Download MIDPHY15_GUIDELINES

2015 HIGHER SCHOOL CERTIFICATE
MID-YEAR EXAMINATION
PHYSICS – MAPPING GRID
Exam
Section
Part A:
Multiple
Choice
Part B:
Free
Response
Question
Marks
Syllabus/Course
Outcomes
Content
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
5
5
7
3
5
7
6
6
4
7
H6, H9
H9
H6, H9
H9
H11
H9
H1
H2, H9
H2, H9
H2, H9
H2, H9
H9
H7, H9
H7, H9
H9, H10
H9, H10
H10
H10
H10
H9
H9
H2, H9
H6
H6
H9
H1, H2, H9
H4, H7, H9
H2, H9, H10
H2, H9
H3, H4, H9
9.2.1
9.2.1
9.2.1
9.2.2
9.2.2
9.2.2
9.2.4
9.3.1
9.3.1
9.3.2
9.3.2
9.3.2
9.3.3
9.3.4
9.4.1
9.4.1
9.4.2
9.4.3, 9.4.2
9.4.3
9.4 .4
9.2 2
9.2.2
9.2.2
9.2.4
9.3.1
9.3.2
9.3.1, 9.3.5
9.4.1
9.4.3
9.4.4
Targeted
Performance
Bands
3-5
4-6
2-4
3-5
3–5
2–4
3–5
2–4
4–6
2–4
3-5
3–5
2–4
3–5
3–5
4–6
2–4
4–6
3–5
3–5
3–5
3–5
4–6
2–4
4–6
3–5
3–5
2–5
2–5
3–5
Answer
B
C
D
C
B
D
A
C
B
B
B
D
A
C
B
A
C
D
A
B
Disclaimer Every effort has been made to prepare this Examination in accordance with the Board of Studies documents. No guarantee or warranty is
made or implied that the Examination paper mirrors in every respect the actual HSC Examination question paper in this course. This paper does not
constitute ‘advice’ nor can it be construed as an authoritative interpretation of Board of Studies intentions. No liability for any reliance, use or purpose
related to this paper is taken. Advice on HSC examination issues is only to be obtained from the NSW Board of Studies. The publisher does not accept
any responsibility for accuracy of papers which have been modified.
MIDPHY15_GUIDELINES
Page 1
2015 HIGHER SCHOOL CERTIFICATE
MID-YEAR EXAMINATION
PHYSICS – MARKING GUIDELINES
Part A – 20 marks
Questions 1-20 (1 mark each)
Question
Correct Response
Outcomes Assessed
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
C
D
C
B
D
A
C
B
B
B
D
A
C
B
A
C
D
A
B
H6, H9
H9
H6, H9
H9
H11
H9
H1
H2, H9
H2, H9
H2, H9
H2, H9
H9
H7, H9
H7, H9
H9, H10
H9, H10
H10
H10
H10
H9
MIDPHY15_GUIDELINES
2
Targeted
Performance Bands
3-5
4-6
2-4
3-5
3-5
2-4
3-5
2-4
4-6
2-4
3-5
3-5
2-4
3-5
3-5
4-6
2-4
4-6
3-5
3-5
Part B – 55 marks
Question 21 (7 marks)
21 (a) (2 marks)
Outcomes Assessed: H9
Targeted Performance Bands: 3 – 5


Criteria
Calculates the initial horizontal velocity and initial vertical velocity
Partial calculation
Marks
2
1
Sample answer
ux = u cos  = 42 x cos 50 = 27.0 ms-1 .
uy = u sin  = 42 x sin 50 = 32.2 ms-1 .
21 (b) (2 marks)
Outcomes Assessed: H9
Targeted Performance Bands: 4 – 6


Criteria
Calculates the height and gives the unit
Partial calculation involving the equations of accelerated motion
Marks
2
1
Sample answer
At top vy = 0
v2 = u2 + 2as
0 = 32.22 + 2 x -9.8 x h
h = 52.8 m
21 (c) (2 marks)
Outcomes Assessed: H9
Targeted Performance Bands: 4 – 6


Criteria
Calculates the velocity and gives the angle
Calculates the velocity or the angle
Sample answer
vx = 27.0 ms-1
Marks
2
1
vy = uy + at = 32.2 + -9.8 x 3 = 2.8 ms-1 .
V2 = 27.02 + 2.82
V = 27.1 = 5.9°
21 (d) (1 mark)
Outcomes Assessed: H9
Targeted Performance Bands: 3 – 5

Criteria
States that V is max at initial and final position
Mark
1
Sample answer
The velocity of the cannon ball is maximum at the initial position (out of the cannon) and at
the final position just before impact.
MIDPHY15_GUIDELINES
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Question 22 (5 marks)
22 (a) (2 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 2 – 4


Criteria
Describes how the two masses fall, and gives the reason
Outlines how the two masses fall
Marks
2
1
Sample answer
The two masses, which were the same size but different of weight, fell to the ground in the
same time. This is because gravity accelerates all masses, big and small, at the same rate.
22 (b) (3 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 3 – 5



