Download Presentation by the Chief Assessor

Andrew R. Hansen
Ringwood Secondary College
• Use black or blue pen as stipulated
in instructions. (Q15)
• Show working for questions worth
more than 1 mark, as per
instructions.
• Do not rely on prepared statements
from A3 sheet.
• Check for ridiculous answers.
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the vector nature of momentum
centripetal force
energy conversion and conservation in springs
more complex projectile motion
apparent weightlessness
explaining direction of induced current using Lenz’s law
operation of transformers and how they work in a power
transmission system
understanding of series and parallel circuits
explaining aspects of the photoelectric effect
applying the concept of path difference in interference
patterns
explaining electron and X-ray diffraction patterns
electron energy level diagrams and associated emissions
and absorptions
how the wave nature of matter can explain the electron
energy levels
Unless otherwise indicated, diagrams are not to scale.
Area of study – Motion in one and two dimensions
Question 1 (7 marks)
Block A, of mass 4.0 kg, is moving to the right at a speed of 8.0 m s–1, as shown in Figure 1. It collides with a
stationary block, B, of mass 8.0 kg, and rebounds to the left. Its speed after the collision is 2.0 m s–1.
4.0 kg
A
8.0 m
s–1
8.0 kg
B
Figure 1
a.
Calculate the speed of block B after the collision.
2 marks
mA vi = mA v f + mB v f
m s–1
b.
[4 ´ 8] = [4 ´ (-2)]+[8 ´ v f ]
32 = -8 + 8v f
Explain whether the collision is elastic or inelastic. Include some calculations in your answer.
vf =
32 + 8
= 5m s-1
8
2 marks
Area of study – Motion in one and two dimensions
Question 1 (7 marks)
Block A, of mass 4.0 kg, is moving to the right at a speed of 8.0 m s–1, as shown in Figure 1. It collides with a
stationary block, B, of mass 8.0 kg, and rebounds to the left. Its speed after the collision is 2.0 m s–1.
4.0 kg
A
8.0 m s–1
8.0 kg
B
Figure 1
a.
Calculate the speed of block B after the collision.
2 marks
mA vi = mA v f + mB v f
[4 ´ 8] = [4 ´ 2]+[8 ´ v f ]
m s–1
b.
32 = 8 + 8v f
32 - 8
-1
vf =
= 3m s
8
Explain whether the collision is elastic or inelastic. Include some calculations in your answer.
2 marks
Students must understand the vector nature of momentum.
L
b.
Calculate the magnitude of the force exerted by the track on the car at its lowest point (L). Show your
working.
N
2 marks
mv 2
F=
+ mg
r
2 ´ 62
F=
+ 2 ´10
4
F = 38N
SECTION A – Core studies – continued
L
b.
Calculate the magnitude of the force exerted by the track on the car at its lowest point (L). Show your
working.
2 marks
mv 2
F=
r
2
2´6
F=
4
F = 18N
N
SECTION A – Core studies – continued
Students must understand forces in vertical circular motion.
The golfer hits the ball at a speed of 40 m s–1 and at an angle of 30° to the horizontal. Ignore air resistance.
2015 PHYSICS
EXAM
a. Calculate
8
the maximum height, h, that the ball rises
above its initial position.
2 marks
Question 5 (5 marks)
A golfer hits a ball on a part of a golf course that is sloping downwards away from him, as shown in
Figure 5.
40 m s–1
h
30°
d
G
m
Figure 5
not to scale
The golfer hits the ball at a speed of 40 m s–1 and at an angle of 30° to the horizontal. Ignore air resistance.
The ballthe
lands
at a point
at ah,horizontal
distance
of 173
m from
the hitting-off point, as shown above.
a.b. Calculate
maximum
height,
that the ball
rises above
its initial
position.
2 marks
Calculate the vertical drop, d, from the hitting-off point to the landing point, G.
xH
t=
vH
173
=
40 cos30
= 5.0sec
m
b.
m
1 2
xv = uvt + at
2
x = (20 ´ 5) + (0.5 ´ -10 ´ 25)
x = 100 -125
x = -25m
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173 m
3 marks
The ball lands at a point at a horizontal distance of 173 m from the hitting-off point,
as shown above.
