physics marking guide and solution
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Physics marking guide and solution Sample external assessment 2020
Maths/Science (90 marks)
Assessment objectives This assessment instrument is used to determine student achievement in the following objectives: 1. describe and explain gravity and motion, electromagnetism, special relativity, quantum
theory and the Standard Model2. apply understanding of gravity and motion, electromagnetism, special relativity, quantum
theory and the Standard Model
3. analyse evidence about gravity and motion, electromagnetism, special relativity, quantumtheory and the Standard Model to identify trends, patterns, relationships, limitations oruncertainty
4. interpret evidence about gravity and motion, electromagnetism, special relativity, quantumtheory and the Standard Model to draw conclusions based on analysis.
Note: Objectives 5, 6 and 7 are not assessed in this instrument.
Marking guide and solution
Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Introduction
The Queensland Curriculum and Assessment Authority (QCAA) has developed mock external assessments for each General senior syllabus subject to support the introduction of external assessment in Queensland.
An external assessment marking guide (EAMG) has been created specifically for each mock external assessment.
The mock external assessments and their marking guides were:
β’ developed in close consultation with subject matter experts drawn from schools, subject
associations and universities
β’ aligned to the external assessment conditions and specifications in General senior syllabuses
β’ developed under secure conditions.
Purpose
This document consists of an EAMG and an annotated response.
The EAMG:
β’ provides a tool for calibrating external assessment markers to ensure reliability of results
β’ indicates the correlation, for each question, between mark allocation and qualities at each
level of the mark range
β’ informs schools and students about how marks are matched to qualities in student responses.
Mark allocation
Where a response does not meet any of the descriptors for a question or a criterion, a mark of β0β
will be recorded.
Where no response to a question has been made, a mark of βNβ will be recorded.
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
External assessment marking guide
Paper 1: Multiple-choice
Question Response
1 B
2 A
3 A
4 D
5 A
6 C
7 D
8 D
9 B
10 C
11 B
12 D
13 B
14 C
15 C
16 A
17 A
18 A
19 D
20 C
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Paper 1: Short response
Question Sample response The response
21 A baryon is a composite subatomic particle made up of 3 quarks.
β’ defines a baryon[1 mark]
22 π’π’π¦π¦ = π’π’ sinππ π’π’π¦π¦ = 12 sin 35ππ
Velocity: 6.9 m sβ1 (to 1 decimal place)
β’ identifies thegeometry of thescenario [1 mark]
β’ shows substitutioncorrectlyperformed[1 mark]
β’ provides 6.9 m s-1
as the velocity[1 mark]
23 πΉπΉππππππ =πππ£π£2
ππ
β΄ ππ =πππ£π£2
πΉπΉππππππ
ππ =5 Γ 82
80Radius: 4 m (to the nearest whole
number)
β’ shows correcttransposition[1 mark]
β’ shows substitutioncorrectlyperformed[1 mark]
β’ provides 4 m asthe radius [1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Question Sample response The response
24 An electron and positron annihilate to produce a photon and then pair produce another electron and positron.
β’ identifies the firststep of theinteraction[1 mark]
β’ identifies the finalstep of theinteraction[1 mark]
β’ provides thecorrectcorrespondingFeynman diagram[1 mark]
25 Fnet = πππ£π£ππ sinππ πππ£π£2
ππ= πππ£π£ππ sinππ
ππ =ππππππ sinππ
π£π£ππ = οΏ½4.8 Γ10β19οΏ½(0.02)οΏ½1Γ10β4οΏ½sin90ππ
30
Mass: 3.2 x 10β26 kg (to 1 decimal place)
β’ indicates that themagnetic force isequal to thecentripetal force[1 mark]
β’ shows correcttransposition[1 mark]
β’ shows substitutioncorrectlyperformed[1 mark]
β’ provides 3.2 x 10β
26 kg as the mass[1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Question Sample response The response
26 One piece of evidence is the bright and dark bands. This shows that there is interference which is a property of waves.