Criteria
Identifies that the foam ball will take longer, and gives an explanation
involving increased air resistance on the foam due to its surface
Identifies that the foam ball will take longer, and mentions air resistance
Identifies that the foam ball will take longer OR identifies that air resistance
slows the ball
Marks
3
2
1
Sample answer
The result above would differ because the foam ball would take longer to hit the ground. This
is because it is slowed down by air resistance to a greater degree than the smooth metal ball,
because it has a rougher surface.
Question 23 (7 marks)
23 (a) (2 marks)
Outcomes Assessed: H6
Targeted Performance Bands: 2 – 4


Criteria
Shows how Kepler’s 3 law links period and radius and planet mass
Outlines Kepler’s 3rd law OR links period and radius and planet mass
rd
Marks
2
1
Sample answer
Kepler’s 3rd law states that R3 / T2 = GM/42 which shows that the period of satellite A
depends on its orbital radius, and on the mass of the planet it is orbiting.
MIDPHY15_GUIDELINES
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23 (b) (2 marks)
Outcomes Assessed: H6
Targeted Performance Bands: 4 – 6


Criteria
Correct calculation for radius and orbital period
Partial calculation
Marks
2
1
Sample answer
R3 / T2 = GM/42
R3 = T2 GM/42
R3 = (6 x 60 x 60)2 x 6.67x10-11 x 5.1x1024 /42 = 4.02 x 1021
R = 1.59 x 107 m
Orbital speed v = 2r/T = 2 x 1.59 x 107 /(6 x 60 x 60) = 4 625 ms-1
23 (c) (2 marks)
Outcomes Assessed: H6
Targeted Performance Bands: 3 – 5


Criteria
Identifies that for B radius is less, and gives an explanation using Kepler’s
Laws
Identifies that for B radius is less
Marks
2
1
Sample answer
If satellite B has the same orbital period as satellite A, the orbital radius of satellite B will be
less than the orbital radius of A. Kepler’s law shows that R3 / T2 = GM/42
so R3 / = T2GM/42
for A, RA 3 = T2GM/42
for B, RB 3 = T2G( ½ M )/42 approximately, giving RB 3 = ½ T2GM/42 which shows
that the radius for B is less.
23 (d) (1 mark)
Outcomes Assessed: H6
Targeted Performance Bands: 2 – 4

Criteria
Outline identifies no change in speed
Mark
1
Sample answer
The orbital speed would not change, as there is only one value for orbital speed at a certain
orbital radius.
MIDPHY15_GUIDELINES
5
Question 24 (3 marks)
24 (a) (2 marks)
Outcomes Assessed: H6
Targeted Performance Bands: 3 – 5


Criteria
Outline identifies that no experiment is possible, and links this to relativity and
the absence of an ether
Identifies that no experiment is possible
Marks
2
1
Sample answer
According to Einstein’s special theory of relativity, there is no experiment that the on-board
astronaut can do to be able to measure the rockets’ absolute velocity relative to a stationary
ether. In fact, the ether has not been detected, which would also preclude such a measurement.
24 (b) (1 mark)
Outcomes Assessed: H6
Targeted Performance Bands: 4 – 6
Criteria

Mark
1
Correct substitution
Sample answer
Tv = t0 /[ 1 – v2/c2 ]0.5 = 10/[ 1 – 0.982 ] = 50.3 s
Question 25 (6 marks)
25 (a) (4 marks)
Outcomes Assessed: H9
Targeted Performance Bands: 3 – 5

Criteria
Explanation involving four features;
- no current with open circuit
- copper is non-magnetic
- a closed circuit bringing about a current and a force
- the force on a current is BIL
Explanation involving three features


Outline involving two features
Outline involving one feature

Marks
4
3
2
1
Sample answer
The copper wire will not move when the switch is open, because there is no current flow, and
copper is not magnetic so it will not be attracted to the magnetic poles.
The copper wire will move when the switch is closed, because there is a current flowing due
to the circuit being compete, and a force acts on a current carrying wire in a magnetic field
given by F = BIL.
MIDPHY15_GUIDELINES
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25 (b) (2 marks)
Outcomes Assessed: H9
Targeted Performance Bands: 3 – 5
Criteria


Marks
2
1
Outline involving two features
Outline involving one feature
Sample answer
The copper wire has a magnetic field given by
F = BIL = 55 x 10-3 x 3.9 x 0.12 = 0.026 N down the page.
Question 26 (7 marks)
26 (a) (2 marks)
Outcomes Assessed: H1
Targeted Performance Bands: 2 – 4


Criteria
Identifies Michael Faraday and electromagnetic induction
Identifies Michael Faraday OR electromagnetic induction
Marks
2
1
Sample answer
The early electricity pioneer who carries this out was Michael Faraday and the effect is
electromagnetic induction.
26 (b) (3 marks)
Outcomes Assessed: H1. H2, H9
Targeted Performance Bands: 3 – 5



