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2013 - 2016 VCAA Physics Sample Exam Solutions
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SECTION A
Area of study – Motion in one and two dimensions
Q1a Conservation of momentum: (80 + 40)v = 80 × 4.0 ,
v = 2.7 ms-1.
Q1b The total kinetic energy after collision is different from that
before collision. Hence the collision is inelastic.
1
Before: E k = (80 ) 4.0 2 = 640 J
2
1
After: E k = (80 + 40 ) 2.7 2 ≈ 430 J
2
mv 2 250 × 32.0 2
Q2 Fnet =
=
= 2.56 × 103 N
r
100
(
)
(
17
= 34 m s-2
0.50
1
1
a = + 34 , u = 0 , t = 1.5 , s = ut + at 2 = ×+ 34 × 1.52 ≈ + 38 m
2
2
.: height above the ground ≈ 38 m
Q6c Acceleration =
Q7 At the point of leaving the rails (minimum speed) at A,
v2
= 10 ,
R = 0 , i.e. trolley is in free fall and a = g , ∴
7.0
v ≈ 8.4 m s-1.
Q8a Jason 2 in orbit is in free fall, i.e. it moves under the
gravity of Earth only.
)
Q3a Change in elastic potential energy = change in kinetic energy
1
1
k 0.10 2 = (0.20 ) 5.0 2 , k = 500 Nm-1
2
2
(
)
(
mg
)
Q3b Work done by friction = change in kinetic energy
1
F friction × 2.5 = (0.20) 5.0 2 , F friction = 1.0 N
2
( )
Q4a Constant velocity, zero acceleration, same as when Helen
is at rest.
Apparent weight = reaction force = mg = 60 × 10 = 600 N.
Q4b Take upward as the positive direction.
r
r
r
Fnet = ma , R + − mg = ma , R + − 600 = 60× − 2.0 , R = + 480 N
Apparent weight = 480 N
Q4c The gravitational force on Helen (her weight) remains the
same but Helen experiences apparent weightlessness, i.e. she
feels weightless because there is no reaction force on her while
she is in free fall (falling under gravity only).
Q5a Take upward as the positive direction.
Vertical component: a = − 10 , u = + 60 sin 30°= + 30 , v = 0
v 2 = u 2 + 2as , .: s = + 45 .
Hence the ball is 45 m above the top of the cliff.
Q5b Take upward as the positive direction.
Vertical component: a = − 10 , u = + 30 , t = 9.0 .
1
s = ut + at 2 , .: s = − 135 .
2
Hence the height of the cliff is h = 135 m.
Q6a Resultant force = 22 − 0.50 × 10 = 17 N
Q8b
T 2 4π 2
,T=
=
GM
r3
Q8c v =
4π 2 r 3
≈ 1.53 × 10 4 s.
GM
2πr
≈ 5.54 × 103 m s-1
T
Area of study – Electronics and photonics
Q9a V2 =
R2
30
×V =
× 14 = 6.0 V
R1 + R2
70
2
Q9b V1 = 14 − 6.0 = 8.0 V, P1 =
Q9c Total resistance RT =
I=
1
40
1
+
V1
8.0 2
=
= 1.6 W
R1
40
1
20
+ 30 =
130
Ω,
3
V
14
= 130 ≈ 0.32 A = 320 mA
RT
3
Q10a LED switch-on voltage = 3.0 V
1, 2 and 3 LEDs are in series
voltage across this series = 3.0 × 3 = 9.0 V
∴V2 = 12 − 9.0 = 3.0 V
V
3.0
I2 = 2 =
= 0.050 A = 50 mA
R2 60
Q6b Impulse of the gunpowder force = 22 × 1.5 = 33 N s
2013 -2016 VCAA Physics Sample Exam Solutions
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1
Q10b LED 2 is reverse biased, it stops the current flowing
through the series. LED 4, 5 and 6 are not affected.
LED 1
OFF
LED 2
OFF
LED 3
OFF
LED 4
ON
LED 5
ON
Area of study − Electric power
Q14a
LED 6
ON
∆v o
−9
=
= −300 , minus sign
∆vi 30 × 10 −3
means inverted output. The amplifier is inverting.
Q11a Voltage gain =
+
‒
Q11b
Q14b The magnetic flux is zero because the magnetic field
inside the solenoid is parallel to the plane of the rectangular
loop.
Q14c
9
Q
P
magnetic force F
−9
(
)
Q14d F = nBIL = 3 5.0 × 10 −2 (4.0 )(0.040) = 0.024 N
Q14e The force is zero because QR is parallel to the magnetic
field inside the solenoid.
