2014 VCAA Physics Examination Solutions
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Area of study − Motion in one and two dimensions
Q1a u = 0 , t = 5 , a = + 0.20 , s = ut +
1 2 1
at =
2
2
( 0.20)(5 ) = 2.5
+
2
.: distance travelled = 2.5 m
Q1b Tension in the coupling
= mass of the two trucks × acceleration = 20 000 × 0.20 = 4 000 N
Q4a Mary and Bob feel weightless when they move under gravity
mv 2
only, .: a = mg ,
= mg , v = gr = 10 × 20 ≈ 14 m s-1
r
Q4b At point A they are in free fall and gravity is the only force
on them. There is no contact reaction force on them, .: they cannot
feel their own weights, i.e. they feel weightless.
Q4c
Q1c Total momentum after = total momentum before collision
80 000v = 40 000 + 4.0 , .: v = + 2.0 , .: the speed is 2.0 m s-1
(
)
Q1d Total kinetic energy after =
1
(80 000) 2.0 2 = 160 000 J
2
( )
1
(40 000) 4.0 2 = 320 000 J
2
Kinetic energy is not conserved, .: the collision is inelastic.
( )
Total kinetic energy before =
Q2a From the given table extension is proportional to the force of
gravity, .: the spring follows Hooke’s law and its spring constant is
0.050 × 10
given by the constant of proportionality k =
= 5.0 Nm-1
0.10
Q2b When the 0.20-kg masses are released from the unstretched
position, they move downwards and the gravitational potential
energy decreases. At the lowest point (x metres below the released
position) the gravitational potential energy is converted to elastic
1
potential energy, .: 0.20 × 10 × x = × 5.0 × x 2 , .: x = 0.8 m
2
The extension at the lowest point is 0.8 m.
Q5a a = g ,
M =
4π 2 r GM
= 2
T2
r
(
)
3
4π 2 r 3
4π 2 7.0 × 1010
=
≈ 1.1 × 1031 kg
2
GT 2
6.67 × 10−11 (1200 × 3600)
Q5b All objects around the star at the same orbital radius have the
same period, independent of the mass of the objects. It is not
possible to determine the mass of the planet from the given data.
Q2c The graphs are correct but they do not add to a constant
because kinetic energy of the system is not included in the total
energy.
Area of study − Electronics and photonics
Q2d The speed is maximum when the acceleration and hence the
net force is zero, .: 5.0 x + − 0.20 × 10 = 0 , .: x = 0.40 m
1
The total energy of the system = × 5.0 × 0.82 = 1.6 J
2
At extension 0.40 m extension:
1
1
× 0.20v 2 + × 5.0 × 0.4 2 + 0.2 × 10 × 0.4 = 1.6
2
2
.: v = 2.0 , .: the maximum speed is 2.0 m s-1
Q6b R 1 and R 2 form a voltage divider ,
300
× 8.0 = 2.0 V
Voltmeter reading = V2 =
900 + 300
Q3a Vertical component: u = + 20 sin 30°= + 10 , v = 0 , a = − 10
v 2 = u 2 + 2as , .: 0 = 100 − 20 s , s = + 5.0
.: maximum height reached is 5.0 m.
Q7a Power dissipated in the LED:
P 300 × 10 −3
P = VI , I = =
= 150 × 10 −3 A
V
2
V
9
VR = 11 − 2 = 9 V, .: R = R =
= 60 Ω
I
150 × 10 −3
Q7b VR = 12 − 3.0 = 9.0 V, I R = 0.50 A, PR = 9.0 × 0.50 = 4.5 W
26
Q3b Horizontal component: t =
≈ 1.5011 s
20 cos 30°
1
Vertical component: t = 1.5011 , a = 10 , u = 10 , s = ut + at 2
2
1
.: s = + 10 × 1.5011 + × − 10 × 1.50112 ≈ + 3.74 , .: h ≈ 3.7 m
2
−
Q6a 0 V
Q8a Read from graph: I = 3.6 mA
Q8b Read from graph: V = 1.3 V
+
Q8c When V = 1.0 V, I = 1.6 mA, R =
V
1.0
=
= 625 Ω
I 1.6 × 10 −3
1
2014 VCAA Physics Exam Solutions
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Q14a 0 V
Q9
Q14b
N primary
V
130
Vin
, in =
, Vin = 10 V (rms)
=
Vout N secondary 400 5200
The peak value of Vin = 10 2 ≈ 14.1 V
Q15a Voltage drop (loss) in the wires = 13 − 3.0 = 10 V
Q15b For the light globe, P =
Q15c I =
V 2 3.0 2
=
= 6.0 W
R
1.5
V 3.0
=
= 2.0 A
R 1.5
V 6.0
=
= 4.0 A
R 1.5
For the wires, V = IR = 4.0 × 5.0 = 20 V
.: the DC power supply = 20 + 6.0 = 26 V
Q10a
Q15d For the light globe, I =
Q16 At the generator end of the transmission lines, a step-up
transformer increases the output voltage and thus decreases the
output current to provide the same power supplied by the generator
for transmission, Vout I out = Vin I in .