SECTION
A – Core studies – continued
DO NOT WRITE IN THIS AR
not to scale
2015 PHYSICS
EXAM
The golfer
hits
8 of 30° to the horizontal. Ignore air resistance.
the ball at a speed of 40 m s–1 and at an angle
a. Calculate
the maximum height, h, that the ball rises above its initial position.
Question
5 (5 marks)
A golfer hits a ball on a part of a golf course that is sloping downwards away from him, as shown in
Figure 5.
2 marks
40 m s–1
30°
h
d
G
173 m
m
Figure 5
not to scale
The golfer hits the ball at a speed of 40 m s–1 and at an angle of 30° to the horizontal. Ignore air resistance.
The ball lands
at a pointheight,
at a horizontal
distance
of 173
m from
the hitting-off
a.b. Calculate
the maximum
h, that the
ball rises
above
its initial
position. point, as shown above.
Calculate the vertical drop, d, from the hitting-off point to the landing point, G.
2 marks
3 marks
Common errors involve breaking the flight up into multiple
phases (launch to top, top to init height, final drop) which
led to errors.
Difficult for the assessors to follow.
m
m
SECTION A – Core studies – continued
b.
The ball lands at a point at a horizontal distance of 173 m from the hitting-off point, as shown above.
N O T W R I T E I N T H DI SO AN RO ET A W R I T E I N T H I S A R E A
Figure 5
not to scale
WRITE IN THIS
9
2015 PHYSICS EXAM
J
Question 6 (8 marks)
A mass of 2.0 kg is suspended from a spring, with spring constant k = 50 N m–1, as shown in Figure 6.
Calculate
potential
energy
at spring
its lowest
point.
Itb.is released
fromthe
thespring
unstretched
position
of the
and falls
a distance of 0.80 m. Take the zero of
gravitational potential energy at its lowest point.
2 marks
0
0.40
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AO
R EN
AOT
0.80
J
a.
c.
Figure 6
Calculate
change
energy(maximum
as the massspeed).
moves from the top position to the
Calculatethethe
speedinofgravitational
the mass atpotential
its midpoint
lowest position.
1 mark
3 marks
1 2
1 2
ET = (mgx) + ( kx ) + ( mv )
2
2
16 = (2 ´10 ´ 0.4) + (0.5´ 50 ´ 0.4 2 ) + (0.5 ´ 2 ´ v 2 )
J
b.
16 = 8 + 4 + v 2
Calculate the spring potential energy at its lowest point.
v =4
2
m s–1
J
v = 2m s -1
2 marks
WRITE IN THIS D
AO
R E NA O T W R I T E I N T H I S A R
lowest position.
1 mark
9
2015 PHYSICS EXAM
Question 6 (8 marks)
J
A mass of 2.0 kg is suspended from a spring, with spring constant k = 50 N m–1, as shown in Figure 6.
It is released from the unstretched position of the spring and falls a distance of 0.80 m. Take the zero of
b. Calculate
the spring
potential
at its lowest point.
gravitational
potential
energy
at its energy
lowest point.
2 marks
0
0.40
0.80
J
a.
c.
b.
Figure 6
Calculate the change in gravitational potential energy as the mass moves from the top position to the
Calculate the speed of the mass at its midpoint (maximum speed).
lowest position.
3 marks
1 mark
1 2 1 2
kx = mv
2
2
J
1
mgx = mv 2
Calculate the spring potential energy at its lowest 2
point.
2 marks
m s–1
Students must understand that there are
three energy’s that need to be accounted for.
SECTION A – Core studies – Question 6 – continued
2015 PHYSICS EXAM
Which one of the following graphs (A.–D.) best shows the acceleration of the mass as it goes from the
highest point to the lowest point? Take upwards as positive. Give a reason for your choice.
2 marks
A.
10
5
acceleration
0
(m s–2)
0
–5
0.20
0.40
0.60
0.80
0.20
0.40
0.60
0.80
spring extension (m)
–10
B.
10
5
acceleration
0
(m s–2)
0
–5
spring extension (m)
–10
C.
