Physics (2007) Sample assessment instrument and indicative responses 3 Supervised assessment

Physics (2007)
Sample assessment instrument and indicative responses
Supervised assessment 3
December 2010
Purposes of assessment
1
The purposes of assessment are to:
 promote, assist and improve student learning
 inform programs of teaching and learning
 provide information for those people — students, parents, teachers — who need to know about
the progress and achievements of individual students to help them achieve to the best of their
abilities
 provide information for the issuing of certificates of achievement
 provide information to those people who need to know how well groups of students are
achieving (school authorities, the State Minister for Education and Training and the Arts, the
Federal Minister for Education).
It is common practice to label assessment as being formative, diagnostic or summative, according
to the major purpose of the assessment.
The major purpose of formative assessment is to help students attain higher levels of performance.
The major purpose of diagnostic assessment is to determine the nature of students’ learning, and
then provide the appropriate feedback or intervention. The major purpose of summative
assessment is to indicate the achievement status or standards achieved by students at a particular
point in their schooling. It is geared towards reporting and certification.
Syllabus requirements
Teachers should ensure that assessment instruments are consistent with the requirements,
techniques and conditions of the Physics syllabus and the implementation year 2007.
Assessment instruments
2
High-quality assessment instruments: 3
 have construct validity (the instruments actually assess what they were designed to assess)
 have face validity (they appear to assess what you believe they are intended to assess)
 give students clear and definite instructions
 are written in language suited to the reading capabilities of the students for whom the
instruments are intended
 are clearly presented through appropriate choice of layout, cues, visual design, format and
choice of words
 are used under clear, definite and specified conditions that are appropriate for all the students
whose achievements are being assessed
 have clear criteria for making judgments about achievements (these criteria are shared with
students before they are assessed)
 are used under conditions that allow optimal participation for all
 are inclusive of students’ diverse backgrounds
 allow students to demonstrate the breadth and depth of their achievements
 only involve the reproduction of gender, socioeconomic, ethnic or other cultural factors if
careful consideration has determined that such reproduction is necessary.
1
QSA 2008, P–12 Assessment Policy, p. 2.
2
Assessment instruments are the actual tools used by schools and the QSA to gather information about student achievement, for
example, recorded observation of a game of volleyball, write-up of a field trip to the local water catchment and storage area, a test of
number facts, the Senior External Examination in Chinese, the 2006 QCS Test, the 2008 Year 4 English comparable assessment task.
3
2
QSA 2008, P–12 Assessment Policy, pp. 2–3.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Physics (2007)
Sample assessment instrument and indicative responses
Supervised assessment
Compiled by the Queensland Studies Authority
October 2010
The QSA acknowledges the contribution of the State Review Panel Chair for Physics in the
preparation of this document.
About this assessment instrument
The purpose of this document is to inform assessment practices of teachers in schools. For this
reason, the assessment instrument is not presented in a way that would allow its immediate
application in a school context. In particular, the assessment technique is presented in isolation
from other information relevant to the implementation of the assessment. For further information
about those aspects of the assessment not explained in this document, please refer to the
assessment section of the syllabus.
This sample provides opportunities for students to demonstrate:

reproduction and interpretation of complex and challenging concepts and principles

comparison and explanation of complex processes

linkage and application of algorithms and concepts in complex and challenging situations

