Download HSC Progress Exam 2010 Solutions - Kotara High School

Physics
Progress Examination
SOLUTIONS
HSC Course
2010
General Instructions
Reading time -
5 minutes
Working time -
70 minutes
Total Marks (52)
Board-approved calculators may be used.
This paper has two sections:
Write using blue or black pen.
Section A
Draw diagrams using pencil.
Total marks (10)
Formulae sheets and a Periodic Table are
Attempt questions 1 – 10.
provided with this question paper.
Allow about 14 minutes for this section.
Answer all questions in the spaces
Section B
provided.
Total Marks (42)
Attempt questions 11 – 20.
Allow about 56 minutes for this part.
Physics HSC Progress Exam 2010 Solutions Page 1 : Candidate Number: STAFF
Section A
Multiple Choice Answers
For each question place a cross (X) in the column which matches your choice.
Question
A
B
C
1
X
2
X
3
X
D
X
4
X
5
X
6
X
7
8
X
X
9
10
X
Teacher’s Use Only
Topic Analysis
Space
Questions 1 to 5
…… / 5
11 to 15
…… / 21
…… / 5
16 to 20
…… / 21
Total …… / 26
Motors
Questions 6 to 10
Total …… / 26
Total Mark
…… / 52
Task Ranking ...... / 15
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Section B
Short Answer Questions
(42 marks)
Answer each question or part of a question in the space provided.
Note that marks vary from question to question.
Question 11 (3 marks)
The photograph below indicates the results of an investigation performed to identify the properties of
cathode rays.
Use the space below to draw and label the apparatus used in this investigation, from a side-on view, and
describe ONE safety precaution for this demonstration.
Standard diagram showing induction coil with 6 V DC input and approx.
20 000 V output. Connections to cathode ray tube showing cathode
emitting electrons (negative electrode). Electron path is impeded by
barrier (Maltese Cross).
Induction coil can be a source of high frequency electromagnetic radiation (usually X rays).
Operators should remain a safe distance away – anything from 1 m to 3 m depending on
manufacturer’s specifications.
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Question 12 (5 marks)
The sketch below shows how the acceleration of a rocket changes during the first stage of the launch of a
rocket.
(a)
Describe the trend shown in the graph.
1
Acceleration is increasing at an increasing rate and then suddenly decreases at the end of stage one.
(b)
Explain why the acceleration changes during the first stage of a rocket launch.
1
As the rocket ascends, it uses fuel to provide the thrust. This means that the overall mass of the rocket
is decreasing, but the thrust from the engines remains constant. Hence the acceleration continues to
increase until the fuel is gone.
(c)
Describe the changes to the gravitational potential energy of the rocket during the first stage of the
launch. Explain the changes.
3
The two factors to take into account are the overall mass of the rocket and the distance between the
rocket and the centre of the Earth. As the rocket’s altitude increases, the GPE increases i.e. it
becomes less negative. However as the fuel is used up the mass of the rocket decreases and this
causes a decrease in the GPE. Since the two factors are in opposition, we would need more
information before we could determine the overall change in GPE.
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Question 13 (8 marks)
The Chinese have an ambitious space program, hoping to land taikonauts (Chinese space travellers) on the
Moon by 2019. A student hopes to be in one of their future space shuttles that is designed to achieve an
altitude of 260 km after a 3 minute launch.
(a)
Explain the issues associated with getting taikonauts into orbit in terms of launching rockets.
The main considerations for the taikonauts is that g forces need to be kept at safe levels during the
launch. Humans can survive g forces of 10 g, but only for a few seconds. Maintaining the g forces
around 3 – 4 g is acceptable, so fuel payloads and thrust are a compromise i.e. higher acclerations
and shorter time frames are achievable, but not with live taikonauts.
(b)
Calculate the orbital velocity of a Chinese shuttle in circular orbit at an altitude of 260 km (radius of
Earth 6370 km).
Use v = √GMe/r = 7.9 x 103 ms-1
(c)
On one such future taikonaut mission an observation satellite may be launched from orbit.
Calculate the radius of the satellite orbit if its period is 14 hours.
Use r3 / T2 = GM / 42 = 2.95 x 107
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Question 14 (3 marks)
The twin paradox is an often quoted example of relativity. One twin is travelling in space and the other
remains on Earth. According to the twin on Earth, the travelling twin will age less. However, the paradox
arises when the travelling twin claims that, since the Earth is moving away at the speed of the spacecraft,
time passes more slowly on Earth, and therefore the effects of time dilation will be reversed.
Discuss a solution to this apparent paradox, quoting frames of reference in your answer.
3
The key to this problem is to recognise that the theory of special relativity recognise that there are two
types of frames of reference i.e. non-accelerated frames called inertial frames of reference, and
accelerated frames called non-inertial frames of reference. All physical laws apply only in inertial
frames of reference. The space travelling twin has to accelerate away from Earth and decelerate on
return, so he/she finds him/herself in a non-inertial frame of reference. Thus the conclusions drawn by
the space travelling twin are false.
Question 15 (3 marks)
(a)
Planet Alpha has a mass 1.5 times that of the Earth. The acceleration due to gravity on planet Alpha is
three times that of Earth.
(i)
Calculate the mass of a space explorer setting foot on Alpha if he weighs 800 N on Earth.
1
Mass will be the same on Alpha as it is on Earth i.e. 800 / 9.8 = 80 kg (approx.)
