Andrew R. Hansen Assistant Chief Assessor • Chief Assessor: – Bruce Walsh (Xavier). • Assistant Chief Assessors: – Andrew Hansen (Ringwood SC). – Nicholas Howes.
Download ReportTranscript Andrew R. Hansen Assistant Chief Assessor • Chief Assessor: – Bruce Walsh (Xavier). • Assistant Chief Assessors: – Andrew Hansen (Ringwood SC). – Nicholas Howes.
Andrew R. Hansen Assistant Chief Assessor
• • • • Chief Assessor: – Bruce Walsh (Xavier).
Assistant Chief Assessors: – Andrew Hansen (Ringwood SC).
– – Nicholas Howes (Caulfield Grammar).
Dan O’Keeffe (AIP).
Senior markers: – Around 10-12 for resolution marking.
– Drawn from general markers.
General markers: – Around 100 practicing or retired Physics teachers.
• • • Prior to the exam the assessors are appointed and informed of training day and deadlines.
On the day of the exam the paper is emailed to assessors.
At ~4:00pm on exam day the CAG meets with reps from setting panel.
– The draft marking scheme is discussed.
• • The setting panel decide the marks per question.
The CAG decides (initially) how the marks are to be allocated.
• • • • In General: 1 mark questions: – Mark for correct answer.
2 mark questions: – 1 mark for correct answer.
– 1 mark for correct formula & substitution.
3 mark questions: – 1 mark for correct answer.
– – 1 mark for final formula and substitution.
1 mark for initial derivation or conversion of info.
• • • • The CAG meets for a day prior to Assessors meeting.
An initial 250 scripts is reviewed.
Two training papers and 4 trial papers are produced.
Six validity items per question are selected and annotated.
• • • • • Occurs the Saturday following the exam.
The marking scheme is discussed and the training papers are completed.
The marking scheme is amended.
The four trial papers are completed.
The marking scheme is finalised and printed.
• • • • Each question is marked (initially) by two markers.
Discrepancy results in a third marker.
– Two most concordant results are used provided they are not still discrepant.
– This marking is done by CAG and senior markers.
Marks that are still discrepant go to adjudication.
– Done by CAG.
Most concordant two marks are used and added for final mark.
• • Each set of scores is summed and totals that are discrepant go to 3 rd round of marking.
– Done by CAG.
Final grades that are more than 1 (or 1.5) grades from teacher’s indicative grades are declared anomalous.
– – Anomalous grade marking.
Done by CAG
• Papers that have images problems or where students wrote where they should not have go to pulled paper marking.
– Done by CAG.
• • • • • ~7300 students and 59 markable items = 430,700 items to mark.
Two initial markings = 861,400 markings.
~25% of item go to resolution = 969,000 marking.
~ 5000 items go to adjudication = 974,000 markings ~ 150 papers referred to AGM = 983,000 items marked.
• • Over 900,000 items marked.
~100 markers •
Over 9000 questions marked per person.
•
In 14 days!!!
v u t
define a
= D
t v
\
at
=
v
-
u
\
v
=
u
+
at
=
v
(1) -
t u
Now, the area under a vt graph yields disp.
So:
x
=
rect
+
triangle x
= (
u
*
t
) + ( 1 2
bh
)
x
= (
ut
) + ( 1 2 ´
t
´ (
v
-
u
))
but v
-
u
=
at from
(1)
So x
=
ut
+ 1 2 ´
t
´
at x
=
ut
+ 1 2
at
2 (2)
Now from
:
x
= (
ut
) + ( 1 2 ´
t
´ (
v
-
u
))
x
=
ut
+ 1 2
vt
1 2
ut x
= 1 2
ut
+ 1 2
vt x
= 1 2 (
u
+
v
)
t
(3) Finally, from (1):
v
\ =
t u
+ =
at v
-
u a
Sub into (3):
x
= 2
ax v
+ = 2
u
´
v
-
u a
(
v
+
u
)(
v
-
u
) 2
ax
=
v
2 -
uv
+
uv
-
u
2 2
ax
=
v
2 -
u
2 \
v
2 =
u
2 + 2
ax
(4)
• • • Always show working.
– Allow markers to exercise “reward good physics” clause.
Avoid algebraic rearrangement.
– Make it easier for markers to award the “formula & substitution” mark(s).
Answers in decimals with appropriate sig figs.
– No surds and no excessive decimal places.
• • • Changing units.
– Students can usually cross out the units provided and respond in units of choice.
Layout & legibility.
– Markers are not obliged to mark illegible responses.
Calculators.
– – Use the exponent key.
Check mode (degrees, not radians).
• • • • • Keep them short!!!
– Stay within the space provided.
Use dot points.