systematic analysis of secondary data to identify patterns and trends

analysis and evaluation of complex scientific interrelationships

exploration of scenarios and possible outcomes with justification of conclusions

discriminating selection, use and presentation of scientific data.
To complete this assessment, students would have access to generic formulae, and a table of
physical constants would be allowed.
This sample assessment instrument is intended to be a guide to help teachers plan and develop
assessment instruments for individual school settings.
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Assessment instrument
The student work presented in this sample is in response to an assessment task which involves
students applying and using relevant knowledge and skills to create a response to a problem or
issue.
Question 1
The force of a tennis racket hitting a ball increases the velocity from rest to 5 m s-1 in a distance of
6 metres. If the ball has a mass of 60 g calculate the force acting on the ball.
Question 2
An object is dropped from the top of a cliff 125 m high. Using g = 9.8 m s-2, find the time it takes
for the object to reach the bottom of the cliff.
Question 3
A tennis racket of mass 1.5 kg falls off the shelf in the garage, 2 m above the floor, onto the floor.
a. Calculate the kinetic energy gained by the tennis racket in falling to the floor.
b. Calculate the gravitational potential energy of the tennis racket half way to the floor.
c. At what velocity will the tennis racket hit the floor?
Question 4
A squash ball of mass 0.4 kg hits the back wall in a squash court at an angle of 300 to the normal,
at a speed of 25 m s-1, and rebounds at the same angle and speed. Calculate the change in
momentum of the squash ball.
N
300
300
Question 5
A cricket ball of mass 156 g is held at a height of 1.00 m above the ground.
a. Calculate the gravitational potential energy of the cricket ball at this height.
b. Determine how high above the ground a squash ball of mass 24 g must be positioned in
order for it to have the same gravitational potential energy as the cricket ball.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Question 6
In cricket, the speed of the ball and height it bounces off the grass pitch are affected by various
factors (e.g. type of grass, density of the soil, dryness of the pitch). In order to determine how fast
a cricket ball is likely to travel as it bounces off the pitch, a “drop test” was devised. A cricket ball
is dropped from a height of 3.00 m and the height of the bounce is measured. The height of the
bounce is used to determine the “speed” of the pitch, as shown in the table below.
Mass of a cricket ball = 156 g
Radius of cricket ball = 3.6 cm
Pitch speed
Height of bounce
Very fast
76 cm and greater
Fast
64 cm to 76 cm
Moderately fast
51 cm to 64 cm
Easy paced
38 cm to 51 cm
Slow
Less than 38 cm
a. If a cricket ball has a coefficient of restitution of 0.48 when dropped from 3.00 m onto a
pitch, what is the “pitch speed”?
b. Describe the energy transformations that occur during the “drop test” and hence explain
how the bounce height of the cricket ball can be used to determine how fast the cricket
ball is likely to travel as it bounces off the pitch. Question 7
A spin bowler intentionally causes the ball to spin in order to make it more difficult for the batsman
to hit the ball. One of the results of making the ball spin is that the ball “dips” as it travels through
the air, as shown in the diagram below. Using your knowledge of relevant Physics principles,
justify the ball’s path.
Question 8
Brett Lee was able to bowl a cricket ball at speeds around 150.0 km h-1.
If a bat is in contact with the ball for 1.30 ms, what force is needed from the bat to hit the ball
straight back at a speed of 70.0 km h-1?
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Question 9
Ricky Ponting hit a ball with a speed of 23 m s-1 at an angle of 680 above the horizontal. The ball
goes straight towards Michael Clarke who was standing 67 m away. When the ball was hit, it was
at a height of 1.0 m above the ground. Calculate the range of the ball and hence determine if
Michael Clarke is likely to catch the ball before it bounces, given that he can run at a speed of
6.5 m s-1. (Note: Ignore air resistance.)
Question 10
A golf ball is hit at 40 m s-1 at an initial direction of 35° to the horizontal. Find
a. the maximum height of the ball
b. the velocity of the ball at maximum height
c. the horizontal range (assuming a level golf course) and
d. the direction of motion after three seconds.
Question 11
The following table shows the most recent Olympic records for the men’s track events of 100 m,
200 m and 10 000 m.
(Note: 26:17.563 means 26 minutes and 17.563 seconds)
Athlete
Event
(in m)
Time
(in mins and s)
Usain Bolt
100
9.58
Usain Bolt
200
19.19
Kenenisa Bekele
10000
26:17.53
a. In which race, the 100 m or the 200 m, did Usain Bolt have the fastest average speed?
Justify your answer.
b. What was the average speed for Kenenisa Bekele in the 10,000 m race?
Question continues next page.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Question 11 continued
During the 10,000 m race, the following times were recorded at 1000 m intervals.
1000 m
2000 m
3000 m
4000 m
5000 m
6000 m
7000 m
8000 m
9000 m
2:46.24
5:34.24
8:19.55
11:04.75
13:40.45
16.18.57
18.57.63
21:37.80
24:26.31
c. "The fifth kilometre was the fastest yet, as 13:40.45 meant they were down to 2:35 pace".
Justify this statement by showing the calculations.
What final time would Kenenisa Bekele have been aiming for at this pace?
If he could have run that time, would the time be better than the previous record of
26:22.75 set by Haile Gebrselassie in 1998?
d. Kenenisa Bekele covered the last 400 m in 57.4 seconds. If he had run at this pace for the
whole race what would have been his final time? Justify your answer.
The following table shows the 10 m split times for two 100 m races for Usain Bolt.
Portion of the
race
Usain Bolt
Usain Bolt
9.69 s race (in s)
9.58 s race (in s)
Beijing Olympics
Berlin World
Championships
Reaction time
0.165
0.146
0–10
1.85
1.89
10–20
1.02
0.99
20–30
0.91
0.90
30–40
0.87
0.86
40–50
0.85
0.83
50–60
0.82
0.82
60–70
0.82
0.81
70–80
0.82
0.82
80–90
0.83
0.83
90–100
0.90
0.83
e. What factors have contributed to Bolt having a faster time of 9.58 s for the second race?
Question continues next page.
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Question 11 continued
The following table shows the 100 m Men’s record times under the 10 minute time since 1968.
Electronic timing began in 1977. Maximum wind factor for a time to be recorded is 2.0.
Time
f.
Year
Runner
Wind factor
ms-1
9.58,
9.69, 9.72
2008
Usain Bolt
0.9, 0.0, 1.7
9.74, 9.77
2008, 2005
Asafa Powell
9.78
2002
Tim Montgomery
2.0
9.79
1996
Maurice Greene
0.1
9.84
1996
Donovan Bailey
0.7
9.85
1994
Leroy Burrell
1.2
9.86
1991
Carl Lewis
1.0
9.90
1991
Leroy Burrell
1.9
9.92, 9.93
1988, 1988
Carl Lewis
9.83
1987
Ben Johnson
1.0
9.93
1983
Calvin Smith
1.4
9.95
1968
Jim Hines
0.3
1.7, 1.6
1.1, 1.0, 1.1
It has been stated that other runners have run as fast a speed as Usain Bolt but do not
have as good an overall time for the 100 m.
You are to respond to this statement by comparing and contrasting the race times for the
100 m in both tables. You will need to draw graphs to support your answer.
g. Do you think that the time set by Usain (9.58 s) could be broken in the future? Predict the
fastest time possible for the 100 m in the future. Refer to your graphs to answer this
question.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Instrument-specific criteria and standards
Schools draw instrument-specific criteria and standards from the syllabus dimensions and exit
standards. Schools will make judgments about the match of qualities of student responses with the
standards descriptors that are specific to the particular assessment instrument. While all syllabus
exit descriptors might not be assessed in a single assessment instrument, across the course of
study, opportunities to demonstrate all the syllabus dimensions and standards descriptors must be
provided.
The assessment instrument presented in this document provides opportunities for the
demonstration of the following criteria:

Knowledge and conceptual understanding

Investigative processes

Evaluating and concluding.
This document provides information about how the qualities of student work match the relevant
instrument-specific criteria and standards at standards A and C. The standard A and C descriptors
are presented below. The complete set of instrument-specific criteria and standards is on page 24.
Key:
Qualifier
Characteristic of general objective
Standard A
Knowledge and
conceptual
understanding
Investigative
processes
Evaluating and
concluding
Standard C
 reproduction and interpretation of complex
and challenging concepts and principles in
relation to motion
 comparison and explanation of complex
processes in relation to motion
 linking and application of algorithms and
concepts to find solutions in complex and
challenging scenarios in relation to motion
 reproduction of concepts and principles
in relation to motion
 explanation of simple processes in
relation to motion
 application of algorithms to find solutions
in a simple scenario in relation to motion
 systematic analysis of secondary data to
identify patterns and trends in relation to
motion
 analysis of secondary data to identify
obvious patterns and trends in relation to
motion
 analysis and evaluation of complex
scientific interrelationships between the
research and the scenarios in relation to
motion
 exploration of scenarios and possible
outcomes with justification of conclusions/
recommendations in relation to motion
 discriminating selection, use and
presentation of scientific data and ideas to
make meaning accessible to intended
audiences through innovative graphs
 description of scientific interrelationships
in the research and scenarios in relation
to motion
 description of scenarios and possible
outcomes with statements of conclusion/
recommendation in relation to motion
 selection, use and presentation of
scientific data and ideas to make
meaning accessible in graphs
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Sample indicative responses: Standard A
Note: Exerpts from the exam questions are set bold. Comments on the student response within
the student response column are set italic.
Standard descriptors
Student response A
Question 1
u = 0, v = 5, s = 6, a = 2.08
F=ma
= 0.06 x 2.08
Application of algorithms
to find solutions to simple
situations
= 0.12 N
Question 2
s = ut + ½ at2
125m = 0 + ½ x 9.8 x t2
125
= 4.9 t2
25.5
= t2
t
= 5.05 s
Question 3
a. KE = ∆ PE
= mgh
= 1.5 x 9.8 x2
Linking and application of
algorithms to find
solutions to complex
situations
= 29.4 J
b. PE = ½ mgh
= 14.7J
c. KE = ½ m v2
29.4 = ½ x 1.5 x v2
29.4 = 0.75 x v2
29.4/0.75 = v2
therefore, v = 4.7 m s-1
Interpretation of complex
concepts
Question 4
v = 25 x 0.866
= 21.65 m s-1
Application of algorithms
to find solutions to simple
situations
∆p=m∆v
= 0.4 x 21.65
= 8.66 kgm s-1 
Note:  is positive.  is negative. h = horizontal, v = vertical.
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Standard A (continued)
Standard descriptors
Student response A
Question 5
Linking and application of
algorithms and concepts
to find solutions in a
complex situation
a. mc = 156 g = 0.156 g, hc = 1.00 m
GPEc = m g hc
= 0.156 x 9.8 x 1.00
= 1.53 J
b. ms = 24.0 g = 0.0240 kg
GPEs = ms g hs
hs
= GPEs / ms g
= 1.53/0.024x9.8
= 6.50 m
Question 6
a. e = 0.48
Application of algorithms
to find solutions to simple
situations
h1 = 3.00 m
e = √ h 2 / h1
therefore h2 = e2 x h1
= (0.48)2 x 3.00
= 0.69 m
= 69 cm
Therefore, the pitch would be fast (refer to the table)
Reproduction of concepts
and principles
b. Initially the ball has only GPE (gravitational potential energy) at the
starting height. As the ball falls it gains KE (kinetic energy) until the
instant of impact when GPE has all been converted to KE. As the ball
bounces it “loses” energy as heat and sound.
When it starts moving back up the KE it possesses is now less than
before. As the KE is converted back to GPE, the ball will not bounce as
high. As such the height reached by the ball related to GPE can be
related to the speed with which the ball leaves the pitch. This is
assuming that energy “loses” as the ball moves through the air are