Q12
Location
Waveform
P
B
Q
A
R
C
S
B
Q13a From graph: Rtherm = 5000 Ω at 5°C
Q13b From graph: Rtherm = 500 Ω at 25°C
Voltage divider:
500 Ω
R
∆φ
dφ
, ξ =−
, induced emf equals
∆t
dt
negative of the rate of change of magnetic flux (gradient).
Q15a Graph A, ξ av = −
Q15b A split-ring commutator alternates the contact of the loop
terminals with the two brushes. This happens every half turn of
the loop if the ring is split in halves. It is used in DC generators.
Slip rings maintain the contact of each loop end with the same
brush. It is used in alternators (AC generators).
Q16a
4V
Voltage at Q relative to the other end of R (V)
0
8V
1
2
4
time (s)
Q16b Faraday’s law for the relative magnitude of emf, and
Lenz’s law for the polarity of the voltage at Q relative to the
other end of resistor R.
R
8
= , R = 1000 Ω.
500 4
Q13c From graph: Rtherm > 500 Ω at temperature < 25°C.
8
To keep the same ratio , R > 1000 Ω. ∴ the variable resistor
4
should be increased.
Q16c ξ av = n
∆φ
3.0 × 10 −4
= 120 ×
= 3.0 V
∆t
0.012
Q16d Direction: Q → P
According to Lenz’s law, induced current flows in the direction
so that it generates a magnetic field to oppose the reduction in
the magnetic field of an external source.
Q17a f =
2013 -2016 VCAA Physics Sample Exam Solutions
1
1
=
= 12.5 Hz
T 80 × 10−3
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2
80
Q17b
RMS voltage output
peak-to-peak voltage output
= 2 ≈ 0.35
160
Q21a E =
hc
λ
(6.63 × 10 )(3.0 × 10 ) = 3.43 × 10
−34
=
8
−19
580 × 10 −9
J
Q21b S2X − S1X = 2λ = 1160 nm, .: λ = 580 nm
Q18a There is a drop in voltage in the transmission lines. .: the
voltage across the globe is less than 2.0 V and does not operate
at its optimal power of 5.0 W, .: the globe does not glow as
brightly as expected from a 5.0 W globe.
Vsetting = 12.5 + 2.0 = 14.5 V
2
Q22a Graph D. Intensity has no effects on the energies of
photoelectrons.
Q22b Graph A. Both graphs have the same gradient (same h),
and the ‘y-intercept’ for magnesium ( −3.7 eV) is above that for
selenium ( −5.1 eV).
P 5.0
=
= 2.5 A
V 2.0
= IR = 2.5 × 5.0 = 12.5 V
Q18b I =
Vdrop
S2Y − S1Y = 2.5λ = 1450 nm = 1.45 × 10 −6 m
Q23a λ =
h
6.63 × 10 −34
=
= 4.86 × 10 −11 m
mv 9.1 × 10 −31 1.5 × 10 7
(
)(
)
2
Q18c Ploss = I R = 2.5 × 5.0 ≈ 31 W
Q18d AC is often used so that transformers can be employed to
step up the voltage for transmission and thus lower the current in
the transmission cables. Hence power loss due to heating
Ploss = I 2 R is reduced. Then the voltage is stepped down to the
correct values for household/commercial usage.
Q18e Answer D.
V peak = 2 × 21.25 = 30.052 V
V peak −to − peak = 2 × 30.052 ≈ 60.10 V
Q23b Option A. Higher speed → higher momentum
→ shorter wavelength → diffraction to a lesser extent, i.e.
smaller ring radii.
Q23c X-rays are electromagnetic waves, and electrons have
wave behaviours. .: diffraction can occur for both. If their
wavelengths are equal, they will produce a pattern with similar
spacing.
Q23d Electron energy
= 500 eV = 500 × 1.6 × 10−19 = 8.0 × 10−17 J
(
)
Electron momentum = p = 2mEk
Q18f
(
n
1
=
, n = 146
1460 10
Q18g I sec ondary =
)(
)
= 2 9.1 × 10−31 8.0 × 10 −17 = 1.207 × 10 −23 kg ms-1
Since the pattern are similar, .: same λ and .: same p.
(
)(
Photon energy = E = pc = 1.207 × 10 −23 3.0 × 108
P 5.0
=
= 2.5 A
V 2.0
)
−15
= 3.620 × 10−15 J =
1
I primary = × 2.5 = 0.25 A
10
Ploss = I 2 R = 0.25 2 × 5.0 ≈ 0.31 W
3.620 × 10
1.6 × 10 −19
= 2.262 × 10 4 eV
≈ 23 keV
Q24 E =
Area of study − Interactions of light and matter
Q19 Young’s double-slit experiment produced an interference
pattern of alternating bright and dark bands of light from two
very close parallel slits, instead of just two bright patches of
light as suggested by the particle model. Interference pattern is a
wave phenomenon and can be easily demonstrated using water
waves from two dippers generating coherent circular waves at
the surface of water. .: Young’s experiment supports the wave
model.
