
of heat transfer P and thickness t of thermal insulation is given to be ∆
=kA T
Pt
,
where k is the thermal conductivity of the medium, A is the surface area normal
to direction of heat transfer and ∆T is the difference tempearture Thus, graph
should be an inverse curve.
300
400
200
100
0
x
x
x
x
t / mm
150 200 250 50 100
P / W
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6
(iv) The extrapolation of the graph to 250 mm may not be accurate because:
- 250mm is outside of the range of data collected.
- the behaviour of the graph/relationship may change beyond 200mm.
- there are too few data-points to determine the shape of the curve accurately.
- it is difficult to extrapolate a non-linear graph accurately.
(c) (i) From the graph,
=
= =
= × = =
= = = =
ideal
hheating e
h
heating e
e
COPe
QCOP
W
Q tWP
t COP
mod
mod
0.094
1( ) 10.638
( ) 0.60 10.638 6.3828
/ 5000783 W
( ) 6.3828
(ii) A heat pump requires less power than actually needed,
thus they are more cost effective (although not efficient).
(iii) Refrigerator, air-conditioner
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7
8
Diagram
Arrangement of LDR and electric light bulb:
Electric light bulb circuit:
LDR circuit:
Problem Definition
To investigate how the resistance of the LDR depends on the intensity of the
illumination incident on the LDR.
Dependent variable (R): resistance of LDR
Independent variable (i): intensity of illumination incident on the LDR
Controlled variables: distance between light source and LDR or light intensity
sensor
e.m.f. of dry cells
alignment of light source with LDR or light intensity sensor
black cardboard tube
to LDR circuit to electric light
bulb circuit
LDR / light
intensity sensor
electric light bulb
mA
1.5 V dry cell V
A
V 12.0 V battery
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8
Apparatus and Material
electric light bulb
rheostat
light-dependent resistor
light meter with light intensity sensor
digital ammeter
digital milli-ammeter
digital voltmeter
dry cells
connecting wires
black cardboard tube
sticky tape
metre rule
Procedures
1. Set up the apparatus as shown in the diagrams above. Tape the LDR in
position at one end of the black cardboard tube with adhesive tape. Tape
the electric light bulb at the other end of the black cardboard tube with
adhesive tape, using a meter rule to ensure that it is aligned along the
same horizontal axis as the LDR.
2. Adjust the rheostat in the electric light bulb circuit to maximum resistance.
Close the circuit to switch on the light bulb.
3. Measure and record the p.d reading V on the voltmeter and current
reading I on the milli-ammeter in the LDR circuit.
4. Replace the LDR with the light intensity sensor, connected to a light meter,
at the same position. Record the intensity reading i on the light meter.
5. Increase the resistance of the rheostat and repeat steps 3 and 4 to obtain
at least 6 sets of readings.
6. The resistance of the LDR can be calculated using the equation
VR
I=
7. The distance between the electric light bulb and the LDR or light intensity
sensor is kept at a constant distance throughout the experiment by using
adhesive tape to fix them in position and using a metre rule to measure the
distance between them before each reading to ensure that it is constant.
The e.m.f. of the dry cells is checked by connecting a voltmeter across them
before each reading to ensure that they remain at a constant value. The alignment
and orientation of the light bulb and LDR or light intensity sensor are kept constant
throughout the experiment by using adhesive tape to fix them in position.
Analysis
Assume that
nR k i=
Where i is the intensity of the illumination incident on the LDR,
R is resistance of the LDR , and
k and n are constants
Taking lg on both sides, +lglg lg kR n i=
.
Plot a graph of lg R against lg k.
If the above relationship is true, a straight line graph will be obtained where the
gradient is equal to n and the y-intercept is equal to lg k.
Hence 10ck = where c is the y-intercept.
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9
Safety Precautions
1. Do not look directly at the bright light source. Wear polaroid protective
glasses.
2. Do not touch the hot light source with bare hands. Wear gloves when
handling the light bulb after use.
3. Do not handle electrical circuits with wet hands.
Producing Reliable Results / Additional Details
1. As the resistance of the LDR is quite large, the current reading will be
small. Hence a milli-ammeter should be used to measure the current in the
LDR circuit.