negligible.
GPE = KE
mgh = ½ mv2
gh = ½ v2
Linking and application of
algorithms and concepts
to find solutions in a
complex situation
Therefore the speed of the ball can be calculated, e.g. using the figures
from part (a)
9.81 x 0.6912 = ½ v2
v2 = 6.78/0.5
v = 3.68 m s-1
Therefore the ball would have been travelling at approx 3.68 m s-1 as it
bounced off the pitch.
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Standard A (continued)
Standard descriptors
Reproduction and
interpretation of complex
and challenging concepts
and principles
Exploration of a scenario
with justification of a
conclusion
Student response A
Question 7
Bernoulli’s Principle states that faster moving air creates a region of
lower pressure. As a result a ball with a difference in pressure between
two sides will experience a force towards the area of lower pressure. As
the ball rotates the boundary layer of air moves with the ball.
Above the ball, air pressure is higher as the relative air speed is slower
due to the boundary layer opposing the direction of air flow. Below the
ball air pressure is lower due to an increased air speed as the boundary
layer of air flow are in the same direction. Therefore the ball will
experience a downwards force in addition to its weight and will dip as
shown.
Question 8
v = 150.0 km h-1
= 41.7m s-1
Linking and application of
algorithms and concepts
to find solutions in a
complex situation
m = 156 g = 0.156 kg
pi = m vi
= 0.156 x 41.7
= 6.50 kg m s-1
pf = m vf
= 0.156 x (-70/3.6)
= - 3.03 kg m s-1
∆ p = pf - p i
Analysis and evaluation of
complex scientific
interrelationships
= - 3.03 – 6.5
= -9.53 kg m s-1
∆ p = I = Ft
therefore, F = ∆ p/ t
= - 9.53 / (1.3 x10-3)
= -7.33 x 103 N
ie 7.33 x103 N in the direction of the final velocity
Students need to recognise that they have to calculate pf and pi before ∆
p and hence F.
B standard students may forget to convert km h-1 to m s-1 or forget to
convert ms to s. C standard studenst may try to calculate F by using F =
ma and use “v” as “a”.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Standard A (continued)
Standard descriptors
Student response A
Question 9
Note:  is positive.  is negative. h = horizontal, v = vertical.
Linking of complex and
challenging concepts
Step 1.
uh = 23 cos 680 = 8.6 m s-1
uv = -23 sin 680
= -21 m s-1
sv = 1.0 m
Step 2.
Time to hit the ground
sv = uv t + ½ a t2
1.0 = -21 t + ½ 9.8 t2
0 = 4.9 t2 -21t -1.0
Using the quadratic equation
t = 4.3 s
Linking and application of
algorithms and concepts
to find solutions in a
complex and challenging
situation
Step 3.
Horizontal distance travelled
s h = uh t
= 8.6 x 4.3
37 m
Exploration of a scenario
with justification of a
conclusion
Step 4.
In the same time, assuming Michael Clarke reacts immediately, he
could run
smc = u mc t
= 6.5 x 4.3
= 28 m
If he was initially 67 m away he would end up 67-28 = 39 m away from
where Ricky Ponting hit the ball. This is 2 m further than where the ball
will land and so it is unlikely he will catch the ball.
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Standard A (continued)
Standard descriptors
Student response A
Question 10
a. Initial vertical velocity = u sin θ
= 40 x sin 350
= 22.94 m s-1
Initial horizontal velocity = u cos θ
= 40 x cos 350
= 32.77 m s-1
At maximum height, vertical velocity is zero.
Consider vertical motion
uy= 22.94 m s-1
Linking of concepts in a
complex and challenging
situation
a = g = -9.8 m s-2
vy = 0
vy2 = uy +2as
2
0 = (22.94)2 + (2x-9.8xs)
19.6s = (22.94)2
s = 26.8 m
Therefore, the maximum height is 26.8 m
b.
Linking and application of
algorithms and concepts
to find solutions in a
complex and challenging
situation
The velocity of the ball at maximum height = 32.77 m s-1
The horizontal velocity is unchanged
c. To find the horizontal range the time of flight must be calculated.
The time of flight is the time taken for the ball to reach its final vertical
displacement (sy = 0)
Need to find t when sy = 0, uy= 22.94 m s-1, a = = -9.8 m s-2
s = ut + ½ a t2
0 = 22.94 t – (½ 9.8 t2 )
0 = t (22.94 – 4.9t)
t = 4.68 seconds
Horizontal range = horizontal velocity x time of flight