β’ identifies onepiece of evidence[1 mark]
β’ identifies that thisis a characteristicof wave behaviour[1 mark]
27 cosππ =π’π’π₯π₯π’π’
=4
17Angle: 76o (to the nearest degree)
β’ identifies cos asthe relevanttrigonometric ratio[1 mark]
β’ shows substitutioncorrectlyperformed[1 mark]
β’ provides 76o asthe angle [1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Question Sample response The response
28 m =πΉπΉππππ
=459.8
m = 4.6 kg
β’ provides 4.6 as themass [1 mark]
Fparallel = πΉπΉππ sinππ Fparallel = β45 sin 30ππ
Fparallel = 22.5 ππ
β’ provides 22.5 asFparallel [1 mark]
Fnet = Fparallel β πΉπΉππ Fnet = 22.5 β 12.5
Fnet = 10 ππ
β’ determines netforce [1 mark]
aparallel =Fparallelnet
ππ
aparallel =104.6
ππ = 2.17 m sβ2
β’ determinesacceleration[1 mark]
π π = π’π’π’π’ + 12πππ’π’2
10 =12
Γ 2.17 Γ π’π’2
β’ shows substitutioncorrectly performed[1 mark]
Time: 3 s (to the nearest whole number)
β’ determines time[1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Paper 2: Short response
Question Sample response The response
1a gradient =
riserun
=3.5 Γ 10β18
0.008
gradient = 4.4 Γ 10β16 Nm2
β’ identifies a correctmethod todetermine thegradient [1 mark]
β’ determinesgradient [1 mark]
πΉπΉ = 4.4 Γ 10β16 Γ1ππ2
β’ identifiesrelationshipbetween F and r[1 mark]
1b gradient = πΊπΊπΊπΊππ 4.4 Γ 10β16 = πΊπΊ (2.5 Γ 10β3)2
Experimental value for G = 7.0 x 10β11 N m2 kgβ2 (to one decimal place)
β’ identifies that thegradient isequivalent to G ΓM Γ m [1 mark]
β’ determines G[1 mark]
2 ππ =πππππΌπΌ2ππππ
=(4ππΓ 10β7)(12)
2ππ Γ 0.06
Magnitude = 4 x 10β5 T (to the nearest whole number)
Direction = into the page
β’ shows substitutioncorrectly performed[1 mark]
β’ provides 4 Γ 10β5 Tas magnitude andthe direction of themagnetic field isinto the page[1 mark]
3 Vector resolution method πΉπΉ1π¦π¦ = 300 sin 40ππ
= 193 N
β’ provides 193 asπΉπΉ1π¦π¦ [1 mark]
πΉπΉ1π₯π₯ = 300 cos 40ππ =
230 N
β’ provides 230 asπΉπΉ1π₯π₯ [1 mark]
πΉπΉ2π¦π¦ = 500 sin 20ππ = 171 N
β’ provides 171 asπΉπΉ2π¦π¦ [1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
πΉπΉ2π₯π₯ = βπΉπΉ2 cos 20ππ = β500 cos 20ππ = β470 N
β’ provides β470 asπΉπΉ1π₯π₯ [1 mark]
πΉπΉπ¦π¦ππππππ = πΉπΉ1π¦π¦ + πΉπΉ2π¦π¦ = 193 + 171 = 364 N
β’ determines netvertical force[1 mark]
πΉπΉπ₯π₯ππππππ = πΉπΉ1π₯π₯ + πΉπΉ2π₯π₯ = 230 + β470 = β240 N
β’ determines nethorizontal force[1 mark]
πΉπΉππππππ = οΏ½οΏ½πΉπΉπ¦π¦πππππποΏ½2 + οΏ½πΉπΉπ₯π₯πππππποΏ½
2
= οΏ½3642 + 2402 = 436 N
β’ determines the netforce [1 mark]
ππ = tanβ1 οΏ½364240οΏ½
= 56.6ππ
56.6Β° above π₯π₯ β πππ₯π₯πππ π directed to the left
β’ determinesdirection [1 mark]
Sine/cosine rule method β’ identifies thegeometry of thescenario [1 mark]
ππ2 = ππ2 + ππ2 β 2πππππππ π ππ β’ states the cosinerule [1 mark]
ππ2 = 5002 + 3002 β 2(500)(300)πππππ π 60 β’ shows substitutioncorrectly performed[1 mark]
ππ = 436 N β’ provides 436 asthe net force[1 mark]
ππππππππππ
= ππππππππππ
β’ states the sine rule[1 mark]
ππππππ60436
= ππππππππ300
β’ shows substitutioncorrectly performed[1 mark]
ππ = 36.6ππ β’ provides 36.6Β° asthe angle [1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
Add ΞΈ to 20o to find the angle up from the horizontal.
56.6Β° above π₯π₯ β πππ₯π₯πππ π directed to the left
β’ determinesdirection [1 mark]
4a gradient =riserun
=(4.2 β 0) Γ 10β19
(18β 12) Γ 1014gradient = 7.0 Γ 10β34 J.s
π¦π¦β intercept = β8.0 Γ 10β19 J
Max πΈπΈππ = 7.0 Γ 10β34ππ β 8.0 Γ 10β19
β’ determinesgradient [1 mark]
β’ determines yβintercept [1 mark]
β’ identifies therelationship[1 mark]
4b ππ =8.0 Γ 10β19 J
1.60 Γ 10β19 J/eV= 5 eV
The metal is most likely to be copper, which has a work function of 5.1 eV.
β’ determines workfunction [1 mark]
β’ identifies metal,with a reason[1 mark]
5 πΈπΈ2 β πΈπΈ1 =βππππ
β’ identifies therelationshipbetween differencein energy levelsand wavelength ofthe emitted photon[1 mark]
Transition from β2.7 eV to β5.7 eV gives 3 eV
ππ =(6.626 Γ 10β34)(3 Γ 108)
(3)(1.60 Γ 10β19)= 414 nm
β’ identifies theenergy changecorresponding to413 nm [1 mark]
Transition from β5.7 eV to β8.2 eV gives 2.5 eV
ππ =(6.626 Γ 10β34)(3 Γ 108)
(2.5)(1.60 Γ 10β19)= 497 nm
β’ identifies theenergy changecorresponding to496 nm [1 mark]
Transition from β5.7 eV to β7.8 eV gives 2.1 eV
ππ =(6.626 Γ 10β34)(3 Γ 108)
(2.1)(1.60 Γ 10β19)= 592 nm
β’ identifies theenergy changecorresponding to590 nm [1 mark]
None of the other possible transitions would produce a 520 nm line.
