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2011
SUPERVISOR TO ATTACH PROCESSING LABEL HERE
STUDENT NUMBER
Letter
Figures
Words
PHYSICS
Written examination 1
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Section
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1. Motion in one and two dimensions
2. Electronics and photonics
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1. Einstein’s special relativity 25
2. Materials and their use in structures 25
3. Further electronics
Number of
questions
Number of questions
to be answered
Number of
marks
23
12
23
12
42
24
12
12
12
12
12
12
24
24
24
Total 90
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2
Electronics and photonics .......................................................................................................................................... 18
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2011 PHYS EXAM 1
18
The following information relates to Questions 1–4.
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Y
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Question 1
Show that the total resistance of the circuit between X and Y is 3.3 ohm.
2 marks
Question 2
A 10 V battery is now connected across XY as shown in Figure 2.
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What is the current through resistor B?
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SECTION A – Area of study 2 – continued
N O W RIT ING ALLOWED IN T HIS AREA
Area of study 2 – Electronics and photonics
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The following information relates to Questions 5 and 6.
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Question 5
What is the resistance of the thermistor when the temperature is 20°C?
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SECTION A – Area of study 2 – continued
N O W RIT ING ALLOWED IN T HIS AREA
2011 PHYS EXAM 1
N O W RIT ING ALLOWED IN T HIS AREA
25
2011 PHYS EXAM 1
Question 9
On the axes below, draw the variation in the brightness of the LED with time.
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0
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1
2
3
4
5
6
7
8
time (s)
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Einstein’s special relativity ........................................................................................................................................ 28
Materials and their use in structures ......................................................................................................................... 32
Further electronics .................................................................................................................................................... 38
6(&7,21% – continued
785129(5
2011 PHYS EXAM 1
28
Question 1
An experiment is done where two protons with very high kinetic energy collide in order to try to create a single stationary
‘Higgs’ particle.
Each proton in the reaction has a kinetic energy of 1.1 × 10–6 J. No other particles are produced in the reaction and the
protons will not exist after the production of the Higgs particle.
The proton rest mass is equal to 1.6726 × 10–27 kg.
Which of the following options is the best estimate of the mass of the Higgs particle?
A. 1.2 × 10–23 kg
B. 2.4 × 10–23 kg
C. 3.3 × 10–27 kg
D. 1.7 × 10–27 kg
Question 2
According to the theory of special relativity, the mass, m, of a particle, and its total energy, E, are equivalent. The
relationship is
E = mc2 where m = ȖPo, and
The amount of energy required to increase the speed of an electron from 0.18c to 0.19c is 0.004 J. By contrast, the
amount of energy required to increase this electron’s speed by the same amount (0.01c) from 0.98c to 0.99c is 27.7 J.
Which of the following best explains why much more energy is required to produce the same change in speed for the
electron when it is moving at the higher speed?
A.
The rest mass (m0) of the electron is not a constant but depends on the speed v.
B.
The electron mass-energy is proportional to v, so it is more massive at high speed.
C.
Because E =
D.
The total mass-energy of the electron depends on Ȗ, which increases rapidly as v approaches c.
1
2
mv2, the larger v, the more energy is required to increase it further.
Question 3
When stationary, a proton has a rest mass-energy of 1.50 × 10–10 J.
A proton is accelerated from a speed with Ȗ = 1.05 to a speed with Ȗ = 1.10.
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second speed?
A. 2.9 × 106 J
B. 3.2 × 10–10 J
C. 7.5 × 10–12 J
D. 8.3 × 10–29 J
SECTION B – Detailed study 1 – continued
N O W RIT ING ALLOWED IN T HIS AREA
Detailed study 1 – Einstein’s special relativity
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785129(5
30
Question 7
Which of the following statements best describes what the Michelson–Morley experiment attempted to measure?
A. the speed of Earth through space
B. changes in the speed of Earth through space
C. accuracy obtainable with an optical interferometer
D. differences in the speed of light in different directions
Question 8
One of Einstein’s postulates of special relativity was that ‘the laws of physics are the same in all inertial reference frames’.
Scientists are planning an experiment that must be conducted in an inertial reference frame. They choose a location in
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Which of the following is the best option for them to choose?
A. a laboratory in a stationary slowly rotating spaceship
B. a laboratory in a spaceship hurtling through space at 0.99c
C. a laboratory in a spaceship moving at 0.99c that is gradually slowing down
D. a laboratory in a very slowly moving spaceship that is gaining speed
Question 9
Scientists observe the path of a short-lived elementary particle in a detector. It is created in the detector and exists only
for a short time, leaving a path of length 5.4 mm long. The scientists measure its speed as 2.5 × 108 m s–1, giving Ȗ = 1.81.
What is the proper lifetime of the particle?
A. 5.3 × 10–11 s
B. 3.3 × 10–11 s
C. 1.8 × 10–11 s
D. 1.2 × 10–11 s
Question 10
Which of the following statements best explains why it is impossible to accelerate particles (such as electrons) so that
they are travelling at the speed of light?
A. It is directly forbidden by one of Einstein’s postulates.
B. As particles increase in speed, the rest mass (m0WHQGVWRZDUGVDQLQ¿QLWHYDOXH
C. The kinetic energy of particles, given by EK = (Ȗ – 1)m0c2WHQGVWRZDUGVDQLQ¿QLWHYDOXH
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Question 11
Which of the following statements about proper length is the most accurate?
A. The proper length of an object is always greater than or equal to another measure of the length of the object.
B. The proper length of an object is always less than another measure of the length of the object.
C. The proper length of an object is sometimes less than another measure of the length of the object, and sometimes
greater than or equal to another measure of the length of the object.