= 32.77 x 4.68
=153 m
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Standard A (continued)
Standard descriptors
Student response A
d. The direction of motion is the direction is the direction of the vertical
vector.
Linking and application of
algorithms and concepts
to find solutions in a
complex and challenging
situation
After three seconds, horizontal velocity = 32.77 m s-1
Vertical velocity after 3 s is given by
t = 3 s, vy = ?, uy = 22.94 m s-1, a = -9.8 m s-2
vy = uy + at
= 22.94 +(-9.8)3
= -6.4 m s-1
To find θ
tan θ = 6.46/32.77
θ = tan-1 6.46/32.77
θ = 11.20
Therefore, after three seconds, the direction of motion is 11.20 down
from the horizontal.
Question 11
Application of algorithms
to find solutions in simple
situations
a. For the 100 m race
Speed = distance /time
= 100/9.58
= 10.438 m s -1
For the 200 m race
Speed = distance /time
Application of algorithms
to find solutions in simple
situations
= 200/19.19
= 10.422 m s -1
Fastest average speed was for the 200 m race
b. What was the average speed for Kenenisa Bekele in the race?
Speed = distance /time
= 10000/26:17.563
Systematic analysis of
secondary data in a
complex situation
Comparison and
explanation of a complex
process
= 6.338 m s -1
c. The time at the 4000 m mark was 11:04.75 or 664.75 seconds and
the time at the 5000 m was 13:40.45 or 820.45 seconds. Therefore the
time for the fifth kilometre was 155.7 seconds or 2:35.7 minutes and
seconds. Hence the term 2.35 pace.
Speed = distance /time
= 1000/155.7
= 6.423 m s -1
At this pace the race time would be expected to be approximately
13:40.35 + ( 5x 2:35) = 26:25.45. This would not be better than the
previous record.
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Standard A (continued)
Standard descriptors
Analysis of secondary
data to identify a trend
Student response A
d. Speed = distance /time
= 400/57.4
= 6.968 m s -1
Therefore time for the 10 000 m race would have been 10000 x 6.968
Exploration of a scenario
with discussion of
possible outcomes with
discussion of conclusions
= 1435.13 seconds
= 23:55.13 s
e. The reaction time of 0.146 was less for the second race. This may
have contributed to the fast starting speed. The wind factor was greater
ie 0.9 for the 9.58 s race against 0.0 for the 9.69 s race. The consistent
times of less than 0.85 for longer periods of the race and the time of
0.81 for the 60-70 m section.
f.
Students will need to calculate the speed for the intervals of the race.
Portion of
the race
Systematic analysis of
secondary data to identify
patterns and trends
Reaction
9.58 s race
(in s) Set in
2009
Time to
get to
this
point
Velocity for
this split time
(ms-1)
0.146
time
Discriminating selection,
use and presentation of
data to make meaning
accessible to the intended
audience
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Physics (2007)
0-10
1.89
1.89
5.29
10-20
0.99
2.88
10.10
20-30
0.90
3.78
11.11
30-40
0.86
4.64
11.63
40-50
0.83
5.47
12.05
50-60
0.82
6.29
12.20
60-70
0.81
7.10
12.35
70-80
0.82
7.92
12.20
80-90
0.83
8.75
12.05
90-100
0.83
9.58
12.05
Sample assessment instrument and indicative responses Supervised assessment
Standard A (continued)
Standard descriptors
Student response A
f.
(continued)
Discriminating selection,
use and presentation of
data to make meaning
accessible to the intended
audience
Students may produce different styles of graphs.
The above graph shows the speed that Usain Bolt was travelling at
during the race. The graph shows that Usain Bolt ran consistently above
the 12 m s-1 from the 40m mark. Other runners have reached that speed
during races but have not been able to sustain this speed for as long as
Bolt during a race.
Responses might mention reaction times. Usain Bolt substantially
lowered his response time in the 9.58 race. This may have contributed
to the fast increase in speed during the first 10 m.
Maximum velocity was during then 60 to 70 m split.
Exploration of a scenario
with possible outcomes
and justification of
conclusions
s = 10m, t = 0.81
Speed = distance /time
= 10/0.81
= 12.35 m s -1
The average velocity was the same from 40 to 50 m and from 80 to 100
m.