β’ identifies that noneof the otherpossible transitionswould produce a520 nm line.[1 mark]
The only spectrum that shows the three visible
spectrum lines highlighted above is Element A. β’ identifies Element
A, giving a reason[1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
6 The concept of waveβparticle duality means that light can exhibit either wave or particle properties depending on the situation. The alternating dark and light bands produced in Youngβs double slit experiment can only be explained by referring to the wave properties of light such as interference and diffraction. From the graph showing the luminosity and wavelength of light emitted from a black body, we can see that the experimental data is not consistent with the classical model for light which predicts that πΌπΌ ~ 1
ππ4.
Planck suggested a solution to this that electron energy levels are quantised. As particles move from higher energy to lower energy, energy is released as a photon. These photonsβ energies are also quantised (having discrete values).
β’ describes wave-particle duality[1 mark]
β’ identifiesinterference and/ordiffraction as wavecharacteristicsobservable inYoungβsexperiment[1 mark]
β’ identifies that thewavecharacteristic/s oflight in Youngβsexperiment cannotbe explained by aparticle model[1 mark]
β’ identifies evidenceof particlecharacteristicsfrom the black-body radiationcurve [1 mark]
β’ identifies that theparticlecharacteristic/s oflight in black-bodyradiation cannot beexplained by awave model[1 mark]
7 π’π’π΅π΅πππππππππ‘π‘ = 2 π’π’π΅π΅πππππ₯π₯ βππππππβππ β΄ 4.20 = 2 π’π’π΅π΅πππππ₯π₯ βππππππβππ π’π’π΅π΅πππππ₯π₯ βππππππβππ = 2.10 π π
π£π£π¦π¦ = πππ’π’ + π’π’π¦π¦ 0 = (9.8)(2.1) + π’π’π¦π¦ π’π’π¦π¦ = 20.58 m sβ1
β’ provides 2.1 as thetime taken for B toreach maximumheight [1 mark]
β’ shows substitutioncorrectly performed[1 mark]
β’ provides 20.6 asinitial verticalvelocity [1 mark]
tan 30ππ =20.58π’π’π₯π₯
π’π’π₯π₯ =20.58
tan 30ππ
π’π’π₯π₯ = 35.64 m sβ1
β’ determines initialhorizontal velocity[1 mark]
π π π₯π₯ = π’π’π₯π₯π’π’π΅π΅πππππππππ‘π‘ π π π₯π₯ = 35.64 Γ 4.2
Horizontal displacement = 150 m (to the nearest whole number)
β’ shows substitutioncorrectly performed[1 mark]
β’ determineshorizontaldisplacement[1 mark]
Physics marking guide and solution Sample external assessment 2020
Queensland Curriculum & Assessment Authority
8 From observer Bβs frame of reference, the ladder would be length contracted:
πΏπΏ = πΏπΏπποΏ½1β οΏ½π£π£2
ππ2οΏ½
πΏπΏ = 20οΏ½1β οΏ½(0.9ππ)2
ππ2οΏ½ = 8.7 m
The ladder would be 8.7 m long and would fit into the 15 m barn.
From observer Aβs frame of reference, the barn would be length contracted:
πΏπΏ = 15οΏ½1β οΏ½(0.9ππ)2
ππ2οΏ½
πΏπΏ = 6.5 m The ladder of 20 m long will not fit into the 6.5 m barn.
β’ identifies thatlength contractionoccurs [1 mark]
β’ calculates ladderlength from Bβsframe of referenceand identifies thatthe ladder will fit[1 mark]
β’ calculates barnlength from Aβsframe of referenceand identifies thatthe ladder will notfit [1 mark]
This could be viewed as a paradox because in one frame of reference the 8.7 m long ladder will fit into the 15 m long barn, while in another frame of reference the 20 m long ladder will not fit into the 6.5 m barn.
β’ identifies theparadox [1 mark]
The concept of simultaneity resolves this problem. This concept states that two events that appear to be simultaneous in one frame of reference will not be in another frame of reference. In observer Bβs frame of reference, the doors open and close at the same instant. However, in observer Aβs frame of reference, the doors do not open and close at the same instant. The key is what happens while the ladder is partially inside the barn. The door in front of the ladder opens before the ladder reaches it, while the door behind the ladder only closes after the back of the ladder is inside the barn. The conclusion is that the ladder does not make contact with the barn doors
β’ identifies thatsimultaneity isrequired to resolvethe paradox[1 mark]
β’ identifies that thedoors are notclosedsimultaneously inAβs frame ofreference [1 mark]
β’ identifies theposition of theladder in relation tothe doors in Aβsframe of reference[1 mark]
β’ concludes that theladder does notmake contact withthe barn doors[1 mark]