10
5
acceleration
0
(m s–2)
0
–5
0.20
spring extension (m)
0.40
0.60
0.40
0.60
0.80
–10
D.
10
5
acceleration
0
(m s–2)
0
–5
–10
0.20
spring extension (m)
0.80
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d.
10
At x=0 there is no spring force
and a=g (-10). At the mid
point the spring force = the
gravitational force and a=0. At
the bottom the spring
provides an upwards force
giving a=10.
While many students were able to identify C, they were
unable to explain why in a coherent way. Those that
selected responses other than C gave supporting
arguments that were not realistic.
Springs make up 20% of the marks in motion and
students need a more thorough understanding of the
underlying Physics.
DO NOT WR
m
c.
Would an astronaut in this spacecraft feel weightless? Explain your answer.
3 marks
Yes.
The astronauts experience apparent weightlessness as they are still
in a gravitational field, and therefore have weight, but they
experience no normal reaction force.
SECTION A – Core studies – continued
TURN OVER
DO NOT WR
m
c.
Would an astronaut in this spacecraft feel weightless? Explain your answer.
3 marks
A number of students said “no they would not feel
weightless” and then went on to explain that they
would feel “apparently weightless”.
There were also a number of students who believe that
weightlessness (of any kind) can only exist in a region
without gravity.
SECTION A – Core studies – continued
TURN OVER
2015 PHYSICS EXAM
12
Area of study – Electronics and photonics
Question 8 (5 marks)
a. You are provided with four resistors, each of 2.0 Ω.
Show how to connect them to produce an effective resistance of 3.0 Ω, using four or fewer resistors.
Draw in the space below, so that points A and B are at either end of the effective resistance.
Use this symbol for a resistor:
Label the resistors in your diagram R1, R2, R3 and R4. If you used fewer resistors, use fewer labels.
A
B
2 marks
2015 PHYSICS EXAM
12
Area of study – Electronics and photonics
Question 8 (5 marks)
a. You are provided with four resistors, each of 2.0 Ω.
Show how to connect them to produce an effective resistance of 3.0 Ω, using four or fewer resistors.
Draw in the space below, so that points A and B are at either end of the effective resistance.
Use this symbol for a resistor:
Label the resistors in your diagram R1, R2, R3 and R4. If you used fewer resistors, use fewer labels.
2 marks
A
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There were a number of poorly designed circuits, some
suggesting that students had never studied electronics at all.
B
The most problematic was the “short circuiting” by trying to
The resistor network you have drawn is now constructed and connected correctly to a 9 VDC power
create
supply. a circuit, as shown.
b.
Calculate the voltage drop across each of the resistors and write the value in the table below.
If you used fewer than four resistors, leave the unused resistor box(es) blank.
3 marks
Vout
R2
=
Vin R1 + R2
10
15
=
60 R1 +15
Þ R = 75kW
Vout
R2
=
Vin R1 + R2
10
R2
=
R’s are mixed up.
60 15 + R2
Students need to know which resistor is which in the
voltage divider equation and how to use this formula.
2015 PHYSICS EXAM
16
In the second trial, the input signal has a peak voltage of 200 mV, as shown in Figure 11a. Figure 11b shows
the corresponding output signal.
0.005
0.010
0.015
0.020
0.025
0.020
0.025
t (seconds)
Figure 11a
10.0
8.0
6.0
4.0
2.0
Vout (V) 0
–2.0
–4.0
–6.0
–8.0
–10.0
0
0.005
0.010
0.015
t (seconds)
Figure 11b
b.
Explain why the output signal in the second trial has the shape shown in Figure 11b.
3 marks
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250
200
150
100
50
Vin (mV)
0
0
–50
–100
–150
–200
–250
The amplifier is an inverting amplifier and the waveform
demonstrates clipping which is where the output of the amplifier
exceeds the maximum supply voltage. (or similar)
2015 PHYSICS EXAM
16
In the second trial, the input signal has a peak voltage of 200 mV, as shown in Figure 11a. Figure 11b shows
the corresponding output signal.
0.005
0.010
0.015
0.020
0.025
0.020
0.025
t (seconds)
Figure 11a
10.0
8.0
6.0
4.0
2.0
Vout (V) 0
–2.0
–4.0
–6.0
–8.0
–10.0
0
0.005
0.010
0.015
t (seconds)
Figure 11b
b.