D. The proper length of an object can only be measured by an observer who is moving relative to the object.
SECTION B – Detailed study 1 – continued
N O W RIT ING ALLOWED IN T HIS AREA
2011 PHYS EXAM 1
N O W RIT ING ALLOWED IN T HIS AREA
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785129(5
2011 PHYS EXAM 1
32
In this section, take the acceleration due to gravity, g, as 10 m s–2.
The following information relates to Questions 1–6.
Engineers are testing samples of different types of steel to determine the best type to use in construction of a bridge.
The samples they are testing have a cross-section area of 2.0 × 10–5 m2. They are all 0.10 m long.
The samples are under tension.
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× 103 N
8
6
P
4
2
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5
10
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Figure 1
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Question 1
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B. 4.0 × 107 N m–2
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D. 4.0 × 109 N m–2
Question 2
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A. 5.0 × 10–7
B. 5.0 × 10–4
C. 5.0 × 10–3
D. 5.0
SECTION B – Detailed study 2 – continued
N O W RIT ING ALLOWED IN T HIS AREA
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35
2011 PHYS EXAM 1
The following information relates to Questions 7 and 8.
A school crossing sign is supported by a rigid rod, AC, smoothly hinged to an upright pole at point A, and a cable, BC,
as shown in Figure 3.
not to scale
B
2.0 m
C
2.0 m
2.8 m
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A
Figure 3
Length of rod AC = 2.8 m
Length of cable BC = 2.0 m
Length AB = 2.0 m
Mass of sign = 40 kg
Ignore mass of rod and all cables.
Question 7
Which of the following best describes the stress in AC and BC?
A. AC and BC are both in tension.
B. AC and BC are both in compression.
C. AC is in tension, BC in compression.
D. AC is in compression, BC in tension.
Question 8
Which of the following best gives the magnitude of the force in the cable BC?
A.
40 N
B.
60 N
C. 400 N
D. 560 N
SECTION B – Detailed study 2 – continued
TURN OVER
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785129(5
40
Question 5
Mary wants to build a regulated power supply powered by a 9 V battery. The required output voltage of the power supply
has to be 5.5 V to suit the music player she bought overseas.
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a 5.5 V Zener diode.
The circuit is shown in Figure 2a.
The voltage-current characteristics of the Zener diode are depicted in Figure 2b.
I (mA)
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She measured the output voltage VXY to be 3.5 V.
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B. 7KHNȍUHVLVWRULVWRSURWHFWWKH=HQHUGLRGHIURPH[FHVVLYHFXUUHQWVDQGWKHYDOXHRIWKHUHVLVWRULVWRRORZ
C. The battery has too high a voltage to allow 5.5 V across XY.
D. The Zener diode is installed incorrectly, with its polarity reversed.
SECTION B – Detailed study 3 – continued
N O W RIT ING ALLOWED IN T HIS AREA
2011 PHYS EXAM 1
3+<6(;$0
N O W RIT ING ALLOWED IN T HIS AREA
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785129(5
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The following information relates to Questions 6–8.
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N O W RIT ING ALLOWED IN T HIS AREA
2011 PHYS EXAM 1
N O W RIT ING ALLOWED IN T HIS AREA
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2011 PHYS EXAM 1
Question 8
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SECTION B – Detailed study 3 – continued
TURN OVER
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2011
PHYSICS
Written examination 1
Tuesday 14 June 2011
Reading time: 11.45 am to 12.00 noon (15 minutes)
Writing time: 12.00 noon to 1.30 pm (1 hour 30 minutes)
FORMULA SHEET
Directions to students
‡ A question and answer book is provided with this formula sheet.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic
devices into the examination room.
© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2011
2011 PHYSICS EXAM 1
1
2
v=
velocity; acceleration
Δx
Δv
; a=
Δt
Δt
v = u + at
1
2
x = ut + at 2
2
equations for constant acceleration
v 2 = u 2 + 2ax
1
2
x=
(v + u )t
Ȉ) = ma
3
Newton’s second law
4
circular motion
5
Hooke’s law
6
elastic potential energy
1 2
kx
2
7
gravitational potential energy near the
surface of the Earth
mgh
8
kinetic energy
1
mv 2
2
9
Newton’s law of universal gravitation
v 2 4π 2 r
= 2
r
T
a=
) = –kx
F
M 1M 2
G
g
r2
G
10
JUDYLWDWLRQDO¿HOG
11
stress
σ=
12
strain
ε=
13
Young’s modulus
E
14
transformer action
V1
V2
15
AC voltage and current
16
voltage; power
VRMS =
1
2 2
Vp − p
V = RI
M
r2
F
A
ΔL
L
stress
strain
N1
N2
I RMS =
1
2 2
P = VI = I2R
I p−p
3
2011 PHYSICS EXAM 1
17
resistors in series
RT = R1 + R2
18
resistors in parallel
1
1
1
=
+
RT R1 R2
19
capacitors
20
Lorentz factor
21
time dilation
22
length contraction
L = Lo/Ȗ
23
relativistic mass
m = moȖ
24
total energy
25
universal gravitational constant
26
mass of Earth
ME = 5.98 × 1024 kg
27
radius of Earth
RE = 6.37 × 106 m
28
mass of the electron
me = 9.1 × 10–31 kg
29
charge on the electron
e = –1.6 × 10–19 C
30
speed of light
c = 3.0 × 108 m s–1
time constant : IJ = RC
γ=
1
1 − v2 / c2
t = toȖ
Etotal = Ek + Erest = mc2
G = 6.67 × 10–11 N m2 kg–2
3UH¿[HV8QLWV
p = pico = 10–12
n = nano = 10–9
ȝ = micro = 10–6
m = milli = 10–3
k = kilo = 103
M = mega = 106
G = giga = 109
t = tonne = 103 kg
END OF FORMULA SHEET