Q20 Observation 2: Einstein used Planck’s photon (particle)
model to explain the photoelectric effect in his equation
KE max = hf − W , where f is the frequency of the light and W the
work function for the metal. Intensity of the incident light does
not appear in the equation. Clearly the energy of emitted
electrons depends on the frequency of the light and is
independent of its intensity. Only light, with frequency higher
than certain value depending on the metal (called the threshold
frequency), can cause the emission of photoelectrons.
hc
(4.14 × 10 )(3.0 × 10 ) = 2.6 eV
−15
=
8
−9
λ
478 × 10
En=4 − En=2 = 12.8 − 10.2 = 2.6 eV
n=4
n=3
n=2
Detailed study 1 − Einstein’s special relativity
1
C
2
A
3
C
4
B
5
B
6
B
Q1 Time taken = distance/speed =
7
A
8
D
9
B
10
B
1000
= 3.3 × 10−6 s.
3.0 × 108
11
A
C
Q2 Length is contracted when it is measured from a moving
frame. L = Lo 1 −
2013 -2016 VCAA Physics Sample Exam Solutions
v2
= 1.00 × 1 − 0.9 2 = 0.44 km
c2
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A
3
Q3 Vicky and Susanna are at the same location when Susanna
sends simultaneous flashes of light. .: Vicky also sees the flashes
of light sent simultaneously by Susanna.
C
Q7 200 MPa is below the elastic limit, .: strain energy per unit
1
1
B
volume = σε = 200 × 10 6 1.25 × 10 − 3 ≈ 1.3 × 105 J m-3
2
2
Q4
Q8 τ = Fd = (100 × 10)(1.20) = 1200 N m
D
Q9 Equate clockwise and anti-clockwise torques:
1200 + 200 × 0.60 = T (0.80 sin 30°) , T = 3300 N
D
Q10 The lower half is in tension and requires placement of
reinforcing steel rods.
C
Q11 All the blocks are in compression.
C
B
Q5




1
2
− 1
Work done = ∆Ek = Ek , f − Ek , i = Ek , f = mo c 
2
 1− u

c





2 
1
.: 1.0754 × 10 − 9 = 6.64424 × 10− 27 × 3.0 × 108 
− 1
 1− u 2

c


.: u ≈ 0.9339c
B
()
(
)
()
B
Q7 The robot measures the proper time
t
t 0 = = 784 × 1 − 0.85 2 = 413 microseconds.
A
) (
)
γ
D
Q9
B
Q10 Etotal = mc 2 , m =
Etotal 3.38 × 10 −11
=
= 3.76 × 10 −28 kg
8 2
c2
3.0 × 10
(
)
3.76 × 10 −28
Mass of a single muon =
= 1.88 × 10 −28 kg B
2
Q11
A
Detailed study 2 − Materials and their use in structures
1
D
Q1
2
D
3
A
4
D
5
D
6
B
7
B
8
D
9
D
10
C
11
C
Q1 Soft steel has greater strain beyond its elastic limit than hard
steel has.
D
Q2
D
Q3 The area under the graph is greater for soft steel than hard
steel.
A
(
)
Q4 ∆l = εl = 1.000 × 10 (10.000) = 0.0100 m
Length under stress ≈ 10.000 + 0.0100 ≈ 10.010 m
−3
D
Q5 Young’s modulus = gradient of the linear section
)
2
A
3
B
4
C
5
A
6
C
7
C
8
C
9
A
10
B
N 1 V1
240 2400 4800
=
=
=
=
N 2 V 2 10.5 105
210
11
C
D
Q2 V peak = 2 × V RMS = 2 × 10.5 ≈ 15 V, i.e. 3 cm
T=
Q8
)(
Detailed study 3 − Further electronics
1
D
Q6
Distance = speed × time
= 0.85 × 3.0 × 108 × 784 × 10 −6 = 199920 m ≈ 200 km
(
(
1
1
=
= 0.02 s = 20 ms, i.e. 4 cm
f 50
Q3 V av ≈ 10 V, I av ≈
A
Vav
10
=
= 0.02 A,
R L 500
Pav ≈ Vav I av = 0.2 W
B
Q4 V pp = 13.6 − 5 ≈ 8.6 V
C
Q5 The Zener diode ensures 6 V maximum, and the input to the
voltage regulator circuit dips to 5 V at times. Hence small
ripples of 1 V approximately.
A
Q6 Read the graph to find τ ≈ 3 s for 63% charging. RC = τ ,
τ
3
C
C= =
= 3 × 10 −3 = 3000 × 10 −6 F = 3000µ F
R 1000
Q7 After 60 s, the capacitor is fully charged to 10 V approx.
