Download Chief Assessor`s Report

2016 Examination: Chief
Assessor’s Report.
Andrew R. Hansen
Ringwood Secondary College
Areas requiring improvement included:
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resultant forces on free bodies
the vector nature of momentum
converting to and from scientific notation
energy relationships in springs
energy considerations in projectile motion
current flow in series and parallel circuits
the function of diodes in circuits
modulation
the function of DC motors, including application of forces,
the role of the commutator and the factors affecting
efficiency
the process of electromagnetic induction
transformers and line loss
graphing data and interpreting graphical data
wave properties of matter particles.
Question 1b.
Expected solution
a = 0.1 m s-1
T
10 x 103 kg
𝐹 = 𝑚𝑎
2000 N
𝑇 − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎
𝑇 − 2000 = 10 × 103 × 0.1
𝑇 = 3000𝑁
A free body diagram would help here.
Question 1b.
Common errors
𝐹 = 𝑚𝑎
𝐹 = 𝑚𝑎
𝑇 = 𝑚𝑎
𝑇 = 𝑚𝑎 + 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑇 = 10 × 103 × 0.1
𝑇 = 1000𝑁
𝑇 = 10 × 103 × 0.1 + 2000
𝑇 = 5000𝑁
Use free body diagrams
Question 5b.
Expected solution
A
The kinetic energy is highest at the beginning and end of the
flight and lowest (not zero) in the middle.
Question 5b.
Common errors
The kinetic energy is zero at the top of the flight.
The ball slows down before reaching the top and speed up on
the way down.
Make sure you know what the question is asking.
Question 6a.
Expected solution
The orbit must be above the equator.
The orbital period must be 24 hours.
The orbital radius can be found using
𝑟3
𝑇2
𝐺𝑀
= 4𝜋2 or similar.
The orbital direction must be the same as the rotation of the earth.
Question 6a.
Common errors
The orbital period of 24 hours was most commonly seen.
The orbit must be above the equator was seen but rarely.
“For the orbit to be geostationary the satellite has to stay stationary over a fixed
point on the earth.”
Restating the stem is a common error
Question 8b.
If one LED fails the current in that arm will fall to zero.
The current in the other arm will double.
The current through the resistor / battery will remain the same.
Expected solution
Question 8b.
Common errors
If one LED fails the rest of the LEDs in that arm will fail.
The current will go through the other side.
The current through the circuit will decrease.
Be careful not to make ridiculous statements or overly vague statements.
Question 14a.
A, B
Expected solution
If XY is not horizontal there will be an applied force.
The force will be co-linear with the axis of rotation.
Question 14a.
Common errors
There was no single common error. Students did not seem
to have a solid understanding of forces on current carrying
wires.
Question 14b.
The best orientation is horizontal.
When the loop is horizontal the force on the loop is maximal.
Expected solution
Question 14b.
Common errors
There was no single common error. Students often gave vertical as the orientation.
Those that gave horizontal struggled to explain force or torque.
Question 14c.
Expected solution
Increasing the battery voltage will increase
the current in the loop which will increase the
force on the loop.
A, B
Increasing the number of turns will also
increase the force on the loop.
Question 14c.
Common errors
The most common errors were to only
identify one improvement or to be unable to
explain how their chosen improvements
worked.
Question 15b.
Expected solution
The initial flux is to the left and decreasing.
The induced flux must be to the left and increasing.
The right hand grip rule / right hand solenoid rule indicates that current will flow to
the left in the resistor.
Question 15b.
Common errors
A number of students simply stated Lenz’s law which did not answer the question.
By far the most common error was being unable to logically explain the initial change in
flux, the induced flux, the induced current.
Students continue to find this type of question difficult.
Question 16b.
Expected solution
Question 16b.
Expected solution
𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛
𝑅1
𝑅1 + 𝑅2
𝑉 18
𝐼= =
= 1.5𝐴
𝑅 12
𝑉𝑜𝑢𝑡 = 18
3
12
𝑉𝑑𝑟𝑜𝑝 = 𝑅𝐼 = 3 × 1.5 = 4.5𝑉
𝑉𝑜𝑢𝑡 = 4.5𝑉
𝑅𝑇𝑜𝑡𝑎𝑙 = 12Ω
OR
Question 16b.
No single common error here beyond lack of understanding.
Attempted solutions were nonsensical.
Common errors
Question 16c.
Expected solution
𝑉𝑔𝑙𝑜𝑏𝑒 = 𝑉𝑠𝑢𝑝𝑝𝑙𝑦 − 𝑉𝑙𝑜𝑠𝑠
𝑉𝑔𝑙𝑜𝑏𝑒 = 18 − 4.5 = 13.5𝑉
Then
𝑉 2 13.52
𝑃=
=
= 20𝑊
𝑅
9
Question 16c.
Common errors
The most common error was to use an incorrect value for the voltage across the globe.
There was also a range of arithmetic errors.
Question 16d.
18V
Expected solution
Step
up
Step
down
Question 16d.
Common errors
The most common error was to draw the circuit with a DC supply.
Other errors were to draw transformers like resistors, place transformers in
series with the globe, draw circuits that with components that were not
identifiable.
Students must spend more time understanding transformer circuits.
Question 17c.
Expected solution
Question 17c.
Expected solution
The EMF is maximum when the coil is horizontal.
When the loop is horizontal the rate of change of flux is maximal.
Question 17c.
Common errors
The most common error was to state that the EMF is maximal when the loop is
vertical because the area is maximal.
Students did not seem to understand the relationship between EMF and rate of
change of flux.
To my knowledge, no student used graphs to help their explanation.
Question 19c.
Expected solution
Question 19c.
Expected solution
The data gives information about the maximum kinetic energy of the photoelectrons.
Question 19c.
Common errors
Most common: Gives information about the kinetic energy of the photoelectron.
Other common responses simply referred to results of the PE effect experiment.
Many responses seemed to be copied directly from the A3 sheet.
Question 20b.
Expected result
ℎ𝑐
𝐸=
𝜆
4.14 × 10−15 × 3 × 108
𝐸=
0.36 × 10−9
𝐸 = 3450𝑒𝑉
Question 20b.
Common errors
ℎ𝑐
𝐸=
𝜆
ℎ𝑐
𝐸=
𝜆
4.14 × 10−15 × 3 × 108
𝐸=
0.36
6.63 × 10−34 × 3 × 108
𝐸=
0.36 × 10−9
𝐸 = 3.45 × 10−6 𝑒𝑉
OR
𝐸 = 5.525 × 10−16 𝑒𝑉
Question 20c.
Expected result
The amount of diffraction depends on wavelength.
The similarity of the patterns implies that the wavelengths are the same.
Question 20c.
Common errors
Most students indicated the importance of wavelength.
The most common errors were to link wavelength to energy or velocity.
Becoming an assessor
In 2016 we had 168 applications for 104 positions.
Selection is based on both teaching experience and previous marking experience.
Teaching experience is just as valuable as marking experience.
Last year thirteen of the assessors were first time assessors.
If this is your third successive year teaching year 12, please consider applying.
2016 Examination: Chief
Assessor’s Report.
Andrew R. Hansen
Ringwood Secondary College