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Standard A (continued)
Standard descriptors
Student response A
g. Students may produce different graphs for this answer, depending
on what data has been chosen to graph. The graph should show a
downward trend although the slope may be different depending on the
data graphed.
Discriminating selection,
use and presentation of
data to make meaning
accessible to the intended
audience
The scale for the horizontal axis may be in years since 1968 or by
runner.
Exploration of a scenario
and possible outcomes
with a justified conclusion
If the downward trend in the times for the 100m continues the limit of 0
seconds would eventually be reached. This would be impossible. The
use of hand timers pre 1977 also makes the slope of 7.7 milliseconds
fall per year seem greater than it really is.
The greatest decrease in time has occurred during 2008 when Usain
Bolt’s times went from 9.72 to 9.58.
Predictions using logistic fit are that the limit is 9.48 in 500 years. There
are many factors which students might mention eg wind factor, type of
track surface, starting blocks, running spikes, coaching methods.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Sample indicative responses: Standard C
Note: Exerpts from the exam questions are set bold. Comments on the student response within
the student response column are set italic.
Standard descriptors
Student response C
Question 1
u = 0, v = 5, s = 6, a = 2.08
F=ma
= 0.06 x 2.08
Application of algorithms to find
solutions to simple situations
= 0.12 N
Question 2
s = ut + ½ at2
125m = 0 + ½ x 9.8 x t2
125
= 4.9 t2
25.5
= t2
t
= 5.05 s
Question 3
Application of algorithms to find
solutions to simple situations
a. KE = ∆ PE
= mgh
= 1.5 x 9.8 x2
= 29.4 J
Application of algorithms to find
solutions to simple situations
b. PE = ½ mgh
= 14.7 J
c. A C response might get part way through finding a solution or
forget to square a figure or to take the square root of another.
Question 4
Interpretation of complex
concepts
v = 25 x 0.866
= 21.65 m s-1
No change in momentum calculated
Application of algorithms and
concepts to find solutions in a
simple situation
Question 5
a. mc = 156 g = 0.156 g, hc = 1.00 m
GPEc = m g hc
= 0.156 x 9.8 x 1.00
= 1.53 J
b. No attempt at this part of the question.
Sample assessment instrument and indicative responses Supervised assessment
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Standard C (continued)
Standard descriptors
Student response C
Question 6
a. e = 0.48
h1 = 3.00 m
Application of algorithms to find
solutions to simple situations
e = √ h 2 / h1
therefore h2 = e2 x h1
= (0.48)2 x 3.00
= 0.69 m
= 69 cm
Therefore, the pitch would be fast (refer to the table).
b. During the drop test some energy transferred. The energy is
transferred from gravitational potential energy to kinectic energy.
Reproduction of concepts and
principles
Description of a scenario with
statements of a conclusion
Question 7
The bowler caused the ball to spin forwards, therefore according to
Bernoulli’s principle the air flowing over the top of the ball is moving
much quicker than the air following on the under side of the ball
therefore causing it to dip and shortening the path of the ball. Without
spin the air flowing on either side of the ball is equal causing it to only
dip as it loses speed.
Question 8
v = 150.0 km h-1
= 41.7 m s-1
m = 156 g = 0.156 kg
Linking and application of
algorithms and concepts to find
solutions in a complex situation
pi = m vi
= 0.156 x 41.7
= 6.50 kg m s-1
pf = m vf
= 0.156 x (-70/3.6)
= - 3.03 kg ms-1
No attempt at this part of the question.
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Standard C (continued)
Standard descriptors
Student response C
Question 9
Analysis of a complex scientific
interrelationships
v = 23 m s-1
θ = 680
h=1m
Assume Michael Clarke catches the ball at the same height as it was
hit from ie 1.0 m
vh = 23 cos 680
assume a = -9.8 m s-2
when ball reaches max height, v = 0 m s-1
v = u + at
0 = 23 + -9.8 t
-23 = -9.8 t
t = 2.35 s
Total time = 2 t
= 4.69 s
No attempt to complete the question.
Question 10
a. Initial vertical velocity = u sin θ
= 40 x sin 350
= 22.94 m s-1