When the switch is moved to position Q, the capacitor starts to
V
10
discharge, I = ≈
= 0.01 A = 10 mA
C
R 1000
Q8 It is a full-wave rectifier circuit.
Q9 A multimeter set on AC volts measures RMS voltage.
V peak
10
V RMS =
=
≈ 7V
2
2
Q10 V Zener = 6 V, ∴V2 = 10 − 6 = 4 V
C
A
B
6
≈
400 × 10
≈ 1.7 × 1011 N m-2
2.4 × 10 − 3
(
)(
D
Q11 I 2 =
)
Q6 F = σA = 200 × 106 8.0 × 10 −5 = 1.6 × 10 4 N
B
V2
V
4
6
=
= 0.04 A. I 1 = 1 =
= 0.002 A
R 2 100
R1 3000
I Zener = I 2 − I 1 = 0.038 A = 38 mA
2013 -2016 VCAA Physics Sample Exam Solutions
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C
4
Detailed study 4 − Synchrotron and its applications
1
D
2
C
3
B
4
C
5
B
6
B
7
A
8
A
9
C
10
B
11
D
 1.38 
Q5 ic = sin −1 
 = 73.4°
 1.44 
D
Q6
C
B
Q1
D
Q7
Q2
C
Q3
B
Q8 The refractive index of water (1.33) is higher than that of the
plastic rod (1.20). Total internal reflection cannot now occur.
C
Q4 eV =
1 2
mv
2
(
Q9
)(
mv 2
9.1 × 10 − 31 4.6 × 107
V=
=
2e
2 1.6 × 10 −19
Q5 B =
(
(
)
)
)(
2
≈ 6017 V
C
)
B
(
)
)(
)(
)
Q6 F = evB = 1.6 × 10 −19 4.6 × 10 7 5.0 × 10 −4 ≈ 3.7 × 10 −15 N
B
Q7 Same wavelength → same energy → elastic scattering
Q8 λ =
(
)
2d sin θ 2 0.314 × 10 −9 sin 15°
=
≈ 0.163 × 10 − 9 m
n
1
nλ
nλ
= sin θ , θ < 90° , sin θ < 1 , .:
< 1,
2d
2d
n 0.200 × 10 −9
< 1 , n < 3.14
2 0.314 × 10 − 9
6
C
7
B
8
C
Q1
Q2 λ =
)(
9
D
10
C
11
B
)
hc 6.63 × 10 −34 3.0 × 108
=
≈ 6.22 × 10 − 7 m = 622 nm
−19
Ve
2.0 1.6 × 10
C
(
)
Q4
If =
If
Ii
=
ri 2
rf
2
6
C
7
A
8
C
9
C
10
A
I
1.7 × 10 −3
= 10 × log10
≈ 92.3 dB
−12
10
10 −12
11
A
B
B
C
ri 2
rf 2
× Ii =
20 2
× 1.7 × 10 − 3 ≈ 1.9 × 10 − 4 W m-2
60 2
Q5 The point, 10 000 Hz at 60 dB, is on the 40 phon curve.
Q7 f1 =
B
B
VR
10
=
= 0.025 A = 25 mA
R 400
B
v
v
333
, L=
=
≈ 0.33 m
4 f1 4 × 256
4L
A
Q8 f 3 = 3 × f1 = 3 × 256 = 768 Hz
C
Q9 It will sound louder.
C
Q10 L =
Q3 Vdiode = 2.0 V, VR = 12 − 2.0 = 10 V,
I ammeter = I R =
5
B
v 330
=
= 1.5 m
f 220
C
(
4
B
Q6 Sounds of different frequencies on the same phon curve will
be perceived of equal loudness by human hearing.
C
Detailed study 5 − Photonics
5
D
3
C
Q2 λ =
D
4
C
2
B
A
C
Q11
3
B
1
B
Q3 L = 10 × log10
)
)
2
C
Detailed study 6 − Sound
Q1 Sound waves in the air are longitudinal.
Q10 The vertical magnetic fields cause the electron beam to
zigzag horizontally.
B
1
C
B
A
Q9
(
(
Q10 Add up the attenuation due to Rayleigh scattering and
absorption. At 1200 nm the sum is about 1.6 watt/km which is
the least signal loss.
C
Q11
9.1 × 10 −31 4.6 × 107
mv
=
≈ 6.5 × 10 − 4 T
re
0.40 1.6 × 10 −19
(
D
λ
2
=
0.325
≈ 0.163 m
2
Q11 High frequency sounds do not diffract much.
hc
, blue light has shorter wavelength .: Vdiode is
eλ
higher .: VR is lower and hence I diode = I R is reduced.
C
Q4 Vdiode =
2013 -2016 VCAA Physics Sample Exam Solutions
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A
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