General Instructions
Reading time - 5 minutes
Working time - 70 minutes
Board-approved calculators may be used.
Write using blue or black pen.
Draw diagrams using pencil.
Formulae sheets and a Periodic Table are provided with this question paper.
Answer all questions in the spaces provided.
Total Marks (50)
This paper has two sections:
Section A
Total marks (12)
Attempt questions 1 – 12.
Allow about 18 minutes for this section.
Section B
Total Marks (38)
Attempt questions 13 – 22.
Allow about 52 minutes for this part.
Physics HSC Progress Exam 2009 Solutions Page 1 : Candidate Number: .....................................................

For each question place a cross (X) in the column which matches your choice.
Motors
Question
1
2
3
4
5
6
7
8
9
10
11
Teacher’s Use Only
12
Topic Analysis
Space Questions 1 to 3
Questions 4 to 6
Questions 7 to 9
Questions 10 to 12
A
X
X
X
…… / 3
…… / 3
…… / 3
…… / 3
B
X
X
X
13 to 15
16 to 18
19 to 20
21 to 22
Total Mark
X
X
…… / 10
…… / 8
…… / 50
C
X
…… / 10
…… / 10
Total …… / 26
Total …… / 24
X
X
X
D
X
Physics HSC Progress Exam 2009 Solutions Page 2 : Candidate Number: .....................................................

Answer each question or part of a question in the space provided.
Note that marks vary from question to question.
Question 13 (4 marks)
(a) Discuss the effect of the Earth’s orbital and rotational motion on the launch of a rocket. 2
The Earth orbits the Sun and also spins on its axis, from west to east; the fastest velocity occurs at the equator. A rocket takes less energy to launch if the effects of these are taken into account. The launch must consider the position of the Earth in relation to the destination of the rocket, in which case it is best to launch when the Earth is in the closest location to the destination. To assist the rocket speed, the rocket is launched near the equator and to the east, as this location gives the maximum assist from the rotational motion of the Earth.
(b) Given the radius of the Earth is 6400 km, the rocket places a satellite into a stable orbit of altitude
6500 km. Calculate the period of the satellite.
Us r 3 / T 2 = GM / 4

2
With r
E
= 6.4 x 10
6
m
Altitude of satellite above Earth = 6.5 x 10
6
m
So r = 1.29 x 10
7
m
G = 6.67 x 10
-11
Gives T = 14552 seconds (or equivalent in hours and minutes)
2
Physics HSC Progress Exam 2009 Solutions Page 3 : Candidate Number: .....................................................

Question 14 (3 marks)
A satellite has a mass of 250 kg and orbits the Earth at an altitude of 3000 km. The Earth’s equatorial diameter is 12 756 km.
(a) Determine its orbital speed. v =

(GM e
/r e
) =

(6.67 x 10
-11
x 6.0 x 10
24
/ 6.37 x 10
6
) = 6.5 x 10
3
ms
-1
(b) Compare the orbital speed of a 500 kg satellite in an identical orbit.
The orbital speed is the same, as orbital speed is independent of the mass of the satellite.
Question 15 (3 marks)
Three planets orbiting a distant star have the following properties:
Property
Diameter (km)
Mass (kg)
Orbital Period
(days)
Planet X Planet Y Planet Z
4000
200
6000
2 x 10
23
8 x 10
23
1.2 x 10
24
300
16000
800
Rotational Period
(days)
200 1.50 1.67
Predict which planet would have the greatest gravitational acceleration at it surface. Justify your answer.
3
3 marks for relating mg = GMm / d 2 and then correctly using this to identify planet Y
2 marks for correct equations with error(s) in calculation
1 mark for correct equation, but inability to proceed
2
1
Physics HSC Progress Exam 2009 Solutions Page 4 : Candidate Number: .....................................................

Question 16 (3 marks)
Explain the idea of the relativity of simultaneity, giving an example to support your answer. 3
The principle of simultaneity states that two events which appear to be simultaneous in one frame of reference may not appear to be simultaneous in a different frame of reference. For example, two exploding stars may be observed to appear in the night sky at the same instant of time by an astronomer on Earth, but an observer in a different part of the galaxy may notice that the star furthest from Earth actually exploded many thousands of years before the star closest to earth.
Question 17 (5 marks)
(a)
Explain what is meant by the term “equivalence between mass and energy”.
2
Einstein’s famous equation E = mc 2
(where c is the speed of light in a vacuum) shows that mass is merely energy in a different form. The equation can be used to calculate the energy equivalent of mass or vice versa.
(b) Most celestial objects outside our Solar System are too distant to reach at current maximum speeds.
Discuss the prospect of near-light speed space travel to such destinations with reference to relativistic changes in mass, time and length. 3
In order to travel such large distances, we would need to do it very quickly or we would not live long enough to get there. This is where the problem arises.
Einstein showed that at very fast speeds time dilates. This means that the faster you travel, the slower time progresses, which is an advantage. Length at these speeds is contracted, which would shorten the distance, which is also an advantage.
The biggest problem would be that of mass and energy. At the speeds we would need to travel, mass increases by very large amounts. The more massive an object is, the more energy is needed to move it.
Additional energy input is further converted to mass, making the problem worse. The energy required makes it very difficult to reach the speeds needed.
Physics HSC Progress Exam 2009 Solutions Page 5 : Candidate Number: .....................................................