Linking of complex and
challenging concepts
Initial horizontal velocity = u cos θ
= 40 x cos 350
= 32.77 m s-1
At maximum height, vertical velocity is zero.
Consider vertical motion
uy= 22.94 m s-1
a = g = -9.8 m s-2
vy = 0
vy2 = uy +2as
2
0 = (22.94)2 + (2x-9.8xs)
19.6s = (22.94)2
s = 26.8 m
Therefore, the maximum height is 26.8 m
No attempt at other parts of the question.
Queensland Studies Authority
December 2010
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Standard C (continued)
Standard descriptors
Student response C
Question 11
a. For the 100 m race
Application of algorithms and
concepts to find solutions in a
simple situation
Speed = distance /time
= 100/9.58
= 10.438 m s -1
For the 200 m race
Speed = distance /time
= 200/19.19
= 10.422 m s -1
Fastest average speed was for the 200 m race
b. What was the average speed for Kenenisa Bekele in the race?
Speed = distance /time
= 10000/26:17.563
= 6.338 m s -1
Application of concepts in a
simple situation
c. The time at the 4000 m mark was 11:04.75 or 664.75 seconds
and the time at the 5000 m was 13:40.45 or 820.45 seconds.
Therefore the time for the fifth kilometre was 155.7 seconds or 2:35.7
minutes and seconds. Hence the term 2.35 pace.
Speed = distance /time
= 1000/155.7
Explanation of a simple
process
= 6.423 m s -1
A C response may not give an answer to this part of the question.
d. Speed = distance /time
= 400/57.4
= 6.968 m s -1
A C response may not give an answer to this part of the question.
e. A C response may not give as many factors for this part of the
question.
The reaction time of 0.146 was less for the second race. This may
have contributed to the fast starting speed. The wind factor was
greater, i.e. 0.9 for the 9.58 s race against 0.0 for the 9.69 s race. The
consistent times of less than 0.85 for longer periods of the race and
the time of 0.81 for the 60 to70 m section.
Application of concepts in a
simple situation
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Physics (2007)
Sample assessment instrument and indicative responses Supervised assessment
Standard C (continued)
Standard descriptors
Student response C
f. A C response may not calculate the speed for the intervals of the
race as would an A response and so not produce a graph such as the
one below.
Application of algorithms and
concepts to find solutions in a
simple situation
Students may produce different styles of graphs.
The above graph shows the speed that Usain Bolt was travelling at
during the race. The graph shows that Usain Bolt ran consistently
above the 12 m s-1 from the 40m mark. Other runners have reached
that speed during races but have not been able to sustain this speed
for as long as Bolt during a race.
This may be calculated. Maximum velocity was during then 60 to 70 m
split.
s = 10 m, t = 0.81
Speed = distance /time
Application of algorithms to find
solutions in simple situations
= 10/0.81
= 12.35 m s -1
The average velocity was the same from 40 to 50 m and from 80 to
100 m.
g. Students may produce different graphs for this answer, depending
on what data has been chosen to graph. A C response may produce a
graph, which should show a downward trend, although the slope may
be different depending on the data graphed.
The scale for the horizontal axis may be in years since 1968 or by
runner.
A comment may be made about the predicted time in the future but
possibly won’t be justified by a graph.
Queensland Studies Authority
December 2010
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Instrument-specific criteria and standards
Key:
Qualifier
Characteristic of general objective
Standard A
Knowledge and
conceptual
understanding
 reproduction and interpretation of
complex and challenging concepts
and principles in relation to motion
 comparison and explanation of
complex processes in relation to
motion
Standard B
 reproduction and interpretation
of complex or challenging
concepts and principles in
relation to motion
 comparison and explanation of
processes in relation to motion
Standard C
Standard D
Standard E
 reproduction of concepts and
principles in relation to
motion
 reproduction of simple
ideas and concepts in
relation to motion