2. The voltmeter should be placed to measure the potential difference across
both the ammeter and LDR when the resistance of the LDR is high. This
will ensure that the ammeter will measure the small current through LDR.
3. The electric light bulb and LDR or light intensity sensor are placed in a
black cardboard tube to minimise light from the surroundings from
reaching the LDR or light intensity sensor.
4. Wait for intensity and multimeter meter readings to stabilise before
recording.
Other accepted variations of answer
1. The multimeter can be used as an ohmmeter to measure the resistance of
the LDR (note that in such a case, the ohmmeter is connected directly to
the LDR, no e.m.f. source is required).
2. Intensity of the light incident on LDR can be varied by varying the distance
between light source and LDR. Power of the light source must be kept
constant in such a case.
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1
Physics Prelim Paper 3 Marking Scheme
SECTION A
1 (a) It is the motion of a body where its acceleration is directly proportional to its
displacement from a fixed point and is always directed towards that fixed point.
(b) (i) B is being forced to oscillate because P is heavier.
(ii) 1. Increases the damping of the rod B.
2. Decreases the natural frequency of rod B.
3. Increases the coupling between pendulum P and rod B.
Increases the amplitude of the driving force on rod B.
(iii)
2 (a) - The electrical resistance R of a conductor is defined as the ratio of the p.d. V
across it to the current I through it.
[No definition of electrical resistance – minus 1M]
- The electrical resistivity ρ of a material is the constant of proportionality relating
the electrical resistance R to the dimensions of the material (length and area).
OR word definition of L
RA
ρ= or A
RL
ρ =
OR R depends on dimensions while ρ is a material characteristic/property
(b)
( )
( )
22
2 2
2 2
2
2
4
4
OR
0.350.00.15
30.0
BC BCAB AB AB
BC BC BCAB AB
L L LR
A dd
d dR L L
R L Ld d
d
d
ρρ ρ
ππ= = =
= × ×
= × =
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2
Physics Prelim Paper 3 Marking Scheme
(c) (i) Terminal p.d. for the 2.00 V battery:
2.00 1.00 V2
r BJ
rV V
r= × = = (null deflection)
Since balance length1
2BJ BCL L= , 2 2 2.00 VBC BJ rV V V= = = .
(ii) Method 1
From part (c)(i),
2.00 VBCV =
Method 2
From part (c)(i),
2.005.00
0.400
BCBC
VR = = = Ω
I
From part (b), 0.15
AB AB
BC BC
R V
R V= =
2.00 0.15 0.30 V
0.30 2.00 2.30 V
AB
AC
V
V
= × =
= + =
OR
2.00 1.15 2.30 VACV = × =
( )2.50 2.30 0.400
0.500
ACE V IR
R
R
= +
= +
= Ω
From part (b), 0.15AB
BC
R
R=
5.00 0.15 0.75
0.75 5.00 5.75
AB
AC
R
R
= × = Ω
= + = Ω
OR
1.15 5.00 5.75 ACR = × = Ω
2.50
0.400
0.500
T ACR R R
R
= = +
= Ω
(d) Over-estimate. 0.20 V is actually the p.d. across R as well as that of the ammeter.
( )
0.500 0.500
AC A
A
E V I R R
R R R
− = +
+ = Ω ⇒ < Ω
To show that the calculated R is an overestimate, there must be some statement
relating the resistance of the ammeter to either
- pd calculated for R actually includes the pd across the ammeter
- resistance calculated of R actually includes the resistance of the ammeter
3 (a) (i) E1 Ground state
E2 Metastable state
E3 Excited state
B2 for any 2 correct answers
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3
Physics Prelim Paper 3 Marking Scheme
(ii) Electrons are pumped up / excited from E1 to E3.
Electrons at E3 will quickly spontaneously decay from E3 to E2. Since E2 is
the metastable state, a population inversion is created between E2 and E1.
When one electron falls spontaneously from E2 to E1, a photon will be
emitted. This photon will go on to stimulate emission of electrons at E2.