Lowering the current in the transmission lines reduces loss of
power in the lines since Ploss = I 2 R , where I is the current in the
lines and R is the total resistance of the lines.
At the user’s end of the transmission lines, a step-down
transformer decreases the high voltage in the lines to suitable
voltages for home or industrial uses.
Q10b
Q17a
Side
WX
XY
YZ
ZW
Q11 On/off voltage VC = 4 V, VLDR = 9 − 4 = 5 V
From graph, at 300 lux, RLDR = 0.5 k Ω
RV
4
.:
= , .: RV = 400 Ω
0.5 × 103 5
Area of study − Electric power
Force direction
down
no force
up
no force
Q17b F = nBIL = 75 × 0.020 × 2.0 × 0.40 = 1.2 N
Q17c The forces on WX and YZ produce a turning effect on the
rectangular coil.
Q17d
Q12a A
Position
before the vertical position
at the vertical position
after the vertical position
Q12b E
Current in side WX
from W to X
no current
from X to W
Q13a C and D
Q13b ξ av =
∆φ 0.050 × 0.080
0.40
= 20 Ω
=
= 0.40 V, R =
∆t
10 × 10 −3
0.020
Q13c
Q18a Before the coil turns to the vertical position, the magnetic
flux to the right increases and thus a current flowing in the
clockwise direction (viewed from above) is induced according to
Lenz’s law. As the coil continues to rotate past the vertical
position, magnetic flux to the right decreases and an induced
current flows in the opposite direction. The induced current
changes directions every half turn of the coil, .: AC
Q18b Period T = 2 × 25 ms = 50 ms , .: f =
As the downward magnetic field decreases, according to Lenz’s
law, there is an induced current flowing in the clockwise direction
(viewed from above) to generate a downward magnetic field inside
the loop to make up for the decrease in magnetic flux.
2014 VCAA Physics Exam Solutions
1
1
=
= 20 Hz
T 50 × 10 −3
Q18c When the period of rotation of the coil decreases, the
magnitude of the voltage output of the generator increases because
∆φ
1
| ξ av |= n
, i.e. | ξ av |∝
for constant n and ∆φ .
∆t
∆t
2
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Area of study − Interactions of light and matter
Q19a P is on the second bright band from the central bright band,
.: S1P − S2 P = 2λ = 2 × 420 = 840 nm
Q19b Now P is on the second dark band from the central bright
3λ
band, .: S1P − S2 P =
= 840 nm, .: λ = 560 nm
2
Q20a
Work function φ = hf threshold = 6.63 × 10 −34 1.8 × 1015 ≈ 1.2 × 10 −18 J
(
)(
)
Detailed study 1 − Einstein’s special relativity
1
B
2
D
3
C
Q3 L = Lo 1 −
4
B
5
A
6
D
7
B
Q10 Etotal = mc 2 = mo c 2
Q21b smaller than
Q11 Ek + E rest = mc 2 , E k + mo c 2 = γmo c 2 ,
λ
. If the same light beam is
w
used, λ remains constant, .: the extent of diffraction is inversely
proportional to the diameter of the aperture w, i.e. the pattern
spreads out less when the aperture diameter increases.