reproduction of isolated
facts in relation to
motion
 explanation of simple
processes in relation to
motion
 description of simple
processes in relation to
motion

recognition of isolated
simple phenomena in
relation to motion
 linking and application of algorithms
and concepts to find solutions in
complex and challenging scenarios in
relation to motion
 linking and application of
algorithms and concepts to find
solutions in complex or
challenging scenarios in
relation to motion.
 application of algorithms to
find solutions in a simple
scenario in relation to motion.
 application of algorithms in
relation to motion

application of simple
given algorithms in
relation to motion
Investigative
processes
 systematic analysis of secondary data
to identify patterns and trends in
relation to motion
 analysis of secondary data to
identify patterns and trends in
relation to motion
 analysis of secondary data to
identify obvious patterns and
trends in relation to motion
 identification of obvious
patterns in relation to
motion

recording of data in
relation to motion
Evaluating and
concluding
 analysis and evaluation of complex
scientific interrelationships in the
scenarios in relation to motion
 analysis of complex scientific
interrelationships in the
scenarios in relation to motion
 explanation of scenarios and
possible outcomes with
discussion of conclusions/
recommendations in relation to
motion
 identification of simple
scientific interrelationships
in the scenario in relation
to motion

 exploration of scenarios and possible
outcomes with justification of
conclusions/ recommendations in
relation to motion
 description of scientific
interrelationships in the
scenarios in relation to
motion
 description of scenarios and
possible outcomes with
statements of conclusion/
recommendation in relation to
motion
 identification of scenarios
or possible outcomes in
relation to motion
identification of obvious
scientific
interrelationships in the
scenario in relation to
motion

statements about
outcomes in relation to
motion

presentation of
scientific data or ideas.
 discriminating selection, use and
presentation of scientific data and
ideas to make meaning accessible to
intended audiences through
innovative graphs.
 selection, use and presentation
of scientific data and ideas to
make meaning accessible to
intended audiences in graphs.
 selection, use and
presentation of scientific data
and ideas to make meaning
accessible in graphs.
 presentation of scientific
data or ideas in graphs.
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December 2010
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