Explain why the output signal in the second trial has the shape shown in Figure 11b.
3 marks
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250
200
150
100
50
Vin (mV)
0
0
–50
–100
–150
–200
–250
The most common error was to focus on the clipping and fail to
comment on the inversion.
2015 PHYSICS EXAM
20
The model is now set up as a DC generator, with the output connected to a voltmeter and oscilloscope via
a commutator, as shown in Figure 13, with the same coil of side length 4.0 cm and 10 turns, and a uniform
magnetic field of 2.0 × 10–3 tesla.
The shaft is rotated by hand.
F
E
G
H
S
V
Figure
21 13
2015 PHYSICS EXAM
c. b. The
shaft one
and of
coil
two complete
revolutions
second.
Which
themake
following
graphs (A.–D.)
best per
shows
the voltage output as viewed on the
oscilloscope as the coil rotates steadily? (At t = 0, the coil is horizontal, as shown in Figure 13.)
Calculate the magnitude of the average voltage as shown on the voltmeter during one-quarter
revolution. Show your working.
A.
B.
V
V
n´B´ A
t
t
2
100 ´ (2 ´10-3 ) ´ (0.04)
x=
D.
0.125
V
x = 0.26mV
x=
t
0
C.
mV
A
V
t
0
t
1 mark
3 marks
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N
2015 PHYSICS EXAM
20
The model is now set up as a DC generator, with the output connected to a voltmeter and oscilloscope via
a commutator, as shown in Figure 13, with the same coil of side length 4.0 cm and 10 turns, and a uniform
magnetic field of 2.0 × 10–3 tesla.
The shaft is rotated by hand.
F
E
H
S
V
Figure
21 13
2015 PHYSICS EXAM
c. b. The
shaft one
and of
coil
two complete
revolutions
second.
Which
themake
following
graphs (A.–D.)
best per
shows
the voltage output as viewed on the
oscilloscope as the coil rotates steadily? (At t = 0, the coil is horizontal, as shown in Figure 13.)
Calculate the magnitude of the average voltage as shown on the voltmeter during one-quarter
revolution. Show your working.
A.
B.
V
V
0
C.
EA
V
d.
mV
The most common errors were:
t
t
0
• omission
of n.
• failure
to convert cm to m
D.
V
• incorrect
value for t
t generator0into an AC generator.
0
The students
wish to convert this DC
t
1 mark
3 marks
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N
G
mV
d.
The students wish to convert this DC generator into an AC generator.
Describe the change or changes the student would have to make to achieve this. Explain your answer.
3 marks
• Replace the commutator with slip rings.
• The slip rings maintain a constant connection with the
loop.
• As the loop rotates an AC current will be generated in
the loop which is then transmitted to the oscilloscope.
mV
d.
The students wish to convert this DC generator into an AC generator.
Describe the change or changes the student would have to make to achieve this. Explain your answer.
The most common error was to rely on stock
answers from the A3 sheet regarding the
difference between slip rings and a split ring
commutator.
Stock descriptions did not respond to the
question being asked.
3 marks
Over half the responses scored no marks. Most diagrams
were nonsensical and showed that students had a poor
understanding of electromagnetic induction.
TE IN THIS AREA
23
c.
2015 PHYSICS EXAM
Determine the direction of the current through the voltmeter as the loop enters the magnetic field.
Write X to Y or Y to X in the answer box below. Explain how you determined this in terms of Lenz’s
Law.
4 marks
• The initial flux is down and increasing.
• Lenz’s Law states that the induced current will give rise to
a change in flux that opposes the change that induced it.
• The induced flux will be up and increasing.
• The right hand grip rule shows a current from y-x in the
loop will give an upwards flux.
• Therefore the current will flow from x to y through the
voltmeter.
ITE IN THIS AREA
23
c.
2015 PHYSICS EXAM
Determine the direction of the current through the voltmeter as the loop enters the magnetic field.
Write X to Y or Y to X in the answer box below. Explain how you determined this in terms of Lenz’s
Law.
4 marks
Marks awarded for this question varied.