These photons form the laser light.
(iii)
3
2
4
4 (3.0 10 )
26.53
minimum 26.5
E t
hE
t
hh f
f Hz
E f fE hf E hf h f
E f f
f Hz
π
π −
∆ ∆ ≥
∆ ≥∆
∆ ≥×
∆ ≥
∆ ∆ ∆ = ⇒ = ⇒ ∆ = = ∆
∆ =
(b)
(i) Occurs when the most energetic electrons are stopped completely and all
their kinetic energy is converted to photon energy.
(ii) Since
min
hc
e Vλ =
∆, minimum wavelength will increase when the accelerating
potential is decreased.
The intensity at all wavelengths will decrease because the speeds / kinetic
energy of the electrons are decreased.
minλ
0
Intensity
Kβ
K
α
Wavelength
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4
Physics Prelim Paper 3 Marking Scheme
4 (a) In the magnetic field,
2
mv mv
Bqv rr Bq
= ⇒ =
Since Tv
=2πr
,
2mvT T
v Bq Bq
mπ ∴ = ⇒ =
2π
(b) (i) 2T
π∴ = =
-27
-7
-19
)1.54x10 s
0.85 x 2 x 1.6 x 10
(6.68 x 10
(ii) In order for the nucleus to accelerate when it crosses the gap, freq. of the
alternating voltage = orbital freq. of the nucleus
1f∴ = =
6
-76.49 x 10 Hz
1.54 x 10
(iii) KE after one revolution = Work done by e-field on helium nucleus = 2qV = 2(2e)V = 4eV
(iv) The gain in KE after each rev. = 4eV
The gain in KE after five rev. = 20eV
2120
2mv eV∴ =
5406 57 10 -1
m s.eV
vm
= = ×
Comments:
(b)(ii) Common mistake is the failure to realize that in order for the nucleus to accelerate
when it crosses the gap, freq. of the alternating voltage = orbital freq. of the nucleus
because the ion crosses the gap twice in one revolution. A handful of students halved
the period of the ion and took the reciprocal to calculate the frequency of the voltage
supply which is incorrect.
(b)(iii) Common mistake 1: KE = 2eV.
Common mistake 2: KE = 2e (450 V) = 900 eV (eV is not electron-volt !)
[B1 – for sub]
[A1 for final answer]
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5
Physics Prelim Paper 3 Marking Scheme
SECTION B
5 (a) N = W cos θ
= (3.00 x 9.81) cos 30.0o
= 25.5 N
(b) a// = g sin θ
= 9.81 sin 30.0o
= 4.91 m s-2
(c) h/s// = sin θ
s// = h / sin θ
=0.500 / sin 30.0o
= 1.00 m
v//
2 = u//
2 + 2 a// s//
= 02 + 2 x 4.91 x 1.00
= 9.82 or 9.81
v// = √9.82 = 3.13 (shown)
(d) sy = uy t + ½ ay t2
2.00 = (3.13 sin 30.0o) t + ½ (9.81) t
2
4.905 t2 + 1.565 t – 2.00 = 0
t = 0.499 s
sx = ux t + ½ ax t2
= (3.13 cos 30.0o) (0.499) + 0
= 1.35 m
(e) (i) Impulse = area under F-t graph
= ½ (0.200) (360)
= 36.0 Ns
(ii) vy = uy + ay t
= (3.13 sin 30.0o) + (9.81) (0.499)
= 6.46 m s-1
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6
Physics Prelim Paper 3 Marking Scheme
∆py = m [vy – uy]
36.0 = 3.00 [vy – (–6.46)]
vy = 5.54 m s-1
(iii) Since the KE or speed after collision is smaller, the collision is inelastic.
During collision, the KE of the ball is converted into sound energy, thermal
energy and/or elastic PE as the ball deforms.
6 (a) Any two:
1. The waves must be coherent.
2. The waves must have approximately the same amplitude.
3. The waves must be unpolarised or polarised in the same plane (for
transverse waves).
4. The waves must interfere to give regions of maxima (constructive
interference) and minima (destructive interference).