Question 18 (2 marks)
Einstein’s Theory of Special Relativity made predictions about time, length and mass measurements for objects travelling at speeds approaching the speed of light.
Calculate the mass of a proton when it is travelling at 75% the speed of light.
2 marks for correct answer including units
2
1 mark for obtaining mass of proton and speed of light from data sheet and attempts to use correct equation (dilation of mass) OR
1 mark for obtaining erroneous answer but still heavier than rest mass of proton
Relativistic mass of proton = 2.59 x 10
-27
kg
Physics HSC Progress Exam 2009 Solutions Page 6 : Candidate Number: .....................................................

Question 19 (4 marks)
Below is a photograph of a hand-operated AC / DC generator.
(a) Identify the function of the split-ring commutator in this generator. 1
The split-ring commutator is used to ensure that the current continues to travel in the same direction through the external circuit rather than changing direction with each turn of the coil.
(b) Compare a DC generator to a DC motor. 3
In a motor the relationship is that a current-carrying conductor in a magnetic field experiences a force and this causes the motion of the conductor, whereas in a generator the motion of the conductor in a magnetic field creates an induced emf which is used to drive a current.
Both a DC motor and a DC generator make use of a split-ring commutator that in a motor maintains direction of rotation and in a generator ensures current flow in one direction.
Both motor and generator have a coil of wire placed within the vicinity of a magnetic field.
Physics HSC Progress Exam 2009 Solutions Page 7 : Candidate Number: .....................................................

Question 20 (4 marks)
The following graph (for a simple DC motor) shows the variation of torque as the angle between the plane of the coil and the uniform magnetic field changes.
(a) A current of 600 mA is travelling through the windings when the coil is parallel to the magnetic field.
The coil consists of 200 turns of wire and has an area of 1.5 x 10
-3
m
2
. Use this information and the
2 graph above to calculate the magnetic field strength.
Use

= nBIAcos

Re-arrange for B gives B =

/ nIAcos

Gives B = 1.67 tesla
(b) Sketch the variation of torque with angle if all other variables are kept the same as in part (a) and magnetic field strength is halved.
Physics HSC Progress Exam 2009 Solutions Page 8 : Candidate Number: .....................................................

Question 21 (4 marks)
“If someday they say of me that in my work I have contributed something to the welfare and happiness of my fellow man, I shall be satisfied.” George Westinghouse
Assess the impact on the environment of the AC generator to determine if George Westinghouse would be satisfied. 4
In his work Westinghouse fought and won a battle against Edison to make AC the main power supply.
It has lead to the development of the electricity supply that we know today which uses AC generators.
AC generators have mainly had a negative effect on the environment. As most generators are run on fossil fuels they impact on the environment by releasing greenhouse gases such as carbon dioxide which are commonly accepted as causing climate change, raising the Earth’s temperature and harming the environment. The AC generator has made possible the cheap supply of electricity. This easy access to electricity has lead to industrialisation, leading to waste e.g. it is cheaper to buy a new computer than to repair an old one. This leads to not only the overuse of finite resources such as metals and petrochemicals but also to increased landfill, further harming the Earth’s environment.
From an environmental point of view Westinghouse has little reason to be happy.
Physics HSC Progress Exam 2009 Solutions Page 9 : Candidate Number: .....................................................

Question 22 (6 marks)
In the picture below, the transformer has 50 turns in its primary coil and 15 in the secondary coil.
(a) If the initial primary voltage is 240V, calculate the output voltage of this transformer.
Use n p
/ n s
= V p
/ V s
Gives V s
= 72 volts
1
(b) Explain why it is necessary to transmit electricity at high voltages.
The transmission of electricity along power lines is hampered by energy loss due to heating of the
2 wires. To reduce the effect of this energy loss it is advantageous to transmit electricity at the lowest possible current, thus requiring a high voltage at the source. Stepping the voltage up at the source causes a step down in current which enables heat losses to be minimised.
(c) Outline where and how a transformer similar to the one above could be used to assist in the transmission of high voltages (approximately 35 kV) to a home requiring only a 240 V supply.
2
Local area substations and suburban transformers are used to reduce the high transmission voltages to voltages suitable for household consumption. Step down transformers with appropriate ratios of turns in the primary and secondary are used to achieve this. The first transformation is from 35 kV to
11 kV which requires a ratio of approx. 3 : 1. The final transformation from 11 kV to 240 V is achieved using a ration of approx. 50 : 1.
(d) Identify ONE other location in which this type of transformer is commonly used within a household.
1
Step down transformers are used for many household appliances such as computers, printers, games consoles etc.
Physics HSC Progress Exam 2009 Solutions Page 10 : Candidate Number: .....................................................