Criteria
Explanation with all three features;
- broken current flow producing a changing flux
- changing flux inducing an emf
- a changing induced current produce a changing magnetic force and a
flickering magnetic needle
Outline with two features
Identifies 1 feature
Marks
3
2
1
Sample answer
When the switch was constantly changed from on to off a broken current was made to flow in
the 1st circuit. This meant that a changing magnetic field was produced. The changing
magnetic field cut across the wire in the 2nd circuit and produced an emf, due to induction.
The induced current was an alternating current, and this produced an alternating magnetic
field in the solenoid coil with the compass needle, making the compass needle move and
flicker.
MIDPHY15_GUIDELINES
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26 (c) (2 marks)
Outcomes Assessed: H1. H2
Targeted Performance Bands: 2 – 4


Criteria
Outline involves verifying results and consistency
Outline involves verifying results OR consistency
Marks
2
1
Sample answer
In order for the results here to become accepted by the scientific community, the experimenter
should repeat the experiment many more times and verify that the effect is real, and he should
make sure that the results are consistent.
Question 27 (6 marks)
Outcomes Assessed: H4, H7, H9
Targeted Performance Bands: 3 – 6
Criteria
Marks
All six features present ;
 similarities given
 differences given
 details on structure
 details on benefits
 details on drawbacks
 appropriate table with headings
 Five features present
 Four features present
 Three features present
 Outline involving a comparison
 One fact given
6
5
4
3
2
1
Sample answer
Structure
Benefits
Drawbacks
MIDPHY15_GUIDELINES
The AC coil motor
The AC induction motor
Similarity – both use an
external voltage source, a
magnetic field, and a
rotor with a current
carrying coil.
Similarity – both convert
electrical energy into
mechanical energy.
If the motor gets stuck
the current will quickly
rise which may cause a
burn out / fire.
Difference- the induction motor’s rotor
coil is not connected to the external
voltage supply and runs on induced
current.
Difference- the induction motor is more
reliable since there is no wearing down
of brushes.
Is slow to get to optimal speed because
of lower torque.
8
Question 28 (7 marks)
28 (a) (2 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 2 – 4


Criteria
Outline links large voltage to acceleration of cathode rays
Mentions large voltage OR acceleration of cathode rays
Marks
2
1
Sample answer
He used an induction coil to produce a large voltage across the ends of the tube: this was
needed to accelerate the cathode rays across the tube.
28 (b) (3 marks)
Outcomes Assessed: H2. H9
Targeted Performance Bands: 3 – 5



Criteria
Explanation involving the cathode rays bending downward, the electric field E
= V/d, and the force
Explanation involving the cathode rays bending downward, and the electric
field OR the force
One true fact about the cathode rays or the plates
Marks
3
2
1
Sample answer
The cathode rays, when made to pass through electric plates, will bend downward and
produce a curved path. This is because there is an electric field E = V/d between the plates
which will create a force on the negatively charged cathode rays, attracting them to the
positive plate.
28 (c) (2 marks)
Outcomes Assessed: H2. H9
Targeted Performance Bands: 3 – 5


Criteria
Two inferences about the canal rays
One inference about the canal rays
Marks
2
1
Sample answer
The canal rays are charged particles, sub- atomic in size, and oppositely charged to the
cathode rays. They too are formed in the cathode ray tube but travel in the opposite direction
of the electric field that drives the cathode rays from cathode to anode.
MIDPHY15_GUIDELINES
9
Question 29 (4 marks)
29 (a) (2 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 2 – 4


Criteria
Outline shows that the n-type silicon formed from impurities of an identified
element and the p-type silicon formed from impurities of an identified element
Outline mentions the n-type silicon formed from impurities of an identified
element OR the p-type silicon formed from impurities of an identified element
Marks
2
1
Sample answer
To make p-type silicon, atoms of an element with fewer valence electrons are added as
impurities to the silicon. E.g. aluminium. To make n-type silicon, atoms of an element with
more valence electrons are added as impurities to the silicon. E.g. phosphorus.
29 (b) (2 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 3 – 5


Criteria
Explanation mentions the junction field and the small positive voltage
producing free conduction for conventional current
Outline mentions the junction field OR the small positive voltage producing
free conduction for conventional current
Marks
2
1
Sample answer
At the p-n junction electrons on the n-type jump over to the holes on the p-type, forming an
electric field and depletion region in between. This junction field will push negative charges
back towards the n-type or right on the diagram, and positive charges are pushed back
towards the p-type or left.
When a small positive voltage is applied to the p-type, the junction field is nullified and a
positive charge (conventional current) will move freely across from left to right, so the diode
is effectively a conductor.
MIDPHY15_GUIDELINES
10
Question 30 (3 marks)
Outcomes Assessed: H3, H4, H9
Targeted Performance Bands: 3 – 5



Criteria
Positives and negatives are both outlined, and an overall conclusion is given
Positives are outlined, and an overall conclusion is given OR negatives are
outlined, and an overall conclusion is given
Outline mentions positives or negatives
Sample answer
High temperature superconductors can be used in electric motors.
Positives;
- stronger magnetic fields can be produced
- higher torque
- more efficient use of energy as far less heat is generated
Negatives;
- more costly to manufacture
- needs extra systems and cryogenics to keep temperatures cold
Conclusion;
Beneficial overall as efficiency and speed are increased, but very costly.
MIDPHY15_GUIDELINES
11
Marks
3
2
1