(ii)
Calculate the weight of the explorer on Alpha.
1
Weight will be three times greater than on Earth i.e. 2 400 N
(iii) Calculate the radius of Alpha compared to the radius of the Earth (RE).
Use g = GM / r2 for both Alpha and Earth, then compare.
Gives ralpha = rearth / √2
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Question 16 (4 marks)
The diagram below is a cross-section of a loudspeaker.
An applied voltage supplies a pulse of DC (I) to the loudspeaker coil as shown.
(a)
Predict the direction of movement of the speaker cone.
1
To the left
(b)
The resistance of the wire in the speaker coil is 1 ohm.
3
When the external voltage is applied, sensitive measurements show the size of the current pulse is less
than that predicted by Ohm’s Law.
The movement of the coil is an example of the motor effect. Using Lenz’s Law, describe how a back
emf could explain this observation.
Once the coil moves it behaves as a conductor cutting lines of magnetic flux. This induces an emf in
the coil. According to Lenz’s law, this emf must oppose the motion which produced it i.e. it will try to
drive a current in the opposite direction. This has the effect of limiting, or reducing, the current in the
coil.
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Question 17 (6 marks)
In the early twenty-first century, small DC generators running from fuel cells became widely available,
allowing consumers more choice of how to provide electricity to their households. This is slowly taking
over from the standard model of large, centralised power stations delivering AC power to cities.
Justify the use of AC generators in nation-wide electricity grids in the twentieth century.
6
A power station generates typically 50 V AC at high current. This is unsuitable for transmission due
to the fact that high current causes significant heat losses according to P = I2R. It is not feasible to
rectify the AC as this would not allow voltages to be transformed easily.
To transmit electrical energy over large distances, the 50 V at high current needs to be transformed to
330 kV at lower current (typically hundreds of amps). This is achieved with the use of high efficiency
transformers with typical losses of 3 – 5%. This high voltage is now suitable for cross-country
transmission.
When these high voltage cross-country power lines get near their destination, they are fed into substations that step down the voltage to approx. 10 000 V at higher current. This incurs more energy
losses as heat. The losses are minimised by the use of high efficiency transformers using laminated
soft iron cores and oil coiled systems. Typically a sub-station serves approx. 500 households.
At individual households, the single phase consumer needs 240 V. This further step down is a
achieved by local vicinity transformers, once again incurring heat losses.
Nation wide electricity grids are particularly suitable when most consumers are concentrated in a few
large cities at a considerable distance from energy sources.
Physics HSC Progress Exam 2010 Solutions Page 8 : Candidate Number: STAFF
Question 18 (2 marks)
The diagram below is of the apparatus used by J.J. Thomson to measure some of the properties of cathode
rays. The cathode ray is fired at high velocity across both magnetic and electric fields, and the deflection is
measured on a fluorescent screen.
(a)
On the diagram above, indicate the direction of the force experienced by the cathode rays, when only
the magnetic field is applied.
1
Down (using RHPR)
(b)
In one of his experiments, J.J. Thomson deflected a cathode ray travelling at 0.01c, by passing it
through a magnetic field of 0.01 T.
Calculate the force acting on the cathode ray due to the magnetic field.
Use F = Bqv
where q = 1.6 x 10-19 C (stream of electrons)
Hence F = 4.8 x 10-15 N
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Question 19 (3 marks)
The diagram below is of a transformer.
(a)
A student claims that this is a step-up transformer. Justify the claim.
1
More turns in the secondary coil than the primary coil
(b)
An input voltage of 27 V is applied to the transformer.
By counting the number of primary turns (Np) and secondary turns (Ns), calculate the output voltage
and explain why this value is likely to be different in the real world.
Np = 7 and Ns = 10
Therefore Vs = 27 x 10/7 = 38.6 V
Students are given credit for correct method, regardless of number of turns counted.
Physics HSC Progress Exam 2010 Solutions Page 10 : Candidate Number: STAFF
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Question 20 (6 marks)
A current carrying wire of length 3.5 cm is placed deep inside a solenoid where the magnetic field is
unform. The wire is kept perpendicular to the magnetic field. Looking down, this is shown below.
As the current in the wire is increased, the wire downward force on the wire is measured.
The results are shown in the table below.
Current in wire (A)
Downward force (x 10-3 N)
0.00
0.00
0.50
2.5
1.0
6.0
1.5
9.5
2.0
12.0
Question 20 continues on the next page.
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Question 20 (continued)
(a)
Draw a line graph of these results on the grid below.
2
Line of best fit
Suitable scales on each axis
(b)
Calculate the gradient of your line graph.
1
Gradient = rise / run = 6 x 10-3 NA-1
(c)
Using these results, a student claims that they can calculate the magnitude of the magnetic field inside
the solenoid. Justify this claim.
2
Using F = BIlsin can be re-arranged to give
B = (F/Il)
Since F/I is gradient and l is known, B can be calculated ( = 90o)
i.e. B 0.17 T
Incorrect numerical answers are achieved if students have wrong gradient
no deduction of marks.
(d)
Without performing another experiment, describe how you could predict the downward force on the
wire when it carries a current of 3.5 A.
Extrapolate the graph to include I = 3.5 A
1
OR
Substitute I = 3.5A into the equation, using F/I = 6 x 10-3
Physics HSC Progress Exam 2010 Solutions Page 12 : Candidate Number: STAFF