λX-ray =
hc
≈ 4 × 10 −11 m, λelectron =
EX-ray
h
2mEelectron
λ
. If the same aperture is used, w
w
remains constant, .: the extent of diffraction is directly proportional
to the wavelength , i.e. the pattern spreads out more for the X-ray
beam than for the electron beam.
Q22a Atoms can absorb or emit photons with energy
corresponding to the difference between two energy levels. There
is a level (the second excited state, 6.7 eV) 1.8 eV higher than the
first excited state of 4.9 eV, but there is no level 1.8 eV lower than
the first excited state.
Q22b The low values 0.9 eV, 1.5 eV and 2.2 eV suggest that the
energy levels are close to the third excited state (x eV).
Let 9.8 − x = 0.9 , x = 8.9
The value x = 8.9 gives the other two emissions, 10.4 − 8.9 = 1.5
and 8.9 − 6.7 = 2.2 .
h
6.63 × 10 −34
Q23a λ =
=
≈ 0.36 × 10 −9 m, i.e. 0.36 nm
mv 9.1 × 10 −31 2.0 × 10 6
(
)(
(
mo × 3.0 × 108
)
Q23b Electrons in stable orbits can be considered as circular
standing waves. A circular standing wave can be sustained if the
circumference is a whole number of wavelength. A different whole
number of wavelengths corresponds to a different energy level.
This explains the quantised energy levels in atoms.
)
2
C
10.0
≈ 42 × 10 −9 s = 42 ns
3.0 × 108
A
= 4.5 × 10 −11 , total mass mo ≈ 5.0 × 10 −28 kg
.: the rest mass of a pion ≈
5.0 × 10 −28
= 2.5 × 10 −28 kg
2
B
Work done = Ek = (γ − 1)mo c 2 = (3.00 − 1)mo c 2 = 4.5 × 10 −11 J A
Detailed study 2 − Materials and their use in structures
2
C
≈ 7 × 10 −12 m
i.e. λX-ray > λelectron
The extent of diffraction ∝
Q9 t = γto = 1.25 ×
1
A
Q21d Let EX-ray = Eelectron = 5 × 10 −15 J.
C
B
Lo 5.00
=
= 4.00 m
γ 1.25
hf is the energy of a particle (photon). There will be no emission
of electrons if the photon energy is lowered than the work function
of the metal.
hc 4.14 × 10 −15 × 3.0 × 108
Q21a λ =
=
≈ 3.0 × 10 −7 m
E
4.1
Q21c The extent of diffraction ∝
11
A
Q5 Distance = ct = 3.0 × 108 × 0.0100 = 3.0 × 10 6 m = 3000 km A
Q8 L =
)
10
B
v2
v2
,
, v = 0.10c
2
.
00
−
0
.
010
=
2
.
00
1
−
c2
c2
i.e. max Ek depends on the frequency of the light and the type of
metal used. It does not depend on the intensity of the light. Thus
the results of the second experiment provide evidence for the
particle-like nature of light. In the equation, max Ek = hf − φ ,
) (
9
A
Q4 Classical (non-relativistic) treatment
Q20b According to the particle model of light, max Ek = hf − φ ,
(
8
C
3
C
4
D
5
A
6
B
7
B
8
B
9
B
10
D
11
D
Q1 ε =
∆x 2.0 × 10 −3
=
= 1.0 × 10 −3
l
2.0
A
Q2 σ =
F mg 0.20 × 10
= 50 × 106 Pa = 50 MPa
=
=
A
A
4.0 × 10 −8
C
Q3 Young’s modulus for nichrome
200 × 10 6
= gradient =
= 2.0 × 1011 Pa (or N m-2)
1.0 × 10 −3
Q4 Toughness of tungsten = area under σ - ε graph
1
= (2.0 + 2.5) 10 −3 400 × 10 6 = 9.0 × 105 J m-3
2
( )(
(
)
)(
C
D
)
Q6 F = σA = 400 × 10 6 1.0 × 10 −8 = 4 N
4
Maximum number of 50 g masses =
=8
0.050 × 10
Q7 Sum of torque about X = 0
T × 0.40 sin 30° − 0.050 × 10 × 0.40 = 0 , .: T = 1.0 N
B
B
Q8 Let the vertical component of the force of the wall on the beam
at point X be FX . Vertically,
FX + T sin 30° − 0.050 × 10 = 0 and T = 1.0 .: FX = 0
B
Q9 Torque on the beam about point X due to the 150 g mass
B
= 150 × 10 −3 × 10 × 0.80 = 1.2 N m
(
)
3
2014 VCAA Physics Exam Solutions
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Detailed study 5 − Photonics
Detailed study 3 − Further electronics
1
D
2
B
3
C
4
A
5
B
6
A
7
D
8
B
9
C
10
D
N s Vs
6 1
=
=
=
N p Vp 24 4
Q2
11
A
B
Q3 Time constant τ = time taken to charge up the capacitor by
−3
63% . From graph, τ ≈ 20 × 10 s.