(b) Intensity required at 12 km away,
( )
( )
11
7 -2
4
2.5 105.0 10 W m
0.50 10
eye
eye
PI
A
−
−
−
×= = = ×
×
Consider light guide,
.
light
light
PI
A
P I A
=
=
( ) ( )27
5.0 10 4 12000π− = ×
905 W=
(c) (i) Shorter wavelengths means the anti-nodal lines will be closer to one another.
Hence, aircrafts may “lock on” to the wrong line of maxima / difficult to identify
the central line of maxima / difficult to differentiate the lines of maxima.
(ii) Since the two radio waves are in phase, along centre-line, path difference is
always zero / phase difference is always zero / P & Q are equidistant from
any point on the centre-line. Hence constructive interference occurs.
(d) (i) P is nearer to the aircraft.
Hence intensity (or amplitude) of signal should be higher.
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7
Physics Prelim Paper 3 Marking Scheme
(ii) From Fig. 6.3, phase difference
2
π= rad
(iii) Since phase difference of signals
2
π= ,
path difference 4
λ=
Distance from P to plane 2 2 2
4800 180 480 4827.30 m= + + =
Distance from Q to plane 2 2 2
4800 230 480 4829.42 m= + + =
Hence ( )4829.42 4827.304
λ− =
8.49 mλ =
635.3 10 Hz
cf
λ= = ×
Note:
Do not accept if student uses formula D
xa
λ= as it is not a 2-D problem.
Accept if student choose path difference as 5
4
λ,
9
4
λ, etc
(e) (i) If aircraft is on the central anti-nodal line,
it should detect maximum signals from both frequencies / the maximum
signal will be stronger
OR
If aircraft is on wrong anti-nodal line,
only one of the frequencies will show a strong signal.
(ii) If the ratio is an integer ratio,
It means that higher orders of maxima will still coincide/overlap
Hence the aircraft could still detect maximum signals from both frequencies
even though it is not on the central anti-nodal line.
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8
Physics Prelim Paper 3 Marking Scheme
(f) Advantage:
Can still work under low visibility conditions / Use of detector to align aircraft is
more accurate than using visual inspection.
Disadvantage:
Possible interference of signals from other sources (e.g. radio stations,
telecommunication base stations, etc) / It is more costly to install the emitters and
receivers on every airplane.
7 (a) Binding energy is defined as the amount of energy needed to split a nucleus into
its individual nucleons.
(b) (i) 4 3 1
2 2 0He He n→ +
(ii)
( ) ( )
Difference in total BE
4 6.8465 3 2.2666
20.5862 MeV
Q =
= −
=
(iii) ( ) ( ) ( )
( ) ( )
He-3 He-4931.494
He-4 He-3
20.58621.0087u
931.494
0.9866u
Qm n m m
m m
+ − =
−
= −
=
(iv) Energy is supplied in order to conserve mass-energy.
(c) (i) Helium-2 is unstable and cannot exist in a bound state.
(ii) Large coulomb repulsion between the protons.
(d) (i) Half life is the time taken for a sample of radioactive atoms to decay to half
its initial number.
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9
Physics Prelim Paper 3 Marking Scheme
(ii) ( )
( )
( ) ( )
.......... 1
........... 2
1 / 2 :
ln2
9ln2
6.149
0.682
P P
S S S
P P
S S S
P
S
S
P
P S
S
A N
A N
A N
A N
NT
NT
TN
T N
N N
λ
λ
λ
λ
=
=
=
=
= ×
=
(iii) 0
ln2
6.14
0
ln2
0
ln2
6.14
0
ln2
0
ln2 ln2
6.14
Let the total initial activity be
0.1
0.9
0.19
0.9
81
1 1ln2 ln81
6.14
ln81108 days
1 1ln2
6.14
P
P
P
P
P P
tT
S
tT
P
tT
S
tP T
t tT T
P P
P P
A
A A e
A A e
A A e
AA e
e
tT T
t
T T
−
−
−
−
−
=
=
= =
=
− =
= =
−
(iv) It can be used to trace a plant’s fertilizer uptake.
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