Clause, because. vs. Clause.
Supporting equations.
Avoid verbal diarrhea.
– Marks can be deducted for contradictions.
June 2008
t e s t a i t e What i u th o r o i You m s t sh w you work ng.
s a n oat 5 r avell ng a 2. 0 m s –1 ?
m s –2 @2.0
m s
1 ,
F friction
= 2 ´ 10 After a ti e, the t ugb at and hi p r e tr velli g at a co stan s eed.
4
N
What is this constant speed?
å
F
= 9 ´ 10 4 2 ´ 10 4 (
from graph
) = 7 ´ 10 4
N a
=
F m
= 7 ´ 10 4 100 ´ 10 4 = 0.07
m s
2 2 marks m s –1 2 marks
SECTI ON A – f st dy 1 V TURN O ER
t e s t a i t What i u th o e r o i You m s t sh w you work ng.
s a 5 n oat t r avell ng a 2. 0 m s –1 ?
m s –2 m After a ti e, the t ugb at and hi p r e tr velli g at a co stan s eed.
What is this constant speed?
v cont
Þ å
F
= 0 Þ
F friction
= 9 ´ 10 4
N
2 marks Þ
v
= 9
m s
1 m s –1 2 marks
SECTI ON A – f st dy 1 V TURN O ER
June 2009 11 2009 PHYS EXAM 1
The following information relates to Questions 13 and 14.
The
Jason 2
satellite reached its operational circular orbit of radius 1.33 × 10 7 m on 4 July 2008 and then began mapping the Earth’s oceans. mass of the Earth = 5.98 × 10 24 kg mass of
Jason 2
= 525 kg
G
= 6.67 × 10 –11 N m 2 kg –2
Question 13
On the fi gure below, draw one or more labelled arrows to show the direction of any force(s) acting on
Jason 2
as it orbits Earth. You can ignore the effect of any astronomical bodies other than the Earth.
direction of motion Earth Fc
Jason 2
2 marks
Question 14
What is the period of orbit of the
Jason 2
satellite?
s 3 marks
END OF AREA OF STUDY 1 SECTI ON A
– continued
TURN OVER
June 2010
Correct solution N F c θ mg Incorrect solution mg θ N F c tan q tan q =
O A
=
O
´ 1
A
=
mv
2
r
´ 1
mg
q q = = tan 6 ° 1 (
v
2
rg
) (5.71
° ) = tan 1 ( 10 100 ´ 2 10 ) = tan 1 (0.1) sin q tan q =
O H
=
mv
2
r
=
O
´ 1
H
´ 1
mg
q q = sin 1 (
v
2
rg
) = = 6 ° (5.74
° ) sin 1 ( 10 2 100 ´ 10 ) = sin 1 (0.1)
Nov 2013
June 2009
25 °
C
= 500 W
V out
=
V in
´
R Th R V
+
R V
12 ´ 8 =
R V
500
R V
+
R V
= 1000 W
INCREASE Lower temp means higher R th .
Ratio R v :R th must stay constant.
Therefore Rv must increase.
June 2011
Nov 2010 Lines must travel from R to L inside solenoid.
Lines must not touch or cross.
Nov 2010
e e = -
n
DF D
t
=
neg gradient
Nov 2011 As the magnet moves away the loop experiences a decrease in flux to the left.
Lenz’s law states that the induced current will give rise to a change in flux that opposes the change in flux that induced it.
Therefore the induced current will cause an increase in flux to the left, Using the right hand grip rule an anticlockwise current will cause a flux to the left.
Nov 2011 Young’s experiment demonstrated an interference pattern. Interference is a wave phenomenon.
Einstein concluded that light is made up of photons with energy proportional to their frequency.
The existence of a cutoff frequency showed that photons with low frequencies did not have enough energy to produce photoelectrons.
• Know what the wave predictions of the PE effect are: – All light is energy therefore all frequencies should be able to produce photoelectrons.
– – Increasing the intensity of the incident light increases the delivered energy and should increase the energy of the photoelectrons.
Decreasing the intensity of the incident light decreases the rate at which the energy is delivered and should result in a time delay.
• • • • • • • • • • • • Elastic collisions.
Apparent weightlessness.
Projectiles.
Diodes.
Amplifiers.
Modulation / demodulation.
Forces on current carrying wires.
Faraday’s law.
Transformers.
Diffraction and the ratio λ :a.
Constructive / destructive interference and path difference.
Planck’s constant as grad of E k vs f graph.
• • • • • Read the Chief Examiner’s Report’s.
Know where the opposition is weak.
Do lots of practice papers.
Keep an eye on the time.
Have a strategy for catching silly mistakes.
Andrew R. Hansen Assistant Chief Assessor