Most students were able to identify some of the
required points but most were unable to provide
a thorough response.
Students should practice responding to these
high scoring questions.
• A real situation would be a power station supplying power
to a town.
• For a constant power delivery, transformers can be used
to increase the transmission voltage and decrease the
transmission current.
• Since power loss = I2R, reducing the current will
significantly reduce the power loss.
Marks awarded for this question varied.
Most students were able to identify some of the required points
but most were unable to provide a thorough response.
The most commonly omitted point was the need to maintain the
delivered power while varying the supply parameters.
Students should practice responding to these high scoring
questions.
EK max (eV) 2
1
O
d.
1
2
3
4
5 6 7 8 9 10 11 12 13
frequency (×1014 Hz)
In experiment 3, metal A is replaced with metal B that has a work function 50% larger than that of
metal A. The original light intensity is used. A dotted line shows the results of experiment 1.
On the graph below, draw a solid line to sketch the graph of maximum kinetic energy versus frequency
for experiment 3.
2 marks
3
EK max (eV) 2
1
O
1
2
3
4
5 6 7 8 9 10 11 12 13
frequency (×1014 Hz)
• Same gradient.
• Intercept at 7.5
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EK max (eV) 2
1
O
d.
1
2
3
4
5 6 7 8 9 10 11 12 13
frequency (×1014 Hz)
In experiment 3, metal A is replaced with metal B that has a work function 50% larger than that of
metal A. The original light intensity is used. A dotted line shows the results of experiment 1.
On the graph below, draw a solid line to sketch the graph of maximum kinetic energy versus frequency
for experiment 3.
2 marks
3
EK max (eV) 2
1
O
1
2
3
4
5 6 7 8 9 10 11 12 13
frequency (×1014 Hz)
The most common errors were the wrong gradient or an
intercept of 10.
What evidence do we have that electrons can behave like waves? Explain how this evidence supports a
wave model of electrons.
2 marks
• Electrons diffract when they interact with things
like crystals or narrow slits.
• Diffraction is a wave phenomenon.
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b.
What evidence do we have that electrons can behave like waves? Explain how this evidence supports a
wave model of electrons.
2 marks
Many students refered to Young’s Double Slit
experiment.
Young experimented with light not electrons so the
argument is not valid.
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b.
THIS AREA
35
2015 PHYSICS EXAM
Question 21 (5 marks)
a. Use the model of quantised states of the atom to explain why only certain energy levels are allowed.
b.
• Electrons exhibit a wave behavior.
• Allowed orbits are where the circumference is a
whole multiple of the electron wavelength and a
standing wave can be formed.
Illustrate your
answer with an appropriate
diagram.different whole multiples
• Different
orbits have
of the electron wavelength.
3 marks
2 marks
35
2015 PHYSICS EXAM
Question 21 (5 marks)
a. Use the model of quantised states of the atom to explain why only certain energy levels are allowed.
3 marks
Students had trouble linking the concept of a standing wave
to the integer multiple of the electron wavelength.
b.
Students
seem
think that
electrons follow a sinusoidal
Illustrate your answer
withto
an appropriate
diagram.
path around the nucleus.
2 marks
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35
2015 PHYSICS EXAM
Question 21 (5 marks)
a. Use the model of quantised states of the atom to explain why only certain energy levels are allowed.
3 marks
b.
2 marks
Illustrate your answer with an appropriate diagram.
Question 22 (2 marks)
Electrons (of mass 9.1 × 10–31 kg) have a de Broglie wavelength of 1.0 × 10–11 m.
Calculate the speed of these electrons.
35
2015 PHYSICS EXAM
Question 21 (5 marks)
a. Use the model of quantised states of the atom to explain why only certain energy levels are allowed.
3 marks
b.
2 marks
Illustrate your answer with an appropriate diagram.
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Students who tried to draw a sinusoidal pattern on
a strip of paper had trouble demonstrating the
standing wave nature.
Question 22 (2 marks)
Electrons (of mass 9.1 × 10–31 kg) have a de Broglie wavelength of 1.0 × 10–11 m.
Calculate the speed of these electrons.
Andrew R. Hansen
Ringwood Secondary College