τ 20 × 10−3
C= ≈
= 20 × 10 −6 F (or 20 µF )
R 1.0 × 103
Q4 Frequency =
C
1
1
=
= 100 Hz
period 10 × 10 −3
A
Q8 Power loss (heat) = input power − output power
= 11 × 5 − 6 × 5 = 25 W
2
A
Q1 E =
3
B
4
D
5
A
6
C
7
B
8
D
9
B
10
C
F 7.2 × 10 −14
=
= 4.5 × 105 N C-1 (or V m-1)
q 1.6 × 10 −19
(
)(
(
)
11
C
B
)
(6.63 × 10 )(2.30 × 10 ) − (6.63 × 10 )(2.22 × 10 )
−34
18
8
3.0 × 10
−
−24
≈ 10 × 10 kg m s-1
18
3.0 × 10
10 × 10 −24
≈ 1.1 × 10 7 m s-1
9.1 × 10 −31
−9
 nλ 
−1  1 × 0.082 × 10
Q9 θ = sin 
 = sin 
−10
 2d 
 2 × 2.7 × 10
−1
7
A
8
A
9
D
10
D
11
A
 ncladding 
 , smaller ncladding , smaller θ c .
Q2 θ c = sin −1 
 ncore 
D
Q4 ∆E =
hc
λ
(4.14 × 10 )(3.0 × 10 ) ≈ 1.6 eV
−15
=
8
B
800 × 10 −9
 1.0 
Q8 θ = tan −1 
 ≈ 5.7° , a wide fibre collects more light
 10 
A
1
A
2
C
3
C
L − 12
Q2 I = 10 10
4
D
5
D
6
C
7
B
8
D
9
B
10
A
= 10 − 6
11
B
C
Q3 When distance is doubled, intensity becomes a quarter of the
original, .: level drops by 6 dB from 60 dB to 54 dB.
C
v
λ
=
340
= 100 Hz
2 × 1.7
C
3
340
λ = 1.7 , λ = 2.267 , f =
= 150 Hz
4
2.267
B
Q9 Without a baffle board the lower frequencies from the back of
the loudspeaker diffract more and interfere destructively with those
from the front of the loudspeaker. The baffle board stops this
process and John will hear the lower frequencies more loudly than
without it.
B
8
Electron: ∆p ≈ 10 × 10 −24 kg m s-1
.: speed ≈
6
C
D
Q7 Photon: ∆p = pfinal − pinitial
−34
5
B
B
Q7
=
4
B
Q1 1.40 × sin (90 − θ c )° = 1.00 × sin 15° , θ c ≈ 79°
Q6 f =
8 2
v
2.5 × 10
Q4 r =
≈
≈ 5.3 × 10 −3 m
a
1.2 × 1019
−
3
C
Detailed study 6 − Sound
A
qvB 1.6 × 10 −19 2.5 × 108 (0.27 )
Q3 a =
=
≈ 1.2 × 1019 m s-2
m
9.1 × 10 −31
2
2
D
B
Detailed study 4 − Synchrotron and its applications
1
A
1
B
B

 ≈ 8.7°

B
Please inform physicsline@itute.com re conceptual,
mathematical and/or typing errors
 nλ 
Q10 θ = sin −1 
 , .: θ is greater when d is smaller and .: less
 2d 
intensity peaks.
C
4
2014 VCAA Physics Exam Solutions
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